berkeley problems in mathematics
DESCRIPTION
problems and solution from different areas of MathematicsTRANSCRIPT
PAULO NEY DE SOUZAJORGE-NUNO SILVA
Berkeley Problems inMathematics
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Problem Books in Mathematics
Edited by K.A. BencsáthP.R. Halmos
Problem Books in Mathematics
Series Editors: K.A. Bencsáth and P.R. Halmos
Pell's Equationby Edward J. Barbeau
Polynomialsby Edward J. Barbeau
Problems in Geometryby Marcel Berger, Pierre Pansu, Jean-Pic Berry, and Xavier Saint-Raymond
Problem Book for First Year Calculusby George W. Bluman
Exercises in Probabilityby T. Cacoullos
Probability Through Problemsby Marek Capiñski and Tomasz Zastawniak
An Introduction to Hilbert Space and Quantum Logicby David W. Cohen
Unsolved Problems in Geometryby Hallard T. Croft. Kenneth J. Falconer, and Richard K Guy
Berkeley Problems in Mathematics (3rd ed.)by Paulo Ney de Souza and Jorge-Nuno Silva
Problem-Solving Strategiesby Arthur En gel
Problems in Analysisby Bernard R. Gelbaum
Problems in Real and Complex Analysisby Bernard R. Gelbaum
Theorems and Counterexamples in Mathematicsby Bernard R. Gelbaum and John MH. Olmsted
Exercises in Integrationby Claude George
(continued after index)
Paulo Ney de Souza Jorge-Nuno Silva
Berkeley Problemsin Mathematics
Third Edition
Springer
Paulo Ney de Souza Jorge-Nuno SilvaDepartment of Mathematics Departarnento de MatemáticaUniversity of California at Berkeley Faculdade de Ciências de LisboaBerkeley, CA 94720-3840 1700 LisboaUSA [email protected] jnsilva@ lmc.fc.uLpt
Series Editors:Katalin A. Bencsáth Paul R. HalmosMathematics Department of MathematicsSchool of Science Santa Clara UniversityManhattan College Santa Clara, CA 95053Riverdale, NY 10471 USAUSA [email protected]@manhattan.edu
Cover illustration: Tracy Hall, University of California at Berkeley
Mathematics Subject Classification (2000): 26-06, 26Bxx, 34Axx, 3OAxx
Library of Congress Cataloging-in-Publication Datade Souza, Paulo Ney.
Berkeley problems in mathematics I Paulo Ney de Souza, Jorge Nuno-Silvap. cm.— (Problem books in mathematics)
Includes bibliographical references and index.ISBN 0-387-20429-6 (alk. paper) — ISBN 0-387-00892-6 (pbk: alk. paper)1. Mathematics—Problems, exercises, etc. 2. University of California. Berkeley. Dept. ofMathematics—Examinations. I. Silva, Jorge-Nuno. II. Title. III. Series.QA43.D38—20045 l0'.76—dc22 2003063347
ISBN 0-387-20429-6 (hardcover) PrInted on acid-free paper.ISBN 0-387-00892-6 (softcover)
© 2004,2001, 1998 All text by Paulo Ney de Souza and Jorge-Nuno Silva.© 2004,2001, 1998 All problems by California Regents.All rights reserved. This work may not be translated or copied in whole or in part without thewritten permission of the publisher (Springer Science+Business Media, Inc., 233 Spring Street,New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarlyanalysis. Use in connection with any form of information storage and retrieval, electronicadaptation, computer software, or by similar or dissimilar methodology now know or hereafterdeveloped is forbidden.The use in this publication of trade names, trademarks, service marks and similar terms, even ifthe are not identified as such, is not to be taken as an expression of opinion as to whether or notthey are subject to proprietary rights.
Printed in the United States of America. (EB)
9876543springer.com
para Aline, Laura, Manuel & Stephanie
Preface
In 1977 the Mathematics Department at the University of California, Berkeley,instituted a written examination as one of the first major requirements toward thePh.D. degree in Mathematics. This examination replaced a system of standardizedQualifying Exams. Its purpose was to determine whether first-year students in thePh.D. program had mastered basic mathematics well enough to continue in theprogram with a reasonable chance of success.
Historically, any one examination is passed by approximately half of the stu-dents taking it and students are allowed three attempts. Since its inception, theexam has become a major hurdle to overcome in the pursuit of the degree and,therefore, a measure of the minimum requirements to successful completion ofthe program at Berkeley. Even though students are allowed three attempts, mostwould agree that the ideal time to complete the requirement is during the firstmonth of the program rather than in the middle or end of the first year. This bookwas conceived on this premise, and its intent is to publicize the material and aidin the preparation for the examination during the undergraduate years, when oneis deeply involved with the material that it covers.
The examination is now offered twice a year in the second week of each semester,and consists of 6 hours of written work given over a 2-day period with 9 problemseach (10 before 1988). Students select 6 of the 9 problems (7 of 10 before 1988).Most of the examination covers material, mainly in analysis and algebra, thatshould be a part of a well-prepared mathematics student's undergraduate training.This book is a compilation of the more than 1000 problems which have appearedon the Prelims during the last few decades and currently make up a collectionwhich is a delightful field to plow through, and solutions to most of them.
When Berkeley was on the Quarter system, exams were given three times ayear: Spring, Summer, and Fall. Since 1986, the exams have been given twice ayear, in January and September.
From the first examination through Fall 1981, the policy was: two attemptsallowed; each examination 6 hours; total 14/20 problems. From Winter 1982through Spring 1988, the policy was: two attempts allowed; each examination8 hours; total 14/20 problems. Starting Fall 1988, the policy was: three attemptsallowed; each examination 6 hours; total 12118 problems. In all cases, the exami-nation must be passed within 13 months of entering the Ph.D. program.
The problems are organized by subject and ordered in increasing level of dif-ficulty, within clusters. Each one is tagged with the academic term of the examin which it appeared using abbreviations of the type Fa87 to designate the examgiven in the Fall semester of 1987. Problems that have appeared more than oncehave been merged and show multiple tags for each exam. Sometimes the mergerequired slight modifications in the text (a few to make the problem correct!), butthe original text has been preserved in an electronic version of the exams (seeAppendix A). Other items in the Appendices include the syllabus, passing scoresfor the exams and a Bibliography used throughout the solutions.
Classifying a collection of problems as vast as this one by subjects is not an easytask. Some of the problems are interdisciplinary and some have solutions as variedas Analysis and Number Theory (1.1.18 comes to mind!), and the choices areinvariably hard. In most of these cases, we provide the reader with an alternativeclassification or pointers to similar problems elsewhere.
We would like to hear about other solutions to the problems here and commentson the existing ones. They can be sent by e-mail to the authors.
This project started many years ago, when one of us (PNdS) came to Berke-ley and had to go through the lack of information and uncertainties of the examand got involved with a problem solving group. First thanks go to the group'smembers: Dino Lorenzini, Hung The Dinh, Kin-Yin Li, and Jorge Zubelli, andthen to the Prelim Workshop leaders, many of whose names escape us now butthe list includes, besides ourselves, Matthew Wiener, Dmitry Gokhman, KeithKearnes, Geon Ho Choe, Mike May, Eliza Sachs, Ben Lotto, Ted Jones, DavidCruz-Uribe, Jonathan Walden, Saul Schleimer and Howard Thompson; and alsoto the many people we have discussed these problems with like Don Sarason,George Bergman, Reginald Koo, D. Popa, C. Costara, JOzsef Elton Hsu,Enlin Pan, Bjorn Poonen, Assaf Wool and Jin-Gen Yang. Many thanks to Deb-bie Craig for swift typesetting of many of the problems and to Janet Yonan forher help with the archeological work of finding many of the old and lost problemsets, and finally to Nefeli's for the best coffee west of Rome, we would not havesurvived without it!
We thank also the Department of Mathematics and the Portuguese Studies Pro-.gram of UC Berkeley, MSRJ, University of Lisbon, CMAF, JNICT, PRAXISXXI, FEDER, project PRAXIS/212.1/MAT/125/94 and FLAD, which supportedone of the authors on the Summers of 96, 97, 2000 and 2002, and CNPq grant20.1553/82-MA that supported the other during the initial phase of this project.
This is a project that could not have been accomplished in any typesettingsystem other than Tp7(. The problems and solutions are part of a two-prongeddatabase that is called by sourcing programs that generate several versions (work-ing, final paper version, per-exams list, and the on-line HTML and PDF versions)from a single source. Silvio Levy's support and counseling was a major re-source backing our efforts and many thanks also to Noam Shomron for help withnon-standard typesetting.
Berkeley, California Paulo Ney de SouzaAugust 2003 desouza@math . berkeley . edu
Jorge Nuno [email protected]
Contents
Preface vii
I Problems 1
1 Real Analysis 31.1 Elementary Calculus 3
1.2 LirnitsandContinuity . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Sequences, Series, and Products 10
1.4 DifferentialCalculus . . 14
1.5 IntegralCalculus . . . . 18
1.6 SequencesofFunctions ..................... 221.7 FourierSeries. . . . . . 27
1.8 Convex Functions . . . 29
2 Multivariable Calculus 31
2.1 LimitsandContinuity ............... 31
2.2 Differential Calculus . . 32
2.3 Integral Calculus . . . . 40
3 Differential Equations 433.1 First Order Equations. . 43
3.2 Second Order Equations 47
HigherOrderEquations . . . . . . . . . . . . . .
SystemsofDifferentialEquations . . .
4 Metric Spaces4.1 . . . . . .
1.2 'I'heory .
4.3 FixedPointTheorem . . . .
S Complex Analysis5.1 ComplexNumbers . . . . .
5.2 Series and Sequences of Functions. . . . . . . . . . . . .
5.3 ConformalMappings5,4 FunctionsontheUnitDisc . . - . . . . . . . . .
5.5 GrowthConditions .
5.6 Analytic and Meromorphic Functions . . . . . . . . . .
5.7 Cauchy'sTheorem5.8 ZerosandSingularities5.9 HarinonicFunctions . . . . .
5.10 Theory . . . . . . . . . . . . . . . . . . .
5.11 IntegralsAlongtheRealAxis . . . . . . . . . . . . . . .
6 Algebra
6.1
6.2
6.36.46.56.66.76.86.9
6.106.116.12
6.13
Examples of Groups and General TheoryHomomorphisms and Subgroups. . . .
Cyclic (roups . . . * . . . . .
Normality, Quotients, and Homomorphisms. . .
I
I)irect Products . . . . . . . . . . . . . . . . . .
Free Groups, Generators, and RelationsFinite Ciroups .
Rings and Their HomomorphismsIdealsPolynomials . .
Fields and TheirExtensionsElementary Number Theory
97* • • •
917
• I S • 99
104• . . . . 106
106
107
111
112
• . . . . 118
7 Linear7.17.27.37.4
7.57.67.7
7.8
123• 123
.125
.129
.129
.134
• 138
.144
.146
3.3
3.4
4950
57576062
65
6567707174
75
8082
8687
93
Algebra1/ëctor Spaces . . . . . . . . . . . .
RankandDeterrninantsSystemsofEquations
LinearTransforrnations. ..Eigenvalues and Eigenvectors . . - . . . . . . . .
Forins . . .
Similarity . • •
Bilinear, Quadratic Forms, and Inner Product Spaces . .
7.9 GeneralTheoryofMatrices . . . . . .149
II Solutions 155
1 Real Analysis1.1 Elementary Calculus1.2 Limits and Continuity1.3 Sequences, Series, and Products1.4 DifferentialCalculus . . .
1.5 Integral Calculus1.6 Sequences of Functions .
1.7 Fourier Series1.8 Convex Functions .
2 Multivariable Calculus2.1 Limits and Continuity2.2 Differential Calculus2.3 Integral Calculus . *
3 Differential Equations3.1 First Order Equations.3.2 Second Order Equations3.3 Higher Order Equations . .
3.4 Systems of Differential Equations
4 Metric Spaces4.14.2 General Theory4.3 Fixed Point Theorem
5 Complex Analysis5.1 Complex Numbers .
5.2 Series and Sequences of Functions.5.3 Conformal Mappings5.4 Functions on the Unit Disc5.5 Growth Conditions5.6 Analytic and Meromorphic Functions5.7 Cauchy's Theorem5.8 Zeros and Singularities5.9 HarmonicFunctions5.10 Residue Theory5.11 Integrals Along the Real Axis .
6 Algebra6.1
. I •
• . . . S S S S S
157. . . 157
• • . . . . 173
. . 178
• . . . 191
• . . . . 200213228232
235235237258
265265274
• . . . . • . . 278
• . . . . . . . . . . . . . . . 280
289• • . . .289
• . . . . 297300
305305
• . . . 309316
• • 321
• • . . 329
• . • • • S 33.4
. . . . • . . 347
• . . • . . 356372
• • . S • 373390
423Examples of Groups and General Theory 423
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.116.126.13
. . . 429.433435
• • S S S S • S S • i434.Ø
. . . . 443• . 445• . 450• S • • • I I S • I 456• . . . 4.60
• . . 463. . . 473
• , 480
7 Linear7.17.27.37.4
7.57.6
7.7
7.8
7.9
Algebra\'ector Spaces . . . •
Rank andDeterminants. . . . . •
SystemsofEquations .
LinearTransforinations. . . . . . . . .
Eigenvalues and Eigenvectors • • . • . . • . •
Canonical Forms . • * . . • . . . • . . . . •
Similarity. S S S • S •
Bilinear, Quadratic Forms, and Inner Product SpacesGeneralTheoryofMatrices
489
• S • I I S • 489• 495
• S I • S S S 501• • S • • • . 503
. . 514• I S S S • • 525
540
• . S S I S 545
ifi Appendices 569
A How to Get the ExamsA.1 On—line . . . • . . . . . . • • • •
A.2 Off—line,theLastResort .
B Passing Scores 577
C The Syllabus 579
References 581
Index 589
Homomorphisms and Subgroups.CyclicGroups . • • . . . . •
Normality, Quotients, and Homomorphisms.,sIn, .
DirectProducts . . •
Free Groups, Generators, and Relations(3roups • . . . . . • . . . . . . . . .
Rings and Their Homomorphisms • . .
Ideals . . . . . • . .
Polynomials . . . . . . . • •
Fields and Their Extensions . . . . • •
Elementary Number Theory . . . . . .
571.571.571
Part I
Problems
1
Real Analysis
1.1 Elementary Calculus
Problem 1.1.1 (Fa87) Prove that (cos 9)P cos(p 0) for 0 0 andO<p<l.Problem 1.1.2 (Fa77) Let f : [0, 1] —÷ R be continuously differentiable, withf(O) = 0. Prove that
sup f(x)I (f'(x))2 dx.0
Problem 1.13 (Sp81) Let f(x) be a real valued function defined for all x 1,
satisfying f(1) = 1 and
f'(x) = x2 +f(x)2Prove that
urn f(x)X—) 00
exists and is less than 1 +
Problem 1.1.4 (Sp95) Let f, g: [0, 1] —÷ [0, co) be continuous functions satis-fying
sup f(x) = sup g(x).
Prove that there exists t E [0, 1] with f(t)2 + 3f(t) = g(t)2 + 3g(t).
Problem 1.1.5 (Fa86) For f a real valued function on the real line, define the
function by = f(x + 1) — f(x). For n 2, define recur-
sively by = 1f). Prove that f = 0 if and only if f has the formf(x) = ao(x)+ai(x)x +" where ao, ai, ... , are periodic
functions of period 1.
Problem 1.1.6 (FaOO) Suppose that f : R R is a nonconstant function such
that f(x) f(y) whenever x y. Prove that there exist a E R and c > 0 such
that f(a + x) — f(a — x) cx for all x E [0, 1].
Problem 1.1.7 (Fa81) Either prove or disprove (by a counterexample) each ofthe following statements:
1. Let f : R -÷ g : R -+ R be such that
urn g(t) = b and urn f(t) = c.
Thenurn f (g(t)) = c.
2. 1ff : IR —+ R is continuous and U is an open set in R, then f(U) is anopen set in R.
3. Letfbe of class C°° on the interval (—1, 1). Suppose that lforall n 1 and all x in the interval. Then f is real analytic; that is, it has a
convergent power series expansion in a neighborhood of each point of theinterval.
Problem 1.1.8 (Fa97) Prove thatfor all x > 0, sin x > x —
Problem 1.1.9 (Su81) Let
y(h)=l—2sin2(2irh), f(y)= 2
Just jfy the statement
f (y(h)) = 2— 4v%r IhI + 0(h2)
where
hrnsup '2'Problem 1.1.10 (Fa82) 1. Prove that there is no continuous map from the
closed interval [0, 1] onto the open interval (0, 1).
2. Find a continuous surjective map from the open interval (0, 1) onto theclosed interval [0, 1].
3. Prove that no map in Part 2 can be bijective.
Problem 1.1.11 (Sp99) Suppose that f is a twice differentiable real functionsuch that f"(x) > 0 for all x E [a, bl. Find all numbers c E [a, bJ at whichthe area between the graph y = f(x), the tangent to the graph at (c, 1(c)), andthe lines x = a, x = b, attains its minimum value.
Problem 1.1.12 (Fa94, Sp98) Find the maximum area of all triangles that canbe inscribed in an ellipse with semiaxes a and b, and describe the triangles thathave maximum area.Note: See also Problem 2.2.2.
Problem 1.1.13 (Fa93) Let f be a continuous real valued function on [0, oo).Let A be the set of real numbers a that can be expressed as a =for some sequence in [0, oo) such that = 00. Prove that if Acontains the two numbers a and b, then contains the entire interval with endpointsa andb.
Problem 1.1.14 (Su81) Show that the equation
x>0, s>0,
has, for each sufficiently small e > 0, exactly two solutions. Let x(e) be thesmaller one. Show that
1. x(e)—÷Oase--÷0+;
yetforanys > 0,
2. —÷ oo as e —÷ 0+.
Problem 1.1.15 (Sp82) Suppose that f(x) is a polynomial with real coefficients
and a is a real number with f(a) 0. Show that there exists a real polynomialg(x) such that if we define p by p(x) = f(x)g(x), we have p(a) = 1, p'(a) = 0,
and p"(a)=O.
Problem 1.1.16 (Su84) Let p(z) be a nonconstant polynomial with real coeffi-cients such that for some real number a, p(a) 0 but p'(a) = p"(a) =0. Provethat the equation p(z) = 0 has a nonreal root.
Problem 1.1.17 (Fa84, Fa97) Let f be a C2 function on the real line. Assume fis bounded with bounded second derivative. Let
A = sup If(x)I, B = sup If"(x)I.XER XER
Prove thatsup If'(x)I (
XGR
Problem 1.1.18 (Fa90) Find all pairs of integers a and b satisfying 0 <a <band a" =
Problem 1.1.19 (Sp92) For which positive numbers a and b, with a > 1, does
the equation loge x = x" have a positive solution for x?
Problem 1.1.20 (Sp84) Which number is or
Problem 1.1.21 (Sp94) For which numbers a in (1, oo) is it true that aX
for alix in (1, co)?
Problem 1.1.22 (Sp96) Show that a positive constant t can satisfy
forall x>0if and only if t <e.
Problem 1.1.23 (Su77) Suppose that f(x) is defined on [—1, 1], and that f"(x)is continuous. Show that the series
(n (f G) — f — 2f"(O))
converges.
Problem 1.1.24 (Fa96) 1ff is a C2 function on an open interval, prove that
urnf(x + h) — 2f(x) + f(x —h) = f"(x).
Problem 1.1.25 (Su85) 1. For 0 8 show that
sinG
2. By using Part 1, or by any other method, show that <1, then
urn RA Pe_RsmOdO=O.R—>oo J0
Problem 1.1.26 (Su78) Letf : R JR be continuous. Suppose that JR containsa countably infinite subset S such that
f(x)dx=0Jp
if p and q are not in S. Prove that f is identically 0.
Problem 1.1.27 (Fa89) Let the function f from [0, 1] to [0, 1] have thefollowingproperties:
• I is of class C1;
• f(O)=f(1)=O;
• f' is nonincreasing (i.e., f is concave).
Prove that the arclength of the graph off does not exceed 3.
Problem 1.1.28 (Sp93) Let f be a real valued C1 function on [0, oo) such thatthe improper integral f1°° If' (x ) Idx converges. Prove that the infinite series
f(n) converges and only if the integral f(x)dx converges.
Problem 1.1.29 (Su82) Let E be the set of all continuous real valued functionsu : [0, 1] -+ R satisfying
Iu(x)—u(y)l 1, u(0)=0.
Let E -÷ R be defined by
plço(u) = I (u(x)2 — u(x)) dx.
JO
Show that ço achieves its maximum value at some element of E.
Problem 1.1.30 (Fa87) Let S be the set of all real C1 functions f on [0, 1] suchthat f(O) = 0 and
p1
IJO
Definepl
J(f)= I f(x)dx.JO
Show that the function J is bounded on S. and compute its supremum. Is there afunction fo E S at which J attains its maximum value? If so, what is fo?
Problem 1.1.31 (Fa82, Fa96) Letfbe a real valued continuous nonnegative func-tion on [0, 1] such that
f(t)2 1 +2 f f(s) ds
fortE [0, 1].Showthatf(t) 1 +tfort E [0,1].
Problem 1.1.32 (Sp96) Suppose is a C1 function on R such that
a and ço'(x) —÷ b as x —÷ 00.
Prove or give a counterexample: b must be zero.
Problem 1.1.33 (Su77) Show that
çir/2 dxF(k)= I
Jo VU—kcos2x
0 k < 1, is an increasing function of k.
Problem 1.1.34 (Fa79) Given that
1' 00
Jdx =
-00
find f'(t) explicitly, where
f(t) = t > 0.
Problem 1.1.35 (Fa80) Define
tCOS X
F(x)= IJsinx
Compute F'(O).
Problem 1.1.36 (Fa95) Let f : R -÷ R be a nonzero C00 function such that
f(x)f(y) = j (vfx2 + y2) for all x andy and that f(x) —÷ 0 as IxI -÷ 00.
1. Prove that f is an even function and thatf(0) is 1.
2. Prove that f satisfies the differential equation f'(x) = f"(O)xf(x), andfind the most general function satisfying the given conditions.
Problem 1.1.37 (FaOl) Let S be the set of continuous real-valued functions on[0, 1] such that f(x) is rational whenever x is rational. Prove that S is uncount-able.
1.2 Limits and Continuity
Problem 1.2.1 (Sp02) Let f be a bounded, continuous, real-valued function on1R2. Define the function g on JR by
g(x) £ dt.
Prove that g is continuous.
Problem 1.2.2 (Fa90) Suppose that f maps the compact interval I into itself andthat
Jf(x)—f(y)t
for all x, y 1, x y. Can one conclude that there is some constant M < Isuch that for allx, y I,
$f(x)—f(y)I
Problem 1.2.3 (Sp90) Let the real valuedfunction f on [0, 1] have the followingtwo properties:
• If[a, b] C [0, 1], then f ([a, b]) contains the interval with endpoints f(a)and f(b) (i.e., f has the Intermediate Value Property).
• For each c E R, the set f1 is closed.
Prove that f is continuous.
Problem 1.2.4 (FaOO) Letf : R -÷ IR be uniformly continuous with f(O) = 0.Prove: there exists a positive number B such that If(x)I1 + B!xI, for all x.
Problem 1.2.5 (Sp83, SpOl) Suppose that f is a continuous function on JR whichis periodic with period 1, i.e., f(x + 1) = f(x). Show:
1. The function f is bounded above and below and achieves its maximum andminimum.
2. The function f is uniformly continuous on JR.
3. There exists a real number xo such that
f(xo+ir)= f(xo).
Problem 1.2.6 (Sp77) Let h : [0, 1) -+ JR be a function defined on the half-open interval [0, 1). Prove that if h is uniformly continuous, there exists a uniquecontinuous function g: [0, 1] —÷ JR such that g (x) = h (x) for all x E [0, 1).
Problem 1.2.7 (Fa99) Let f be a continuous real valued function on [0, oo) suchthat tim f(x) exists (finitely). Prove that f is uniformly continuous.
x-+oo
Problem 1.2.8 (Sp84) Prove or supply a counterexample. If the function f fromJR to JR has both a left limit and a right limit at each point of JR. then the set of
discontinuities off is, at most, countable.
Problem 1.2.9 (Fa78) Let f : JR —÷ JR satisfy f(x) f(y) for x y. Prove
that the set where f is not continuous is finite or countably infinite.
Problem 1.2.10 (Su85, Fa96) A function f : [0, 11 -+ R is said to be uppersemicontinuous given x E [0, 1] and s > 0, there exists a 8 > 0 such thatif — xI < 8, then f(y) < f(x) + e. Prove that an upper semicontinuouSfunction f on [0, 1] is bounded above and attains its maximum value at somepoint p E [0, 11.
Problem 1.2.11 (Su83) Prove that a continuous function from JR to JR whichmaps open sets to open sets must be monotonic.
Problem 1.2.12 (Fa91) Let f be a continuous function from JR to JR such thatJf(x) — x — alix andy. Prove that the range off is all of JR.Note: See also Problem 2.1.8.
Problem 1.2.13 (Fa81) Let f be a continuous function on [0, 1]. Evaluate thefollowing limits.
1.
ciurn I x'2f(x)dx.
n 1 x?Zf(x)dx.n.-+oo J0
Problem 1.2.14 (FaSS, Sp97) Let f be a function from [0, 11 into itself whosegraph
G1={(xf(x)) xE[0,1])is a closed subset of the unit square. Prove that f is continuous.Note: See also Problem 2.1.2.
Problem 1.2.15 (Sp89) Let f be a continuous real valued function defined on[0, 1] x [0, 1]. Let the function g on [0, 1] be defined by
g(x) =max(f(x,y) I yE [0,1]).Prove that g is continuous.
Problem 1.2.16 (FaOl) Let the function f : JR JR be bounded on boundedsets and have the properly that (K) is closed whenever K is compact. Provef is continuous.
1.3 Sequences, Series, and Products
Problem 13.1 (Su85) Let A1 ?' A2 ... ? 0. Evaluate
urnn-+oo
Note: See also Problem 5.1.11.
Problem 13.2 (Sp96) Compute
L=liml—n-+oo
Problem 133 (Sp92) Let xo = 1 and
Prove that = urn exists, and find its value.n-+oo
Problem 13.4 (Fa97) Define a sequence of real nunthers by
1XO = 1, Xn+t = for n 0.
Show that (Xv) converges, and evaluate its limit.
Problem 1.33 (Fa89, Sp94) Let a be a number in (0, 1). Prove that any se-quence of real numbers satisfying the recurrence relation
Xn+I =
a limit, and find an expression for the limit in terms of a, xo andxj.
Problem 1.3.6 (Fa92) Let k be a positive Determine those real numberscfor which every sequence of real numbers satisfying the recurrence relation
1 + i) =
has period k (i.e., Xfl+k = for all n).
Problem 13.7 (Sp84) Let a be a positive real Define a sequence (xc) by
Find a necessary and sufficient condition on a in order that afinite limit urnshould exist.
Problem 1.3.8 (Sp03) Let be a sequence of real numbers so that
urn (2Xn+1 — = x.n—,.oo
Show that urn = X.
Problem 13.9 (SpOO) Let a andxo be positive numbers, and define the sequencerecursively by
if axn=—Ixn_1+
Prove that this sequence converges, and find its limit.
Problem 1.3.10 (Fa95) Let be a real numbe, 0 <xi < 1, and define a se-quence by Xn+i = — Show that > 0.
Problem 1.3.11 (Fa80) Let f(x) — + x — x2. For any real number x, definea sequence by x and = f(Xn). If the sequence converges, let
x the sequence is bounded and nondecreasing and findX00 = A.
2. Find all y R such that = A.
Problem 1.3.12 (Fa81) The Fibonacci nwnbers f', f2, . .. are defined recursivelybyfi=1,
limn—÷oo
exists, and evaluate the limit.Note: See also Problem 7.5.20.
Problem 13.13 (Fa79) Prove that
urn ( +n-+oo\n+1 n+2 2nj
Problem 13.14 (Sp02) For n a positive integei let denote the partial sumof the harmonic series
Let k> 1 be an integei Prove that
<logk (n=1,2,...),
where log k is the natural logarithm of k, and C is a constant.
Problem 1.3.15 (Sp90) Suppose xi, X2, X3, ... is a sequence of nonnegative realnumbers satisfying
Xn+1n
n 1. Prove that urn exists.fl—300
Problem 1.3.16 (Sp93) Let (ar) and (8,i) be sequences of positive numbers. As-sume that = 0 and that there is a number k in (0, 1) such thatan+1 + for every n. Prove that urn = 0.
n-+oo
Problem 1.3.17 (Fa83) Prove or disprove (by giving a counterexample), the fol-lowing assertion: Every infinite sequence xJ , ... of real numbers has either anondecreasing subsequence or a nonincreasing subsequence.
Problem 13.18 (Su83) Let bi b2, .. be positive real numbers with
urn = 00 and urn = 1.n-+oo
Assume also that b1 <b2 <b3 <••-. Show that the set of quotientsis dense in (1, oo).
Problem 1.3.19 (Sp81) Which of the following series converges?
1.
(2n)!(3n)!L_ n!(4n)!
2.
00
n=I
Problem 13.20 (Fa91) Let ai, ... be positive numbers.
I. Prove that <00 implies <00.
2. Prove that the converse of the above statement is false.
Problem 13.21 (Su80, Sp97) For each (a, b, c) E consider the series
n=3
Determine the values of (a, b, c) for which the series
1. converges absolutely;
2. converges but not absolutely;
3. diverges.
Problem 13.22 (Sp91) For which real numbers x does the infinite series
n=1
converge?
Problem 13.23 (Fa94) For which values of the real number a does the series
Co (1
converge?
Problem 1.3.24 (Sp91) Let A be the set of positive integers that do not containthe digit 9 in their decimal expansions. Prove that
—<00;aEA
that is, A defines a convergent subseries of the harmonic series.
Problem 1.3.25 (Sp89) Let a i, a2, ... be positive numbers such that
<00.
Prove that there are positive numbers ... such that
tim =00 and <00.n—+oo
n=1
Problem 13.26 (Fa90) Evaluate the limit
ir irlim cos cos • cos —.
n—+oo 2L
1.4 Differential Calculus
Problem 1.4.1 (Su83) Outline a proof, starting from basic properties of the realnumbers, of the following theorem: Let f : [a, b] -+ IR be a continuous functionsuch that f'(x) = Oforallx E (a, b). Then f(b) = f(a).
Problem 1.4.2 (Sp84) Let f(x) = x log(1 + x'), 0 < x <00.
1. Show that f is strictly monozonically increasing.
2. Compute urn f(x) as x -+ 0 and x -+ 00.
Problem 1.4.3 (Sp85) Let f(x), 0 x <00, be continuous, differentiable, withf(0) = 0, and that f'(x) is an increasing function ofxforx ? 0. Prove that
{ f(x)/x, x>0gx f'(O), x=0is an increasing function of x.
Problem 1.4.4 (Su79, Fa97) I. Give an example of a differentiable functionf : JR —+ JR whose derivative f' is not continuous.
2. Let f be as in Part 1. If f'(O) <2 < f'(I), prove that f'(x) = 2 for someXE[O,1].
Problem 1.4.5 (Sp90) Let y : JR —+ JR be a C°° function that satisfies the dWer-ential equation
y"+y'—y=Ofor x [0, L], where L is a positive real Suppose that y(O) y(L) =0.Prove that y 0 on [0, L].
Problem 1.4.6 (Su85) Let u(x), 0 x I, be a real valued C2 function whichsatisfies the differential equation
u"(x) = eXu(x).
1. Show that (f 0 < 1, then u cannot have a positive local maximum atSimilarly, show that u cannot have a negative local minimum at xo.
2. Now suppose that u(0) = u(1) = 0. Prove that u(x) 0, 0 x 1.
Problem 1.4.7 (Sp98) Let K be a real constant. Suppose that y(t) is a positivedifferentiable function 'satisfying y'(t) Ky(t) for t 0. Prove thaty(t) fort ) 0.
Problem 1.4.8 (Sp77, Su82) Suppose f is a the re-als into the reals. Suppose f'(x) > f(x) for all x E JR. and f(xo) = 0. Provethatf(x) > Oforallx > xO.
Problem 1.4.9 (Sp99) Suppose that f is a twice dWerentiable real-valued func-tion on JR such that f(0) =0, f'(O) > 0, and f"(x) ? f(x)forallx 0. Provethatf(x) > Oforallx > 0.
Problem 1.4.10 (Sp87) Show that the equation aeX = 1 + x + x2/2, where a isa positive constant, has exactly one real root.
Problem 1.4.11 (Su78, Fa89) Suppose f : [0, 1] -÷ JR is continuous withf(0) = 0, and for 0 < x < 1, f is differentiable and 0 f'(x) 2f(x).Prove thatf is identically 0.
Problem 1.4.12 (Sp84) Let f : [0, 1] —÷ JR be continuous function, withf(0) = f(1) =0. Assume that f" exists on 0< x < 1, with f/I + 2f' + f ? 0.Showthatf(x) OforallO x ( I.
Problem 1.4.13 (Sp85) Let vi and V2 be two real valued continuous functions on
JR such that (x) < v2(x) for all x JR. Let ç'i (t) and co2(t) be, respectively,
solutions of the equations
dx dxand
fora <t <b. If coi(to) = co2(to) for some tO E (a,b), showthatcoi(t)for all t (ta, b).
Problem 1.4.14 (Fa83, Fa84) Prove or supply a counterexample: 1ff and g are
C1 real valued functions on (0, 1), if
urn f(x) = urn g(x) =0,x—)O x—O
ifg and g' never vanish, and if
• f(x)lim = c,x-÷O g(x)
then• f'(x)
urn =c.x-+O g'(x)
Problem 1.4.15 (Fa00) Let f be a real-valued differentiablefunction on (—1, 1)such that f(x)/x2 has a finite limit as x —÷ 0. Does it follow that f"(O) exists?Give a proof or a counterexample.
Problem 1.4.16 (Sp90, Fa91) Let f be an infinitely differentiable functionfromR to IL Suppose that, for some positive integer n,
f(1) = f(0) = f'(O) = f"(O) = ..• = = 0.
Prove that = Oforsomex in (0, 1).
Problem 1.4.17 (SpOO) Let f be a positive function of class C2 on (0, oo) suchthat f' 0 and 1" is bounded. Prove that urn f'(t) = 0.
t—+ 00
Problem 1.4.18 (S p86) Let f be a positive differentiable function on (0, oo).Prove that
f(x) )exists (finitely) and is nonzero for each x.
Problem 1.4.19 (Sp88) Suppose that f(x), —00 <x <00, is a continuous realvalued function, that f'(x) exists forx 0, and that urn f'(x) exists. Prove that
f'(O) exists.
Problem 1.4.20 (Sp88) For each real value of the parameter t, determine thenumber of real roots, counting multiplicities, of the cubic polynomialp,(x) = (1 + t2)x3 — 3t3x + t4.
Problem 1.4.21 (Sp91) Let the real valued function f be defined in an open in-
terval about the point a on the real line and be differentiable at a. Prove that if
(Xe) is an increasing sequence and is a decreasing sequence in the domainoff, and both sequences converge to a, then
urn—
= f'(a).
f be a continuous real valued function on [0, 1] suchrhat,foreachxo E [0,1),
f(x)—f(xo)tim sup 0.x_+xo+ x —
Prove that f is nondecreasing.
Problem 1.4.23 (Sp84) Let I be an open interval in IR containing zero. Assumethat f' exists on a neighborhood of zero and f"(O) exists. Show that
f(x) = f(0) + f'(O)sinx + + o(x2)
(o(x2) denotes a quantity such that0(x)
—÷ 0 as x —÷ 0).
Problem 1.4.24 (Sp84) Prove that the Taylor coefficients at the origin of thefunction
f(z)=
are rational numbers.
Problem 1.4.25 (Sp79) Give an example of a function f : IR -÷ IR having all
three of the following properties:
• f(x) = Oforx <0 andx > 2,
• f'(l) = 1,
• fhas derivatives of all orders.
Problem 1.4.26 (Sp99) Prove that if n is a positive integer and s are realnumbers with s > 0, then there is a real function f with derivatives of all orders
such that
1. efork = 0, 1, . . . , n — 1 and alix E IR,
2. f(k)(0)OforkO,1,...,n_1,
3.
Problem 1.4.27 (Su83) Let f : IR —* ]R be continuously differentiable, periodicof period 1, and nonnegative. Show that
dl f(x) \—I i-÷0 (asc—÷oO)dx \1+cf(x)J
unjformly in x.
Problem 1.4.28 (Su81) Let I c IR be the open interval from 0 to 1.
Let f: I —÷ C be C1 (i.e., the real and imaginary parts are continuouslyferentiable). Suppose that f(t) -÷ 0, f'(t) —÷ C 0 as t -+ 0+. Show thatthe function g(t) = If(t)t is C1 for sufficiently small t > 0 and that lirng'(t)exists, and evaluate the limit.
Problem 1.4.29 (Fa95) Let f : IR -÷ JR be a C°° function. Assume that 1(x)has a local minimum at x =0. Prove there is a disc centered on the y axis whichlies above the graph off and touches the graph at (0, f(0)).
1.5 Integral Calculus
Problem 1.5.1 (Fa98) Let f be a real function on [a, b}. Assume that f is differ-entiable and that f' is Riemann integrable. Prove that
b
f f'(x)dx = f(b) —
Problem 1.5.2 (Sp98) Using the properties of the Rienzann integral, show thatif f is a non-negative continuous function on [0, 11, and 101 f(x)dx = 0, thenf(x) = Oforallx E [0, 1].
Problem 1.5.3 (Fa90) Suppose f is a continuous real valued function. Show that
,1 II
JO 3
for E [0, 1].
Problem 1.5.4 (Sp77) Suppose that f is a real valued function ofone real vari-able such that
urn f(x)X-9'C
existsfor all c [a, b]. Show that f is Riemann integrable on [a, b].
Problem 1.5.5 (Sp78) Let f: [0, 1] -+ JR be Riemann integrable over [b, 1]forall b such that 0 <b 1.
1. 1ff is bounded, prove thatf is Riemann integrable over [0, 1].
2. What iff is not bounded?
Problem 1.5.6 (Su81) Let f : R —÷ R be continuous, with
coo
I lf(x)Idx<oo.J—oo
Show that there is a sequence (xv) such that -÷ 00, -÷ 0, andf (—xv) —+ 0 as n —+ oo.
Problem 1.5.7 (Su85) Let
f(x) = ex2/2f°°e_2/2dt
forx > 0.
1. Showthat0 < f(x) <L
2. Show that f(x) is strictly decreasing for x > 0.
Problem 1.5.8 (Su84) Let ço(s) be a C2 function on [1,2] with ço and ci! vanish-ing at s = 1, 2. Prove that there is a constant C > 0 such that for any A> 1,
Problem 1.5.9 (Fa85) Let 0 a 1 be given. Determine all nonnegative con-tinuous functions f on [0, 1] which satisfy the following three conditions:
'I'I f(x)dx=1,
Jo
I"I xf(x)dx=a,
Jo
j x2f(x)dx=a2.Jo
Problem 13.10 (Fa85, Sp90) Let f be a differentiablefunction on [0, 11 and let
sup If'(x)I = M <00.O<x<1
Let n be a positive Prove that
f(j/n)— f
Problem 15.11 (Fa83) Let f: [0, oo) —÷ R be a uniformly continuous functionwith the properly that
urn I f(x)dxb-÷ooJO
exists (as a finite limit). Show that
urn f(x) = 0.x-+00
Problem 1.5.12 (Fa86) Let f be a real valued continuous function on [0, oo)such that
(1(x) + f(t)dt)
exists. Prove thaturn f(x) =0.
00
Problem 1.5.13 (Sp83) Let f R+ R+ be a monotone decreasing function,defined on the positive real numbers with
f00I f(x)dx<oo.
Jo
Show thaturn xf(x) = 0.
X—+ 00
Problem 15.14 (Fa90, Sp97) Let f be a continuous real valued function satis-fying f(x) 0, for all and
f00f(x)dx<oo.
JO
Prove that
— xf(x)dx 0nJoasn —÷ 00.
Problem 1.5.15 (Sp87) Evaluate the integral
fSIflXJo x
to an accuracy of two decimal places; that is, find a number J* such thatIl—Il <0.005.
Problem 1.5.16 (Fa87) Show that the following limit exists and is finite:
If' dxlimii\Jo (x4 + t4)
Problem 1.5.17 (Fa95) Let f and f' be continuous on [0, oo) and f(x) = Oforx 1010. Show that
Problem 1.5.20 (Fa99) Let f and g be continuous real valued functions on Rsuch that f(x) = 0 and Ig(x)fdx <oo. Define the function h on
Problem 1.5.21 (Sp88) Prove that the integrals
converge.
too
Jo
Problem 13.22 (Fa85) Let f(x), 0 x 1, be a real valued continuous func-tion. Show that
,1lim(n+1) I
n-*oo Jo
Problem 1.523 (Su83, Sp84, Fa89) Compute
where a > 0 is a constant.
fOO logxI dx
Jo x2+a2
x2f (x)2dxro
f'(x)2dxjf(x)2dx
Problem 15.18 (Fa88) Let f be a continuous, strictly increasing function from[0, oo) onto [0, oo) and let g = f'. Prove that
I f(x)dx+Jo Jo
for all positive numbers a and b, and determine the condition for equality.
Problem 1.5.19 (Sp94) Letfbe a continuous real valued function on IR such thatthe improper Riemann integral If(x)I dx converges. Define the function gon by
'Cog(y)= I f(x)cos(xy)dx.
J—oo
P rove that g is continuous.
'Coh(x)= I
J_oo
Prove that urn h(x) = 0.IxI-÷oo
f(x — y)g(y)dy.
Co
0cosx2 dx and sinx2 dx
Problem 1.5.24 (Sp85) Show that
fir1= 1 log(sinx)dx
Jo
converges as an improper Riemann integral. Evaluate 1.
f°°sinxProblem 1.5.25 (Sp02) Prove that dx converges as an improper Rie-
Jo ,tIx(°° Isinxl
mann integral, but that I dx = 00.Jo,fi
1.6 Sequences of Functions
Problem 1.6.1 (Fa84) Prove or supply a counterexample: If f is a nondecreas-ing real valuedfrnction on [0, 1], then there is a sequence of continuous functionson [0, 11, {f,j, such thatfor each x E [0, 11,
urnfl—) 00
Problem 1.6.2 (Fa77, Sp80) Let : IR -* R be differentiable for eachn = 1,2,... with lforalln andx. Assume
urnn-+o0
for all x. Prove that g : R -÷ IR is continuous.
Problem 1.63 (Fa87) Suppose that (f,j is a sequence of nondecreasing func-tions which map the unit interval into itself Suppose that
litu
f is a continuous function. Prove that (x) —÷ f(x) uni-formly as n —k 00, 0 ( x 1. Note that the functions are not necessarilycontinuous.
Problem 1.6.4 (Fa85) Let f and n = 1,2,..., be functions from R to R.Assume that f (x) as n —+ 00 whenever -÷ x. Show that f iscontinuous. Note: The functions are not assumed to be continuous.
Problem 1.6.5 (Sp99) Suppose that a sequence offunctions : R -÷ R con-verges uniformly on IR to a function f : R -÷ R, and that = lim f,, (x) exists
X-+ 00for each positive integer n. Prove that urn and urn f(x) both exist and are
x—+00equal.
Problem 1.6.6 (Sp81) 1. Give an example of a sequence of C' functions
fk:[O,00)—÷IR, k=O,1,2,...
such that fk(O) = Ofor all it; and fk'(x) -+ f01(x)for all x as k -÷ 00, butfk(x) does not converge to fo(x)forallx ask —÷ 00.
2. State an extra condition which would imply that fk(x) -÷ fo(x)for all xask —+ 00.
Problem 1.6.7 (Fa84) Show that if f is a homeomorphism of [0, 1] onto itselfthen there is a sequence (pn), n = 1, 2, 3,... of polynomials such that Pn f
on [0, 1] and each is a homeomorphism of[0, 1] onto itself
Problem 1.6.8 (Sp95) Let [0, 1] [0, oo) be a continuous function, forn = 1,2,.... Supposethat one has
(*) )f3(x) for alix E [0,1].
Let f(x) = tim (x) and M = sup f(x).n-+oo
1. Prove that there exists t [0, 1] with f(t) = M.
2. Show by example that the conclusion of Part 1 need not hold if instead of(*) we merely know thatfor each x E [0, 1] there exists such thatfor alln one has
Problem 1.6.9 (Fa82) Let fi, ... be continuousfunctions on [0, 1] satisfying
Ii f2 and such that (x) = Ofor each x. Must the sequenceconverge to 0 uniformly on [0, 1]?
Problem 1.6.10 (Sp78) Let k 0 be an integer and define a sequence of maps
2n=1,2,....x+n
For which values of k does the sequence converge unifonnly on R? On everybounded subset of R?
Problem 1.6.11 (Sp81) Let f : [0, 1] —* R be continuous and k E N. Provethat there is a real polynomial P(x) of degree k which minimizes (for all such
polynomials)sup — P(x)J.
Problem 1.6.12 (Fa79, Fa80) Let be a sequence of real polynomials of de-
gree D, afixed Suppose that 0 pointwise for 0 x ( 1.
Prove that 0 uniformly on [0, 1].
Problem 1.6.13 (Su85) Let f be a real valued continuousfunction on a com-pact interval [a, b] Given 0, show that there is a polynomial p such thatp(a) = f(a), p'(a) 0, and Ip(x) — f(x)I <eforx [a, b].
Problem 1.6.14 (Sp95) For each positive integer n, define : R —÷ R by
(x) = cos nx. Prove that the sequence offunctions {f,j has no uniformly con-
vergent subsequence.
Problem 1.6.15 (Fa03) Let C(O,l] denote the space of continuous functions on[0, 1]. Define
11 If(x)—g(x)Id(f,g)
= Jo 1 + If(x) —g(x)Idx.
1. Show that d is a metric on
2. Show that (C[o,lJ, d) is not a complete metric space.
Problem 1.6.16 (Fa86) The Arzelà—Ascoli Theorem asserts that the sequence {f,jof continuous real valued functions on a metric space isprecompact (i.e., hasa uniformly convergent subsequence) if
(i) is compact,
(ii) sup <oo(where = X E
(iii) the sequence is equicontinuous.
Give examples of sequences which are not precompact such that: (i) and (ii) holdbut (iii) fails; (i) and (iii) hold but (ii) fails; (ii) and (iii) hold but (i) fails. Take
to be a subset of the real line. Sketch the graph of a typical member of thesequence in each case.
Problem 1.6.17 (SpOl) Let the functions : [0, 1] [0, 11 (n = 1, 2,...)satisfy Ifn(x) — Ix—yl whenever — yj 1/n. Prove that the sequence
has a uniformly convergent subsequence.
Problem 1.6.18 (Fa92) Let be a sequence of real valued C1 functions on[0, 1] such that, for all n,
1
IJo
Prove that the sequence has a subsequence that converges uniformly on [0, 1].
Problem 1.6.19 (Fa96) Let M be the set of real valued continuous functions fon [0, 11 such that f' is continuous on [0, 1], with the norm
hf II = sup If(x)I + sup lf'(x)I.
Which subsets of M are compact?
Problem 1.6.20 (Su80) Let (an) be a sequence of nonzero real numbers. Provethat the sequence offunctions : JR —÷ JR
I(x) = — + cos(x + an)
has a subsequence converging to a continuous function.
Problem 1.6.21 (Sp82) be a sequence ofcontinuousfunctionsfrom [0, 1]to JR. Suppose that (x) -÷ 0 as n —÷ 00 for each x E [0, 1] and also that, forsome constant K, we have
j11
I fn(X)dXJo
for all n. Doesfl
urn Ifl-+OOJo
Problem 1.6.22 (Sp82, Sp93) Let (ga) be a sequence of twice differentiablefunc-tions on [0, 1] such that = = 0 for all n. Suppose also that
1 for all n and all x E [0, 1]. Prove that there is a subsequence of(ga) which converges uniformly on [0, 1].
Problem 1.6.23 (Fa93) Let K be a continuous real valued function defined on[0, 1] x [0, 1}. Let F be the family offunctions f on [0, 1] of the form
plf(x)= I g(y)K(x,y)dy
Jo
with g a real valued continuous function on [0, 1] satisfying 1 everywhere.
Prove that the family F is equicontinuous.
Problem 1.6.24 (Fa78) Let (ga) be a sequence of Riemann integrable functionsfrom [0, 1] into JR such that 1 for all n, x. Define
IJo
Prove that a subsequence of converges uniformly.
Problem 1.6.25 (Su79) Let be a sequence of continuous real functions de-fined [0, 1] such that
pII dy 5
Jo
for all n Define : [0, 1] -÷ JR by
i'I
_____
gn(x) = f -./i+Jo
1. Finda constant K 0 such that Kforalln.
2. Prove that a subsequence of the sequence tEn) converges uniformly.
Problem 1.6.26 (Su81) Let be a sequence of continuous functions definedfrom [0, 1] -+ such that
p1
I as n,m -+Jo
Let K : [0, 1] x [0, 11 -+ R be continuous. Define : [0, 1] —÷ by
gn(x) = I K(x,Jo
Prove that the sequence } converges uniformly.
Problem 1.6.27 (Fa82) Let ... , be nonnegative continuous func-tions on [0, 1] such that the limit
I"urn I xkcon(x)dx
n-+00Jo
exists for every k 0, 1, .... Show that the limit
Jim In-+00
exists for every continuousfunctionfon [0, 11.
Problem 1.6.28 (Sp83) Let A1, A2, .. ., As,... be real numbers. Show that theinfinite series
00e
converges unzfonnly over R to a continuous limit function f : R -÷ C. Show,furthei that the limit
1urn — f(x)dx
T—)OO 2T J--Texists.
Problem 1.6.29 (Sp85) Define the function by
Prove that is defined and has continuous derivatives of all orders in theinterval I <x <00.
Prove that converges uniformly to a limit on every finite interval [a, b].
Problem 1.631 (Sp87) Let f be a continuous real valued functionon R satisfy-ing
If(x)I< C
2'1+xwhere C is a positive constant. Define the function F on R by
F(x)=>2f(x+n).
1. Prove that F is continuous and periodic with period 1.
2. Prove that if G is continuous and periodic with period 1, then
(I tooI F(x)G(x)dx f(x)G(x)dx.
JO J—oo
Problem 1.632 (Sp79) Show that for any continuous function f: [0, 1] -÷ JR
and s > 0, there is a function of the form
g(x)=
forsomen EQ and Ig(x)—f(x)I <eforallxin[0,1].
1.7 Fourier Series
Problem 1.7.1 (Sp80) Let f : JR -÷ JR be the unique function such that
1. Prove that the Fourier series off is
n+1
nn=I
2. Prove that the series does not converge uniformly.
3. For each x JR. find the sum of the series.
Problem 1.7.2 (Su81) Let f : ]R -÷ IR be the function of period 2ir such that
f(x) = x3for—ir x
1. Prove that the Fourier series for f has the form sin nx and write an
integralformula for (do not evaluate it). 1
2. Prove that the Fourier series converges for all x.
3. Prove2jr6
Problem 1.7.3 (Su82) Let f : [0, n] -÷ IR be continuous and such that
p,rI f(x)sinnxdx=0
Jo
for all integers n 1. Is f identically 0?
Problem 1.7.4 (Sp86) Letf be a continuous real valued function on IR such that
for all x. Prove thatf is constant.
Problem 1.7.5 (Sp88) Does there exist a continuous real valued function f(x),0 1, suchthat
(I fII xf(x)dx = 1 and I x'1f(x)dx = 0
Jo Jo
forn = 0,2,3,4 2 Give an example ora proof that no such f exists.
Problem 1.7.6 (Fa80) Let g be 2ir-periodic, continuous on [—sr, it] and haveFourier series
ao
n=1
Let f be 2ir -periodic and satisfy the djfferential equation
f"(x) + kf(x) g(x)
where k n2, n = 1,2,3,.... Find the Fourier series off and prove that itconverges everywhere.
Problem 1.7.7 (Su83) Let f be a twice differentiable real valued function on[0, 2jr}, with f(x)dx = 0 = f(2ir) — f(0). Show that
p2ir
J (f(x))2 dx I (f'(x))2 dx.0 JO
Problem 1.7.8 (Fa81) Let f and g be continuous functions on R such thatf(x + 1) = f(x), g(x +1) =g(x),forallx ER. Provethat
iirnf f(x)g(nx)dx= f f(x)dx f g(x)dx.
1.8 Convex Functions
Problem 1.8.1 (Sp81) Let f: [0, 1] -÷ R be continuous with f(O) = 0. Showthere is a continuous concave function g : [0, 1] —÷ R such that g (0) = 0 andg(x) f(x)forallx E [0, 1].
Note: A function g : I —÷ R is concave
g(tx+(1 —t)y) —t)g(y)
1.
Problem 1.8.2 (Sp82) Let f: I -÷ R (where I is an interval of R) be such thatf (x) > 0, x E I. Suppose that f (x) is convex in I for every real number c.Show that log f(x) is convex in 1.Note: A function g : I -+ R is convex if
g (tx + (1 — t)y) tg(x) + (1 — t)g(y)
for 1.
Problem 1.83 (Sp86) Let f be a real valued continuous function on R satisfyingthe mean value inequality below:
1
f(y)dy, xER, h>0.2h Jx—h
Prove:
1. The maximum of f on any closed interval is assumed at one of the end-points.
2. f is convex.
2
Multivariable Calculus
2.1 Limits and Continuity
Problem 2.1.1 (Fa94) Let the function f : -+ satisfy the following twoconditions:
(i) 1(K) is compact whenever K is a compact subset of 1W2.
(ii) If (Ku) is a decreasing sequence of compact subsets of R'2, then
fProve that is continuous only if its
graph is closed in 1W2 x Is the converse true?Note: See also Problem 1.2.14.
Problem 2.1.3 (Su79) Let U C W2 be an open set. Suppose that the maph : U -÷ 1W2 is a homeomorphism from U onto which is uniformly con-tinuous. Prove U =
Problem 2.1.4 (Sp89) Let f be a real valued function on R2 with the followingproperties:
1. For each in R, the function x i—+ f(x, Yo) is Continuous.
2. For each xo in R, the function y i-÷ f(xo, y) is continuous.
3. f(K) is compact whenever K is a compact subset ofR2.
Prove that f is continuous.
Problem 2.13 (Sp91) Let f be a continuous function from the ball= (x E R" I
II x < 1) into itself. (Here, denotes the Euclidean norm.)
Assume IIf(x)II < lix for all nonzero x E Let xo be a nonzero point ofand define the sequence (xk) by setting Xk = f(xk_1). Prove that tim xk =0.
Problem 2.1.6 (Su78) Let N be a norm on the vector space Rlt; that is,N : R'1 —÷ R satisfies
N(x) 0 and N(x) = 0 only if x =0,N(x)+N(y),
N(Ax) = IMN(x)
for allx,y€IR'1 andAElR.
1. Prove that N is bounded on the unit sphere.
2. Prove that N is continuous.
3. Prove that there exist constants A > 0 and B > 0, such that for allx E R'7, AflxlI N(x) BIlxII.
Problem 2.1.7 (Fa97) A map f : Rm -÷ R'1 is proper if it is continuous andf'(B) is compactfor each compact subset B ofR'3; f is closed if it is continuousand f(A) is closed for each closed subset A of R"1.
1. Prove that every proper map f : —÷ R'1 is closed.
2. Prove that every one-to-one closed map f : is proper.
Problem 2.1.8 (Sp83) Suppose that F : —÷ 1W1 is continuous and satisfies
IF(x) — F(y)Il —
for all x, y E 1W1 and some A> 0. Prove that F is one-to-one, onto, and has acontinuous inverse.Note: See also Problem 1.2.12.
2.2 Differential Calculus
Problem 2.2.1 (Sp93) Prove thatX y eX+Y2 for x 0, y 0.
Problem 2.2.2 (Fa98) Find the minimal value of the areas of hexagons circum-scribing the unit circle in 1R2.Note.- See also Problem 1.1.12.
Problem 2.2.3 (Sp03) Define f : 1R2 by f(x, 0) 0 and
I x2\f(x,y)=
yl,
for
1. Show that f is continuous at (0, 0).
2. Calculate all the directional derivatives off at (0, 0).
3. Show that f is not differentiable at (0,0).
Problem 2.2.4 (Fa86) Letf : -÷ R be defined by:
f(x, ) = f x4/3 sin(y/x) if x 0if x=O.
Determine all points at which f is differentiable.
Problem 2.2.5 (SpOO) Let F, with components F1, ..., be a differentiablemap of 1W1 into 1W1 such that F(0) = 0. Assume that
j.k=1ax,,
Prove that there is a ball B in IR'1 with center 0 such that F(B) C B.
Problem 2.2.6 (Fa02) Let p be a polynomial over R of positive degree. Definethe function f : R2 -+ R2 by f(x, y) = (p(x + y), p(x — y)). Prove that thederivative Df(x, y) is invertible for an open dense set of points (x, y) in 1R2.
Problem 2.2.7 (Fa02) Find the most general continuously differentiable functiong : -+ (0, oo) such that the function h(x, y) = g(x)g(y) on R2 is constant oneach circle with center (0, 0).
Problem 2.2.8 (Sp80, Fa92) Let f —÷ 1W1 be continuously differentiable.Assume the Jacobian matrix has rank n everywhere. Suppose f isproper; that is, f'(K) is compact whenever K is compact. Prove f(1W1) = 1W1.
Problem 2.2.9 (Sp89) Suppose f is a continuously differentiable map of 1R2 intoAssume that f has only finitely many singular points, and that for each posi-
tive number M, the set (z E R2 If(z)I M} is bounded. Prove that f mapsonto
Problem 2.2.10 (Fa81) Let f be a real valued function on of class C2. Apointx E W1 is a critical point offlf all the partial den vatives off vanish atx; acritical point is nondegenerate if the n x n matrix
is nonsingulaiLet x be a nondegenerate critical point off Prove that there is an open neigh-
borhood of x which contains no other critical points (i.e., the nondegenerate crit-ical points are isolated).
Problem 2.2.11 (Su80) Let f: RPZ -÷ R be afunction whose partial derivativesof order 2 are everywhere defined and continuous.
1. Let a E be a cniti cal point off (i.e.., = 0, i = 1, . .. , n). Prove
that a is a local minimum provided the Hessian matrix
is positive definite at x = a.
2. Assume the Hessian matrix is positive definite at all x. Prove thatf has, atmost, one critical point.
Problem 2.2.12 (Fa88) Pmve that a real valued C3 function f on R2 whoseLaplacian,
a2fax2
+ a2'is everywhere positive cannot have a local maximum.
Problem 2.2.13 (FaOl) Let the function u on R2 be harmonic, not identically 0,and homogeneous of degree d, where d> 0. (The homogeneity condition meansthat u(tx, ty) = y)for t > 0.) Prove that d is an
Problem 2.2.14 (Su82) Let f : R3 -+ R2 and assume that 0 is a regular valueoff (i.e., the differential of f has rank 2 at each point of f'(0)). Prove thatR3 \ (0) is arcwise connected.
Problem 2.2.15 (Sp87) Let the transformation T from the subsetU = ((u, v) I u > v) ofR2 into R2 be defined by T(u, v) = (u + v, u2 + v2).
1. Prove that T is locally one-to-one.
2. Determine the range of T, and show that T is globally one-to-one.
Problem 2.2.16 (Fa91) Let f be a C' function from the interval (—1, 1) into R2such that f(0) = 0 and f'(O) 0. Prove that there is a number e in (0, 1) suchthat Ilf(t)iI is an increasing function oft on (0, s).
Problem 2.2.17 (Fa80) For a real 2x2 matrix
yt
let OXII = x2 + y2 + z2 + t2, and define a metric by d(X, Y) = lix — }'Il. LetE = fX
Idet(X) = 0). Let
(1 0
A to E and exhibit an S E E that achieves thisminimum.
Problem 2.2.18 (Su80) Let S C R3 denote the ellipsoidal surface defined by
2x2+(y— 1)2+(z— 10)2=1.
Let T C JR3 be the surface defined by
IZ x2+y2+1
Prove that there exist points p E S, q E T, such that the line is perpendicularto Satp and to T atq.
Problem 2.2.19 (Sp80) Let P2 denote the set of real polynomials of degree 2.Define the map J: P2 —* IR by
'IJ(f)= I f(x)2dx.
JO
Let Q = ff E P2 f(1) = 1). Show that J attains a minimum value on Q anddetermine where the minimum occurs.
Problem 2.2.20 (Su79) Let X be the space of orthogonal real n x n matrices.Let E JR'3. Locate and describe the elements of X, where the map
f : X —÷ R, f(A) = (Do, Avo)
takes its maximwn and minimum values.
Problem 2.2.21 (Fa78) Let W C JR'2 be an open connected set and f a realvalued function on W such that all partial derivatives off are 0. Prove that f isconstant.
Problem 2.2.22 (Sp77) In 1R2, consider the region A defined by x2 + y2 > 1.
Find differentiable real valued functions f and g on A such that af/ax = dg/dybut there is no real valued function h on A such that f = dh/dy and g = dh/dx.
Problem 2.2.23 (Sp77) Suppose that u(x, t) is a continuous function of the realvariables x and t with continuous second partial derivatives. Suppose that u andits first partial derivatives are periodic in x with period 1, and that
d2u 82u
ox2 0t2
Prove that
E(t)= ij' +
dx
is a constant independent oft.
Problem 2.2.24 (Su77) Let f(x, t) be a C1 function such that Of/Ox = Of/at.Suppose that f(x, 0) > Ofor all x. Pmve that f(x, t) > Ofor all x and t.
Problem 2.2.25 (Fa77) Let f : —÷ IR have continuous partial derivativesand satisfy
ax3
for all x = (xl, ... , j = 1, . . . , n. Prove that
If(x) — .fiiKlIx —
(where hull = + • . +
Problem 2.2.26 (Fa83, Sp87) Let f : \ {0) IR be a function which iscontinuously differentiable and whose partial derivatives are uniformly bounded:
Of
dxi
for all that thenfcan be extended to acontinuousfunction defined on all of W'. Show that this is false ifn = 1 by givinga counterexample.
Problem 2.2.27 (Sp79) Let f : \ (0) -÷ IR be differentiable. Suppose
ofurn (x)x-+O
existsforeachj= 1...,n.1. Can f be extended to a continuous map from to R?
2. Assuming continuity at the origin, is f differentiable from to R?
Problem 2.2.28 (Sp82) Let f : —÷ R have directional derivatives in all di-rections at the origin. Is f differentiable at the origin? Prove or give a counterex-ample.
Problem 2.2.29 (Fa78) Let f : R have the following properties: f isdifferentiable on \ (0), f is continuous at 0, and
urnp-÷O ax,
for i = 1, . . ., n. Prove that f is differentiable at 0.
Problem 2.2.30 (Su78) Let U C R" be a convex open set and f: U —÷ R'2 adifferentiable function whose partial derivatives are uniformly bounded but notnecessarily continuous. Prove that f has a unique continuous extension to theclosure of U.
Problem 2.2.31 (Fa78) 1. Show that if u, v : 1R2 —÷ JR are continuously dif-
ferentiable and = then u = v = some f : JR2 ...+ JR
2. Prove there is no f: JR2 \ (0) —+ JR such that
—y andX
8x x2+y2 ay x2+y2
Problem 2.2.32 (Su79) Let f : JR3 -÷ JR be such that
f'(0)=(vER3I llvII=1).
Suppose f has continuous partial derivatives of orders ( 2. is there a p E JR3
with 1 such that
a2fax ay az
Problem 2.2.33 (Sp92) Let f be a dzfferentiablefunctionfivm R" to JR'2. Assumethat there is a differentiable function gfrom IR'2 to JR having no critical points suchthat g o f vanishes identically. Prove that the Jacobian determinant off vanishesidentically.
Problem 2.2.34 (Fa83) Let f, g : JR —÷ JR be smooth functions with f(0) = 0and f'(O) 0. Consider the equation f(x) = tg(x), t E JR.
1. Show that in a suitably small interval It I <8, there is a unique continuousfunction x(t) which solves the equation and satisfies x(0) =0.
2. Derive the first order Taylor expansion of x(t) about t = 0.
Problem 2.2.35 (Sp78) Consider the system of equations
3x+y—z+u4=0x —y+2z+u =0
2x+2y—3z+2u=0
1. Prove that for some e > 0, the system can be solved for (x, y, u) as afunction of z E [—e, s], with x(0) = y(O) = u(0) = 0. Are such functionsx(z), y(z) and u(z) continuous? Differentiable? Unique?
2. Show that the system cannot be solved for (x, y, z) as a function ofU E [—8,8],forallS > 0.
Problem 2.2.36 (Sp81) Describe the two regions in (a, b)-space for which thefunction
fa,b(X, y) = ay2 + bx
restricted to the circle x2 + y2 = 1, has exactly two, and exactly four criticalpoints, respectively.
Problem 2.2.37 (Fa87) Let u and v be two real valued C' functions on JR2 suchthat the gradient Vu is never 0, and such that, at each point, Vv and Vu arelinearly dependent vectors. Given = (xo, yo) E JR2 show that there is a C1function F of one variable such that v(x, y) = F (u(x, y)) in some neighborhoodofpo.
Problem 2.238 (Fa94) Let f be a continuously djfferentiable function from JR2into JR. Prove that there is a continuous one-to-one function gfrom [0, 1] into JR2such that the composite function f o g is constant.
Problem 2.239 (Su84) Let f : JR —÷ JR be C' and let
u=f(x)v = —y +xf(x).
If f'(xo) 0, show that this transformation is locally invertible near (xo, yo) andthe inverse has the form
x=g(u)y =—v+ug(u).
Problem 2.2.40 (Su78, Fa99) Let denote the vector space of real nxn ma-trices. Define a map by f(X) = X2. Find the derivative off.
Problem 2.2.41 (Su82) Let M2x2 be the four-dimensional vector space of all2x2 real matrices and define f : M2x2 -÷ M2x2 by f(X) = X2.
1. Show that f has a local inverse near the point
?)•
2. Show that f does not have a local inverse near the point
Problem 2.2.42 (Fa80) Show that there is an s > 0 such that A is any real2x2 matrix satisfying efor all entries a,) of A, then there is a real 2x2matrix X such that X2 + X' = A, where Xt is the transpose of X. Is X unique?
Problem 2.2.43 (Sp96) Let M2x2 be the space of2x2 matrices over identifiedin the usual way with Let the function Ffmm M2x2 into be defined by
F(X)=X+X2.
Prove that the range ofF contains a neighborhood of the origin.
Problem 2.2.44 (Fa78) Let denote the vector space of n x n real matrices.Prove that there are neighborhoods U and V in of the identity matrix suchthat for every A in U, there is a unique X in V such that X4 = A.
Problem 2.2.45 (Sp79, Fa93) Let denote the vector space of n x n realmatrices for n 2. Let det: -+ R be the determinant map.
1. Show that det is C°°.
2. Show that the derivative of det at A E is zero if and only if A hasrank
Problem 2.2.46 (Fa83) Let F(t) = (f,j (t)) be an nxn matrix of continuouslydjfferentiable functions f1,: R —÷ R, and let
u(t) = tr(F(t)3).
Show that u is differentiable and
u'(t) =
Problem 2.2.47 (Fa81) Let A = be an n x n matrix whose entries a'f arereal' valued differentiable functions defined on R. Assume that the determinantdet(A) of A is everywhere positive. Let B = be the inverse matrix of A.Prove the formula
d7log(det(A)) = dt
by,.i.f=1
Problem 2.2.48 (Sp03) 1. Prove that there is no continuously djfferentiable,measure-preserving bijective function f: R —+ R+.
2. Find an example of a continuously differentiable, measure-preserving bi-jective function f: JR x JR -÷ JR x IR÷.
2.3 Integral Calculus
Problem 2.3.1 (Sp78) What is the volume enclosed by the ellipsoid
— —a b c
Problem 23.2 (Sp78) Evaluate
ft 22e X )' dxdy,
JJA
where A = {(x, y) E 1R2 I x2 + y2 1).
Problem 2.3.3 (Sp98) Given the fact that f e_X2 dx = evaluate the inte-
gral
= L L dy.
Problem 2.3.4 (Fa86) Evaluate
ff(x3 — 3xy2) dxdy,
where
Problem 2.3.5 (Fa98) Let ço(x, y) be a function with continuous second orderpartial derivatives such that
1. + çoyy + cox = 0 in the punctured plane ]R2 \ (0),
2. rcox2irr 2irr
—÷ 0.
Let CR be the circle x2 + y2 = R2. Show that the line integral
I dx + cox dy)JCR
is independent of R, and evaluate it.
Problem 2.3.6 (SpSO) Let S = ((x, y, z) E R3 I x2 + y2 + z2 = 1) denote theunit sphere in JR3. Evaluate the surface integral over 5:
fj(x2+y+z)dA.
Problem 2.3.7 (Sp81) Let 7, j, and be the usual unit vectors in 1R3. Let F de-note the vector field
1x2 + y — 4)7 + 3xyj + (2xz +
1. Compute V x F (the curl of F).
2. Compute the integral of V x F over the surface x2 + y2 + z2 = 16, z 0.
Problem 2.3.8 (Sp91) Let the vector field F in 1R3 have the form
F(r) = g(IIrII)r (r (0, 0, 0)),
where g is a real valued smooth function on (0, oo) and fl denotes the Euclideannorm. (F is undefined at (0,0,0).) Prove that
IJc
for any smooth closed path C in R3 that does not pass through the origin.
Problem 2.3.9 (Fa91) Let 23 denote the unit ball of 1R3, 13= fr E WI lint 11.
Let J = (J1, J21J3) be a smooth vector field on 1R3 that vanishes outside of 13and satisfies J = 0.
1. For f a smooth, scalar-valued function defined on a neighborhood of 13,prove that
J (Vf) • Jdxdydz =0.13
2. Prove that
J J1 dxdydz =0.13
Problem 2.3.10 (Fa94) Let V denote the open unit disc in R2. Let u be an eigen-function for the Laplacian in V; that is, a real valued function of class C2 defined
in 1), zero on the boundary of V but not identically zero, and satisfying the djf-
ferential equation82u 82u
2+ 2=Au,ax ay
where A is a constant. Prove that
(*) 'IDIgrad u12 dxdy + xff u2dxdy =0,
and hence that A <0.
Problem 23.11 (Fa03) Let A, a e R, with a < 0. Let u(x, y) be an infinitelydifferentiable function defined on an open neighborhood of closed unit disc Vsuch that
a2u a2u2+ 2=Au inint(D)
dxindV.
Here denotes the directional derivative of u in the direction of the outwardunit normal. Prove that if u is not identically zero in the interior of V then A <0.
Problem 23.12 (Sp92) Let f be a one-to-one C1 map of JR3 into JR3. and let Jdenote its Jacobian determinant. Prove that if xo is any point of JR3 anddenotes the cube with center xo, side length r, and edges pa rallel to the coordinateaxes, then
IJ(xo)I = urn urn supIIf(x) —
r—÷O IIX—xolI
Here, • II is the Euclidean norm in JR3.
3
Differential Equations
3.1 First Order Equations
Problem 3.1.1 (Fa87) Find a curve C in JR2. passing through the point (3,2),with the following property: Let L (xo, be the segment of the tangent line toC at (xo, which lies in the first quadrant. Then each point (xo, Yo) of C is themidpoint of L (xo, Yo)-
L(xo, YO)
C
(xo, yo)
Pmblan 3.1.2 (Su78) Solve the differential equatlong' = 2g, g(O) = a, wherea isa real constant.
Problem 3.1.3 (flfl, Fa77) Len be an integer Miser than 1. Is there aon [0, cc) whose derivative equaLc Its power and whose valUe
at the origin is positive?
Pmblen 3.1.4 (Sp78) 1. For which real numbers a > 0 does the differential—on
x(0)=O,
have a solution on some interval [0, bJ, b> 0?
2. For which values of a are there intervaLt on which two solutions are Sfined?
PrOblem 3.1.5 (Sp78) Consider the differential equation
x(0)=I.
L Prove thatfor some b> 4 there isa solution definedfort E (0, bJ.
2. FInd an explicit value of b having the property in Pan 1.
3. Findac ,Osuththatthere is no solution on [0,4
Pieblan 3.1.6 (kfl) Solve the differential equation
y(0)=1.
Pmbkm 3.1.7 (Sp93) Prove that every solution x(t) Ct >0) of the d(fferential
2 6—x
withx(0) > = I.
Problan 3.1.8 (SpN) Consider the differential equation
X3—X
1+9I. Find all its constant solutions
2. DIscuss lirnx(t), where x(t) Is the solution such thatx(0) =
Problem 3.1.9 (Su77, Su80, Sp82, Sp83) Prove that the initial value problem
dx= 3x +85 cosx, x(O) =77,
has a solution x(t) defined for all t E JR.
Problem 3.1.10 (Fa82) Let f : JR -÷ JR be a continuous nowhere vanishingfunction, and consider the differential equation
dy—=f(y).dx
1. For each real number c, show that this equation has a unique, continuouslydifferentiable solution y = y(x) on a neighborhood of 0 which satisfies theinitial condition y(O) = c.
2. Deduce the conditions on f under which the solution y exists for all x E JR.
for evety initial value c.
Problem 3.1.11 (Sp79) Find all solutions to the differential equation
y(O)=O.
Problem 3.1.12 (Sp83) Find all solutions y : JR -+ JR to
— = (y — 2), y(O) =0.dx
Problem 3.1.13 (Su83) Find all real valued C' solutions y(x) of the differentialequation
dyx—+y=x (—1<x<1).dx
Problem 3.1.14 (Sp84) Consider the equation
dy— = ydx
Show that there is an e> 0 such that if lyol <e, then the solution y = f(x) with
f(0) = satisfiesurn f(x)=0.
X—)' —00
Problem 3.1.15 (Fa84) Consider the equation
dy y—=3xy+dx 1+y2
Pmve
1. For each n = 1,2,..., there is a unique solution y = (x) defined for0 x 1 such that (0) = 1/n.
2. urnn—+00
Problem 3.1.16 (Fa85) Let y(t) be a real valued solution, defined for0 <t <00, of the differential equation
= — +dtShow that y(t) —+ +00 as t -+ +00.
Problem 3.1.17 (SpOl) Consider the differential-delay equation given byy'(t) = —y(t — ta). Here, the independent variable t is a real variable, the func-tion y is allowed to be complex valued, and tO is a positive constant. Prove that0 < tO then every solution of the form y(t) = with complez tendstoOast —+ +00.
Problem 3.1.18 (Fa86) Prove the following theorem, or find a counterexample:lip and q are continuous real valued functions on IR such that Ip(x)Ifor all x, and if every solution f of the differential equation
f'+qf =0satisfies urn f(x) =0, then every solution f of the differential equation
x-++0o
f'+pf =0satisfies urn f(x) =0.
+00
Problem 3.1.19 (Fa86) Discuss the solvability of the differential equation
(?sin y)(y')3 + (eX cos y)y' + tan x = 0
with the initial condition y(O) = 0. Does a solution exist in some interval about0? If so, is it unique?
Problem 3.1.20 (Fa92) Let f and g be positive continuous functions on R, withg f everywhere. Assume the initial value problem
x(0)=0,
has a solution defined on all of JR. Prove that the initial value problem
x(0)=0,
also has a solution defined on all of JR.
Problem 3.1.21 (Sp95) Let f : JR -+ JR be a bounded continuously differentiablefunction. Show that every solution of y'(x) = f (y(x)) is monotone.
3.2 Second Order Equations
Problem 3.2.1 (Sp97) Suppose that f"(x) = (x2 — 1)f(x) for all x E R, andthatf(0) = 1, f'(O) = 0. Show that f(x) -+ 0 as x 00.
Problem 3.2.2 (Sp77) Find the solution of the differential equation
y"—2y'+y =0,
subject to the conditions
y(O)=l, y'(O)=l.
Problem 3.23 (Fa77) Find all solutions of the differential equation
d2x dx—2— +x = sintdt2 dt
subject to the condition x(0) = I and x'(O) = 0.
Problem 3.2.4 (Su79) Let x : R —+ R be a solution to the differential equation
5x"+ lOx'+6x = 0.
Prove that the function f : R —÷ R,
2
— 1 +x(t)4
attains a maximum value.
Problem 3.2.5 (Su84) Let x(t) be the solution of the differential equation
x"(t) + 8x'(t) + 25x(t) = 2cost
with initial conditions x(0) = 0 and x'(O) = 0. Show that for suitable constantsa and8,
urn (x(t) — a cos(t — = 0.t-*oo
Problem 3.2.6 (Fa79, Su81, Fa92) Let y = y(x) be a solution of the differentialequation y" = with —00 <X <00, y(O) = 1 and y'(O) =0.
1. Show that y is an even function.
2. Show that y has exactly one zero on the positive real axis.
Problem 3.2.7 (Fa95) Determine all real numbers L > 1 so that the boundaiyvalue problem
x2y"(x)+y(x)=O, 1 L
y(1)=y(L)=0has a nonzero solution.
Problem 3.2.8 (Fa83) For which real values of p does the dWerential equation
y"+2py'+y =3
admit solutions y = f(x) with infinitely many critical points?
Problem 3.2.9 (Sp87) Letp, q and r be continuous real valued functions on IR,with p > 0. Prove that the differential equation
p(t)x"(t) + q(z)x'(t) + r(t)x(t) = 0
is equivalent to (i.e., has exactly the same solutions as) a differential equation ofthe form
(a(t)x'(t))' + b(t)x(t) = 0,
where a is continuously djfferenciable and b is continuous.
Problem 3.2.10 (Fa93) Let the function x(t) (—oo <t <oo) be a solution ofthe differential equation
d2x dx—2b—+cx =0dt2 dt
such that x(0) = x(1) = 0. (Here, b and c are real constants..) Prove thatx(n) = Ofor every integern.
Problem 3.2.11 (Sp93) Let k be a positive integer. For which values of the realnumber c does the differential equation
d2x dx-2c—+x=0dt2 dt
have a solution satisfying x(0) = x(2nk) = 0?
Problem 3.2.12 (FaOl) Consider the second-order linear differential equation
d2x dx(*)
Here, the independent variable t varies over R, the unknown function x is as-sumed to be real valued, and p and q are continuous functions on R. Assumethat the solutions of(*) are defined for all t (which is actually guaranteed by thetheory), and that the solution set is translation invariant: if f is a solution and sis a real numbe, then the function g(t) = f(t + s) is also a solution. Prove thatp and q are constant.
Problem 3.2.13 (Sp85) Let h > 0 be given. Consider the linear difference equa-tion
y((n+2)h)—2y((n+1)h)+y(nh)=—y(nh), n=0,1,2,...
(Note the analogy with the differential equation y" = —y.)
1. Find the general solution of the equation by trying suitable exponentialsubstitutions,
2. Find the solution with y(O) 0 and y(h) = h. Denote it by Sh (nh),n = 1,2
3. Letx befixedandh = Show that
lim I — I = sinx.n-+oo
3.3 Higher Order Equations
Problem 3.3.1 (Su78) Let E be the set offunctions f : R -÷ R which are solu-tions to the differential equation 1" + f" — 2f = 0.
1. Prove that E is a vector space and find its dimension.
2. Let Eo C E be the subspace of solutions g such that urn g(t) =0. FindIg E Eosuchthatg(0) =Oandg(0)=2.
Problem 3.3.2 (Fa98) Find a function y(x) such that y(4) + y = Ofor x 0,y(0) =0, y'(O) = 1 and urn y(x) = urn y'(x) =0.
x-+oo x-÷oo
Problem 3.3.3 (Sp87) Let V be afinite-dimensional linear subspace of C°°(R)(the space of complex valued, infinitely differentiable functions). Assume that V isclosed under D, the operator of differentiation (i.e., f e V Df e V). Provethat there is a constant coefficient differential operator
L=>akDk
such that V consists of all solutions of the differential equation Lf = 0.
Problem 3.3.4 (Fa94) 1. Find a basis for the space of real solutions of thedifferential equation
(*)
2. Find a basis for the subspace of real solutions of(*) that satisfy
lirn x(t)=O.
Problem 33.5 (Sp94) 1. Suppose the functions sin t and sin 2t are both solu-tions of the differential equation
)JCkdk=O,
where cO, ..., are real constants. What is the smallest possible order ofthe equation? Write down an equation of minimum order having the givenfunctions as solutions.
2. Will the answers to Part 1 be different if the constants co, ..., are al-lowed to be complex?
Problem 33.6 (Sp95) Let y : IR -÷ R be a three times differentiable functionsatisfying the differential equation y" — y = 0. Suppose that y(x) = 0.
Find real numbers a, b, c, and d, not all zero, such that ay(0)+y' (0)+cy" (0) = d.
3.4 Systems of Differential Equations
Problem 3.4.1 (Sp79) Consider the system of differential equations:
dx =y+tzdy—=z+t'xdtdz
=x+e y.
Prove there exists a solution defined for all t E [0, 11, such that
/1 2 3\ /x(0)\ /05 61 Iy(°)1=I°
\7 8 9J \z(0))and also
f' (x(t)2 + y(t)2 + z(t)2) dt = 1.
Problem 3.4.2 (Su79, Fa79, Fa82, Su85) Find all pairs of C°° functions x(t)and y(t) on IR satisfying
x'(t) = 2x(t) — y(t), y'(t) = x(t).
Problem 3.43 (Su80) Consider the djfferential equations
dx dy= --x + y, = log(20 + x) — y.
Let x(t) and y(t) be a solution defined for alit ? 0 with x(0) > 0 and y(0) > 0.Prove that x(t) and y(t) are bounded.
Problem 3.4.4 (Sp81) Consider the system of differential equations
dx=y+x(1 —x2—y2)
dy=—x+y(1—x2—y2).
1. Show that for any xo and yo, there is a unique solution (x(t), y(t)) definedfor alit IR such thatx(0) = y(O) = Yo.
2. Show that if xo 0 and yo 0, the solution referred to in Part 1 ap-proaches the circle x2 + y2 = 1 as t -÷ 00.
Problem 3.4.5 (Sp84) Show that the system of differential equations
dfx\ 10 1 0\fxj-IYI=12 0 OflytkzJ \O 0 3J\z
has a solution which tends to oo as t —÷ —oo and tends to the origin as t -÷ +00.
Problem 3.4.6 (Sp91) Let x(t) be a nontrivial solution to the system
dxAx,
dt
where /1 6 1
A=I—4 4 11
\—3 —9 8
Prove that lix (t) II is an increasing function oft. (Here, denotes the Euclideannonn.)
Problem 3.4.7 (Su84) Consider the solution curve (x(t), y(t)) to the equations
dx 12.siny
dt
with initial conditions x (0) = 0 and y(0) = 0. Prove that the solution must crossthe linex = 1 inthexy plane bythetimet =2.
Problem 3.4.8 (Fa80) Consider the differential equation x" + x' + x3 = 0 andthe function f(x, x') = (x + x')2 + (x')2 + x4.
1. Show that f decreases along trajectories of the differential equation.
2. Show that ifx(t) is any solution, then (x(t), x'(t)) tend,s to (0,0) as t -÷ 00.
Problem 3.4.9 (Fa84) Consider the differential equation
dx dy 35—=y, —=—ay—x —x , where a>O.dt dt
1. Show thaty2 x4 x6
F(x,y)=
decreases along solutions.
2. Show that for any e > 0, there is a 5 > 0 such that whenever(x(0), y(O)) fl < 5, there is a unique solution (x(t), y(t)) of the given
equations with the initial condition (x(O), y(O)) which is defined for allt 0 and satisfies (x(t), y(t)) <e.
Problem 3.4.10 (Fa83) 1. Let u(t) be a real valued differentiable function ofa real variable t which satisfies an inequality of the form
u'(t) au(t), t 0, u(0) b,
where a and b are positive constants. Starting from firs: principles, derivean upper bound for u(t)for t > 0.
2. Let x(t) = (xi(t), x2(t), ... , be a differentiable function from R towhich satisfies a differential equation oftheform
x'(t) = f(x(t)),
where f : —÷ is a continuous function. Assuming that f satisfies thecondition
y) 11y112, y E
derive an inequality showing that the norm IIx(t)H grows, at most, expo-nentially.
Problem 3.4.11 (Fa81) Consideran autonomous system of differential equations
dx1=
F = (F1, .... : -+ IWZ is a C1 vector field.
1. Let U and V be two solutions on a <t <b. Assuming that
(DF(x)z, z) 0
for z in R'2, show that IIU(t) — V(t)112 is a decreasing function of t.
2. Let W(t) be a solution defined fort > 0. Assuming that
z) —liz 112,
show that there exists C E JR'1 such that
urn W(t) = C.1—400
Problem 3.4.12 (Fa81) Let V : JR'2 -÷ JR be a C1 function and consider thesystem of second order differential equations
x'(t)_—f,(x(t)), 1
whereavfi=-—.
Let x(t) = (Xl (t), ... , (t)) be a solution of this system on a finite intervala<t<b.
1. Show that the function
H(t) + V(x(t))
is constant for a <t <b.
2. Assuming that V(x) M > —ooforallx E R'2, show thatx(t), x'(t), andx"(t) are bounded on a <t <b, and then prove all three limits
urn x(t), urn x'(t), lim x"(t)z-+b
exist.
Problem 3.4.13 (Sp86) For A a real number, find all solutions of the integralequations
ço(x) =? + A dy,
Jo
Problem 3.4.14 (Sp86) Let V be a finite-dimensional vector space (over C) ofC°° complex valued functions on JR (the linear operations being defined point-wise). Prove that if V is closed under differentiation (i.e., f'(x) belongs to Vwhenever f(x) does), then V is closed under translations (i.e., f(x + a) belongsto V whenever f(x) does, for all real numbers a).
Problem 3.4.15 (Fa99) Describe all three dimensional vector spaces V of C00complex valued functions on JR that are invariant under the operator of differen-tiation.
Problem 3.4.16 (Fa88) Let the real valued functions fi, ..., on R satisfythe system of differential equations
k=1,...n— —(n +
Prove that for each k,
urn fk(t)=0.1—400
Problem 3.4.17 (Fa91) Consider the vector differential equation
dx(t) = A(t)x(t)
where A is a smooth n x n function on JR. Assume A has the properly that(A(t)y, y) cIIyII2 for all y in JR" and all t, where c is a fixed realProve that any solution x (t) of the equation satisfies lix (t) Ii Ii x (0)11 for allt>0.
Problem 3.4.18 (Sp94) Let W be a real 3x3 antisymmetric matrix; i.e.,
= —W. Let the function
fxi(t)X(t) = I
\x3(t)
be a real solution of the vector differential equation = WX.
1. Prove that IIX(t)Il, the Euclidean norm of X(t), is independent oft.
2. Prove that if v is a vector in the null space of W, then X (t). v is independentoft.
3. Prove that the values X(t) all lie on afixed circle in JR3.
Problem 3.4.19 (SpSO) For each t E JR. let P(t) be a symmetric real n x n matrixwhose entries are continuous functions oft. Suppose for all t that the eigenvaluesof P (t) are all —1. Let x (t) = (Xi (t), ... , (t)) be a solution of the vectordifferential equation
= P(t)x.
Prove that
urn x(t)=O.l-400
Problem 3.4.20 (Sp89) Let
/0 0 0 o\ /0 1 0 0110001 10010A—10
1 0 01' B=100 0 1\ooio) k0000
Find the general solution of the matrix differential equation = AX B for theunknown 4x4 matrix function X(t). t
4Metric Spaces
4.1 Topology of
Problem 4.1.1 (Fa02) Let S be a subset of IR. Let C be the set of points x in Rwith the property that S fl (x — x + 5) is uncountable for every ö> 0. Pmvethat S \ C is finite or countable.
Problem 4.1.2 (Sp03) Let A ç IR be uncountable.
1. Show that A has at least one accumulation point.
2. Show that A has uncountably many accumulation points.
Problem 4.1.3 (Fa02) Let (x(i, j) I i, j E N) be a doubly indexed set in a com-plete metric space (X, p) such that
1 11 11 11 1p(x(i,j),x(k,t)) li' ,maxj-, jProve that the iterated limits urn urn x(i, j), urn urn x(i, j) exist and are
i—+ooJ--+oo J—*ooL—*ooequaL
Problem 4.1.4 (SpOl) Let T0 be the interior of a triangle in R2 with verticesA, B, C. Let Ti be the interior of the triangle whose vertices are the midpoints ofthe sides ofTo, T2 the interior of the triangle whose vertices are the midpoints ofthe sides ofT1, and so on. Describe the set
Problem 4.1.5 (Sp86, Sp94, Sp96, Fa98) Let K be a compact subset of R'1 and(By) a sequence of open balls that covers K. Prove that there is a positive number
such that each €-ball centered at a point of K is contained in one of the balls
Problem 4.1.6 (Su81) Prove or disprove: The set Q of rational numbers is theintersection of a countable family of open subsets of R.
Problem 4.1.7 (Fa77) Let X c IR be a nonempy connected set of real numbers.If every element of X is rational, prove X has only one element.
Problem 4.1.8 (Su80) Give an example of a subset of having uncountablymany connected components. Can such a subset be open? Closed?
Problem 4.1.9 (FaOO) Let : > be continuous (n = 1,2, . . .). LetK be a compact subset of Rk. Suppose f uniformly on K. Prove that
S = f(K) U (K) is compact.
Problem 4.1.10 (Sp83) Show that the interval [0, 1] cannot be written as a count-ably infinite disjoint union of closed subintervals of [0, 1].
Problem 4.1.11 (Su78, Sp99, Sp03) Let X and I be nonemply subsets of a met-ric space M. Define
d(X, I) = inf{d(x, y) I x E X, y E I).
1. Suppose X contains only one point x, and I is closecL Prove
d(X, I) = d(x, y)
for some y E I.
2. Suppose X is compact and I is closed. Prove
d(X, I) = d(x, y)
for somex E X, y El.
3. Show by example that the conclusion of Part 2 can be false if X and I areclosed but not compact.
Problem 4.1.12 (Sp82) Let S C be a subset which is uncountable. Prove thatthere is a sequence of distinct points in S converging to a point of S.
Problem 4.1.13 (Fa89) Let X C W' be a closed set and r a fixed positive realLet I = {y E I Ix — = r for some x X). Show that I is closed.
Problem 4.1.14 (Sp92, Fa99) Show that every infinite closed subsetof a countable set.
Problem 4.1.15 (Fa86) Let (U1, U2...) be a cover of R'1 by open sets. Provethat there is a cover fV1, V2. ...) such that
1. C Uj for each j;
2. each compact subset of IWZ is disjoint from all but finitely many of the Vi.
Problem 4.1.16 (Sp87) A standard theorem states that a continuous real valuedfunction on a compact set is bounded. Prove the converse: JfK is a subset of R'7and if every continuous real valued function on K is bounde4, then K is compact.
Problem 4.1.17 (Su77) Let A C be compact, x E A; let (xe) be a sequencein A such that every convergent subsequence of(x1) converges to x.
1. Prove that the entire sequence (x1) converges.
2. Give an example to show that A is not compact, the result in Part 1 is notnecessarily true.
Problem 4.1.18 (Fa89) Let X C be compact and let f : X —÷ R be continu-ous. Given s > 0, show there is an M such that for all x, y E X,
Jf(x) MIx—yI+s.
Problem 4.1.19 (Su78) Let (Sa) be a family of connected subsets of 1R2 all con-taining the origin. Prove that Ua Sa is connected.
Problem 4.1.20 (Fa79) Consider the following properties of a functionf:IR'2
1. f is continuous.
2. The graph off is connected in x R.
Prove or disprove the implications 1 2, 2 1.
Problem 4.1.21 (SpOl) Let U be a nonemply, open subset of R". Con-struct a function f : —÷ R that is discontinuous at each point of U andcontinuous at each point \U.
Problem 4.1.22 (Sp82) Prove or give a counterexample. Every connected, lo-cally pathwise connected set in is pathwise connected.
Problem 4.1.23 (Sp81) The set of real 3 x 3 symmetric matrices is a real, finite-dimensional vector space isomorphic to ]R6. Show that the subset of such matricesof signature (2, 1) is an open connected subspace in the usual topology on 1R6.
Problem 4.1.24 (Fa78) Let be the vector space of real n x n matrices,identified with Let X C be a compact set. Let S C C be the set ofall numbers that are eigenvalues of at least one element of X. Prove that S iscompact.
Problem 4.1.25 (Su81) Let 50(3) denote the group of orthogonal transforma-tions ofR3 of determinant 1. Let Q C S0(3) be the subset of symmetric transfor-mations I. Let P2 denote the space of lines through the origin in R3.
1. Show that P2 and S0(3) are compact metric spaces (in their usual topolo-gies).
2. Show that P2 and Q are homeomorphic.
Problem 4.1.26 (Fa83) Let m and n be positive integers, with m <n. Let Mm xnbe the space of linear transformations of into R" (considered as n x m matri-ces) and let L be the set of transformations in Mm )(fl which have rank m.
1. Show that L is an open subset 01Mm xn•
2. Show that there is a continuous function T : L -÷ Mm xn such thatT(A)A 'm for all A, where Im is the identity on Rm.
Problem 4.1.27 (Fa91) Let be the space of real nxn matrices. Regard itas a metric space with the distance function
d(A, B) = —J
(A = B =i,j=1
Prove that the set of nilpotent matrices in x is a closed set.
Problem 4.1.28 (SpOO) Let S be an uncountable subset of R. Prove that thereexists a real number t such that both sets S fl (—oo, t) and S fl (t, oo) are un-countable.
4.2 General Theory
Problem 4.2.1 (Sp02) Let (X, p) and (Y, a) be metric spaces. Assume that:
1. f, fi, 12, ... are bijectivefunctionsofX onto Y with inverses g,
g is uniformly continuous
3. f uniformly as n —+ 00.
Prove that g unjformly as n —+ oo.
Problem 4.2.2 (Fa99) Let Ei, E2, ... be nonempty closed subsets of a completemetric space (X, d) with C for all positive integers n, and such thaturn diam (En) = 0, where diam (E) is defined to be
fl—) 00
E E}.
ProvethatflEn
Problem 4.2.3 (FaOO) Let A be a subset of a compact metric space (X, d). As-sume that, for every continuous function f : X -+ R, the restriction of f to Aattains a maximum on A. Prove that A is compact.
Problem 4.2.4 (Fa93) Let X be a metric space and a convergent sequencein X with limit xo. Pmve that the set C = (xo, XI, X2, ...) is compact.
Problem 4.2.5 (Sp79) Prove that every compact metric space has a countabledense subset.
Problem 4.2.6 (FaSO) Let X be a compact metric space and f : X —÷ X anisometry. Show that f(X) = X.
Problem 4.2.7 (Sp97) Let M be a metric space with metric d. Let C be a nonemplyclosed subset of M. Define f : M —+ R by
f(x) = inf(d(x, y) I y E C).
Show that f is continuous, and that f(x) = 0 and only if x E C.
Problem 4.2.8 (Su84) Let C"3 be the set of real valued functionsf on the closedinterval [0, II such that
1. f(O)=0;
2. 11111 isfinite, where by definition
IIf II = sup {If(x)—f(y)I
i x y].
Verify that V II is a norm for the space C"3, and prove that C"3 is complete withrespect to this norm.
Problem 4.2.9 (SpOO) Let be a uniformly bounded equicontinuous se-quence of real-valued functions on the compact metric space (X, d). Define thefunctions : X —÷ R, for n E N by
gn(x) =max{f,(x),...,fn(x)}.
Prove that the sequence
Problem 4.2.10 (Sp87) Let F be a uniformly bounded, equicontinuousfamily ofreal valued functions on the metric space (X, d). Prove that thefunction
g(x) = supff(x) f E F)
is continuous.
Problem 4.2.11 (Fa91) Let X and Y be metric spaces and f a continuous map of
X into Y. Let K1, K2,... be nonempty compact subsets of X such thatK = fl Prove that f(K) = fl
Problem 4.2.12 (Fa92) Let (Xi, di) and (X2, d2) be metric spaces andf : X1 -÷ X2 a continuous surjective map such that q) d2(f(p), f(q))
for every pair of points p, q in X1.
1. If Xi is complete, must X2 be complete? Give a proof or a counterexample.
2. If X2 is complete, must Xj be complete? Give a proof or a counterexample.
4.3 Fixed Point Theorem
Problem 43.1 (Fa79) An accurate map of California is spread out flat on a tablein Evans Hall, in Berkeley. Prove that there is exactly one point on the map lyingdirectly over the point it represents.
Problem 4.3.2 (Fa87) Define a sequence of positive numbers as follows. Letxc, > 0 be any positive and let = (1 + Prove that thissequence converges, and find its limit.
Problem 433 (Su80) Let f : R —÷ R be monotonically increasing (perhapsdiscontinuous). Suppose 0 < f(0) and f(100) <100. Prove f(x) = x for somex.
Problem 43.4 (Su82, Sp95) Let K be a nonempty compact set in a metric spacewith distance function d. Suppose that K —÷ K satisfies
d(ço(x), ço(y)) <d(x, y)
for all x y in K. Show there exists precisely one point x E K such thatx=(p(x).
Problem 43.5 (Fa82) Let K be a continuous function on the unit square0 x, y 1 satisfying IK(x, < 1 for all x and y. Show that there is acontinuous function f(x) on [0, 1] such that we have
'I'f(x) + K (x, y) f (y) dy = eX2.
Jo
Can there be more than one such function f?
Problem 43.6 (Fa88) Let g be a continuous real valued function on [0, 1]. Provethat there exists a continuous real valued function f on [0, 1] satisfying the equa-tion
f(x)— f f(x — t)e_t2 dt = g(x).
Problem 43.7 (Su84) Show there is a unique continuous real valued functionf: [0, 1] -÷ JR such that
f(x)=sinx+f dy.
Problem 43.8 (Fa85, Sp98) Let (M, d) be a nonempty complete metric space.Let S map M into M, and write S o S (S(x)). Supposethat is a strict contraction; that is, there is a constant A < I such that for allpoints x, y E M, d (S2(x), S2(y)) Xd(x, y). Show that S has a unique fixedpoint in M.
5
Complex Analysis
5.1 Complex Numbers
Problem 5.1.1 (Fa77) If a and b are complex numbers and a 0, the set abconsists of those complex numbers c having a logarithm of the form ba, for somelogarithm a of a. (That is, = c and = a for some complex number a.)Describe set Qb when a = 1 andb = 1/3 + i.
Problem 5.1.2 (Su77) Write all values of in the form a + bi.
Problem 5.13 (Sp85) Show that a necessary and sufficient condition for threepoints a, b, and c in the complex plane to form an equilateral triangle is that
a2+b2+c2 =bc+ca+ab.
Problem 5.1.4 (Fa86) Let the points a, b, and c lie on the unit circle of the com-plex plane and satisfy a + b + c = 0. Prove that a, b, and c form the vertices ofan equilateral triangle.
Problem 5.1.5 (Sp77) 1. Evaluate (1), where (x) is the polynomial
f—ix—1
2. Consider a circle of radius 1, and let be the vertices of aregular n-gon inscribed in the circle. Join Qi to by seg-ments of a straight line. You obtain (n—i) segments oflengths A2, A3, ... ,
Show that
=n.
Problem 5.1.6 (Sp03) Prove that for each integer n 0 there is a polynomial(x) with integer coefficients such that the identity
holds for all z.
2cosnz =
Problem 5.1.7 (Sp90) Let Zi, Z2, ... , Zn be complex numbers. Prove that thereexists a subset J C (1, 2, .. ., n} such that
Zj lZj I.jEJ j=1
Problem 5.1.8 (Sp94) Let ai, ... , be complex numbers. Prove that thereis a point x in [0, 1] such that
Problem 5.1.9 (Fa82) Let a and b be complex numbers whose real partsare neg-ative or 0. Prove the inequality lea — e'i Ia —
Problem 5.1.10 (Fa95) Let A be a finite subset of the unit disc in the plane, andlet N(A, r) be the set of points at distance r from A, where 0 < r < 1. Show
that the length of the boundary N(A, r) is, at most, C/r for some constant Cindependent of A.
Problem 5.1.11 (Su82) For complex numbers aj, a2, ... , prove
k1/n
limsup =
Note: See also Problem 1.3.1.
5.2 Series and Sequences of Functions
Problem 5.2.1 (Fa95) Show that
(1+z+z2+.• +z9Xl+z'°+z20+.• +z90X1+z'°°+z20°+' +z900)• •
1
1 —z
forlzl <1.
Problem 5.2.2 (Fa94) Suppose the coefficients of the power series
>2 an?
are given by the recurrence relation
1, at = —1, 3a, + i — an....2 =0, n =2,3
Find the radius of convergence of the series and the function to which it convergesin its disc of convergence.
Problem 5.2.3 (Fa03) Show that the differential equation
f"(z) = zf(z), f(0) = 1, f'(O) = 1
has an unique entire solution in the complex plane.
Problem 5.2.4 (Fa93) Describe the region in the complex plane where the infi-nite series
001 ( nz
converges. Draw a sketch of the region.
Problem 5.2.5 (Su77) Let f be an analytic function such that
f(z)=1+2z+3z2+ for Izt<1.
Define a sequence of real numbers ao, ai, a2, ... by
f(z) >an(z
What is the radius of convergence of the series
Problem 5.2.6 (Sp77) Let the sequence ao, ai,... be defined by the equation
(O<x< 1).
Find
n-+oo
Problem 5.2.7 (Su78) Suppose the power series
converges for Izi <R where z and the are complex nwnbers. If E C aresuch that all n, prove that
converges for Izi <R.
Problem 5.2.8 (Sp99) Let b1, b2, .. . be a sequence of real numbers such thatbk+1 for all k and bk = 0. Prove that the power series
converges for all complex numbers z such that Izi 1 and z 1.
Problem 5.2.9 (Sp79) For which z E C does
converge?
Problem 5.2.10 (Su79) Show that
00Z
(i+z2)'1
convergesfor all complex numbers z exterior to the lemniscate
I
I = 1.
Problem 5.2.11 (Su82) Determine the complex numbersz for which the powerseries
and its term by term derivatives of all orders converge absolutely.
Problem 5.2.12 (Su84) Suppose
f(z) =
has radius of convergence R > 0. Show that
is entire and that for 0 < r <R, there is a constant M such that
Ih(z)I Me14".
Problem 5.2.13 (SpOl) Let the power series with positive radius ofconvergence R, represent the function f in the disk Izi <R. Fork = 0, 1, ... letSk be the k-th partial sum of the series (z) = Prove that
>1f(z)—sk(z)I <00
foreachzinthedisklzl <R.
Problem 5.2.14 (Sp85) Let R > 1 and let f be analytic on Izi < R except atz = 1, where f has a simple pole. If
1(z) = (IzI <1)
is the Maclaurin series for f, show that urn exists.n-+00
Problem 5.2.15 (Fa95) Find the radius of convergence R of the Taylor seriesabout z = 1 of thefunction f(z) = 1/(1 + z2 + z4 + z6 + z8 + z'°). Express theanswer in terms of real numbers and square roots only.
Problem 5.2.16 (Sp78) Prove that the uniform limit of a sequence of complexanalytic functions is complex analytic. Is the analogous theorem true for realanalytic functions?
Problem 5.2.17 (Su79) Let gn (z) be an entire function having only real zeros,n = 1,2,.... Suppose
urn gn(z) = g(z)n—*00
uniformly on compact sets in C, with g not identically zero. Prove that g (z) hasonly real zeros.
Problem 5.2.18 (Sp86) Let f, ... be entire functions. Assume that
j. (0)1f(k) for all n and k;
2. urn (0) exists for all k.fl—+ 00
Prove that the sequence converges uniformly on compact sets and that itslimit is an entire function.
Problem 5.2.19 (Sp92) Find a Laurent series that converges in the annulus
1 <IzI <2 to a branch of the function log(z(2_:))
5.3 Conformal Mappings
Problem 53.1 (Fa77) Consider the following four types of transformations:
liz, zt-÷kz
(where
z a variable complex number and the other letters denote constant com-plex numbers. Show that each transformation takes circles to either circles orstraight lines.
Problem 53.2 (Fa78) Give examples of confonnal maps as follows:
1. from fz I tzI < 1) onto {z I <0),
2. from Izt <1) onto itself with f(0) =0 and f(112) = i,/2,
Problem 5.3.3 (Sp83) A fractional linear transfor,nation maps the annulusr < IzI < 1(when?r > =and Izi = 1. Findr.
Problem 5.3.4 (SpSO) Does there exist an analytic function mapping the annulus
onto the annulusB=[zI
and taking Ci Ci, C4 -÷ C2, where Cr is the circle of radius r?
Problem 53.5 (Su80) Exhibit a confonnal map from {z C I IzI < 1, Iflz > 0)ontoD=(z€C IIzI<1).
Problem 53.6 (Sp90) Find a one-to-one conformal map of the semidisc
1 11onto the upper half-plane.
Problem 5.3.7 (Fa97) map the region inside the disc given by(Z E C Iz — ii 1) and outside the disc fz C I Iz — onto theupper half-plane.
Problem 53.8 (Sp95) Prove that there is no one-to-one conformal mapof the punctured disc G = (z E C I 0 < fzI < 1) onto the annulusA=(zECI1 <IzI<2).
Problem 53.9 (Fa02) Let n be a positive Find a group of linear frac-tional transformations of C thatfixes the two points z = I and z = —1 and hasorder n.
5.4 Functions on the Unit Disc
Problem 5.4.1 (Fa02) Let a be a number in (0, ir/2). Prove that the functionf(z) = is uniformly continuous in Sa = (z 1 0 < IzI 1, IArgzi a),a sector of the complex plane.
Problem 5.4.2 (Fa82) Let a and b be nonzero complex numbers andf(z) = az + bz1. Determine the image under f of the unit circle fz I Izi = 1).
Problem 5.4.3 (Su83, Fa96) Let f be analytic on and inside the unit circleC = (z I Iz I 1). Let L be the length of the image of C under f. Show thatL 2irjf'(O)I.
Problem 5.4.4 (Sp80) Let
1(z) =
be analytic in the disc D = (z E C I Izi < 1). Assume f maps D one-to-one ontoa domain G having area A. Prove
A
Problem 5.4.5 (Su83) Compute the area of the image of the unit disc (zi lzI<1}under the map 1(z) = z ÷ z2/2.
Problem 5.4.6 (Sp80) Let
1(z) =
an analytic function in the open unit disc D. Assume that
n=2
with
Pmve that f is infective.
Problem 5.4.7 (Sp03) Let f(z) be afunction that is analytic in the unit disc 1D' =
tIzI < 1). Suppose that If(z)I 1 in D. Prove that 1(z) has at least two fixedpoints Zi and Z2. then 1(z) = zfor all z E D.
Problem 5.4.8 (Su85) For each k > 0, let Xk be the set of analytic functions1(z) on the open unit disc D such that
sup {(1 — IzIYc If(z)I}
isfinite. Show thatf E Xk if and only jff' E Xk+1.
Problem 5.4.9 (Sp88) Let the functionf be analytic in the open unit disc of thecomplex plane and real valued on the radii [0, 1) and [0, Prove thatf isconstant.
Problem 5.4.10 (Fa91) Let the function f be analytic in the disc I zI < 1 of thecomplex plane. Assume that there is a positive constant M such that
ç23r
I If'(re'°)IdO M, (0 r < 1).Jo
Pmve that
f lf(x)ldx(0,l)
Problem 5.4.11 (Fa78) Suppose h(z) is analytic in the whole plane, h(O) = 3+4i, and Ih(z)I 5 < 1. What is h'(O)?
Problem 5.4.12 (Fa98) Let f be analytic in the closed unit disc, withf(— log 2) = 0 and If(z)I 1e9 for all z with IzI = 1. How large can If(log2)jbe?
Problem 5.4.13 (Fa79, Fa90) Suppose that f is analytic on the open upper half-plane andsatisfies If(z)I lforallz, f(i) =0. How large can If(2i)I be underthese conditions?
Problem 5.4.14 (Fa85) Let f(z) be analytic on the right half-planeH = (z I > 0) and suppose If(z)l 1 for z E H. Suppose also thatf(1) = 0. What is the largest possible value of If'(l)I?
Problem 5.4.15 (Su82) Let f(z) be analytic on the open unit disc D={zIIzI.<1).Prove that there is a sequence (Zn) in 11) such that —÷ 1 and isbounded.
Problem 5.4.16 (Sp93) Let f be an analytic function in the unit disc, Izi < 1.
1. Prove that there is a sequence (Zn) in the unit disc with 1
and f (Zn) exists (finitely).
2. Assume f nonconstant. Prove that there are two sequences (Zn) and (Wn)in the disc such that IZnI = limn...÷c,o IWnI = 1, and such that bothlimits and f(Wn) exist (finitely) and are not equal.
Problem 5.4.17 (Fa81, Sp89, Fa97) Let f be a holomorphic map of the unit disc
= IZ IIzI < 1) into itself, which is not the identity map f(z) z. Show that f
can have, at most, one fixed point.
Problem 5.4.18 (Fa87) If f(z) is analytic in the open disc Izi < 1, andIf(z)I 1/(1 — IzI), show that
/ 1\n<e(n+1).nj
Problem 5.4.19 (Sp88) 1. Let f be an analytic function that maps the openunit disc, lED, into itself and vanishes at the origin. Prove thatIf(z) + f(—z)I ( 2Iz12 in TED.
2. Prove that the inequality in Part 1 is strict, except at the origin, unless fhas the form f(Z) with A a constant of absolute value one.
Problem 5.4.20 (Sp91) Let the function f be analytic in the unit disc, with
If(z)I 1 and f(0) 0. Assume that there is a number r in (0, 1) such thatf(r) = f(—r) = 0. Prove that
— r2If(z)I Izi _r2Z2
Problem 5.4.21 (SpSS) Let f(z) be an analytic function that maps the open disc
IzI <1 into itself Show that lf'(z)J I /(1 — 1z12).
Problem 5.4.22 (Sp87, Fa89) Letf be an analytic function in the open unit discof the complex plane such that If(z) I C/ (I — I z I) for all z in the disc, where Cis a positive constant. Prove that If'(z)I 4C/(1 — IzD2.
5.5 Growth Conditions
Problem 5.5.1 (Sp03) Let f be an entire functi on such that —2 for all
z C. Show that f is constant.
Problem 5.5.2 (Fa90) Let the function f be analytic in the entire complex plane,and suppose that f(z)/z —÷ 0 as Izi —* 00. Prove that f is constant.
Problem 53.3 (Fa99) Let the rational function f in the complex plane have nopoles for 0. Prove that
supflf(z)I I = sup(If(z)I I = 0).
Problem 5.5.4 (Fa97) Let f be an entire function such that, for all z, If(z)I =I sin Prove that there is a constant C of modulus 1 such that f(z) = C sin z.
Problem 53.5 (Fa79, SuSi) Suppose f andg are entire functions with If(z)IIg(z)I for all z. Prove that f(z) = cg(z) for some constant c.
Problem 5.5.6 (Sp02) Let p and q be nonconstant complex polynomials of thesame degree whose zeros lie in the open disk IzI < 1. Prove that if Ip(z)I = Iq(z)Ifor IzI 1 then q = Apfor a unimodular constant A.
Problem 53.7 (Fa98) Let f be an entire function. Define = C \ (—oo, 0],the complex plane with the ray (—oo, 0] removed. Suppose that for all z E c�,
If(z)I I Jog zI, where log z is the principal branch of the logarithm. What canone conclude about the function f?
Problem 53.8 (Sp97) Let f and g be two entire functions such that, for allz C, k (independent of z). Showthat there are constants a, b such that
1(z) =ag(z)+b.
Problem 5.5.9 (Sp02) Let the continuous real—valued function ço on the complexplane satisfy the Sub—Mean—Value Property: for any point zo,
p2irço(zo) — I 60(zo + re'6)dO (0 <r < I).2irj0
Prove that ço obeys the Maximum Modulus Principle: the maximum ofço over anycompact set K is attained on the boundaiy of K.
Problem 5.5.10 (Su78) Let f : C —÷ C be an entire function and let a > 0 andb> 0 be constants.
1. If If(z)I bforallz, prove that f isa constant.
2. What can one prove about f jf
If(z)I
for all z?
Problem 5.5.11 (Fa90) Let the function f be analytic in the entire complex planeand satisfy
p2ir
I I f(re'0 ) I dO r17'3Jo
for all r > 0. Prove that f is the zero function.
Problem 5.5.12 (Fa96) Does there exist a function f, analytic in the puncturedplane C \ (0), such that
If(z)I
for all nonzero z?
Problem 5.5.13 (Fa91) Let the function f be analytic in the entire complex planeand satisfy the inequality If(z)I /2 off the imaginary axis. Prove thatfis constant.
Problem 5.5.14 (FaOl) Let the nonconstant entire function f satisfy the condi-tions
1. f(O)=O;
2. for each M > 0, the set {z I If(z)I <M) is connected.
Prove that f(z) = c and positive integer n.
5.6 Analytic and Meromorphic Functions
Problem 5.6.1 (Sp96) Let f = u + iv be analytic in a connected open set D,where u and v are real valued. Suppose there are real constants a, b and c suchthata2+b2
au+bv=cin D. Show that f is constant in D.
Problem 5.6.2 (Sp88) True or false: A function f(z) analytic on Iz —a I
r extends, for some 8 > 0, to a function analytic onlz — a <r + 8? Give a pro of or a counterexample.
Problem 5.6.3 (Fa80) Do there exist functions 1(z) and g (z) that are analytic atz = 0 and that satisfy
1. f(1/n) = f(—1/n) = 1/n2, n = 1,2,...,
2. g(1/n) =g(—1/n)= 1/n3,n = 1,2,...?
Problem 5.6.4 (Fa02) Let k be an integer larger than 1. Find all entire functionsf that satisfy f(zk) = (f(z)Y'.
Problem 5.6.5 (FaOl) Let F be a polynomial over C of positive degree d, andlet S be its zero set. Prove that every rationalfunction R whose finite poles lie inS can be written uniquely as
where m and n are integers, m n, and the ak are polynomials whose degreesare less than d.
Problem 5.6.6 (1?a99) Let A = fOJU fl/n I n Z, n> 1), and letD be the openunit disc in the complex plane. Prove that every bounded holomorphic function onID) \ A extends to a holomorphic function on II).
Problem 5.6.7 (FaOO) Let U be a connected and simply connected open subsetof the complex plane, and let f be a holomorphic function on U. Suppose a is apoint of U such that the Taylor series off at a converges on an open disc D thatintersects the complement of U. Does it follow that f extends to a holomorphicfunction on U U D? Give a proof or a counterexample.
Problem 5.6.8 (Su78) 1. Suppose f is analytic on a connected open setU C C andf takes only real values. Prove thatf is constant.
2. Suppose W C C is open, g is analytic on W, and g'(z) Ofor all Z E W.Show that
(Iftg(z) + I z WI c R
is an open subset of R.
Problem 5.6.9 (Sp78) Let f : C -÷ C be a nonconstant entire function. Provethat 1(C) is dense in C.
Problem 5.6.10 (Fa03) Let L be a line in C, and let f be an entire functionsuch that f(C) 1) L = 0. Prove that f is constant. (Do not use the theorem ofPicard that the image of a nonconstant entire function omits at most one complexnumbei)
Problem 5.6.11 (SpOl) 1. Prove that an entire function with a positive realpart is constant.
2. Prove the analogous result for 2 x 2 matrix functions: If F(z) = (fjk (z))is a matrix function in the complex plane, each entry being entire, andif F(z) + F(z)* is positive definite for each z, then F is constant. (Here,F(Z)* is the conjugate transpose of F(z).)
Problem 5.6.12 (Su82) Let s(y) and t(y) be real differentiable functions of y,<00, such that the complex function
f(x + iy) = eX (s(y) + it(y))
is complex analytic with s(0) = I and t(0) =0. Determine s(y) andt(y).
Problem 5.6.13 (Sp83) Determine all the complex analytic functions f definedon the unit disc ID) which satisfy
fFlQ.)+f(!) =0
forn = 2,3,4
Problem 5.6.14 (FaOO) Assume the nonconstant entire function f takes real val-ues on two intersecting lines in the complex plane. Prove that the measure ofeither angle formed by the lines is a rational multiple of it.
Problem 5.6.15 (Su83) Let be an open subset of 1R2, and let f : —÷
be a smooth map. Assume that f preserves orientation and maps any pair oforthogonal curves to a pair of orthogonal curves. Show that f is holomorphic.Note: Here we identify 1R2 with C.
Problem 5.6.16 (Sp87) Letf be a complex valued function in the open unit disc,ID), of the complex plane such that the functions g = f2 and h = f3 are bothanalytic. Prove that f is analytic in ID).
Problem 5.6.17 (Fa84) Prove or supply a counterexample: If f is a continuouscomplex valued function defined on a connected open subset of the complex planeand if f2 is analytic, then f is analytic.
Problem 5.6.18 (Sp88) 1. Let G be an open connected subset of the complexplane, fan analytic function in G, not identically 0, and n a positive integer.Assume that f has an analytic root in G; that is, there is an analyticfunction g in G such that gfl = f. Prove thatf has exactly n analyticroots in G.
2. Give an example of a continuous real valued function on [0, 1] that hasmore than two continuous square roots on [0, 1].
Problem 5.6.19 (Sp99) 1. Prove that if f is holomorphic on the unit disc ID)and f(z) Ofor all z ID), then there is a holomorphic function g onsuch that f(z) for all z ID).
2. Does the conclusion of Part 1 remain true if 11) is replaced by an arbitraryconnected open set in C?
Problem 5.6.20 (Fa92) Let the function f be analytic in the region Izi> I of thecomplex plane. Prove that if f is real valued on the interval (1, oo) of the realaxis, then f is also real valued on the interval (—oo, —I).
Problem 5.6.21 (Fa94) Let the function f be analytic in the complex plane, realon the real axis, 0 at the origin, and not identically 0. Prove that if f maps theimaginary axis into a straight line, then that straight line must be either the realaxis or the imaginary axis.
Problem 5.6.22 (FaOl) Let the entire function f be real valued on the lines= 0 and £iz = it. Prove that f has 2iri as a period: f(z + 2iri) = f(z)
for allz.
Problem 5.6.23 (Fa87) Let f(z) be analytic for z 0, and suppose thatf(1/z) = f(z). Suppose also that f(z) is real for all z on the unit circle Izi = 1.Prove that f(z) is real for all real z 0.
Problem 5.6.24 (Fa91) Let p be a nonconstant complex polynomial whose zerosare all in the half-plane > 0.
1. Prove that > 0 on the real axis.
2. Find a relation between deg p and
dx.p(x)
Problem 5.6.25 (Sp92) Let f be an analytic function in the connected open sub-set G of the complex plane. Assume that for each point z in G, there is a positiveinteger n such that the nih derivative off vanishes at z. Prove thatf is a polyno-mial.
Problem 5.6.26 (Sp92) Let the function f be analytic in the entire complex plane,real valued on the real axis, and ofpositive imaginary part in the upper half-plane.Prove f'(x) > Oforx real.
Problem 5.6.27 (Sp94) 1. Let U and V be open connected subsets of thecomplex plane, and let f be an analyticfunction in U such that f(U) C V.Assume is compact whenever K is a compact subset of V. Provethatf(U) = V.
2. Prove that the last equality can fail if analytic is replaced by continuous inthe preceding statement.
Problem 5.6.28 (Sp94) Let f = u + iv and g = p + iq be analytic functions de-fined in a neighborhood of the origin in the complex plane. AssumeIg'(O)I < If'(O)I. Prove that there is a neighborhood of the origin in whichthe function h = f + is one-to-one.
Problem 5.6.29 (Sp87) Prove or disprove. If the function f is analytic in theentire complex plane, and iff maps every unbounded sequence to an unboundedsequence, then f is a polynomial.
Problem 5.630 (Fa88, Sp97) Determine the gmup Aut(C) of all one-to-oneanalytic maps of C onto C.
Problem 5.631 (Sp77) Let f(z) be a nonconstantmeromorphicfunction. A com-plex nwnber w is called a period off f (z)for all z.
1. Show that if Wi and w2 are periods, so are +n2w2 for all integers fliandn2.
2. Show that there are, at most, afinite number of periods off in any boundedregion of the complex plane.
Problem 5.6.32 (Sp91) Let the function f be analytic in the punctured disc0 < izI <ro, with Laurent series
f(z) =
Assume there is a positive number M such that
r4 If(re'°)12d0 <M, 0 <r <ro.Jo
Prove that = 0for n <—2.
Problem 5.6.33 (Sp98) Let a > 0. Show that the complex function
1+z+az2f(z)=1 —z +az
satisfies If(z)( <1 for all z in the open left half-plane <0.
Problem 5.634 (Fa93) Let f be a continuous real valued function on [0, 1], andlet the function h in the complex plane be defined by
'Iih(z) = I f(t) cos zt dt.
Jo
1. Prove that h is analytic in the entire plane.
2. Prove thath is the zero function only is the zero function.
Problem 5.635 (Su79, Sp82, Sp91, Sp96) Let f be a continuous complex val-ued function on [0, 1], and define the function g by
I"g(z) = / f(t)e(Z dt (z C).
Jo
Prove that g is analytic in the entire complex plane.
Problem 5.636 (Fa84, Fa95) Let f and g be analytic functions in the open unitdisc, and let Cr denote the circle with center 0 and radius r, oriented counter-clockwise.
1. Prove that the integral
1.1 !f(w)g (!) dw2irz c,w w
is independent of r as long as IzI <r < 1 and that it defines an analyticfunction h(z), IzI < 1.
2. Prove or supply a counterexainple: 1ff 0 and g 0, then h 0.
Problem 5.6.37 (Sp84) Let F be a continuous complex valued function on theinterval [0, 1]. Let
f(z)F(t)
dt,o t—z
for z a complex number not in [0, 1].
1. Prove that f is an analytic function.
2. Express the coefficients of the Laurent series of f about 00 in terms of F.Use the result to show that F is uniquely determined by f.
Problem 5.638 (Fa03) Let f (z) be a meromorphic function on the complex plane.Suppose that for every polynomial p(z) C [z] and every closed contour Favoiding the poles off, we have
I p(z)2f(z)dz = 0.Jr
Prove that f(z) is entire.
5.7 Cauchy's Theorem
Problem 5.7.1 (Fa85) Evaluate
j27ree'° dO.
Problem 5.7.2 (Su78) Evaluate
I —10) dO.Jo
Problem 5.7.3 (Sp98) Let a be a complex number with IaI < 1. Evaluate theintegral
['IzII Iz—a12
Problem 5.7.4 (Fa99) For 0 <a <b, evaluate the integral
1f23r 1
= 2ir Jo laS° —
Problem 5.7.5 (Sp77, Sp82) Prove the Fundamental Theorem of Algebra: Everynonconstant polynomial with complex coefficients has a complex root.'
Problem 5.7.6 (Su77) Letf be continuous on C and analytic on (zI
0).Prove that f must be analytic on C.
Problem 5.7.7 (FaOO) Let f(z) be the rational function p(z)/q(z), where p(z)and q (z) are nonzero polynomials with complex coefficients, such that the degreeof p(z) is less than the degree of q(z), and such that q(z) has no complex ze-ros with nonnegative imaginary part. Prove that if zo is a complex number withpositive imaginary part, then
p +00 f(t)f(zo) dt.
2irzj_00 t—ZO
Problem 5.7.8 (Fa78, Su79) Let f(z) = ao + a complexpolynomial of degree n > 0. Prove
I f2iri JIzI=R
Problem 5.7.9 (SpOl) Let f be an entire function such that
p2,r
I If(re'°)12 dO Ar2k (0 <r <oo),Jo
where k is a positive integer and A is a positive constant. Prove that f is a con-stant multiple of the function
Problem 5.7.10 (Fa95) Let f(z) = u(z) + iv(z) be holomorphic on IzI < 1, uand v real. Show that
p2g
I u(re'0 )2d0 = / v(re'0 )2d0Jo JO
forO < r < I if u(0)2 =
Problem 5.7.11 (Su83) Let f : C -÷ C be an analytic function such that
(1 + iz,k)' dmf
is bounded for some k and m. Prove that d!7f/dzIz is identically zero for suffi-ciently large n. How large must n be, in terms of k and m?
Problem 5.7.12 (Su83) Suppose is a bounded domain in C with a boundaryconsisting of a smooth Jordan curve y. Let f be holomorphic on a neighborhoodof the closure of and suppose that f (z) 0 for z E y. Let z 1, ..., Zk be thezeros off in and let be the order of the zero off at Zj (for j = 1, ... , k).
1. Use Cauchy's integral formula to show that
2iri £dz
= k
2. Suppose that f has only one zero in with multiplicity nj = 1. Find aboundary integral involving f whose value is the point z i.
Problem 5.7.13 (Fa88) Let f be an analytic function on a disc D whose centeris the point zo. Assume that If'(z) — f'(zo)I < If'(zo)I on D. Prove that fone-to-one on D.
Problem 5.7.14 (Fa89) Let f(z) be analytic in the annulus = (1 <IzI <2}.Assume that f has no zeros in Show that there exists an integer n and ananalytic function g in such that,for all z E f(z) =
Problem 5.7.15 (Sp02) Define the function f on C \[0, 1] by
f(z)=I di'.Jo t—Z
Prove that f is analytic, and find its Laurent series about cc.
Problem 5.7.16 (Sp90) Let the function f be analytic and bounded in the com-plex half-plane IRz > 0. Prove that for any positive real number c, the function fis uniformly continuous in the half-plane c.
5.8 Zeros and Singularities
Problem 5.8.1 (Fa77, Fa96) Let C3 denote the set of ordered triples of complexnumbers. Define a map F : C3 —÷ C3 by
F(u,v,w)=(u+v+w,uv+vw+wu,uvw).
P rove that F is onto but not one-to-one.
Problem 5.8.2 (Fa79, Fa89) Prove that the polynomial
p(z) = z47 — z23 + 2z11 — z5 + 4z2 + 1
has at least one root in the disc IzI <1.
Problem 5.83 (Fa80) Suppose that f is analytic inside and on the unit circleI z = .1 and satisfies If (z) I < 1 for I z I = 1. Show that the equation 1(z) = z3
has exactly three solutions (counting multiplicities) inside the unit circle.
Problem 5.8.4 (Fa81) 1. How many zeros does the function f(z) =have inside the unit circle (counting multiplicities)?
2. Are the zeros distinct?
Problem 5.8.5 (Fa92) 1. How many roots does the polynomial defined byp(z) = 2z5 + 4z2 + 1 have in the disc IzI <1?
2. How many roots does the same polynomial have on the real axis?
Problem 5.8.6 (Su80) How many zeros does the complex polynomial
3z9 + 8z6 + z5 + 2z3 + I
have in the annulus 1 <IzI <2?
Problem 5.8.7 (Fa83) Consider the polynomial
p(z)=z5+z3+5z2+2.
How many zeros (counting multiplicities) does p have in the annular region1 < IzI <2?
Problem 5.8.8 (Sp84, Fa87, Fa96) Find the number of roots of
z7—4z3— 11=0
which lie between the two circles Izi = 1 and IzI =2.
Problem 5.8.9 (F'aOl) Let e be a positive Prove that the polynomialp(z) = ez3 — z2 — 1 has exactly two roots in the half-plane <0.
Problem 5.8.10 (Sp96) Let r < 1 <R. Show thatfor all sufficiently small e > 0,the polynomial
p(z) = ez7 + z2 + 1
has exactly five roots (counted with their multiplicities) inside the annulus
rs115 <IzI <Rs"
Problem 5.8.11 (Sp86) Let the 3x3 matrixfunction A be defined on the complex
plane by14z2 1 —1
A(z)=: —1 2z2 00 1
How many distinct values of z are there such that IzI <1 and A(z) is not invert-ible?
Problem 5.8.12 (Fa85) How many roots does the polynomial z4 + 3z2 + z + Ihave in the right half z-plane?
Problem 5.8.13 (Sp87) Prove that if the nonconstant polynomial p(z), with com-plex coefficients, has all of its roots in the half-plane Iitz > 0, then all of the rootsof its derivative are in the same half-plane.
Problem 5.8.14 (SpOO) Let f be a nonconstant entire function whose values onthe real axis are real and nonnegative. Prove that all real zeros of f have evenorder.
Problem 5.8.15 (Sp92) Let p be a nonconstant polynomial with real coefficientsand only real roots. Prove that for each real number r, the polynomial p — rp'has only real roots.
Problem 5.8.16 (Sp79, Su85, Sp89) Prove that if I <A <00, the function
fx(z)= Z+A_ez
has only one zero in the half-plane <0, and that this zero is real.
Problem 5.8.17 (Fa85) Prove that for every A> 1, the equation = I hasexactly one root in the disc IzI < 1 and that this root is real.
Problem 5.8.18 (Sp85) Prove that for any a E C and any integer n 2, theequation 1 + z + = 0 has at least one root in the disc Izi 2.
Problem 5.8.19 (Sp98) Prove that the polynomial z4 + z3 + 1 has exactly oneroot in the quadrant (z = x + iy I x, y> 0).
Problem 5.8.20 (Sp98) Let f be analytic in an open set containing the closedunit disc. Suppose that If (z) I > m for Iz I
= 1 and If <m. Prove that f(z)has at least one zero in the open unit disc IzI < 1.
Problem 5.8.21 (Su82) Let 0 < ao Prove that the equation
=0
has no roots in the disc IzI < 1.
Problem 5.8.22 (Fa86) Show that the polynomial p(z) — 6z + 3 has fivedistinct complex roots, of which exactly three (and not five) are reaL
Problem 5.8.23 (5p90) Let CO. . . ., be complex numbers. Prove that allthe zeros of the polynomial
zn+cn_lzn—l +w+cIZ+CO
lie in the open disc with center 0 and radius
+
Problem 5.8.24 (Sp95) Let P(x) be a polynomial with real coefficients and withleading coefficient 1. Suppose that P(0) = —1 and that P(x) has no complexzeros inside the unit circle. Prove that P(1) = 0.
Problem 5.8.25 (Su81) Prove that the number of roots of the equationz7-" + a natural numbe, a and fi real, nonzero) that havepositive real part is
1. n is even, and
2. n—lifnisodd.
Problem 5.8.26 (Su84) Let p > 0. Show that for n large enough, all the zeros of
++ 1
z 2!z
lie in the circle IzI <p.
Problem 5.8.27 (Fa88) Do the functions f(z) = ez + z and g(z) = Zez + 1 havethe same number of zeros in the strip <
Problem 5.8.28 (Sp93) Let a be a complex number and s a positive numberProve that the function 1(z) = sinz + has infinitely many zeros in the strip
Problem 5.8.29 (S p99) Suppose that f is holomorphic on some neighborhoodof a in the complex plane. Prove that either f is constant on some neighborhoodof a, or there exist an integer n > 0 and real numbers 8,8> 0 such that for eachcomplex number b satisfying 0 < Jb — f(a)I < e, the equation 1(z) = b hasexactly n roots in (z E C I I z — a I <6).
Problem 5.8.30 (Su77, SpSl) Let a0 + +• + be a polynomial having2 as a simple root. Show that there is a continuous function r: U C, whereU is a neighborhood of(âo, ... , in such that r(ao, . ..,a root of ao +alz + . and r(âo, .. . , = 2.
Problem 5.8.31 (Su85) Let
f(z) = >anztz
where all the are nonnegative reals, and the series has radius of convergence 1.Prove that f(z) cannot be analytically continued to afunction analytic in a neigh-borhoodofz = 1.
Problem 5.832 (Su80) Let f be a meromorp hi c function on C which is analyticin a neighborhood of 0. Let its Maclaurin series be
with all a pole of modulus r > 0 and no pole hasmodulus < r. Pmve there is a pole at z = r.
Problem 5.833 (Sp82) Decide, without too much computation, whether a finitelimit
((tan zY2 — z_2)
exists, wherez is a complex variable, and jf yes, compute the limit.
Problem 5.8.34 (Sp89, SpOO) Let f and g be entire functions such thatlirnf(g(z)) = 00. Prove thatf and g are polynomials.
Problem 5.8.35 (Fa98) Let ... , Zn be distinct complex numbers, and let, be nonzero complex numbers such that 5, = = 0 for
p = 0, 1, .. . , m — 1 but Sm 0. Here 1 m n — 1. How many zeros does therationalfunction 1(z) have in C? Why is m n impossible.
5.9 Harmonic Functions
Problem 5.9.1 (Fa77, Fa81) Let u : -+ IR be the function defined byu(x, y) = — 3xy2. Show that u is harmonic and find v : 1R2 —÷ IR suchthat thefunction f : C —+ C defined by
f(x+iy)=u(x,y)+iv(x,y)
is analytic.
Problem 5.9.2 (Fa80) Let 1(z) be an analytic function defined for IzI 1 andlet
u(x,y)=IItf(z), z =x+iy.Prove that
fau dtiI —dx——dy=0Jc8y dx
where C is the unit circle, x2 + y2 = 1.
Problem 5.93 (Fa83) 1. Let f be a complex function which is analytic on anopen set containing the disc Izi 1, and which is real valued on the unitcircle. Prove that f is constant.
2. Find a nonconstant function which is analytic at every point of the complexplane except for a single point on the unit circle Izi = 1, and which is realvalued at every other point of the unit circle.
Problem 5.9.4 (Fa92) Let s be a real numbe, and let the function u be definedin C\(—oo, 01 by
u(re'8) = rS cos sO (r > 0, —it <0 <it).
Prove that u is a harmonic function.
Problem 5.9.5 (Fa87) Let u be a positive harmonic function on R2; that is,
d2u a2u
ax2+
ay2=0.
Show that u is constant.
Problem 5.9.6 (Sp94) Let u be a real valued harmonic function in the complexplane such that
u(z)
a and b are positive constants. Prove that u is constant.
Problem 5.9.7 (Fa03) Let D = (z E C : Izi 1) \ {1, —1). Find an explicitcontinuous function f: D -+ R satisfying all the following conditions:
• f is harmonic on the interior of D (the open unit disk),
•
•
5.10 Residue Theory
Problem 5.10.1 (Fa83) Let rI, r2, ... , be distinct complex nwnbers. Show thata rationalfunction of the form
bo + biz + + +f(z)= (z—rl)(z—r2)"(z—rn)
can be written as a sum
A1 A2+ +•"+
z—r2 z—rn
for suitable constants A1, ... ,
Problem 5.10.2 (Fa82) Let
cot(irz)=
>
be the Laurent expansion for cot(Jrz) on the annulus I < <2. Compute the<0.
Problem 5.103 (Sp78) Show that there is a complex analytic function defined onthe set U = (z E C I Izi > 4) whose derivative is
z
(z—1)(z—2)(z—3)
Is there a complex analytic function on U whose derivative is
2z9
(z — I)(z — 2)(z — 3)
Problem 5.10.4 (Fa88) Let n be a positive Prove that the polynomial
i 2 n
=
in R[x] has n distinct complex zeros, Z2, ... and that they satisfy
for
Problem 5.10.5 (Sp79, Sp83) Let P and Q be complex polynomials with the de-gree of Q at least two more than the degree of P. Prove there is an r > 0 suchthat if C is a closed curve outside Izi = r, then
f P(z) dz=0.Jc Q(z)
Problem 5.10.6 (Sp93) Prove that for any fixed complex number
1p2ir °°
—,2irJ0n—O
Problem 5.10.7 (Fa99) Evaluate the integral
1 f (z+2)22jri 114=1 z2(2z
where the direction of integration is counterclockwise.
Problem 5.10.8 (Sp80) Let a > 0 be a constant 2. Let Ca denote the positivelyoriented circle of radius a centered at the origin. Evaluate
I
_______
z2(z — 2)dz.
Problem 5.10.9 (Sp02) Let r be a positive number and a a complex numbersuch that laI r.Evaluate the integral
.dzIzI=r a — z
where the circle IzI = r has the counterclockwise orientation.
Problem 5.10.10 (Su80) Let C denote the positively oriented circle Izt = 2,z E C. Evaluate the integral
where the branch of the square root is chosen so that 4122 — 1 > 0.
Problem 5.10.11 (Fa00) Evaluate the integral
2iri sin4z
where the direction of integration is counterclockwise.
Problem 5.10.12 (Fa02) Evaluate the integrals
1' 1
Jfl 1 • -dz, n=O,1,...,Jcn Z SIflZ
where is the circle IzI = (n + with the counterclockwise orientation.
Problem 5.10.13 (Su81) Compute
1 fdz2jri Ic
where C is the circle IzI = positively oriented.
Problem 5.10.14 (Su84) 1. Show that there is a unique analytic branch out-side the unit circle of the function 1(z) = + z + 1 such that f(t) ispositive when t> 1.
2. Using the branch determined in Part 1, calculate the integralif dz
2lriJC 41z2+Z+1
where C,. is the positively oriented circle Izi r and r> 1.
Problem 5.10.15 (Sp86) Let C be a simple closed contour enclosing the points0, 1, 2, . .. , k in the complex plane, with positive orientation. Evaluate the inte-grals
1kf — — , k=0,1,...,CZ(Z 1) (z k)
f (z—1)••(z—k)Jk I dz, k=0,1
Jc Z
Problem 5.10.16 (Sp86) Evaluate
I + 1)_2 dzJ 14=1
where the integral is taken in counterclockwise direction.
Problem 5.10.17 (Fa86) Evaluate
1 f zi'2iri JIzI=1 12z'2 4z9 + 2z6 — 4z3 + 1
dz
where the direction of integration is counterclockwise.
Problem 5.10.18 (Sp89) Evaluate
I (2z —Jc
where C is the circle Izi = 2 with counterclockwise Orientation.
Problem 5.10.19 (Fa90) Evaluate the integral
dz— 2tri Ic (z — 2)(I + 2z)2(1 — 3z)3
where C is the circle IzI = I with counterclockwise orientation.
Problem 5.10.20 (Fa91) Evaluate the integral
1 r1= dz,2irz c3z"—l
where n is a positive integei and C is the circle IzI = 1, with counterclockwiseorientation.
Problem 5.10.21 (Fa92) Evaluate
Ic z(2z+ 1)2dz,
where C is the unit circle with counterclockwise orientation.
Problem 5.10.22 (FaOl) Let a= 1 + b = —
1+i•Let F be the polygonal
path in the complex plane with successive vertices —1, —1 + i, 1, 1 + i, —1.Evaluate the integral
= Jr (z — a)2(z — b)3dz.
Problem 5.10.23 (Fa93) Evaluate the integral f, 1(z) dz for the function
f(z) = z2(1 — z2YleZ and the curve y depicted by
Problem 5.10.24 (Sp81) Evaluate
f2 dz
JcZ (z—l)where C is the closed curve shown below:
Problem 5.10.25 (Sp95) Let n be a positive integer and 0 <0 <ir. Prove that
1 f sinn0
2iri JIzI=2 1 — 2z cos 0 + —
where the circle I z I = 2 is oriented counterclockwise.
Problem 5.10.26 (Su77, Fa84, Sp94, Sp96) Use the Residue Theorem to evalu-ate the integral
1(a)=f dO
o a+cosOwhere a is real and a > 1. Why the formula obtained for 1(a) is also valid forcertain complex (nonreal) values of a?
Problem 5.10.27 (Fa78) Evaluate
(27r dO
Jo 1—2rcos0+r2
where r2 1.
Problem 5.10.28 (Sp87) Evaluate/1
C0540dO.
j0 1 + cos2O
Problem 5.10.29 (Fa87) Evaluate the integral
1—cos23O
dOJo 5—4cos2O
Problem 5.10.30 (SpOO) Evaluate
1=f COS3ZdZ14=:! Z
where the direction of integration is counterclockwise.
Problem 5.10.31 (Sp95) Let n be a positive integel Compute
1—cosn0
Problem 5.10.32 (Fa94) Evaluate the integrals
sinnOI dO, n=1,2
sinO
Problem 5.10.33 (Sp88) For a> 1 and n =0, 1,2, ..., evaluate the integrals
JC0SnO
dO, =sinnO
dO.a — cosO a — cosO
5.11 Integrals Along the Real Axis
Problem 5.11.1 (Sp86) Let the complex valued functions n E 7L, be definedon IR by
(x—i)'1= j)fl+l
Prove that these functions are orthonormal; that is,
(001 1 if m=njfm(x)fn(x)clx=j if
Problem 5.11.2 (Fa85) Evaluate the integral
1 —cosaxI dx
x2
fora E IR.
Problem 5.11.3 (SpOO) Evaluate the integrals
fOOJ(t)= •2dx, —oo<t<oo.
j_00(x+i)
Problem 5.11.4 (FaOl) Evaluate the integrals F(t)= f dx
—00 < t < 00.
Problem 5.11.5 (Sp78, Sp83, Sp97) Evaluate
(00dx.
J—oo
Problem 5.11.6 (Fa82, Sp92) Evaluate
(00 sin3x dx.J—0o
Problem 5.11.7 (Sp93) Evaluate
(°° x3sinxI dx.
Problem 5.11.8 (Sp81) Evaluate
(°° xsinxI dx.J_00(1+x2)2
Problem 5.11.9 (Sp90, Fa92) Let a be a positive real Evaluate the im-proper integral
f°° sinxI dx.
Jo x(x2+a2)
Problem 5.11.10 (Sp91) Prove that
fR sinxlimI dxx —31
exists and find its value.
Problem 5.11.11 (Sp83) Evaluatecoo sinxI dx.
Problem 5.11.12 (Fa02) Evaluatecoo cosxI dx.
(1 +x2)3
Problem 5.11.13 (Fa97) Evaluate the integral
(00 coskxI dxf—co 1+x+x2
where k 0.
Problem 5.11.14 (Fa82) Evaluate(00 cosirxI dx.J_004x21
Problem 5.11.15 (Sp77, Fa81, Sp82) Evaluate
P°° cosnxI dx.
J_ooX4+1
Problem 5.11.16 (Sp79) Evaluate
f°°x2+lI dx.
Jo
tOO cProblem 5.11.17 (Sp02) Evaluate the integrals I (a) = 2 2dxj_ooa +x
fora > 0.
Problem 5.11.18 (Su84) Evaluate
(00 xsinxdx.
J_oox2+4x+20
Problem 5.11.19 (Fa84) Evaluate
f°°x—sinxI dx.
x3
Problem 5.11.20 (Fa84) Evaluate
(00 dx
(1 +x +x2)2
Problem 5.11.21 (Fa79, Fa80, Sp85, Su85, Fa98, Sp99) Prove that
f00 Jrdx= •
J0 1+x sinna
What restrictions must be placed on a?
Problem 5.11.22 (Fa96) Evaluate the integral
fOO__1=1 dx.
Jo 1+x2
Problem 5.11.23 (SpOl) Evaluate
/'ooI dx.
J0 1+x5
Problem 5.11.24 (Fa77, Su82, Fa97) Evaluate[00 dxI
J—0O
where n is a positive
x2Problem 5.11.25 (Fa03) Evaluate dx, where n 4 is an even inte-
j_00x?2+l
Problem 5.11.26 (Fa88) Prove that
x fl.2I
Jo 8
Problem 5.11.27 (Fa93) Evaluate
I dx.x2-2x+4Problem 5.11.28 (Fa86) Evaluate
logx
Problem 5.11.29 (Fa94) Evaluate
[00dx.
Jo x2+1
Problem 5.1130 (Fa83) Evaluate
J(sech x)2 cos Ax dx
where A is a real constant and
2sechx=
ex +
Problem 5.11.31 (Sp85) Prove that
J cos(2bx) dx =
What restrictions, if any, need be placed on b?
Problem 5.11.32 (Sp03) Evaluate
1000 e_X2 cosx2 dx.
Problem 5.11.33 (Sp97) Prove that
00
1100
is independent of the real parameter y.
6
Algebra
6.1 Examples of Groups and General Theory
Problem 6.1.1 (Sp77) Let G be the collection of 2 x 2 real matrices with nonzerodeterminant. Define the product of two elements in G as the usual matrix product.
1. Show that G is a group.
2. Find the center Z of G; that is, the set of all elements z of G such thataz = zaforalla E G.
3. Show that the set 0 of real orthogonal matrices is a subgroup of G(a matrix is orthogonal if = I, where A' denotes the transpose ofA). Show by example that 0 is not a normal subgroup.
4. Find a nontrivial homomorphism from G onto an abelian group.
Problem 6.1.2 (Fa77) Let G be the set of3x3 real matrices with zeros below thediagonal and ones on the diagonal.
1. Prove G is a group under matrix multiplication.
2. Detennine the center of G.
Problem 6.1.3 (Su78) For each of the following either give an example or elseprove that no such example is possible.
1. A nonabelian group.
2. A finite abelian group that is not cyclic.
3. An infinite group with a subgroup of index 5.
4. Two finite groups that have the same order but are not isomorphic.
5. A group G with a subgroup H that is not normal.
6. A nonabelian group with no normal subgroups except the whole group andthe unit element.
7. A group G with a normal subgroup H such that the factor group G/H isnot isomorphic to any subgroup of G.
8. A group G with a subgroup H which has index 2 but is not normal.
Problem 6.1.4 (Fa80) Let R be a ring with multiplicative identity 1. Call x E Ra unit ifxy = yx = I for some y R. Let G(R) denote the set of units.
1. Prove G(R) is a multiplicative group.
2. Let R be the ring of complex numbers a + bi, where a and b are integers.Prove G(R) is isomorphic to Z4 (the additive group of integers modulo 4).
Problem 6.1.5 (Sp83) In the triangularnetwork in R2 depicted below, the pointsPo, Pi, P2, and P3 are respectively (0, 0), (1, 0), (0, 1), and (1, 1). Describe thestructure of the group of all Euclidean transformations of R2 which leave thisnetwork invariant.
\%\
7N/ N ...
..S
S
7 / NN S
S F SF
S S
Problem 6.1.6(Fa90) Does the setG = {a ]R a > 0,a 1)forma groupwith the operation a * b =
Problem 6.1.7 (Sp81) Let G be a finite group. A conjugacy class is a set of theform
C(a)={bab'IbEG}for some a C.
1. Prove that the number of elements in a conjugacy class divides the order ofG.
2. Do all conjugacy classes have the same number of elements?
3. If G has only two conjugacy classes, prove G has order 2.
Problem 6.1.8 (Sp91) Let G be a finite nontrivial group with the properly thatfor any two elements a and b in G different from the identity, there is an elementc in G such that b = Prove that G has order 2.
Problem 6.1.9 (Sp99) Let G be a finite group, with identity e. Suppose that forevery a, b E G distinct from e, there is an automorphism a of G such thata(a) = b. Prove that G is abelian.
Problem 6.1.10 (S p84) For a p-group of order p4, assume the center of G hasorder p2. Determine the number of conjugacy classes of G.
Problem 6.1.11 (Sp85) In a commutative group G, let the element a have or-der r, let b have order s (r, s < oo), and assume that the greatest commondivisor of r and s is 1. Show that ab has order rs.
Problem 6.1.12 (Fa85) Let G be a group. For any subset X of G, define its cen-tralizer C(X) to be {y G xy = yx, for all x E X}. Prove the following:
1. If X C Y, then C(Y) C C(X).
2. XcC(C(X)).
3. C(X) = C (C (C(X))).
Problem 6.1.13 (5p88) Let D be a group of order 2n, where n is odd, with asubgroup H of order n satisfying xhx1 = h1 for all h in H and allx in D \ H.Prove that H Is commutative and that every element of D \ H is of order 2.
6.2 Homomorphisms and Subgroups
Problem 6.2.1 (Fa78) How many homomorphisms are there fivm the groupZ2 x Z2 to the symmetric group on three objects?
Problem 6.2.2 (Fa03) 1. Let G be a finite group and let X be the set of pairsof commuting elements of G:
X ={(g,h)€ G x G
Prove that IXI = cIGI where c is the number of conjugacy classes in G.
2. Compute the number of pairs of commuting permutations on five letters.
Problem 6.2.3 (Fa03) The set of 5 x 5 complex matrices A satisfying A3 = A2
is a union of conjugacy classes. How many conjugacy classes?
Problem 6.2.4 (Sp90) Let C * be the multiplicative group of nonzero complexnumbers. Suppose that H is a subgroup offinite index of C *• Prove that H = C *•
Problem 6.23 (Su80) Let G be a finite group and H C G a subgroup.
1. Show that the number of subgroups of G of the form x Hx1 for some x E G
is the index ofHin G.
2. Prove that some element of G is not in any subgroup of the formXEG.
Problem 6.2.6 (Su79) Prove that the group of automorphisms of a cyclic groupof prime order p is cyclic and find its orde.
Problem 6.2.7 (Su81) Let G be a finite group, and let q be an automorphism ofG which leaves fixed only the identity element of G.
1. Show that every element of G may be written in theform
2. if q has on'Jer 2 (i.&, ço. q = id) show that q is given by the formulag i—f g1 and that G is an abelian group whose order is odd.
Problem 6.2.8 (Fa79, Sp88, Fa91) Prove that every finite group of order> 2has a nontrivial automorphism.
Problem 6.2.9 (Fa90) Let A be an additively written abelian group, andu, v: A —÷ A homomorphisms. Define the group homomorphisms f, g : A Aby
f(a) = a — v (u(a)), g(a) = a — u (v(a)) (a E A).
Prove that the kernel off is isomorphic to the kernel ofg.
Problem 6.2.10 (Su81) Let G be an additive group, and u, v: G G homo-morphisms. Show that the map f: G —* G, f(x) = x — v (u(x)) is surf ective ifthe maph : G G, h(x) =x —u(v(x)) issurjective.
Problem 6.2.11 (Sp83) Let H be the group of integers mod p. under addition,where p is a prime numbei Suppose that n is an integer satisfying I n p,and let G be the group H x H x ••• x H (n factors). Show that G has noautomorphism of order p2.
Problem 6.2.12 (Sp03) 1. Suppose that H1 and H2 are subgroups ofa groupG such that H1 U H2 is a subgroup of G. Prove that either H1 c F!2 orH2çH1.
2. Show that for each integer n 3, there exists a group G with subgroupsH1, H2, ..., such that no H1 is contained in any othei and such that
Problem 6.2.13 (Fa84) Let G be a group and H a subgroup of index n <00.Prove or disprove the following statements:
1. If a E G, then E H.
2. Ifa E G, rhenforsomek, 1 we have a' E H.
Problem 6.2.14 (Fa78) Find all automorphisms of the additive group of rationalnumbers.
Problem 6.2.15 (SpOl) Let a and $ be real numbers such that the subgroup I'of ]R. generated by a and is closed. Prove that a and $ are linearly dependentover Q.
Problem 6.2.16 (SpOO) Prove that the group G = Q/Z has no proper subgroupoffinite
Problem 6.2.17 (Fa87, Fa93) Let A be the group of rational numbers under ad-dition, and let M be the group of positive rational numbers under multiplication.Determine all homomorphisms ço: A M.
Problem 6.2.18 (SpOO) Suppose that H1 and Fl2 are distinct subgroups of a groupG such that [G : Hj] = [G : H2] = 3. What are the possible values of[G:HlflH2]?
Problem 6.2.19 (Fa92) Let G be a group and H and K subgroups such that Hhas a finite index in G. Prove that K fl H has a finite index in K.
Problem 6.2.20 (Fa94) Suppose the group G has a nontrivial subgroup H whichis contained in every nontrivial subgroup of G. Prove that H is contained in thecenter of G.
Problem 6.2.21 (FaOO) Show that for each positive integer k there exists a posi-tive integer N such that there are at least k nonisomorphic groups oforder N.
Problem 6.2.22 (Fa95) Let G be a group generated by n elements. Find an upperbound N(n, k)for the number of subgroups H of G with the index [G: H] = k.
6.3 Cyclic Groups
Problem 6.3.1 (Su77, Sp92) 1. Prove that every finitely generated subgroupof Q, the additive group of rational numbers, is cyclic.
2. Does the same conclusion hold for finitely generated subgroups of Q /Z,where Z is the group of integers?
Note: See also Problems 6.6.3 and 6.7.2.
Problem 6.3.2 (Sp98) Let G be the group Q /Z. Show that for every positiveinteger t, G has a unique cyclic subgroup of order t.
Problem 6.3.3 (Fa99) Show that a group G is isomorphic to a subgroup of theadditive group of the rationals if and only if G is countable and every finite subsetof G is contained in an infinite cyclic subgroup of G.
Problem 63.4 (Su85) 1. Let G be a cyclic group, and let a, b G be ele-ments which are not squares. Prove that ab is a square.
2. Give an example to show that this result is false if the group is not cyclic.
Problem 6.3.5 (Sp82) Prove that any group of order 77 is cyclic.
Problem 63.6 (Fa91) Let G be a group of order 2p, where p is an odd prime.Assume that G has a normal subgroup of order 2. Prove that G is cyclic.
Problem 63.7 (Fa97) A finite abelian group G has the property that for eachpositive integer n the set fx E G I x" = 1) has at most n elements. Prove that Gis cyclic, and deduce that every finite field has cyclic multiplicative group.
Problem 63.8 (FaOO) Let G be a finite group of order n with the property thatfor each divisor d of n there is at most one subgroup in G of order d. Show G iscyclic.
6.4 Normality, Quotients, and Homomorphisms
Problem 6.4.1 (Fa78) Let H be a subgroup of a finite group G.
1. Show that H has the same number of left cosets as right cosets.
2. Let G be the group of symmetries of the square. Find a subgroup H suchthatxH
Problem 6.4.2 (Fa80) Let G be the group of orthogonal transformations of R3to R3 with determinant 1. Let v E R3, lvi = 1, and let {T E G I Tv = v).
1. Show that is a subgroup of G.
2. Let = {T G I T is a rotation of 1800 about a line orthogonal to v).Show that is a coset of in G.
Problem 6.43 (Su84) Show that if a subgroup H of a group G has just one leftcoset different from itself then it is a normal subgroup of G.
Problem 6.4.4 (Su85) Let G be a group of order 120, let H be a subgroup oforder 24, and assume that there is at least one left coset of H (other than H itself)which is equal to some right coset of H. Prove that H is a normal subgroup of G.
Problem 6.4.5 (Fa02) Let G be a group of order 112. Prove that G has a non-trivial normal subgroup.
Problem 6.4.6 (Fa02) List all groups of order 6 up to isomorphism. Find agroup of order 120 that contains as a subgroup an isomorphic copy of each ofthem. Prove that no group of order < 120 has the preceding property.
Problem 6.4.7 (Sp89) For G a group and H a subgroup, let C(G, H) denote thecollection of left cosets of H in G. Prove that if H and K are Iwo subgroups of Gof infinite index then G is not afinite union of cosets from C(G, H) U C(G, K).
Problem 6.4.8 (Fa82, Fa92) Let
N is a normal subgroup of G and prove that C/N is isomorphictoR.
2. Find a normal subgroup N' of G satisfying N C N' C G (where theinclusions are proper), or prove that there is no such subgroup.
Problem 6.4.9 (Sp86) Let Z2 be the group of lattice points in the plane (orderedpairs of integers, with coordinatewise addition as the group operation). Let Hjbe the subgroup generated by the two elements (1,2) and (4, 1), and H2 the sub-group generated by the two elements (3,2) and (1,3). Are the quotient groupsCi = Z2/Hi and C2 = Z2/H2 isomorphic?
Problem 6.4.10 (Sp78, Fa81) Let C be a group of order 10 which has a nonnalsubgroup of order 2. Prove that C is abelian.
Problem 6.4.11 (Sp79, Fa81) Let C be a group with three normal subgroups N1,
N2, and N3. Suppose fl = fe) and = Gfor all i, j with i j. Showthat C is abelian and is isomorphic to for all i, j.
Problem 6.4.12 (Fa97) Suppose H1 is a normal subgroup of a group G forI i k, such that fl = (1)for i j. Prove that G contains a subgroup
isomorphic to Hi x H2 x• x Hk if k =2, but not necessarily if k 3.
Problem 6.4.13 (Sp80) G is a group of order n, H a proper subgroup of order m,and (n/rn)! < 2n. Prove G has a proper normal subgroup different from theidentity.
Problem 6.4.14 (Sp82, Sp93) Prove that is a group containing no subgroupof index 2, then any subgroup of index 3 is normal.
Problem 6.4.15 (Sp03) Suppose G is a nonabelian simple group, and A is itsautomorphism group. Show that A contains a normal subgroup isomorphic to G.
Problem 6.4.16 (Sp89) Let G be a group whose order is twice an odd numberFor g in G, let Ag denote the permutation of G given by ).8(x) = gxfor x E G.
1. Let g be in G. Prove that the permutation Ag is even if and only if the orderof g is odiL
2. Let N = (g E G I order(g) is odd). Prove that N is a normal subgroup ofG of index 2.
Problem 6.4.17 (Fa89) Let G be a group, G' its commutator subgroup, and Na normal subgroup of G. Suppose that N is cyclic. Prove that gn = ng for allg G'andalln EN.
Problem 6.4.18 (Fa90) Let G be a group and N be a normal subgroup of Gwith N G. Suppose that there does not exist a subgroup H of G satisfyingN C H C G and N H G. Prove that the index of N in G is finite and equalto a prime numbem
Problem 6.4.19 (Sp94) Let G be a group having a subgroup A of finite index.Prove that there is a normal subgroup N of G contained in A such that N is offinite index in G.
Problem 6.4.20 (Sp97) Let H be the quotient of an abelian group G by a sub-group K. Prove or disprove each of the following statements:
1. If H is finite cyclic then G is isomorphic to the direct product of H and K.
2. If H is a direct product of infinite cyclic groups then G is isomorphic to thedirect product of H and K.
6.5 Sn, An, Dn,
Problem 6.5.1 (Fa80) Let F2 = (0, 1) be the field with two elements. Let G bethe group of invertible 2 x 2 matrices with entries in F2. Show that G is isomorphicto S3, the group of permutations of three objects.
Problem 6.5.2 (Su84) Let Sn denote the group ofpermutations ofn objects. Findfour different subgroups of S4 isomorphic lo 53 and nine isomorphic to S2.
Problem 6.5.3 (Fa86) Let G be a subgroup 0185, the group of all permutationsoffive objects Prove that if G contains a 5-cycle and a 2-cycle, then G = S5.
Problem 6.5.4 (Fa85) Let G be a subgroup of the symmetric group on six objects,S6. Assume that G has an element of order 6. Prove that G has a normal subgroupH of index 2.
Problem 6.5.5 (Sp79) Let S7 be the group of permutations of a set of seven ob-jects. Find all n such that some element of 57 has order n.
Problem 6.5.6 (Sp80) S9 is the group of permutations of 9 objects.
1. Exhibit an element of S9 of order 20.
2. Prove that no element of S9 has order 18.
Problem 63.7 (Sp88, SpOl) Let S9 denote the group ofpermutations of nine ob-jects and let A9 be the subgroup consisting of all even permutations. Denote by1 E S9 the identity permutation. Determine the minimum of all positive integersm such that every a E S9 satisfies am = 1. Determine also the minimum of allpositive integers m such that every a E A9 satisfies am = 1.
Problem 6.5.8 (Sp92) Let denote the group of permutations of 999 objects,and let G C be an abelian subgroup of order 1111. Prove that there existsi E (1,...,999)suchthatforalla E G,onehasa(i) =i.
Problem 6.5.9 (Sp02) Prove that Sn, the group of permutations of(1, 2, . . .,is isomorphic to a subgroup of the alternating subgroup Of
Problem 6.5.10 (Fa81, Sp95) Let be the group of all permutations of n ob-jects and let G be a subgroup of of order where p is a prime not dividingn. Show that G has a fixed point; that is, one of the objects is left fixed by everyelement of G.
Problem 6.5.11 (Sp80) Let G be a subgroup of the group ofpermutations of nobjects. Assume G is transitive; that is, for any x and y in 5, there is some a E G
witha(x) =y.
1. Prove that n divides the order of G.
2. Suppose n 4. For which integers k I can such a G have order 4k?
Problem 6.5.12 (Su83) Let G be a transitive subgroup of the group ofpermu-tations of n objects (1, ... , n). Suppose that G is a simple group and that is anequivalence relation on {1, ... , n) such that i -s-' j implies that a(i) a(j)forall a G. What can one conclude about the relation ''?
Problem 6.5.13 (Sp89) Let be the dihedral group, the group of rigid motionsof a regular n-gon (n 3). (It is a noncommutative group of order 2n.) Determineits center Z = (c I cx = xc for all x E
Problem 6.5.14 (Fa92) How many Sylow 2-subgroups does the dihedral groupof order 2n have, when n is odd?
6.6 Direct Products
Problem 6.6.1 (Fa83) Let G be afinite group and suppose that G x G has exactlyfour normal subgroups. Show that G is simple and nonabelian.
Problem 6.6.2 (Sp99) Let G be a finite simple group of order n. Determine thenumber of normal subgroups in the direct product G x G.
Problem 6.63 (Sp91) Prove that Q, the additive group of rational numbers,cannot be written as the direct sum of two nontrivial subgroups.Note: See also Problems 6.3.1 and 6.7.2.
Problem 6.6.4 (Su79, Fa93) Let A, B, and C be finite abelian groups such thatA x B and Ax C are isomorphic. Prove that B and C are isomorphic.
Problem 6.65 (Su83) Let Gj, G2, and G3 be finite groups, each of which is gen-erated by its commutators (elements of the form y1). Let A be a subgroupof G1 x G2 x G3, which maps surjectively, by the natural projection map, tothe partial products Gi x G2, G1 x G3 and G2 x G3. Show that A is equal toG1xG2xG3.
Problem 6.6.6 (Fa82) Let A be a subgroup of an abelian group B. Assume that Ais a direct summand of B, i.e., there exists a subgroup X of B such that A1)X 0and such that B = X + A. Suppose that C is a subgroup of B and satisfyingA C C C B. Is A necessarily a direct summandof C?
Problem 6.6.7 (Fa87, Sp96) Let G and H be finite groups of relatively primeShow that Aut(G x H), the group of automorphisms of G x H, is isomorphic
to the direct product of Aut(G) and Aut(H).
Problem 6.6.8 (Fa03) Letp be a prime, and let G be the group x Z,,. Howmany automorphisms does G have?
6.7 Free Groups, Generators, and Relations
Problem 6.7.1 (Sp77) Let Q + be the multiplicative group of positive rationalnumbers.
1. Is Q + torsionfree?
2.
Problem 6.7.2 (Sp86) Prove that the additive group of Q, the rational numberfield, is not finitely generated.Note: See also Problems 6.3.1 and 6.6.3.
Problem 6.7.3 (Fa79, Fa82) Let 0 be the abelian group defined by generatorsx, y, and z, and relations
15x+3y =03x+7y+4z=0
18x + l4y + 8z = 0.
1. Express G as a direct product of two cyclic groups.
2. Express 0 as a direct product of cyclic groups of prime power order.
3. How many elements of 0 have order 2?
Problem 6.7.4 (Sp82, Sp93) Suppose that the group 0 is generated by elementsx andy thatsatisfyx5y3 = x8y5 = 1. Is G the tn vial group?
Problem 6.7.5 (Su82) Let G be a group with generators a and b satisfying
a1b2a = b3, b1a2b = a3.
Is G trivial?
Problem 6.7.6 (Fa88, Fa97) Let the group G be generated by two elements, aand b, both of order 2. Pmve that 0 has a subgroup of index 2.
Problem 6.7.7 (Fa89) Let be the free group on n generators. Show that G2and G3 are not isomorphic.
Problem 6.7.8 (Sp83) Let 0 be an abelian group which is generated by, at most,n elements. Show that each subgroup of G is again generated by, at most, n ele-ments.
Problem 6.7.9 (Sp84) Determine all finitely generated abelian groups 0 whichhave only finitely many automorphisms.
Problem 6.7.10 (Fa89) Let A be a finite abelian group, and m the maximum ofthe orders of the elements of A. Put S = (a E A I lal = m}. Prove that A isgenerated by S.
6.8 Finite Groups
Problem 6.8.1 (Sp91) List, to within isomorphism, all the finite groups whose
orders do not exceed 5.
Problem 6.8.2 (Fa84) Show that all groups of order 5 are commutative. Givean example of a noncommutative group of order 6.
Problem 6.8.3 (Fa80) Prove that any group of order 6 -is isomorphic to either Z6
or 53 (the group of pennutations of three objects).
Problem 6.8.4 (Sp87) 1. Show that, to within isomorphism, there is just onenoncyclic group G of order 4.
2. Show that the group of automorphisms of G is isomorphic to the permuta-tion group S3.
Problem 6.8.5 (Fa88) Find all abelian groups of order 8, up to isomorphism.Then ident4fy which type occurs in each of
1.
2. (7L17)*/(±1),
3. the roots of z8 — 1 in C,
4. Ft.
5.
Fg is the field of eight elements, and Ft is its underlying additive group; R* isthe group of invertible elements in the ring R, under multiplication.
Problem 6.8.6 (Sp90, Fa93, Sp94) Show that there are at least two nonisomor-phic nonabelian groups of order 24, of order 30 and order 40.
Problem 6.8.7 (Fa03) List eight groups of order 36 and prove that they are notisomorphic.
Problem 6.8.8 (Fa97) Prove that if p is prime then every group of order p2 isabelian.
Problem 6.8.9 (Sp93) Classify up to isomorphism all groups of order 45.
Problem 6.8.10 (Sp02) Let G be a group of order 56 having at least 7 elements oforder 7. Prove that G has only one Sylow 2-subgroup P, and that all nonidentityelements of P have order 2.
Problem 6.8.11 (Sp79, Sp97) Classify all abelian gmups of order 80 up to iso-morphism.
Problem 6.8.12 (Fa88) Find (up to isomorphism) all groups of order 2p, wherep isa prime(p 2).
Problem 6.8.13 (Sp87) Prove that any finite group of order n is isomorphic to asubgroup of 0(n), the group of n x n orthogonal real matrices.
Problem 6.8.14 (Fa98) Suppose that G is a finite group such that every Sylowsubgroup is normal and abelian. Show that G is abelian.
Problem 6.8.15 (SpOl) If G is a finite group, must S = {g2 I g E G) be a sub-group? Provide a proof or a counterexample.
Problem 6.8.16 (Fa99) Let G be a finite group acting transitively on a set X ofsize at least 2. Prove that some element g of G acts without fixed points.
Problem 6.817 (Fa02) Let G be a finite non-A belian group of order n. Showthat there exists an integer d satisfying 2 d n/2 and a set P of cardinality d,such that G acts transitively on P.
Problem 6.8.18 (Su80, Fa96) Prove that every finite group is isomorphic to
1. A group of permutations;
2. A group of even permutations.
Problem 6.8.19 (Sp00) Let G be a finite group and p a prime numbei: Supposea and b are elements of G of order p such that b is not in the subgroup generatedby a. Prove that G contains at least p2 — 1 elements of order p.
Problem 6.8.20 (Fa01) Find all finite abelian groups G (up to isomorphism)such that the group of automorphisrns of G has odd
Problem 6.8.21 (Fa03) Let n be a positive Let çb(n) be the Euler phifunction, so çb(n) = Prove that if gcd(n, > 1, then there exists anoncyclic group of order n.
6.9 Rings and Their Homomorphisms
Problem 6.9.1 (Fa80) Let be the ring of real 2x2 matrices and S Cthe subring of matrices of the form
(a —ba
1. Exhibit an isomorphism between S and C.
2. Prove that (0 31
lies in a subring isomorphic to S.
3. Prove that there is an X E M2x2 such that X4 + 13X = A.
Problem 6.9.2 (Sp03) Let denote the ring of 2 x 2 matrices with en-
tries in Q. Let R be the set of matrices in M2x2(Q) that commute with i)•
1. Prove that R is a subring of M2x2(Q).
2. Prove that R is isomorphic to the ring Q [x}/(x2).
Problem 6.9.3 (Sp86) Prove that there exists only one automorphism ofthefieldof real numbers, namely the identity automorphism.
Problem 6.9.4 (Sp86) Suppose addition and multiplication are defined on Ccomplex n-space, coordinatewise, making C into a ring. Find all ring homo-morphisms of C onto C.
Problem 6.9.5 (Fa88) Let R be afinite ring. Prove that there are positive integersm.andn withm > n such thatxm = x'1 for every x in R.
Problem 6.9.6 (Sp89) Let R be a ring with at least two elements. Suppose thatfor each nonzero a in R there is a unique b in R (depending on a) with aba = a.
Show that R is a division ring.
Problem 6.9.7 (Sp91) Letp be a prime number and R a ring with identity con-taining p2 elements. Prove that R is commutative.
Problem 6.9.8 (FaOO) Let R be a ring with identity, ha ving fewer than eight ele-ments. Prove that R is commutative.
Problem 6.9.9 (Sp03) Let R be the set of complex numbers oftheform
a+3bi, a,bEZ.
Prove that R is a subring of C, and that R is an integral domain but not a uniquefactorization domain.
Problem 6.9.10 (Fa93) Let R be a commutative ring with identity. Let G be afi-nite subgroup of R*, the group of units of R. Prove that if R is an integral domain,then G is cyclic.
Problem 6.9.11 (Fa98) Let R be a finite ring with identity. Let a be an elementof R which is not a zero divisor. Show that a is invertible.
Problem 6.9.12 (Fa94) Let R be a ring with identity, and let u be an element ofR with a right inverse. Prove that the following conditions on u are equivalent:
1. u has more than one right inverse;
2. u is a zero divisor;
3. u is not a unit.
Problem 6.9.13 (Su81, Sp93) Show that no commutative ring with identity hasadditive group isomorphic to Q /Z.
Problem 6.9.14 (Sp81) Let D be an ordered integral domain and a E D. Provethat
a2 —a+ 1> 0.
Problem 6.9.15 (Fa95) Prove that Q [x, y]/ (x2 + y2 — 1) is an integral domainand that its field offractions is isomorphic to the field of rational functions Q (t).
Problem 6.9.16 (SpOO) Find the cardinality of the set of all subrings of Q, thefield of rational numbers.
6.10 Ideals
Problem 6.10.1 (Sp98) Let A be the ring ofreal2x2 matrices ofthe formWhat are the 2-sided ideals in A?
Problem 6.10.2 (Fa79, Fa87) Let be the ring of n x n matrices over afield F. Prove that it has no 2-sided ideals except and (0).
Problem 6.10.3 (Fa83, Su85) Let be the ring of nxn matrices over afield F. For n 1 does there exist a ring homomorphism fromonto
Problem 6.10.4 (Sp97) Let R be the ring of n x n matrices over afield. SupposeS is a ring and h: R —÷ S is a homomorphism. Show that h is either infective orzero.
Problem 6.10.5 (Sp84) Let F be a field and let X be a finite set. Let R(X, F)be the ring of allfunctions from X to F, endowed with the pointwise operations.What are the maximal ideals of R(X, F)?
Problem 6.10.6 (FaOO) Suppose V is a vector space over afield K If U and Ware subspaces, let E(U, W) be the set of linear endomorphisms F of V over Kwith the properly that the image of FU in VI W is finite dimensional. Show thatE(U, U) is a subring of the ring of endomorphisms of V with two-sided idealsE(V, U) and E(U, 0).
Problem 6.10.7 (Sp88) Let R be a commutative ring with identity element anda E R. Let n and m be positive integers, and write d = gcd{n, m). Prove that theideal of R generated by a" — 1 and am — 1 is the same as the ideal generated byad_i.
Problem 6.10.8 (Sp89) 1. Let R be a commutative ring with identity contain-ing an element a with a3 = a + 1. let 1 be an ideal of R ofindex
2. Show that there exists a commutative ring with identity that has an element
a with a3 = a + I and that contains an ideal of index 5.
Note: The term index is used here exactly as in group theoiy; namely the index ofin R means the order of R/3.
Problem 6.10.9 (Sp90) Let R be a commutative ring with 1, and R* be its groupof units. Suppose that the additive group of R is generated by (U2 I U E R*
}.
Prove that R has, at most, one ideal lfor which R/J has cardinalily 3.
Problem 6.10.10 (SpOl) Find all commutative rings R with identity such that Rhas a unique maximal ideal and such that the group of units of R is trivial.
Problem 6.10.11 (Fa90) Let R be a ring with identity, and let be the left idealof R generated by (ab — ba I a, b E RI. Prove that 1 is a two-sided ideal.
Problem 6.10.12 (Fa99) Let R be a ring with identity element. Suppose that i,12, ..., are left ideals in R such thatR = (as additive
groups). Prove that there are elements u, E such that for any elements a Ea, and ajuj =0 if j i.
Problem 6.10.13 (Sp95) Suppose that R is a subring of a commutative ring Sand that R is offinite index n in S. Let m be an integer that is relatively primeto n. Prove that the natural map R/mR —÷ S/mS is a ring isomorphism.
Problem 6.10.14 (Sp81) Let M be one of the following fields: JR. C, Q, and F9(the field with nine elements). Let C M[x] be the ideal generated by x4 +2x —2.For which choices of M is the ring M[x]/1 afield?
Problem 6.10.15 (Sp84) Let R be a principal ideal domain and let and 3 benonzero ideals in R. Show that 3 if and only if 3+3= R.
6.11 Polynomials
Problem 6.11.1 (Fa77) Suppose the nonzero complex number a is a root of apolynomial of degree n with rational coefficients. Prove that 1/a is also a root ofa polynomial of degree n with rational coefficients.
Problem 6.11.2 (Su85) By the Fundamental Theorem ofAlgebra, the polynomialx3 +2x2 +7x +1 has three complex roots, ai, anda3. Compute a?
Problem 6.11.3 (Sp85) Let = e be a primitive root of unity. Find a cubicpolynomial with integer coefficients having a = + ç-1 as a root.
Problem 6.11.4 (Sp92, Su77, Fa81) 1. Prove that a = is algebraicover Q, by explicitly finding a polynomial f(x) in Q [x] of degree 4 havinga as a root.
2. Prove that f(x) is irreducible over Q.
Problem 6.11.5 (Fa90) Prove that + is irrational.
Problem 6.11.6 (Su85) Let P(z) be a polynomial of degree < k with complexcoefficients. Let cot, .. . be the kth roots of unity in C. Prove that
k
= P(0).
Problem 6.11.7 (Fa95) Let f(x) E Q [x] be a polynomial with rationalcoefficients. Show that there is a g(x) E Q [x], g 0, such that f(x)g(x) =a2x2 + a3x3 + a polynomial in which only prime exponents
Problem 6.11.8 (Fa91) Let be the ideal in the ring Z[x] generated by x — 7and 15. Prove that the quotient ring Z[x}/1 is isomorphic to Z15.
Problem 6.11.9 (Fa92) Let J denote the ideal in 7L[x], the ring of polynomialswith coefficients in Z, generated by x3 + x + 1 and 5.Is a prime ideal?
Problem 6.11.10 (Su77) In the ring Z[x} ofpolynomials in one variable over theintegers, show that the ideal generated by 5 and x2 + 2 is a maximal ideaL
Problem 6.11.11 (Sp78) let denote the ring of integers modulo n. Letbe the ring of polynomials with coefficients in Let denote the ideal ingenerated by x2 + x + 1.
1. For which values of n, 1 n 10, is the quotient ring afield?
2. Give the multiplication table for
Z be the ring of integers, p a prime, and= 7L/pZ the field of p elements. Let x be an indeterminate, and set
Ri = Fp(x]/(x2 — 2), R2 = — 3). Determine whether the rings R1and R2 are isomorphic in each of the cases p = 2,5, 11.
Problem 6.11.13 (Fa79, Su80, Fa82) Considerthe polynomial ring Z[x] and theideal generated by 7 and x —3.
1. Show that for each r E Z[x], there is an integer a 0 a 6
such thatr —a 1.
2. Finda in the special case r = + 15x14 +x2 + 5.
Problem 6.11.14 (Fa96) Let Z[x] be the ring ofpolynomials in the indeterminatex with coefficients in the ring Z of integers. Let 1 C Z[x] be the ideal generatedby 13 andx —4. Find an integer m such that 0 m 12 and
Problem 6.11.15 (Fa03) Give an example, with proof of a nonconstant irre-ducible polynomial f(x) over Q with the property that f(x) does not factor intolinear factors over the field K = Q[x]/(f(x)).
Problem 6.11.16 (Sp77) 1. In R[x], consider the set ofpolynomials f(x)forwhich f(2) = f'(2) = f"(2) = 0. Prove that this set forms an ideal andfind its monic generatoi
2. Do the polynomials such that f(2) = 0 and f'(3) 0 form an ideal?
Problem 6.11.17 (Sp94) Find all automorphisms of Z[x], the ring of polynomi-als over Z.
Problem 6.11.18 (Su78) Let R denote the ring of polynomials over a field F.Let pi, ..., be elements of R. Prove that the greatest common divisor ofpi, ..., is 1 if and only if there is an n x n matrix over R of determinant Iwhose first row is (pie ', Pn).
Problem 6.11.19 (Sp79) Let f(x) be a polynomial over the field of integersmodp. Let g(x) = — x. Show that the greatest common divisor of f(x) andg(x) is the product of the distinct linear factors of f(x).
Problem 6.11.20 (Su79) Let F be a subfield of a field K. Let p and q be polyno-mials over F. Prove that their greatest common divisor in the ring of polynomialsover F is the same as their gcd in the ring of polynomials over K.
Problem 6.11.21 (Su81, Su82) Show thatx'0+x9 +x8 +. • .+x+1 is irreducibleoverQ. How aboutx1t +x10 + +x + 1?
Problem 6.11.22 (Su84) Let Z be the ring of integers and Z[x] the polynomialring over Z. Show that
x6 +539x5 —511x+ 847
is irreducible in Z[x].
Problem 6.11.23 (Sp82) Prove that the polynomialx4 +x +1 is irreducible overQ.
Problem 6.11.24 (Fa83, Fa86) Prove that if p is a prime then the poly-nomial
f(x) +xP2+...+ 1is irreducible in Q[x].
Problem 6.11.25 (Sp96) Prove that f(x) = x4 +x3 +x2 +6x +1 is irreducibleover Q.
Problem 6.11.26 (SpOl) Prove that the polynomial f(x) = 16x5 — 125x4 +50x3 — 100x2 + 75x + 25 is irreducible over the rationals.
Problem 6.11.27 (Su84) Let Z3 be thefield of integers mod 3 and Z3[xJ the cor-responding polynomial ring. Decompose x3 + x + 2 into irreducible factors inZ3[x].
Problem 6.11.28 (Sp85) Factorx4 + x3 + x +3 completely in Z5[xJ.
Problem 6.11.29 (Fa8S) 1. How many different monic irreducible polynomi-als of degree 2 are there over the field Z5?
2. How many different monic irreducible polynomials of degree 3 are thereover the field Z5?
Problem 6.11.30 (Sp78) Is x4 + 1 irreducible over the field of real numbers? Thefield of rational numbers?A field with 16 elements?
Problem 6.11.31 (Sp81) Decompose x4 —4 andx3 —2 into irreducibles over R,over Z, and over Z3 (the integers modulo 3).
Problem 6.11.32 (Fa84) Let a be an element in afield F and let p be a prim&Assume a is not a power. Show that the polynomial — a is irreducible inF[x].
Problem 6.11.33 (Sp92) Let p be a prime intege? p 3 (mod 4), and letF1, = Z/pZ. If x4 + I factors into a product g(x)h(x) of two quadratic poly-nomials in prove that g(x) and h(x) are both irreducible over
Problem 6.1134 (Fa88) Let n be a positive integer and let f be a polynomial inR[xJ of degree n. Prove that there are real numbers ao, ai, . . . , not all equalto zero, such that the polynomial
n
axi=O
is divisible byf
Problem 6.11.35 (Fa89) Let F be afield,, F[xJ the polynomial ring in one vari-able over F, and R a subring ofF[xl with F C R. Prove that there exists a finiteset f2, ... , f,j of elements ofF[x] such that R = F[f1,
Problem 6.11.36 (Sp87) Let F be a finite field with q elements and let x be anindeterminate. For f a polynomial in F[xJ, let denote the corresponding func-tion ofF into F, defined by ço1(a) = f(a), (a E F). Prove that if is anyfunctionofF into F, then there is an f in F[x} such that q Prove that f is uniquelydetermined by q. to within addition of a multiple of — x.
6.12 Fields and Their Extensions
Problem 6.12.1 (Su78, Fa87, Sp93) Let R be the set of 2 x 2 matrices of the form
(a —ba
where a, b are elements of a given field F. Show that with the usual matrix oper-ations, R is a commutative ring with identity. For which ofthefollowingfields FisRafield:F=Q, C, Z5, Z7?
Problem 6.12.2 (Fa83) Prove that every finite integral domain is afield.
Problem 6.123 (Sp77, Sp78) Let F c K be fields, and a and b elements of Kwhich are algebraic over F. Show that a + b is algebraic over F.
Problem 6.12.4 (Fa78, Fa85) Prove that every finite multiplicative group of com-plex numbers is cyclic.
Problem 6.12.5 (Sp87, Fa95) Let F be a field. Prove that every finite subgroupof the multiplicative group of nonzero elements of F is cyclic.
Problem 6.12.6 (Fa02) Let K be a field such that the additive group of K isfinitely generated as a group. Prove that K is finite.
Problem 6.12.7 (Sp85) Let F = (a + + a, b, C E Q). Prove that Fis afield and each element in F has a unique representation as a + +
c E Q. Find (1 — in F.
Problem 6.12.8 (Sp85) Let F be a finite field. Give a complete proof of the factthat the number of elements ofF is oftheform pr, where p 2 is a prime numberand r is an integer 1.
Problem 6.12.9 (Fa02) Let K be a field and L C K a subfield containing(a2 a E K). Prove that if the characteristic of K is not 2 then L = K andthat the same conclusion holds for finite fields K of characteristic 2, but not ingeneral for fields of characteristic 2.
Problem 6.12.10 (Su85) Let F be afield of characteristic p > 0, p 3. If a isa zero of the polynomial f(x) = — x +3 in an extension field ofF, show thatf(x) has p distinct zeros in the field F(a).
Problem 6.12.11 (Fa99) Let K be the field Q ( Prove that K has degree 10over Q, and that the group of automorphisms of K has order 2.
Problem 6.12.12 (Fa85) Let f(x) = x5 — 8x3 +9x —3 andg(x) = — 5x2 —
6x +3. Prove that there is an integer d such that the polynomials f(x) and g(x)have a common root in the field Q What is d?
Problem 6.12.13 (Fa86) Let F be a field containing Q such that [F: Q} = 2.
Prove that there exists a unique integer m such that m has no multiple primefactors and F is isomorphic to Q
Problem 6.12.14 (Sp96) Exhibit infinitely many pairwise nonisomorphicquadratic extensions of Q and show they are pairwise nonisomorphic.
Problem 6.12.15 (Fa94) Let Q be the field of rational numbers. For 0 a reallet F0 = Q(sin 0) and = Q (sin Show that Eo is an extension field
of Fe, and determine all possibilities for dimF0Ee.
Problem 6.12.16 (Fa98) Show that the field Q(t1, ..., of rational functionsin n variables over the rational numbers is isomorphic to a subfield of IR.
Problem 6.12.17 (Sp99) Let f(x) E Q [x] be an irreducible polynomial of de-gree n 3. Let L be the splitting field off, and let a E L be a zero off. Giventhat[L : Q] =n!,provethatQ(a4) = Q(a).
Problem 6.12.18 (FaOl) Let K be a field. For what pairs of positive integers(a, b) is the subring ti'] ofK[t] a unique factorization domain?
Problem 6.12.19 (Sp95) Let F be a finite field of cardinality p's, with p primeand n > 0, and let G be the group of invertible 2x2 matrices with coefficients inF.
1. Prove that G has order — —
2. Show that any p-Sylow subgroup of G is isomorphic to the additive groupofF.
Problem 6.12.20 (Sp02) Let p be a prime and k, n positive integers. Prove thatthe group of invertible n x n matrices over (the field of p elements)contains an element of order and only
Problem 6.12.21 (FaOl) Let F be afield, and let G be the group of2x2 upper-triangular matrices over F of determinant 1.
1. Determine the commutator subgroup of G.
2. Suppose F is the finite field of order p1. Determine for this case the mini-mum nwnber of generators for the subgroup found in (i).
Problem 6.12.22 (Fa94) Let p be an odd prime and the field of p elements.How many elements of F,, have square roots in How many have cube roots
Problem 6.12.23 (Sp94) Let F be a finite field with q elements. Say that afunc-
tion f : F —+ F is a polynomial function if there are elements ao, ai, .. ., ofFsuch that f(x) = ao + aix + + for all x E F. How many polynomial
functions are there?
Problem 6.12.24 (Sp95) Let F be a finite field, and suppose that the subfield ofF generated by (x3 x E F) is different from F. Show that F has cardinality 4.
Problem 6.12.25 (Sp97) Suppose that A is a commutative algebra with identityover C (i.e., A is a commutative ring containing C as a subring with identity).Suppose further that a2 0 for all nonzero elements a E A. Show that if thedimension of A as a vector space over C is finite and at least two, then the equa-tions a2 = a is satisfied by at least three distinct elements a E A.
Problem 6.12.26 (Sp02) Let A be a commutative ring with 1. Prove that thegroup A* of units of A does not have exactly 5 elements.
Problem 6.12.27 (FaOO) Suppose K is afield and R is a nonzero K-algebra gen-erated by two elements a and b which satisfy a2 = b2 = 0 and (a + b)2 = 1.
Show R is isomorphic to M2(K) (the algebra of 2 x 2 matrices over K).
6.13 Elementary Number Theory
Problem 6.13.1 (Fa86) Prove that if six people are riding together in an EvansHall elevato,; there is either a three-person subset of mutual friends (each knowsthe other two) or a three-person subset of mutual strangers (each knows neitherof the other two).
Problem 6.13.2 (Fa03) Let Um,n be an array of numbers for 1 m N andI n N. Suppose that um,n = 0 when m is I or N, or when n is I or N.Suppose also that
IUm,n = (Um_1,n + Um+I,n + um,n_1 + um,n+i)
whenever 1 <rn <N and 1 <n <N. Show that all the um,n are zero.
Problem 6.13.3 (Sp02) How many functions f: (1,2,3,4,5) -÷ (1,2,3,4,5)have a range of size exactly 3?
Problem 6.13.4 (Sp98) Let m 0 be an integei Let aI, .. . , am be integersand let
f(x)
Show that � 0 is an integer then f(x)d /d! can be expressed in the form
md
where the b1 are integers.
Problem 6.13.5 (Fa03) Let f : [0, 1] -+ [0, 1] be an increasing (not strictlyincreasing) function such that
whenever the are 0 or 2. Prove that there is a constant Co such that
If(x) — Coax —
for alix, y [0, 1].
Problem 6.13.6 (Sp77) Letp be an odd prime. Let Q(p) be the set of integers a,0 a p — 1, for which the congruence
x2—=a (modp)
has a solution. Show that Q(p) has cardinality (p + 1)/2.
Problem 6.13.7 (Su77) Let p be an odd prime. If the congruence x2 —1
(mod p) has a solution, show that p 1 (mod 4).
Problem 6.13.8 (SpSO) Let n 2 be an integer such that + n2 is prime. Provethat
n_=3 (mod6).
Problem 6.13.9 (Fa77) 1. Show that the set of all units in a ring with unityform a group under multiplication. (A unit is an element having a two-sidedmultiplicative inverse.)
2. In the ring of integers modn, show thatk is a unit if and only if k and nare relatively prime.
3. Suppose n = pq, where p and q are primes. Prove that the number of unitsis(p — 1)(q —1).
Problem 6.13.10 (Su79) Which rational numbers t are such that
3t3+ lOt2 —3t
is an integer?
Problem 613.11 (Fa96) Show the denominator of(1/2) is a power of2for all
integers n.
Problem 6.13.12 (Su82) Let n be a positive intege?
1. Show that the binomial coefficient
f2n=
is even.
2. Prove that is divisible by 4 if and only if n is not a power of 2.
Problem 6.13.13 (Sp83) Suppose that n> 1 is an integer. Prove that the sum
1 11+—+...+-2 n
is not an integer.
Problem 6.13.14 (Fa84, Fa96) Let gcd abbreviate greatest common divisor and1cm abbreviate least common multiple. For three nonzero integers a, b, c, showthat
gcd (a, lcm(b, c}) = 1cm fgcd(a, b}, gcdfa, c)).
Problem 6.13.15 (Sp92) Let a2, ... , be integers with 1 a1 25, for1 i 10. Prove that there exist integers ni, n2, ... , n io, not all zero, such that
10
= 1.
Problem 6.13.16 (Su83) The number 21982145917308330487013369 is the thir-teenth power of a positive Which positive integer?
Problem 6.13.17 (Sp96) Determine the rightmost decimal digit of
A =
Problem 6.13.18 (Sp88) Determine the last digit of
232323
in the decimal system..
Problem 6.13.19 (Sp88) Show that one can represent the set of nonnegative inte-gers, as the union of two disjoint subsets N1 and N2 (N1 flN2 = 0, N1 UN2 =Z+) such that neither Nj nor N2 contains an infinite arithmetic progression.
Problem 6.13.20 (Fa89) Let ço be Euler's totient function; so is a positivethen ço(n) is the number of integers m for which 1 m n and
gcdfn, m} 1. Let a and k be two integers, with a > 1, k > 0. Prove that kdivides ço(ak — 1).
Problem 6.13.21 (Sp90) Determine the greatest common divisor of the elementsof the set {n13 — n I n Z).
Problem 6.13.22 (Sp91) For n a positive integei let d(n) denote the number ofpositive integers that divide n. Prove that d(n) is odd and only if n is a perfectsquare.
Problem 6.13.23 (FaOl) Which of the numbers 0,1,2,3,4,5,6,7,8,9 occur as thelast digit of n'1 for infinitely many positive integers n?
Problem 6.13.24 (Sp03) Let N 30030, which is the product of the first sixprimes. How many nonnegative integers x less than N have the property that Ndividesx3 — 1?
7
Linear Algebra
7.1 Vector Spaces
Problem 7.1.1 (Sp99) Let p, q, r and s be polynomials of degree at most 3.Which, if any, of the following two conditions is sufficient for the conclusion thatthe polynomials are linearly dependent?
1. At 1 each of the polynomials has the value 0.
2. At 0 each of the polynomials has the value 1.
Problem 7.1.2 (Sp03) For an analytic function h on C, let denote its i-thderivative, with = h. Suppose that f and g are analytic functions on Csatisfying
+ an_i f(?2I) + • + = 0
=0
for some constants ao, ... , an—i, bo, ... , bm_i E C. Show that the productfunc-tion F = fg satisfies
+ + • • + c0F' = 0
for some constants ... , cmn E C not all zero.
Problem 7.13 (Su79, Sp82, Sp83, Su84, Fa91, Fa98) Let F be afinite field withq elements and let V be an n-dimensional vector space over F.
1. Determine the number of elements in V.
2. Let (F) denote the group of all n x n nonsingular matrices over F.Determine the order of
3. Let (F) denote the subgroup of (F) consisting of matrices withdeterminant 1. Find the order of
Problem 7.1.4 (SpOl) Let F be a finite field of order q, and let V be a two di-mensional vector space over F. Find the number of endomorphisms of V that fixat least one nonzero
Problem 7.1.5 (Sp97) Let GL2(Zm) denote the multiplicative group of invertible2x2 matrices over the ring of integers modulo m. Find the order of GL2(Zpn)for each prime p and positive integer n.
Problem 7.1.6 (FaOO) Let denote the field of p elements (p prime). Let n be apositive Prove that there is a transformation A E (F,,) (the group ofinvertible linear transformations from into itself) which, as a permutationof the nonzero vectors acts as a single cycle of length p" — 1.
Problem 7.1.7 (Sp96) Let G be the group of 2x2 matrices with determinant 1over the four-element field F. Let S be the set of lines through the origin in F2.Show that G acts faithfully on S. (The action is faithful if the only element of Gwhich fixes every element of S is the identity.)
Problem 7.1.8 (Su77) Prove the following statements about the polynomial ringF[x], where F is any field.
1. F[x} is a vector space over F.
2. The subset F [x] of polynomials of degree n is a subspace of dimensionn+IinF{x}.
3. The polynomials 1, x —a, ..., (x —a)" form a basis of any a E F.
Problem 7.1.9 (Sp02) Let U, V, W be finite-dimensional subspaces of a vectorspace. Prove thatdim(U)+dim(V)+dim(W)—dim(U+V+W) max{dim(UflV)7 dim(U fl W), dim(V fl W)}.
Problem 7.1.10 (Su84) Suppose V is an n-dimensional vector space over thefield F. Let W c V be a subspace of dimension r <n. Show that
W = fl (U U is an (n — 1) — dimensional subspace of V and W C U).
Problem 7.1.11 (Sp80, Fa89) Show that a vector space over an infinite field can-not be the union of a finite number of proper subspaces.
Problem 7.1.12 (Fa88) Let A be a complex n x n matrix and let C(A) be thecommutant of A; that is, the set of complex n x n matrices B such thatAB = BA. (it is obviously a subspace of the vector space of all com-plex n x n matrices.) Prove that dim C(A) n.
Problem 7.1.13 (Sp89, Fa97) Let S be the subspace of (the vector spaceof all real n x n matrices) generated by all matrices of the form AB — BA withA and B in Pmve that dim(S) = n2 — 1.
Problem 7.1.14 (Sp90) Let A and B be subspaces ofa finite-dimensional vectorspace VsuchthatA+B = V. Writen = dimV, a = dimA, andb = dimB. LetS be the set of those endomorphisms f ofVfor which f(A) C A and f(B) C B.Prove that S is a subspace of the set of all endomorphisms of V. and express thedimension of S in terms of n, a, and b.
Problem 7.1.15 (Sp81) Let T be a linear transformation of a vector space V intoitself Supposex E V is such that TmX = 0, Ofor some positive integerm. Show that x, Tx, ..., are linearly independent.
Problem 7.1.16 (Fa97) Let aa, . .. , be distinct real numbers. Show thatthe n exponential functions ea2t, ..., are linearly independent over thereal numbers.
Problem 7.1.17 (Su83) Let V be a real vector space of dimension n with a pos-itive definite inner product. We say that two bases (a1) and (b,) have the sameorientation if the matrix of the change ofbasisfmm (a,) to (b1) has a positive de-terminant. Suppose now that (as) and (b1) are orthonormal bases with the sameorientation. Show that (a1 + 2b,) is again a basis of V with the same orientationas (aj).
7.2 Rank and Determinants
Problem 7.2.1 (Sp78, Fa82, Fa86) Let M be a matrix with entries in a field F.The row rank of M over F is the maximal number of rows which are linearly inde-pendent (as vectors) over F. The column rank is similarly defined using columns
instead of rows.
1. Prove row rank = column rank
2. Find a maximal linearly independent set of columns of
103—2212 001—44111 2101 2
taking F = JR.
3. If F is a subfield of K, and M has entries in F, how is the row rank of Mover F related to the row rank of M over K?
Problem 7.2.2 (Fa02) Let the n x n real matrix A be diagonalizable and have aone-dimensional null space. Prove that a nonzero left null vector of A cannot beorthogonal to a nonzero right null vector of A.
Problem 7.23 (Su85, Fa89) Let A be an n x n real matrix and A its transpose,Show that At A and A' have the same range.
Problem 7.2.4 (Sp97) Suppose that P and Q are n x n matrices such thatp2 = P. Q2 = Q, and I — (P + Q) is invertible. Show that P and Q havethe same rank.
Problem 7.2.5 (Fa03) Let A(m, n) be the m x n matrix with entries
ajj=j'where = 1 by definition. Regarding the entries of A(m, n) as representingcongruence classes (mod p), determine the rank of A(m, n) over the finite fieldIF,, =Z,,forallm,n land all primes p.
Problem 7.2.6 (SpOl) Let A be an n x n matrix over afield K. Prove that
rank A2 — rank A3 rank A — rank A2
Problem 7.2.7 (Sp91) Let T be a real, symmetric, n x n, tridiagonal matrix:
b1 0 0 0 0b1 a2 b2 0 0 0o b2 b3 0 0
T=: : :
o o 0 0 ...o 0 0 0
(All entries not on the main diagonal or the diagonals just above and below themain one are zero.) Assume Ofor all f.
Prove:
1.
2. T has n distinct eigenvalues.
Problem 7.2.8 (Sp83) Let A = be an nxn real matrix satisfying the condi-tions:
Show that det(A) > 0.
Problem 7.2.9 (Sp91) Let A = be a square matrix with integer entries.
1. Prove that if an integer n is an eigenvalue of A, then n is a divisor of det A,the determinant of A.
2. Suppose that n is an integer and that each row of A has sum n:
Prove that n is a divisor of det A.
Problem 7.2.10 (FaOl) Let A be a symmetric n x n matrix over R of rank n —1.Prove there is a k in {1, 2, ... , n) such that the matrix resulting from deletion ofthe kth row and kth column from A has rank n — 1.
Problem 7.2.11 (Fa84) Let IR[xj, . . . , be the polynomial ring over the realfield R in the n variables Xi, . .. , x,,. Let the matrix A be the n x n matrix whose
row is (1, x,, .. . , i = 1,..., n. Show that
detA = —x3).i>j
Problem 7.2.12 (Sp77) A matrix of the form
1 aij ...I a1 ...
where the a, are complex numbers, is called a Vandermonde matrix.
1. Prove that the Vandermonde matrix is invertible ao, ai, . .. , are alldifferent.
2. If ao, ai, ... , are all different, andbo, b1, . . . , are complex numbers,prove that there is a unique polynomial f of degree n with complex coeffi-cients such that f(ao) = b0, f(ai) = b1,..., =
Problem 7.2.13 (Fa03) Let A and B be n x n complex unitary matrices. Provethat I det(A + B)I <2's.
Problem 7.2.14 (Sp90) Give an example of a continuous function v : -÷with the property that v(t1), V(t2), and v(t3)form a basis for whenever ti, t2,and t3 are distinct points of lEt
Problem 1.2.15 (Fa95) Let ..., be continuous real valued functionson [a, b]. Show that the set . . . , is linearly dependent on [a, b] if and
only ifb
det (fProblem 7.2.16 (Fa81) Let M2x2 be the vector space of all real 2x2 matrices.Let
and define a linear transformation L : M2x2 -÷ M2x2 by L(X) = AXB. Compute
the trace and the determinant of L.
Problem 7.2.17 (Su82) Let V be the vector space of all real 3x3 matrices andlet A be the diagonal matrix
(1001020.\001Calculate the determinant of the linear T on V defined byT(X) = + XA).
Problem 7.2.18 (Sp80) Let M3x3 denote the vector space of real 3x3 matrices.For any matrix A E M3x3, define the linear operator LA : M3x3 M3x3,
LA(B) = AB. Suppose that the determinant of A is 32 and the minimal polyno-mial is (t — 4)(t — 2). What is the trace of LA?
Problem 7.2.19 (Su81) Let S denote the vector space of real n x n skew-symmetricmatrices. For a nonsingular matrix A, compute the determinant of the linear mapTA:S—+S,TA(X)=AXA.
Problem 7.2.20 (Fa94) Let M7 denote the vector space of real 7 x 7 matrices.Let A be a diagonal matrix in M7 that has + 1 in four diagonal positions and—1 in three diagonal positions. Define the linear transformation T on byT(X) = AX — XA. What is the dimension of the range of T?
Problem 7.2.21 (Fa93) Let F be afield. For m and n positive integers, let Mm xnbe the vector space of m x n matrices over F. Fix m and n, and fix matrices Aand B in Mmxn. Define the linear transformation Tfrom Mnxm to Mmxn by
T(X) = AXB.
Prove that if m n, then T is not invertible.
7.3 Systems of Equations
Problem 73.1 (Su77) Determine all solutions to the following infinite system oflinear equations in the infinitely many unknowns
X1 +X3+X50X2+X4+X60X3 +X5 +X7 =0
How many free parameters are required?
Problem 73.2 (Fa77, Su78) 1. Using only the axioms for a field F, provethat a system of m homogeneous linear equations in n unknowns withm <n and coefficients in F has a nonzero solution.
2. Use Part 1 to show that V is a vector space over F which is spanned by afinite number of elements, then every maximal linearly independent subsetof V has the same number of elements.
Problem 73.3 (Sp88, Sp96) If a finite homogeneous system of linear equationswith rational coefficients has a nontrivial complex solution, need it have a non-trivial rational solution? Give a proof or a counterexample.
Problem 734 (Sp84, Sp87) Let A be a real m x n matrix with rational entriesand let b be an m-tuple of rational numbers. Assume that the system of equationsAx = b has a solution x in complex n-space C Show that the equation has asolution vector with rational components, or give a counterexample.
7.4 Linear Transformations
Problem 7.4.1 (Fa77) Let E and F be vector spaces (not assumed to be finite-dimensional). Let 5: E —÷ F be a linear transformation.
1. Prove S(E) is a vector space.
2. Show S has a kernel (0) if and only ifS is injective (i.e., one-to-one).
3. Assume S is infective; prove : S(E) —+ E is linea,
Problem 7.4.2 (Sp82) Let T : V -÷ W be a linear transformation betweenfinite-dimensional vector spaces. Prove that
dim(ker T) + dim(range T) = dim V.
Problem 7.4.3 (Fa99) Let V and W be finite dimensional vector spaces, let X be
a subspace of W, and let T: V -÷ W be a linear map. Prove that the dimension
of (X) is at least dimV — dimW+dimX.
Problem 7.4.4 (Fa98) Let A and B be linear transformations on a finite dimen-sional vector space V. Prove that dim ker(AB) dim ker A + dim ker B.
Problem 7.4.5 (Fa02) Let V and W be vector spaces over a field K. AssumeA: V —+ W and B: V —÷ W are linear transformations such that A has rank atleast 2, andfor every vector v in V the vectors A v and B v are linearly dependent.Prove that the linear transformations A and B are linearly dependent.
Problem 7.4.6 (Sp95) Suppose that W c V are finite-dimensional vector spacesover a field, and let L: V -÷ V be a linear transformation with L(V) C W.Denote the restriction of L to W by Prove that det(1 — tL) = det(1 —
Problem 7.4.7 (FaOO) Let V be a finite-dimensional vector space, and letf : V —÷ V be a linear transformation. Let W denote the image of f. Provethat the restriction of f to W, considered as an endomorphism of W, has thesame trace as f: V —÷ V.
Problem 7.4.8 (Fa99) Let T: V —÷ V be a linear operator on an n dimensionalvector space V over afield F. Prove that T has an invariant subspace W otherthan fO) and V if and only if the characteristic polynomial of T has a factorfEF[t]withO <degf <n.
Problem 7.4.9 (FaOO) Let -÷ R'2 be a linear transformation, where n>1.Prove that there is a 2-dimensional subspace M R'2 such that T(M) c M.
Problem 7.4.10 (Sp95) Let V be a finite-dimensional vector space over a fieldF, and let L: V —÷ V be a linear transfor,nation. Suppose that the characteristicpolynomial x of L is written as x = where xi and X2 are two relativelyprime polynomials with coefficients in F. Show that V can be written as the directsum of two subspaces V1 and V2 with the property that (L) = Ofor i = 1,2.
Problem 7.4.11 (Su79) Let E be a three-dimensional vector space over Q. Sup-poseT: E —÷ Eisalineartransformation and Tx = y, Ty = z, Tz = x+y,for certain x, y, z E E, x 0. Prove thatx, y, and z are linearly independent.
Problem 7.4.12 (Su80) Let T: V —÷ V be an invertible linear transformationof a vector space V. Denote by G the group of all maps fk,a : V -÷ V wherekEZ,aEV, and forx€V,
fk,a(X) Tkx + a (x E V).
Prove that the commutator subgroup G' of G is isomorphic to the additive groupof the vector space (T — I) V, the image of T — I. (G' is generated by all
g andh in G.)
Problem 7.4.13 (Sp86) Let V be a finite-dimensional vector space and A andB two linear transformations of V into itself such that A2 = B2 = 0 andAB+BA—I.
1. Prove that if NA and NB are the respective null spaces of A and B, thenNA = ANN, NB = BNA, and V = NA NB.
2. Prove that the dimension of V is even.
3. Prove that if the dimension of V is 2, then V has a basis with respect towhich A and B are represented by the matrices
(0 i\ fo o\o) and
Problem 7.4.14 (Su84) Let f : Rm -÷ n 2 be a linear transfor,nationof rank n — 1. Let f(v) = f2(V),...,fn(V)) for v E Rm. Show thata necessary and sufficient condition for the system of inequalities f, (v) > 0,i = 1, ... , n, to have no solution is that there exist real numbers 0, not allzero, such that
=0.
Problem 7.4.15 (Sp95) Let n be a positive integer, and let S C R'7 a finite subsetwith 0 E S. Suppose that p: S -÷ S is a map satisfying
ço(0) =0,
d(ço(s), q?(t)) = d(s, t) for all s, t E S,
where d( , ) denotes Euclidean metric. Prove that there is a linear mapf : whose restriction to S is ço.
Problem 7.4.16 (S p86) Consider be equipped with the Euclidean metricd(x, y) = lix — yll. Let T be an isometry of R2 into itself Prove that T canbe represented as T(x) = a + U(x), where a is a vector in R2 and U is anorthogonal linear transformation.
Problem 7.4.17 (Sp88) Let X be a set and V a real vector space of real val-ued functions on X of dimension n, 0 <n <oo. Prove that there are n pointsX1,X2,...,Xn (f(xi),...,f(xn))ofVtolWZ isanisomorphism.
Problem 7.4.18 (Sp97) Suppose that X is a topological space and V is a finite-dimensional subspace of the vector space of continuous real valued functions onX. Prove that there exist a basis fn)for V and points xi, in Xsuch that ft (xj) =
Problem 7.4.19 (Fa90) Let n be a positive integer and let P2,*' be the vectorspace of real polynomials whose degrees are, at most, 2n + 1. Prove that thereexist unique real numbers Cl, ... , such that,for all p E
n
[ p(x)dx = 2p(O) + Ck(p(k) + p(—k) — 2p(O))J—l k=1
Problem 7.4.20 (Sp94) Let T : -÷ JR'2 be a diagonalizable linear transfor-mation. Prove that there is an orthonormal basis for IR'2 with respect to which Thas an upper-triangular matrix.
Problem 7.4.21 (Fa77) Let P be a linear operator on a finite-dimensional vectorspace over afinitefield. Show that (f P is invertible, then = Iforsomepositiveintegern.
Problem 7.4.22 (Fa82) Let A be an n x n complex mat riz and let B be the Her-mitian transpose of A (i.e., = Suppose that A and B commute with eachothe, Consider the linear transformations a and fi on C'2 defined by A and B.Prove that a and have the same image and the same kernel.
Problem 7.4.23 (Su79, Fa96) Prove that a linear transformation T : JR3 —÷ JR3
has
1. a one-dimensional invariant subspace, and
2. a two-dimensional invariant subspace.
Problem 7.4.24 (Fa83) Let A be a linear transfor,nation on JR3 whose matrix(relative to the usual basis for R3) is both symmetric and orthogonal. Prove thatA is either plus or minus the identity, or a rotation by 1800 about some axis in
or a reflection about some two-dimensional subspace of JR3
Problem 7.4.25 (Fa84) Let 8 and q.' be fixed, 0 8 27r, 0 2jr and letR be the linear transfor,nation from iii? to 1EV whose matrix in the standard basis
7, and is
/1 0 0
( 0 cosO sin8\0 —sinO cosO
Let S be the linear transformation of JR3 to JR3 whose matrix with respect to thebasis
is
/ cOsço sinço 0—sinçi cosq 0
0 1
Prove that T R o S leaves a line invariant.
Problem 7.4.26 (Sp86) Let e = (a, b, c) be a unit vector in JR3 and let T be thelinear transformation on 1EV of rotation by 180° about e. Find the matrix for Twith respect to the standard basis.
Problem 7.4.27 (Su80) Exhibit a real 3x3 matrix having minimal polynomial(t2 + 1)(t — 10), which, as a linear transformation of R3, leaves invariant the lineL through (0,0,0) and (1, 1, 1) and the plane through (0,0,0) perpendicular toL.
Problem 7.4.28 (Su77) Show that every rotation of 1R3 has an axis; that is, givena 3x3 real matrix A such that A = A—1 and detA > 0, prove that there is anonzero vector v such thatAv = v.
Problem 7.4.29 (Sp93) Let P be the vector space of polynomials over IlL Let thelinear transformation E: P —÷ P be defined by Ef = f + f', where f' is thederivative off. Prove that E is invertible.
Problem 7.4.30 (Fa84) Let be the vector space of all real polynomials withdegrees at most n. Let D : —÷ be given by differentiation: D(p) = p'.Let ir be a real polynomial. What is the minimal polynomial of the transformation
Problem 7.4.31 (Su77) Let V be the vector space of all polynomials of degree10, and let D be the differentiation operator on V (i.e., Dp(x) = p'(x)).
1. Show that tr D = 0.
2. Find all eigenvectors of D and eD.
Problem 7.4.32 (Sp02) Let xo, Xl, ... , be distinct points of R. Prove that there
are unique real numbers ao, ai, ... , such that
n
/ p(t)dt = >ajp(xj)JO j=O
for all polynomials of degree n or less.
Problem 7.433 (SpOO) Let ..., be disjoint closed nonemply subintervals
of R.
1. Prove that if p is a real polynomial ofdegree less than n such that
I p(x)dx=0, forj=1,...,nJ!J
then p=0.
2. Prove that there is a nonzero realpolynomial p of degree n that satisfies all
the above equations.
7.5 Eigenvalues and Eigenvectors
Problem 73.1 (Fa77) Let M be a real 3x3 matrix such that M3 = I, M I.
1. What are the eigenvalues of M?
2. Give an example of such a matrix.
Problem 7.5.2 (Fa79) Let N be a linear operator on an n-dimensional vectorspace, n> 1, such that NP? = 0, 0. Prove there is no operator X withX2=N.
Problem 733 (Sp89) Let F be afield, n and m positive integers, and A an n x nmatrix with entries in F such that =0. Prove that A" =0.
Problem 7.5.4 (Fa98) Let B be a 3x3 matrix whose null space is 2-dimensional,and let x (X) be the characteristic polynomial of B. For each assertion below,provide either a proof or a counterexample.
1. X2 is a factor of x (X).
2. The trace of B is an eigenvalue of B.
3. B is diagonalizable.
Problem 7.5.5 (Sp99) Suppose that the minimal polynomial of a linear operatorT on a seven-dimensional vector space is x2. What are the possible values of thedimension of the kernel of T?
Problem 7.5.6 (Sp03) Let L be a real symmetric n x n matrix with 0 as a simpleeigenvalue, and let v E IR".
I. Show that for sufficiently small positive real e, the equation Lx + ex = vhas a unique solution x = x(s)
2. Evaluate exfr) in term$ of v, the eigenvectors of L, and the innerproduct (, ) on
Problem 7.5.7 (Su81, Su82) Let V be a finite-dimensional vector space over therationals Q and let M be an automorphism of V such that M fixes no nonzerovector in V. Suppose that is the identity map on V. where p is a prime nwnbe,Show that the dimension of V is divisible by p — 1.
Problem 73.8 (Fa92) Let F be afield, V afinite-dimensional vector space overF, and T a linear transformation of V into V whose minimum polynomial, isirreducible over F.
1. Let v be a nonzero vector in V and let V1 be the subspace spanned by v andits images under the positive powers of T. Prove that dim V1 = deg
2. Prove that deg divides dim V.
Problem 7.5.9 (Fa02) Suppose A and M are n x n matrices over C, A is in-vertible and AMA—' = M2. Prove the nonzero eigenvalues of M are roots ofunity.
Problem 7.5.10 (Su79, Fa93) Prove that the matrix
0510505015050050has two positive and two negative eigenvalues (counting multiplicities).
Problem 7.5.11 (Fa94) Prove that the matrix
/ 1 1.00001 1
1.00001 1 1.00001
\ 1 1.00001 1
has one positive eigenvalue and one negative eigenvalue.
Problem 7.5.12 (Sp85) For arbitrary elements a, b, and c in a field F, computethe minimal polynomial of the matrix
/0 0 aIlOb.i c
Problem 75.13 (FaOl) Let A be an n x n complex matrix such that the threematrices A + I, A2 + I, A3 + I are all unitary. Prove that A is the zero matrix.
Problem 75.14 (Sp03) Let k be afield, and let n 1. Prove that the followingproperties of an n x n matrix A with entries in k are equivalent:
• A is a scalar multiple of the identity matrix.
• Every nonzero vector V E is an eigenvector of A.
Problem 7.5.15 (Fa85, Sp97, Fa98) Suppose that A and B are endomorphismsof a finite-dimensional vector space V over a field F. Prove or disprove the fol-lowing statements:
1. Every eigenvector of AB is also an eigenvector of BA.
2. Every eigenvalue of AB is also an eigenvalue of BA.
Problem 7.5.16 (Sp78, Sp98) Let A and B denote real nxn symmetric matricessuch that AB = BA. Prove that A and B have a common eigenvector in
Problem 7.5.17 (Sp86) Let S be a nonempty commuting set of n x n complexmatrices (n 1). Prove that the members of S have a common
Problem 7.5.18 (Sp84) Let A and B be complex n x n matrices such thatAB = BA2, and assume A has no eigenvalues of absolute value 1. Prove thatA and B have a common (nonzero) eigenvector.
Problem 73.19 (Su78) Let V be a finite-dimensional vector space over an al-gebraically closed field. A linear operator T : V —+ V is called completelyreducible if whenever a linear subspace E C V is invariant under T, that isT(E) C E, there is a linear subspace F C V which is invariant under T andsuch that V = E F. Prove that T is completely reducible if and only V hasa basis of eigenvectors.
Problem 7.5.20 (Fa79, Su81) Let V be the vector space of sequences (an) ofcomplex numbers. The shift operator S: V -÷ V is defined by
S((aI,a2,a3,...)) = (a2,a3,a4,...).
1. Find the eigenvectors of S.
2. Show that the subspace W consisting of the sequences with
= Xn+1 + is a two-dimensional, S-invariant subspace of V andexhibit an explicit basis for W.
3. Find an explicit formula for the Fibonacci number fn, where
f2f1 1.
Note: See also Problem 1.3.12.
Problem 75.21 (Fa82) Let T be a linear transformation on a finite-dimensionalC -vector space 1' and let f be a polynomial with coefficients in C - If A is aneigenvalue of T, show that f(A) is an eigenvalue of f(T). Is every eigenvalue off(T) necessarily obtained in this way?
Problem 73.22 (Fa83, Sp96) Let A be the n x n matrix which has zeros on themain diagonal and ones everywhere else. Find the eigenvalues and eigenspacesof A and compute det(A).
Problem 7.5.23 (Sp85) Let A and B be two n x n self-adjoint (i.e., Hermitian)matrices over C and assume A is positive definite. Prove that all eigenvalues ofAB are real.
Problem 7.5.24 (Fa84) Let a, b, c, and d be real numbers, not all zero. Find theeigenvalues of the following4x4 matrix and describe the eigenspace decomposi-tion of R4.
aa ab ac adba bb be bdca cb cc cddadbdcdd
Problem 7.5.25 (Sp81) Show that the following three conditions are all equiva-lent for a real 3x3 symmetric matrix A, whose eigenvalues are Ai, A2, and A3:
1. trA is not an eigenvalue of A.
2. (a + b)(b + c) (a + c) 0.
3. The map L : S -÷ S is an isomorphism, where S is the space of3x3 realskew-symmetric matrices and L(W) = AW + WA.
Problem 7.5.26 (Su84) Let
A(ab
be a real matrix with a, b, c, d> 0. Show that A has an eigenvector
ER2\YJ
withx,y >0.
Problem 7.5.27 (Sp90) Let n be a positive and let A = be thenxn matrix with a 1±1 = —1, and a-1 =0 otherwise,- that is,
2 —1 0 0 ••• 0 0 0—1 2 —1 0 •.. 0 0 00 —1 2 —1 0 0 00 0 —1 2 ••• 0 0 0
A = . . . . .
0 0 0 2 —1 00 0 0 0 •-. —1 2 —1
0 0 0 0 ••• 0 —1 2
Prove that eveiy eigenvalue of A is a positive real number.
Problem 7.5.28 (Sp92) Let A be a real symmetric n x n matrix with nonnegativeentries. Prove that A has an eigenvector with nonnegative entries.
Problem 7.5.29 (SpOO) Let be the n x n matrix whose entries ajk are givenby
_J I if Ii—kI=1ajk
— 0 otherwise.
Prove that the elgenvalues of A are symmetric with respect to the origin.
Problem 7.530 (Fa91) Let A = be a real n x n matrix with nonnega-tive entries such that
Pmve that no eigenvalue of A has absolute value greater than 1.
Problem 7.531 (SpOl) Let S be a special orthogonal n x nmatrix satisfying S'S = I and det(S) = 1.
1. Prove that is odd then 1 is an eigenvalue of S.
2. Prove that is even then I need not be an eigenvalue of S.
Problem 7.5.32 (Sp85, Fa88) Let A and B be two n x n self-adjoint (i.e., Her-mitian) matrices over C such that all eigenvalues of A lie in [a, a'] and all eigen-values of B lie in [b, b'}. Show that all eigenvalues of A + B lie in [a + b a' + b'].
Problem 7.5.33 (Fa85) Let k be real, n an integer 2, and let A = (a,j) be then x n matrix such that all diagonal entries a,, = k, all entries immediatelyabove or below the diagonal equal 1, and all other entries equal 0. For example,zfn=5, kl000
1k100A= 0 1 k 1 0
001k10001k
Let Xmin and denote the smallest and largest eigenvalues of A, respectively.Show that Xmjn k — I and Xmax k + 1.
Problem 75.34 (Fa87) Let A and B be real nxn sym,netric matrices with Bpositive definite. Consider the function defined for x 0 by
(Ax,x)G(x)=(Bx,x)
1. Show that G attains its maximum value.
2. Show that any maximum point U for G is an eigenvector for a certain ma-trix related to A and B and show which matrix.
Problem 7.5.35 (Fa90) Let A be a real symmetric n x n matrix that is positivedefinite. Let y E IRE, y 0. Prove that the limit
yfAm+Iylim
m-÷oo ytAmy
exists and is an eigenvalue of A.
7.6 Canonical Forms
Problem 7.6.1 (Sp90, Fa93) Let A be a complex n x n matrix that has finiteorder; that is, = I for some positive integer k. Prove that A is diagonalizable.
Problem 7.6.2 (Sp84) Prove, or supply a counterexample: If A is an invertiblen x n complex matrix and some power of A is diagonal, then A can be diagonal-ized.
Problem 7.63 (Fa78) Let
A=(IExpress A—' as a polynomial in A with real coefficients.
Problem 7.6.4 (Sp81) For x E R, let
fx 1 1 1
11 x 1 1
i 1
1 1 x
1. Prove that = (x — (x + 3).
2. Prove that if x # 1, —3, then A1 = —(x — +
Problem 7.6.5 (Sp88) Compute A10 for the matrix
f3 11A=( 2 42.
—1 1
Problem 7.6.6 (Fa87) Calculate A'°° and A7, where
1 3/2 1/21/2
Problem 7.6.7 (Sp96) Prove or dispmve. For any 2x2 matrix A over C, thereis a 2x2 matrix B such that A = B2.
Problem 7.6.8 (Su85) 1. Show that a real 2x2 matrix A satisfies A2 = —I
if and only
______
—pq
where p and q are real numbers such that pq I and both upper or bothlower signs should be chosen in the double signs.
2. Show that there is no real 2 x 2 matrix A such that
A2_(1 01€
withs >0.
Problem 7.6.9 (Fa96) Is there a real 2x2 matrix A such that
A20_(1 00 —1—eJ
Exhibit such an A or prove there is none.
Problem 7.6.10 (Sp88) For which positive integers n is there a 2 x 2 matrix
A(ab
d
with integer entries and order n; that is, A'2 = I but Ak I for 0 <k <n?Note: See also Problem 7.7.8.
Problem 7.6.11 (Sp92) Find a square root of the matrix
/1 3 —3
1045\oo 9
How many square roots does this matrix have?
Problem 7.6.12 (Sp92) Let A denote the matrix
000000000
For which positive integers n is there a complex 4 x 4 matrix X such that X'2 A?
Problem 7.6.13 (Sp88) Prove or disprove: There is a real n x n matrix A suchthat
A2+2A+51 =0if and only if n is even.
Problem 7.6.14 (Su83) Let A be an n x n Hermitian matrix satisfying the con-dition
A5 +A3 +A =31.Show that A = 1.
Problem 7.6.15 (Su80) Which of the following matrix equations have a real ma-trix solution X? (It is not necessary to exhibit solutions.)
1.
/0x3=(i
oo\0 01,3 01
0100
2.
f3 5 o\2X5+X=(5 91,\090J
X6+2X4+IOX=(? _1)
/340x4=Io 3 0
0 3
Problem 7.6.16 (Sp80) Find a real matrix B such that
/2 0 02 0
—l 1
Problem 7.6.17 (Fa87) Let V be afinite-dimensional vector space and T: V -+ Va diagonalizable linear transformation. Let W C V be a linear subspace whichis mapped into itself by T. Show that the restriction of T to W is diagonalizable.
Problem 7.6.18 (Fa89) Let A and B be diagonalizable linear transformations ofR'2 into itself such that AB = BA. Let E be an eigenspace of A. Prove that therestriction of B to E is diagonalizable.
Problem 7.6.19 (Fa83, Sp87, Fa99) Let V be a finite-dimensional complex vec-tor space and let A and B be linear operators on V such that AB = BA. Pmvethat A and B can each be diagonalized, then there is a basis for V which simul-taneously diagonalizes A and B.
Problem 7.6.20 (Sp80) Let A and B be n x n complex matrices. Prove or dis-prove each of the following statements:
1. If A and B are diagonalizable, so is A + B.
2. If A and B are diagonalizable, so is AB.
3. If A2 = A, then A is diagonalizable.
4. If A is invertible and A2 is diagonalizable, then A is diagonalizable.
Problem 7.6.21 (Fa77) Let
154
Find a real matrix B such that B' AB is diagonal.
Problem 7.6.22 (Su77) Let A : 1R6 R6 be a linear transformation such thatA26 = I. Show thatR6 = where V1. V2, and V3 are two-dimensionalinvariant subspaces for A.
Problem 7.6.23 (Sp78, Sp82, Su82, Fa90) Determine the Jordan CanonicalFormof the matrix f123
A=IO 4 5
Problem 7.6.24 (Su83) Find the eigenvalues, eigenvectors, and the Jordan Canon-ical Form of
f2 1 i\A=I1 2\112/
considered as a matrix with entries in F3 = Z/3Z.
Problem 7.6.25 (Su83) Let A be an n x n complex matrix, and let x and bethe characteristic and minimal polynomials of A. Suppose that
X(x) = — i),
fh(x)2 = X(x)(X2 + 1).
Determine the Jordan Canonical Form of A.
Problem 7.6.26 (Fa78, Fa84) Let M be the n x n matrix over a field F, all ofwhose entries are equal to 1.
1. Find the characteristic polynomial of M.
2. Is M diagonalizable?
3. Find the Jordan Canonical Form of M and discuss the extent to which theJo rdan form depends on the characteristic of the field F.
Problem 7.6.27 (Fa86) Let denote the vector space of complex 2 x2 ma-trices. Let
(0 1
and let the linear transformation T : M2x2 -÷ M2x2 be defined byT(X) XA — AX. Find the Jordan Canonical Form for T.
Problem 7.6.28 (SpOl) Let be the vector space of n x n complex matrices.For A in define the linear transformation of TA on by TA (X) = AX — X A.Prove that the rank of TA is at most n2 — n.
Problem 7.6.29 (FaOl) Let A be an n x n matrix with real entries, let denoteits cha racteristic polynomial, and let g(t) E REt] be a polynomial of degree n — 1
dividing What are the possibilities for the rank of g(A)?
Problem 7.6.30 (Fa88) Find the Jordan Canonical Form of the matrix
100000110000101000100100100010111111Problem 7.6.31 (Sp02) Let A and B be complex matrices ofsizes 3x5 and 5x3,respectively, such that /110
1 0\o 0 —1
Find the Jordan canonicalform of BA.
Problem 7.6.32 (Sp99) Let A = (ajj) be a n x n complex matrix such that0 if i = j + 1 but = 0 if i j + 2. Prove that A cannot have
more than one Jordan block for any eigenvalue.
Problem 7.633 (Fa89) Let A be a real, n n matrix that com-mutes with its transpose. Prove that A is diagonal.
Problem 7.634 (Su78) 1. Prove that a linear operator T : C —÷ C isdiagonalizable iffor all A E C, ker(T — Al)'1 = ker(T — Al), where I isthe n x n identity matrix.
2. Show that T is diagonalizable T commutes with its conjugate transpose(i.e., (T*)jk =
Problem 7.6.35 (Fa79) Let A be an n x n complex matrix. Prove there is a unitarymatrix U such that B = UAU' is upper-triangular: B3k = Ofor j > k.
Problem 7.636 (Sp81) Let b be a real nonzero n x 1 matrix (a column vector).Set M = bb' (an n x n matrix) where bt denotes the transpose of b.
1. Prove that there is an orthogonal matrix Q such that QMQ1 = D isdiagonal, and find D.
2. Describe geometrically the linear M : R" —*
Problem 7.6.37 (Sp83) Let M be an invertible real n x n matrix. Show that thereis a decomposition M = UT in which U is ann x n real orthogonal matrix and Tis upper-triangular with positive diagonal entries. Is this decomposition unique?
Problem 7.6.38 (Su85) Let A be a nonsingular real n x n matrix. Prove that thereexists a unique orthogonal matrix Q and a unique positive definite symmetricmatrix B such that A = QB.
Problem 7.639 (Fa03) Let A be a 2 x 2 matrix with complex entries. Prove thatthe series I + A + A2 +... converges if and only if every eigenvalue of A hasabsolute value less than 1.
Problem 7.6.40 (Sp95) Let A be the 3x3 matrix
11—i 0f—i 2 —1
i
Determine all real numbers afor which the limit tim a7Z A?Z exists and is nonzeron—+oo
(as a matrix).
Problem 7.6.41 (Fa96) Suppose p is a prime. Show that every elementhas order dividing either p2 — 1 or p(p — 1).
7.7 Similarity
Problem 7.7.1 (Fa80, Fa92) Are the matrices give below similar?
/1 oo\ 1110A=f—i 1 i) and B=I0 1 0
\—102) \002Problem 7.7.2 (Fa78, Sp79) Which of the following matrices are similar as ma-trices over
fioo\ fooi\ fioo\(a)1 0 1 01, (b)(0 1 (c)( 1 i of,\oo 1/ \ioo/ \oo 1)
fi 0 o\ /1 1 o\ /0 i 1
(d)11 1 0), (e)I0 of, (f)I° 1 0\0 1 1/ \0 0 1) \i 0 0
Problem 7.73 (SpOO) Are the 4x4 matrices
(1000 1—100010 —1 00 I—i 1 1 —iA=10001 and B=1100
0\o 000 \—ioi 0
similar?
Problem 7.7.4 (Sp79) Let M be an n x n complex Let GM be the set ofcomplex numbers A such that the matrix AM is similar to M.
1. What is GM if foo4\M=IO 0 01?\oooJ
2. Assume M is not nilpotent. Prove GM is finite.
Problem 7.7.5 (Su80, Fa96) Let A and B be real 2 x 2 matrices such thatA2 = B2 = I and AB + BA = 0. Prove there exists a real nonsingular ma-trix T with
TBT-'=(?
Problem 7.7.6 (Su79, Fa82) Let A and B be n x n matrices over a field F suchthat A2 = A and B2 = B. Suppose that A and B have the same rank Prove thatA and B are
Problem 7.7.7 (Fa97) Prove that if A is a 2 x 2 matrix over the integers such that= I for some strictly positive integer n, then 12 =
Problem 7.7.8 (Su78) Let G be a finite multiplicative group of 2 x 2 integer ma-trices.
1. Let A G. What can one prove about
(i) detA?
(ii) the (real or complex) eigenvalues of A?
(iii) the Jordan or Rational Canonical Form of A?
(iv) the order of A?
2. Find all such groups up to isomorphism.
Note: See also Problem 7.6.10.
Problem 7.7.9 (Fa80) Exhibit a set of2 x 2 real matrices with the following prop-erly: A matrix A is similar to exactly one matrix in S provided A is a 2 x 2 invert-ible matrix of integers with all the roots of its characteristic polynomial on theunit circle.
Problem 7.7.10 (Fa81, Su81, Sp84, Fa87, Fa95) Let A and B be two real n xn matrices. Suppose there is a complex invertible n x n matrix U such thatA = UBU'. Show that there is a real invertible n x n matrix V such thatA = V B V — (In other words, if two real matrices are similar over C, then theyare similar over R.)
Problem 7.7.11 (Sp91) Let A be a linear transformation on an n-dimensionalvector space over C with characteristic polynomial (x — Prove that A issimilar to A1.
Problem 7.7.12 (Sp94) Prove or disprove: A square complex matrix, A, is simi-
lar to its transpose, At.
Problem 7.7.13 (Sp79) Let M be a real nonsingular3 x 3 matrix. Prove there arereal matrices S and U such that M = SU = US, all the eigenvalues of U equal1, and S is diagonalizable over C.
Problem 7.7.14 (Sp77, Sp93, Fa94) Find a list of real matrices, as long as pos-sible, such that
• the cha racteristic polynomial of each matrix is (x — 1)5(x + 1),
• the minimal polynomial of each matrix is (x — 1)2(x + 1),
• no two matrices in the list are similar to each other.
Problem 7.7.15 (Fa95) Let A and B be nonsimilar n x n complex matrices withthe same minimal and the same characteristic polynomial. Show that n 4 andthe minimal polynomial is not equal to the characteristic polynomial.
Problem 7.7.16 (Sp98) Let A be an n x n complex matrix with tr A = 0. Showthat A is similar to a matrix with all 0's along the main diagonal.
Problem 7.7.17 (Sp99) Let M be a 3 x 3 matrix with entries in the polynomialftOo\ring R[t] such that M3 = 0 t 0 Let N be the matrix with real entries
\OOt/(0 1 0
obtained by substituting t = 0 in M. Prove that N is similar to 0 0 1\000Problem 7.7.18 (Sp99) Let M be a square complex matrix, and letS = (XMX' J Xis non-singulaij be the set of all matrices similar to M. Showthat M is a nonzero multiple of the identity matrix and only if no matrix in Shas a zero anywhere on its diagonaL
Problem 7.7.19 (SpOO) Let A be a complex n x n matrix such that the sequenceconverges to a matrix B. Prove that B is similar to a diagonal matrix
with zeros and ones along the main diagonal.
7.8 Bilinear, Quadratic Forms, and Inner ProductSpaces
Problem 7.8.1 (Fa98) Let f : P?3 —÷ ]R be a function such that
1. the function g defined by g(x, y) = f(x + y) — f(x) — f(y) is bilinear,
2. for allx E IR'3 andt E IR, f(tx) = t2f(x).
Show that there is a linear transformation A : -÷ such that f(x) =(x, Ax) where (.,.) is the usual inner product on (in other words, f is aquadratic form).
Problem 7.8.2 (Sp98) Let A, B,..., F be real coefficients. Show that thequadratic form
Ax2 +2Bxy+Cy2 +2Dxz+2Eyz+Fz2
is positive definite if and only if
AB ABDA>0, B C >0, B C E >0.DEF
Problem 7.8.3 (FaOO) Find all real numbers t for which the quadratic form Q,on R3, defined by
Qt(xi, X2, X3) = + 34 + 2tX1x2 + 2X1X3,
is positive definite.
Problem 7.8.4 (Fa90) Let be 3-space with the usual inner product, and(a, b, c) E iiV a vector of length 1. Let W be the plane defined by ax+by+cz =0.Fin4 in the standard basis, the matrix representing the orthogonal projection ofiiV onto W.
Problem 7.8.5 (Fa93) Let w be a positive continuous function on [0, 1], n a pos-itive and the vector space of real polynomials whose degrees are atmost n, equipped with the inner product
q) = I p(t)q(t)w(t) dt.Jo
1. Prove that has an orthonormal basis pi, ..., (i.e., (Pj' Pk) = Ifor j = k and Ofor j k) such that deg Pk = kfor each k.
2. Prove that (pk. = Ofor each k.
Problem 7.8.6 (Sp98) For continuous real valued functions f, g on the interval
[—1, 11 define the inner product (f, g) = f (x)g(x)dx. Find that polyn omial
of the form p(x) a + bx2 — x4 which is orthogonal on [—1, 1] to all lowerorder polynomials.
Problem 7.8.7 (Su80, Fa92) Let E be a finite-dimensional vector space over afield F. Suppose B : E x E-+ F is a bilinear map (not necessarily symmetric).Define subspaces
E1 = fx E E IB(x, y) =0 for all y E E),
allxEE)Prove that dim E1 = dim E2.
Problem 7.8.8 (Su82) Let A be a real n x n matrix such that (Ax, x) Oforevery real n-vectorx. Show that Au =0 if and only if AtU =0.
Problem 7.8.9 (Fa85) An n x n real matrix T is positive definite if T is symmetricand (Tx, x) > Ofor all nonzero vectors x E JRiI, where (u, v) is the standardinner product. Suppose that A and B are two positive definite real matrices.
1. Show that there is a basis {vi, V2, ..., of R" and real numbersA1, A2,..., such that,for 1 i, j n.
0
and
2. Deducefrom Part 1 that there is an invertible real matrix U such that UAUis the identity matrix and UBU is diagonal.
Problem 7.8.10 (Sp83) Let V be a real vector space of dimension n, and letS : V x V -÷ JR be a nondegenerate bilinear form. Suppose that W is a lin-ear subspace of V such that the restriction of S to W x W is identically 0. Showthat dim W n/2.
Problem 7.8.11 (Fa85) Let A be the symmetric matrix
if13 —5 —2—1—5 13 —26\_2 —2 10
Denote by v the column vectorfx(y
in JR3. and by xt its transpose (x, y, z). Let liv II denote the length of the vector v.As v ranges over the set of vectors for which v A v = 1, show that v isand determine its least upper bound.
Problem 7.8.12 (Fa97) Define the index of a real symmetric matrix A to be thenumber of strictly positive eigenvalues of A minus the number of strictly nega-tive eigenvalues. Suppose A, and B are real symmetric n x n matrices such thatxt Ax xt Bx for all n x 1 matrices x. Prove the the index of A is less than orequal to the index of B.
Problem 7.8.13 (Fa78) For x, y E C", let (x, y) be the Hermitian inner productLet T bea linear operator on C" such that (Tx, Ty) = 0 zf(x, y) = 0.
Prove that T = kS for some scalar k and some operator S which is unitary:(Sx, Sy) = (x, y)forallx andy.
Problem 7.8.14 (Sp79) Let E denote a finite-dimensional complex vector spacewith a Hermitian inner product (x, y).
1. Prove that E has an orthonormal basis.
2. Let f : E —÷ C be such that f(x, y) is linear in x and conjugate linear iny. Show there is a linear map A : E —* E such that f(x, y) = (Ax, y).
Problem 7.8.15 (Fa86) Let a and b be real numbers. Prove that there are two or-thogonal unit vectors u and v in JR3 such that u = (ui, U2, a) and v = (vi, v2, b)
and only if a2+b2 1.
7.9 General Theory of Matrices
Problem 7.9.1 (FaOl) Prove that a commutative C —algebra of2 x2 complexma-trices has dimension at most 2 over C.
Problem 7.9.2 (Fa81) Pmve the following three statements about real n x n ma-trices.
1. If A is an orthogonal matrix whose eigenvalues are all djfferent from —1,then I + A is nonsingular and S = (I — A)(I + is skew-symmetric.
2. IfS is a skew-symmetric matrix, then A = (I — S)(1 is an orthogonalmatrix with no eigenvalue equal to —1.
3. The correspondence A S from Parts I and2 is one-to-one.
Problem 7.9.3 (Fa79) Let B denote the matrix
faOO\10 b 01\OOc)
where a, b, and c are real and lal, JbI, and id are distinct. Show that there areexactly four symmetric matrices of the form BQ, where Q is a real orthogonalmatrix of determinant 1.
Problem 7.9.4 (Sp79) Let P be a n x n real matrix such that x' Py = —y' Pxfor all column vectors x, y in JRfl• Prove that P is skew-symmetric.
Problem 7.9.5 (Fa79) Let A be a real skew-symmetric matrix (A,3 = —A31).
Prove that A has even rank
Problem 7.9.6 (Fa98) A real symmetric n x n matrix A is called positive semi-definite Ofor all x JR'1. Prove that A is positive semi-definite if andonly if trAB Ofor every real symmetric positive semi-definite nxn matrix B.
Problem 7.9.7 (Fa80, Sp96) Suppose that A and B are real matrices such thatA—A,
forallv E JR" andAB+BA=0.
Show that AB = BA =0 and give an example where neither A nor B is zero.
Problem 7.9.8 (Fa02) Let A and B be n x n matrices over IR. such that A + B isinvertible. Prove that
A(A + BY'B = B(A + BY'A.
Problem 7.9.9 (Sp78) Suppose A is a real n x n matrix.
1. Is it true that A must commute with its transpose?
2. Suppose the columns of A (considered as vectors) form an orthonormal set;is it true that the rows of A must also form an orthonormal set?
Problem 7.9.10 (Sp98) Let M1 == (53 M3 = (53 For
which (if any) i, 1 i 3, is the sequence (M') bounded away from oo? Forwhich i is the sequence bounded away from 0?
Problem 7.9.11 (Su83) Let A be an nxn complex matrix, all of whose eigenval-ues are equal to 1. Suppose that the set I .n = 1,2, . . .} is bounded. Show thatA is the identity matrix.
Problem 7.9.12 (Fa81) Consider the complex 3x3 matrix
/ao al a2\A=(a2 aij,
\ai a2 aoj
where ao, al, E C.
1. Show that A = aoI3 + alE + a2E2, where
/0 1 00 1
2. Use Part 1 to find the complex eigenvalues of A.
3. Generalize Parts 1 and 2 to n x n matrices.
Problem 7.9.13 (Su78) Let A be a n x n real matrix.
1. If the sum of each column element of A is 1 prove that there is a nonzerocolumn vector x such that Ax = x.
2. Suppose that n = 2 and all entries in A are positive. Prove there is anonzero column vector y and a number X > 0 such that Ay =
Problem 7.9.14 (Sp89) Let the real 2n x 2n matrix X have the form
(AB
where A, B, C, and D are n x n matrices that commute with one anothe, Provethat X is invertible jf and only if AD — BC is invertible.
Problem 7.9.15 (Sp03) Let GL2(C) denote the group of invertible 2 x 2 matri-ces with coefficients in the field of complex numbers. Let PGL2(C) denote the
quotient of GL2(C) by the normal subgroup Let n be a posi-
tive and suppose that a, b are elements of PGL2(C) of order exactly n.Prove that there exists c E PGL2(C) such that cac1 is a power of b.
Problem 7.9.16 (Sp89) Let B = be a real 20 x 20 matrix such that
b11=0 for
1 i j.
Prove that B is
Problem 7.9.17 (Sp80) Let
(1 2
Show that every real matrix B such that AB BA has the form sI + tA, whereS, t E ILL
Problem 7.9.18 (Su84) Let A be a 2 x2 matrix over C which is not a scalarmultiple of the identity matrix I. Show that any 2x2 matrix X over C commutingwithAhastheformX =aI+j3A, where a,fi C.
Problem 7.9.19 (Sp02) 1. Determine the commutant of the n x n Jordanmatrix A10...00
0 A 1 ... 0 00 0 A ... 0 0
AB= . . . . •
0 ... A 1
0 0 0 ... 0 A
In particulai determine the dimension of the commutant as a complex vec-
tor space.
2. What is the dimension of the commutant of the 2n x 2n matrix
1.1 o\
Problem 7.9.20 (Fa96) Let
/2—1 0A=I —1 2 —1
\ 0—1 2
Show that every real matrix B such that AB = BA has the form
B =aI+bA+cA2
for some real numbers a, b, and c.
Problem 7.9.21 (Sp77, Su82) A square matrix A is nilpotent Ak = Ofor somepositive integer k.
1. If A and B are nilpotent, is A + B nilpotent?
2. Prove: If A and B are nilpotent matrices and AB = BA, then A + B isnilpotent.
3. Prove: If A is nilpotent then I + A and I — A are invertible.
Problem 7.9.22 (Sp77) Consider the family of square matrices A(8) defined bythe solution of the matrix differential equation
dA(O) = BA(O)
with the initial condition A(0) = I, where B is a constant square matrix.
1. Find a property of B which is necessary and sufficientfor A(O) to be orthog-onalfor all 0; that is, A(9)t = A(9)1, where A(0)' denotes the transposeof A(8).
2. Find the matrices A(0) corresponding to
and give a geometric interpretation.
Problem 7.9.23 (Su77) Let A be an r x r matrix of real numbers. Prove that theinfinite sum
A2 A'7
2 n!
of matrices converges (i.e., for each i, j' the sum of (i, entries converges),and hence that eA is a well-defined matrix.
Problem 7.9.24 (Sp97) Show that
det(exp(M)) =
for any complex n x n matrix M, where exp(M) is defined as in Problem 7.9.23.
Problem 7.9.25 (Fa77) Let T be an n x n complex matrix. Show that
urn T'=Ok-÷oo
and only all the eigenvalues of T have absolute value less than 1.
Problem 7.9.26 (Sp82) Let A and B be nxn complex matrices. Prove that
)tr(AB*)12 tr(AA*)tr(BB*).
Problem 7.9.27 (Fa84) Let A and B be n x n real matrices, and k a positive inte-Find
1.
urn — ((A + tB)k — Ak).t \
2.
Problem 7.9.28 (Fa91) 1. Prove that any real nxn matrix M can be writtenas M = A + S + ci, where A is antisymmetric, S is symmetric, c is aI is the identity matriZ and tr S =0.
2. Prove that with the above notation,
tr(M2) = tr(A2) + tr(S2) + !(trM)2.
Problem 7.9.29 (Fa99) Let A be an n x n complex matrix such that tr = Ofork=1,...,n.ProvethatAisnilpotent.
Problem 7.9.30 (Sp98) Let N be a nilpotent complex Let r be a positiveinteger. Show that there is a n x n complex matrix A with
- K=I+N.Problem 7.931 (Fa94) Let A = be a real n x n matrix such thatajj 1 for all i, and
i#jProve that A is invertible.
Problem 7.9.32 (Fa95) Show that an n x n matrix of complex numbers A satis-
fying
i ( n must be invertible.
Problem 7.9.33 (Sp93) Let A = be an n x n matrix such that I<1
for each i. Prove that I — A is invertible. j=1
Problem 7.9.34 (Sp94) Let A be a real n x n matrix. Let M denote the maximumof the absolute values of the eigenvalues of A.
1. Prove that if A is symmetric, then flAx ii M lix II for all x in R'1. (Here,denotes the Euclidean norm.)
2. Prove that the preceding inequality can fail if A is not symmetric.
Problem 7.935 (SpOO) Let A be an n x n matrix over C whose minimal polyno-mial has degree k.
1. Prove that, if the point A of C is not an eigenvalue of A, then there is apolynomial pj. of degree k — 1 such that = (A — AI)'.
2. Let Xi, ..., be distinct points of C that are not eigenvalues of A. Provethat there are complex numbers ci, ..., such that
k
Problem 7.936 (Sp99) Let lix II denote the Euclidean norm of a vector x. Showthatfor any real m x n matrix M there is a unique non-negative scalar a, and(possibly non-unique) unit vectors u E and v E such that
1. IIMxII ollxfl for allx E W2,
2. Mu = av,
3. MTv = au (where MT is the transpose of M).
Part II
Solutions
1
Real Analysis
1.1 Elementary Calculus
Solution to 1.1.1: Let f(9) = cos p9 — (cos 9)P. We have f(0) = 0 and, for0<9<ir/2,
f'(O) = —psinp9sin0 \
= P + C0SI—P0)
>0
since sin is an increasing function on [0, ir/2] and 9 E (0, 1). We concludethat f(9) 0 for 0 9 ir/2, which is equivalent to the inequality we wantedto establish.
Solution to 1.1.2: Let x E [0, 1]. Using the fact that f(0) = 0 and the Cauchy—Schwarz Inequality [MH93, p. 69] we have,
If(x)I=
'0Jf'(t)I2dt
and the conclusion follows.
Solution to 1.13: As f' is positive, f is an increasing function, so we have, for
t> 1, f(t)> f(1) = 1. Therefore, fort> 1,
/ — 1 <1f(t)— t2+f2(t) t2+l'
so
f(x) = 1 + f'(t)dt
<i+f=1+1;
hence, f(x) exists and is, at most, 1 + The strict inequality holdsbecause
limf(x)=1+ff'(t)dt < 1+f
Solution to 1.1.4: Denote the common supremum off and g by M. Since f and gare continuous and [0, 1] is compact, there exist a, E [0, 1] withf(a) = = M. The function h defined by h(x) = f(x) — g(x) satisfiesh(a) = M — g(a) ? 0, h(fl) = f(J3) — M 0. Since h is continuous, it has azero t [a, 6]. We have f(t) = g(t), so f(t)2 + 3f(t) = g(t)2 + 3g(t).
Solution to 1.15: Call a function of the desired form a periodic polynomial, andcall its degree the largest k such that occurs with a nonzero coefficient.
If a is 1-periodic, then = a/if for any function f, so, by the InductionPrinciple [MH93, p. 7], for all n.
We will use Complete Induction [MH93, p. 32]. For n = 1, the result holds:= Oifandonlyiffis 1-periodic.Assumeitistrueforl, ... ,n — 1.If
is a periodic polynomial of degree, at most, n — 1, then
= + • • +
and the induction hypothesis implies that all the terms vanish except, maybe, thelast. We have = n —2 by theBinomial Theorem [BML97, p. 15]. So the induction hypothesis also implies
1) = 0 and the first half of the statement is established.
For the other half, assume f = 0. By the induction hypothesis, is aperiodic polynomial of degree, at most, n — 2. Suppose we can find a periodicpolynomialg, of degree, at most, n—i, such that = Then, as =0, the function f g will be 1-periodic, implying that f is a periodic polynomialof degree, at most, n — 1, as desired. Thus, it is enough to prove the followingclaim: If h is a periodic polynomial of degree n (n = 0, 1, .. .), then there is aperiodic polynomial g of degree n + I such that = h.
If n = 0, we can take g hx. Assume h has degree n > 0 and, as an inductionhypothesis, that the claim is true for lower degrees than n. We can then, withoutloss of generality, assume h = ax's, where a is 1-periodic. By the BinomialTheorem,
is a periodic polynomial of degree n — 1, so it equals for some periodicpolynomial gi of degree n, and we have h = where
n+1 +gi
as desired.
Solution to 1.1.6: Since f is increasing and nonconstant, there exist ro suchthat f(ro) < f(so). Increasing if necessary, we may assume 8 = — ro > 1.LetA=f(so)—f(ro).
Choose [ri, si] to be either [ro, (ro + or [(re + se], whichevermakes f(si) — larger. (Choose either if they are equal.) Then
f (rj) /2. Similarly choose [r2, s2] as the left or right half of [rj, Si], to max-imize f(S2)— f(r2), and soon, so = 2'8, and f(r1)
Let a = limr,, and set c = Given x E (0, 1], choose the first integer
i 0 such that 2'S x. Then 2'S x/2. Also Er,, s,] contains a and is oflength x, so [ri, s,] [a — x, a + x]. Since f is increasing,
f(a + x) — f(a — x) f(s,) — f(rj) = 2c2_'S cx.
Finally, the inequality f(a + x) — f(a — x) cx is immediate for x =0.
Solution to 1.1.7:1. This is false. For
10 forf(t)=g(t)=11 for t=0
we have Iim,...÷o g(t) = f(t) = 0 but f (g(t)) = 1.
2. This is false. f(t) = maps the open interval (—1, 1) onto [0, 1), which is not
open.
3. This is true. Let x, E (—1, 1). By Taylor's Theorem [Rud87, p. 110], there
is E (—1, 1) such that
n—I k (n)
f(x)=>Jk! n!
(x—xoY' (nEN).
We have. x—xol'1
lim (x — xo) limn-÷oo n!
so00 (k'f' '(xe) k
k!(x—xo)
foranyxo E (—1, 1)andf isreal analytic.
Solution to 1.1.8: Let f(x) — sinx —x +x3/6; then f(0) = f'(O) = f"(O) 0andf"'(x) = 1 —cosx,sof"(x) Oforallx,andf"(x) > OforO <x <2n.Hence, for x > 0, f"(x) = f"(t) dt > 0; similarly f'(x) = f"(t) dt > 0,
and finally f(x) = f'(t) dt >0.
Solution 2. The function f(x) = sin x — x + has derivatives
x2
f"(x) = —sinx+xf"(x) = — cos x + 1.
Now f"(x) 0 with equality at the discrete set of points 2irn, n E Z. Thereforef" is strictly increasing on IR, and since f"(O) = 0, we obtain f"(x) > 0 forx > 0. Therefore f' is strictly increasing on R+; since f'(O) = 0, we obtain
f'(x) > 0 for x > 0. Therefore f is strictly increasing on since f(0) 0we obtain f(x) > 0 for x > 0, as required.
Solution 3. Consider the Taylor expansion of sin x around 0, of order 5, with theLagrange remainder. For x > 0 we have
x3 x5
6!
forsome0 <x.Since >Oand that theresultfollows for positive x.
Solution to 1.1.9: The Maclaurin expansion of sin x of order 3 is
thereforesin2 x = x2 + 0(x4) (x —÷ 0).
So, for h near 0, we have y(h) = I — 8ir2h2 + 0(h4). (Alternatively, observe thaty(h) = cos4pih and use the expansion for cosine.) Thus,
2 2f(y(h))= (h-40).1 + + 0(h4) 1 +
2=2— 2x + 2x2 + 0(x2), we get1+x
f(y(h)) =2— Ihi + 0(h2), (h -+ 0).
Solution to 1.1.10: 1. Suppose f : [0, 1] —' (0, 1) is a continuous surjection.Consider the sequence such that ((0, 1/n)). By the Boizano—Weierstrass Theorem [Rud87, p. 40], [MH93, p. 153], we may assume that (Xv)converges, to x E [0, 1], say. By continuity, we have f(x) = 0, which is absurd.Therefore, no such a function can exist.2. g(x) = I sin 2jrxl.3. Suppose g: (0, 1) -÷ [0, 1] is a continuous bijection. Let xo = andxj = g1(1). Without loss of generality, assume xo < xj (otherwise consider1 —g). By the Intermediate Value Theorem [Rud87, p. 93], we have g([xo, xl]) =[0, 1]. As x1 E (0, 1), g is not an injection, which contradicts our assumption.
Solution to 1.1.11: Let A(c) denote the area the problem refers. The condition onf" implies the convexity of f, 50 the graph of f is always above any tangent toit, and we have
bA(c)
= f (1(x) — f(c) — f'(c)(x — c)) dx.
The derivative of A is given by
b
A'(c)= _f f"(c)(x —c)dx
b2—a2= —f"(c)
2+ (b — a)cf'(c)
f"(c)(b — a)(
a + b)
so the minimum occurs at c = (a + b)/2. As A' is an increasing function, A isconvex, so its minimum in [a, b] corresponds to the only critical point in (a, b).
Solution to 1.1.12: Using the parameterization
x=acost, y=bsint,
a triple of points on the ellipse is given by
(acost1,bsint,), i = 1,2,3.
So the area of an inscribed triangle is given by
1acosti bsinti ab 1 Cost1 Sinti
— 1 a cos t2 b sin 12 = 1 cos t2 sin t22
ii a cos 13 b sin 13 1 COS t3 sin 13
which is ab times the area of a triangle inscribed in the unit circle. In the case ofthe circle, among all inscribed triangles with a given base 2w (0 < w 1), theone of maximum area is an isosceles triangle whose area equals
g(w) = w(1 + — w2).
Using elementary calculus one finds that the maximum of g on the intervalo w 1 occurs at w = corresponding to an equilateral triangle, andequals Alternatively, fixing one side of the triangle as the basis, we easilysee that among all the inscribed triangles the one with the greatest area is isoscelesbecause of the maximum height, showing that the angle at the basis is the same.Fixing another side we see that the triangle is indeed equilateral. Hence, the areais maximal when
2ir 2irand
that is, when the corresponding triangle inscribed in the unit circle is regular.For the ellipse with semiaxes a, b, this corresponds to an inscribed triangle with
maximum area equalsSolution 2. Let f R2 be the stretch function f(x, y) = (ax, by). By awell know lemma in the proof of the Change of Variable for Integration [MH93,p. 524, Lemma 1], or the theorem itself when proven in its whole generality,
vol(f(T)) = Idetfi volA
since the determinant of f is constant, following on the steps of the previousproof, the maximum for the area is achieved over the image of an equilateraltriangle and it is equal to
vol (f(T)) = ab. vol(T) = 3ab—4--•
Solution to 1.1.13: Assume that a and b are in A and that a < b. Supposea <c <b. Let and be sequences in [0, oo) tending to +oo such thata and b = Deleting finitely many terms fromeach sequence, if necessary, we can assume <c and f(yn) > c for everyn. Then, by the Intermediate Value Theorem [Rud87, p. 93], there is for each ii a
point between and such that = c. Since obviously urn Zn = +00,it follows that c is in A, as desired.
Solution to 1.1.14: For each x > 0, the equation x(1 + log(1/s4Ji)) = I maybe solved for s to obtain s = e/(xh/2el/X) that we will call f(x).
Forx > 0 define g(x) = xh/2elIx = e/f(x). Then g'(x) = —2).Thus gis strictly decreasing on (0,2] and strictly increasing on [2,00). Moreover
g(x) = oo = g(x), and g(2) > 0. Thus f(x) = 0 =f(x), f is strictly increasing on (0, 2] and strictly decreasing on [2, oo),
f is continuous on (0,00), and f(2) Denote the restrictions of f to(0,2) and (2,00) by fj, 12 respectively. Then for each 0 < e the givenequation has exactly two solutions, namely x = fI'(s), (e). The smallersolution is then x = fr(s). As fi is strictly increasing and continuous, withfi(0+) =0,wededucethatx(e) —+ Oasc-÷ 0+.
Fix S > 0. Now e3x(s) = = where x =x(s) = fr'(e). As s -÷ 0+, x -÷ 0+. The function x i-* (s/2)es/1 haslimit oo as x —÷ 0+. This proves the second part.
Solution to 1.1.15: Let g be a polynomial,
g(x)=ao+aI(x —a)+a2(x —a)2
Ifwetake
1 f'(a)g(a) f"(a)g(a)+f'(a)g'(a)ao = aj = —
a calculation shows that the requirements on g are met.
Solution to 1.1.16: Suppose that all the roots of p are real and let deg p = n. Weand
= n. By differentiating this expression, we see that the r1 's are roots of p' oforder — 1 when > 1. Summing these orders, we see that we have accountedfor n — k of the possible n — I roots of p'. Now by Rolle's Theorem [MH93, p.200], for each i, 1 i k — 1, there is a point s1, r1 <s1 < Tj+1, such that
= 0. Thus, we have found the remaining k — 1 roots of p'. and they aredistinct. Now we know that a is a root of p' but not of p, so a for all i. Buta is a root of p", so a is a multiple root of p'; hence, a for all i. Therefore,a is not a root of p', a contradiction.
Solution to 1.1.17: Let x E R and h > 0. By the Taylor's Theorem [Rud87, pag.110], there is a w E (x, x + 2h) such that
f(x + 2h) = f(x) + 2hf'(x) ÷ 2h2f"(w),
or rewritingf'(x) = f(x + 2/i) — f(x)
— hf"(w).
Taking absolute values and applying our hypotheses, we get
If'(x)l +hB.
Now making h = the conclusion follows:
If'(x)I
Solution to 1.1.18: Consider the function f(x) = log x/x. We have = 'iff(a) = f(b). Now f'(x) = (1 — log x)/x2, so f is increasing for x <c anddecreasing for x > e. For the above equality to hold, we must have 0 <a <e,so a is either 1 or 2, and b> e. For a = 1, clearly there are no solutions, and fora = 2 and b = 4 works; since f is decreasing for x > e, this is the only solution.
Solution 2. Clearly, a and b have the same prime factors. Let b = a + t, for somepositive integer t. Then = b" which implies that therefore ajb. Wehave then b = ka, for some integer k> 1. Now = = ab implies that kis apowerofa, sob am forsome m> l.Now b" = ama = exactly whenma = am, which can easily be seen to have the unique solution a = m = 2. Soa =2 and b= 22 = 4.
Solution 3. Let b = a(1 + t), for some positive t. Then the equation a1' = isequivalent to any of the following
= (a(1 + t))a1+: = aa (1 + t)a
(aa)' =(1 +t)aa = 1+t.
We have, by the power series expansion of the exponential function, that e> 1 +tfor positive t, so a <e. As a = I is impossible, we conclude a =2. The originalequation now becomes
2b=b2
which, considering the prime decomposition of b, clearly implies b =4.
Solution to 1.1.19: The equation can be rewritten as = x, or
logx =loga.
There is thus a solution for x if and only if log a is in the range of x i-÷ (log x)/xb.Using elementary calculus, we get that the range of this function is (—oo, 1/be].We conclude then that the original equation has a positive solution for x if andonly if log a 1/be, that is, if and only if 1 <a
Solution to 1.1.20: Let f(x) = for x > 0. We have
/ 3x(xlog3_3) 3f(x)= >0 for x>x log3
As3/log3 <3 <ir,wehavef(3)=1 <371•
Solution 2. The same kind of analysis can be performed to the function f(x) =ln(x)/x, which is decreasing forx > e, as well as others likeg(x) = — 3' andh(x) = (3
Solution to 1.1.21: Fix a in (1, cc), and consider the function f(x) on(1, cc), which we try to minimize. Since logf(x) = x loga — a logx, we have
f'(x) a=loga——,f(x) x
showing that f'(x) is negative on (1, iota) and positive on , cc). Hence, fattains its minimum on (1, cc) at the point Xa
aa a log a log
a
a thus has the required property if and only if eloga 1. To seewhich numbers a in (1, cc) satisfy this condition, we consider the functiong(y) = logy on (1, cc). We have
/ 1—logy2y
from which we conclude that g attains its maximum on (1, cc) at y = e, themaximum value being g(e) = Since g(y) < on (1, cc)\(e), we conclude
that < 1 for a in (1, cc), except for a = e. The number a = e is thus theonly number in (1, cc) with the required property.
Solution to 1.1.22: Let g(x) = eX/xt for x > 0. Since g(x) —+ cc as x —+ 0 and
as x -÷ cc, there must be a minimum value in between. At the minimum,
g'(x) = exx_(1 — t/x) = 0,
so the minimum must occur at x = t, where
g(x) = g(t) = et/tt = (e/t).
Thus,xe
and the right-hand side is strictly larger than if and only if t <e.
Solution to 1.1.23: f can be written as its second degree Maclaurin polynomial[PMJ85, p. 127] on this interval:
(3)
f(x) = f(0) + f' (0)x +2
X +6
X
where is between 0 and x. Letting x = in this formula and combining the
results, we get, for n 1,
(i =
for some E [—1, 1]. As f" is continuous, there is some M > 0 such thatIf'"(x)I <M for all x [—1, 1]. Hence,
Solution to 1.1.24: Using Taylor's Theorem [Rud87, p. 110],
f(x + h) — f(x) = f'(x)h + for some z E (x, x + h)
and similarly
f(x — h) — f(x) = —f'(x)h + for some
The result follows by adding the two expressions, dividing by h2, and taking thelimit when h -+ 0.
Solution to 1.1.25: 1. The geometric construction
n(fin)+
+
—2f'(O)
6n2
W €(x—h,x).
1
,r/20
shows that
forJr 2
An analytic proof can be written down from the fact that the sin function is con-cave down (second derivative negative) in the interval 0 0
It can also be seen from the following geometric construction due to József[Sán88] and later rediscovered by Feng Yuefeng [Yue96]:
013= OM+MP OA
2. The integral inequality
20
Jr
J=IJopr/2
JoR dO
ir/2= _ire2Rem0
<It
is called Jordan's Lemma [MH87, p. 301]. Our limit is then
urn IR—+oo J0
sinG dO = lirn IR—oo Jo
< urnR-+oo
Solution 2. We have
RX e_R SinG dO + RA dO.
B
Q
As cos 0 1/2 for 0 6 and sin0 is an increasing function on [0, ir/2},we have
RA dO 2RXfT
cosO dO + dO
= (i — +
=o(1) (R—*oo)
Solution to 1.1.26: Let T = IR \ S. T is dense in IR because each nonemptyinterval contains uncountably many numbers.
Fix p E T and define F : IR IR by
F(x) f(t)dt.Jp
F vanishes on T, so, as it is continuous, F vanishes on IL Therefore, we have
Solution to 1.1.27: f'(c) = 0 for some c E (0, 1), by Rolle's Theorem [MH93,
p. 200]. The concavity of f shows that f is increasing on (0, c) and decreasingon (c, 1). The arc length of the graph off on [0, c] is
L(o,c)= f + dx = urn +
where E (kc/n, (k + l)c/n). By the Mean Value Theorem [Rud87, p. 1081 wecan assume the satisfy
/ I ((k + 1)c/n) — f(kc/n)c/n
We get
L(o,c) = 11rn + (f ((k + 1)c/n) — f(kc/n))2
r >(c/n) + (f ((k + 1)c/n) — f(kc/n))
=c+f(c)since f is increasing. A similar reasoning shows that 1 — c + f(c). SoL[o,11 —c+f(c)
Solution to 1.1.28: The convergence of fr If'(x)Idx implies the convergenceof f1°° f'(x)dx, which implies that f(x) exists. If that limit is not 0,
then f(n) and f1°° f(x)dx both diverge. We may therefore, assume thatf(x) = 0. Then f(x)dx -+ 0 as r —+ oo (where In is the greatest
integer r), implying that ir f(x)dx converges if and only iff' f(x)dx exists (where n here tends to 00 through integer values). In
other words, the convergence of f1°° f(x)dx is equivalent to the convergence of
f(x)dx. It will therefore, suffice to prove that
pn+1f(x)dx—f(n)J <00.
n=1
We have
f(x)dx — f(n) = (f(x) — f(n))dxJn
pn+1
n n n n
n+1
=1j'OO
Hence, J f(x)dx — f(n) J lf'(t)Idt <oo, as desired.n=I I
Solution to 1.1.29: We have
iu(x)I =lu(x)—u(O)I lxi
andIu2ix) — u(x)f = Iu(x)iIu(x) — ii IxKlxI + 1)
so
f — u(x)l f x(x + 1)dx =
Equality can be achieved if (u(x)I = x and Iu(x) — II = x + 1. This is the casefor u(x) = —x which is in E.
Solution to 1.1.31: Letjit
u(t) = 1 +2 I f(s)ds.Jo
We haveu'(t) = 2f(t)
so
therefore,
Solution to 1.132: We will show b must be zero. By subtracting and multiplyingby constants, we can assume a = 0 b. Given e> 0, choose R 1 such that
and
for all x R. By the Fundamental Theorem of Calculus [MH93, p. 209],
ço(x) = ço(R) + f ço'(x)dx,
sorXb
2s ço(x) — ço(R) I —dx = (x — R)b/2.JR2
Forx =5R,we getb
s > 0 was arbitrary, we must have b =0.
Solution to 1.1.33: Let 0 ki <k2 < 1, then for all x E (0, ir/2),
—k1 cos2 x> —k2 cos2 x
cos2x > —k2cos2x1 1
— ki cos2 x — k2 cos2 x
i 1
J dx<! dx.o — k1 cos2 x Jo ,/1 — k2 cos2 x
Solution to 1.1.34: With the change of variables y = xVi, we have
f(t)= f etX2 dx
= f = f eY2 dy =—00 —00 4J1 t
so
f'(t) =
Solution to 1.1.35: Let
G(u, v, x) =L"
Then F(x) = G(cosx, sinx, x), so
aGau aGav 8Gduax Fivax ax
= eU2 U(_ sinx) — cosx +
and1
F'(O) —1 + I te2dt = —(e — 3).2
Solution to 1.136: 1. Let 1(z) 0. Then
f(x)f(z) = f (v'x2 + z2) = f(—x)f(z),
so f(x) = f(—x) and f is even.Also, 1(0)1(z) = 1(z), so f(0) = I.
2. We will show now that = (f(x))'3 for real x and natural n, using theInduction Principle [MH93, p. 7]. The result is clear for n = 1. Assume it holdsfor n = k. We have
lx)
f f(x)= (f(x))k f(x)= (f(x))k+I
Ifp,q €N,then1(p) .f(p2 i)
and
= I ([fl) = (from which follows
(1 = (f(1))P2
• Iff(1) >0, we have
I =
so, by continuity on R,
f(x) =
• If f(1) = 0, then f vanishes on a dense set, so it vanishes everywhere,contradicting the hypothesis.
• To see that f(1) <0 cannot happen, consider p even and q odd. We getf(p/q) > 0, so f is positive on a dense set, and f(1) 0.
Note that we used only the continuity of f and its functional equation.Differentiating, we easily check that f satisfies the differential equation. The
most general function satisfying all the conditions is then
C
withO <C <1.Solution2. l.Letx = y =0.Thenf(0)2 = f(O),sof(0)then 0 = for any x, so, in fact, f(x) = 0 for all x > 0. If f(y) 0for any y, then f(x)f(y) = 0 implies f(x) = 0 for all x, so f(x) = 0 for allx if f(0) = 0. Since we assume f is nonzero, we must have f(0) = 1. Thenevaluating at y = 0 gives f(x) = f is an even function.2. Differentiate with respect to y to get
f(x)f'(y) =
where r = /x2 + y2 and Ty denotes the partial derivative of r with respect to y.Differentiate again to get
f(x)f"(y) = +
Since Ty = y/r and = x2/r3, we get
f'(x) = f"(O)xf(x)
for y = 0. The solution of this differential equation is
f(x) = ef"x2,12
and since f vanishes at infinity, we must have f"(0)/2 = —y < 0. Thus,f(x) for some positive constant y•
Solution to 1.137: Let C be the set of all sequences such thatequals 1 or —1; it is an uncountable set. For 8 in C let f8 be the function on [0,1]such that
• fe(0)=0;
• fe(l/n) = e12/n forn a positive integer;
• on each interval [l/(n + 1), 1/n], is the linear function whose values atthe endpoints agree with those given by (2).
Each function is continuous: continuity at points of (0,11 is obvious, and con-tinuity at 0 follows because )f(x)I x. Each fe takes rational values at ra-tional points: if x is a rational point between a = 1/(n + 1) and b = 1/n, thenx = (1 —:)a+tb with t a rational point of [0,1], so f8(x) = (1 isrational because (a) and f8 (b) are. The functions thus form an uncountablesubset of S. showing that S is uncountable.
1.2 Limits and Continuity
Solution to 1.2.1: Fix We have
fg(x)t) —f(xo,t)I
dx= f+f+f,
where R is any positive number. Fix e > 0. Because of the boundedness of f1
and the convergence of I dt, we can choose R so that the first and lastJ
integrals on the right are each less than sf3. Since f is continuous on R2, it isuniformly continuous on the compact set [xo — 1, xo+ 1] x [—R, R}. Hence, thereis a 8 in (0,1) such that lf(x, t) — f(xo, t)I <s/6R for all tin [—R, R] wheneverIx — xol <8. Hence, for Ix — xol <8, the middle integral is also less than €/3,implying that g(x) — g(xo)l <s. This establishes the continuity of g.
Solution to 1.2.2: Consider f(x) = sinx. The Mean Value Theorem [Rud87, p.108) implies that
f(x) — f(y) = — y) = — y) for some (0, 1),
and since I cos I < 1, this implies
11(x) — <lx — whenever x y.
However, if M < 1 were such that
Jf(x) — f(y)1 <Mix — YI for all x, y I,
then, putting x = 0 and letting y 0, we would get If' (0)1 M < 1, whichcontradicts the fact that f'(O) = 1.
Solution to 1.2.3: Suppose f is not continuous at E [0, 1]. Then, for somes > 0, there is a sequence converging to with — > s for alln. By the first condition, there is a sequence such that lies between and
and =e.Then
s) — s) f' + s) — s)
which contradicts the second condition.
Solution to 1.2.4: There exists 5 > 0 such that Ix — yl ( 8 implies
11(x) — 1. Let B = 1/8. Take any x > 0 and let = Lx/8], thegreatest integer not exceeding x/8. Then
lf(x)l =11(x) - 1(0)1 11(x) — + — f((i — 1)8)1
1+n, 1 +Bx.
The proof for x <0 is similar.
Solution to 1.23: 1. Let 1' be the restriction of f to [0,2]. The ranges of f andfi are the same, by periodicity, so f attains its extrema.2. Let 8 > 0. Ii is uniformly continuous, being a continuous function defined ona compact set, so there is S > 0 such that
lfi (a) — 1' <8 for a, b [0,2], Ia — bI <s.
Let x, y E IR with Ix — <e. Then, there are xi, x2 = +1, yi, Y2 = yi +1[0,2] with f(xi) = f(x2) = f(x), f(yi) = f(Y2) = 1(y)' and — <efor some choice of i, j (1, 2), and the conclusion follows.
3. Let f attain its maximum and minimum at and respectively. Then
+jr) — 0 and
f is continuous, the conclusion follows from the Intermediate Value Theorem[RudS7, p. 93].
Solution to 1.2.6: Let be a sequence of numbers in [0, 1) converging tozero. As h is uniformly continuous, given 8 > 0 we can find e > 0 such thatIh(x) — h(y)I <8 if Ix — <e; therefore, we have
— h(Xm)1 <8
for n and m large enough. is a Cauchy sequence then, so it converges, tosay. If is another sequence with limit zero, a similar argument applied to
shows that lim = The function g: [0, 1] -+ IR givenby
(x) — J h(x) for x E [0, 1)g for x=0
is clearly the unique extension of h to [0, 1].
Solution to 1.2.7: Let 1(x) = a. Given e > 0, let K > 0 satisfy— at <e/2 for x K. If x, y K we have then
if(x)—f(y)I
The interval [0, K] is compact, so f is uniformly continuous there, that is, thereexists S > 0 such that, if x, y E [0, K] and lx — <5, then 11(x)— <e/2.Finally, if x K, y> K verify lx — <5, we have
lf(x) — lf(x) — f(K)l + 11(K) — f(y)l <e,therefore, if x, y 0 satisfy Ix — <8, we have lf(x) — <s, as desired.
Solution to 1.2.8: Let E be the set of discontinuities of f. We have E E1 UE2 U E3 U E4, where
E1 = (x E EIf(x—) =f(x+) <f(x)), E2= {x EJf(x—) > f(x+))
E3 = (x E EIf(x—) = f(x+) > f(x)}, E4= (x E EIf(x—) <f(x+)}.For x E E1, let ax E Q be such that f(x—) < ax < f(x+). Now let's take
Q in such a way
t implies f(t)
Thismapq: E1 —÷ Q3 given byx i—f =(ay, implies f(y) <ax <f(y) forx y. So E1 is, at most, countable.
For x E E2, take ax E Q with f(x—) > a7 > f(x+) and choose c, E Qsuch that <x and
b, <t <x implies f(t) > ax
andt implies f(t) <ax;
this map is an injection E2 —k Q so E2 is, at most, countable.Similar methods lead to analogous results for E3 and E4. As the union of count-
able sets is countable, the result follows.
Solution 2. Define the function a : R -+ IR by
o(x) = max(If(x) — f(x+)I, )f(x) — f(x—)l);
observe that a (x) > 0 if and only if x is a discontinuity of f.For each n E N, let the set be given by
n
It is clear that the set of discontinuities of f is D = We shall provethat each has no accumulation points, so, it is countable. If a E usingthe fact that f(a+) = f(x), we can find S > 0 such that for all x,a <x <a+S,wehave
f(a+)_L <f(x)<f(a+)+—,412 4n
that is, for every point in this interval, o(x) 1/2n. In the same fashion, we canfind an open set a — S <x <a such that no point is in showing that is
made up of isolated points so it is countable, and so is D.
Solution to 1.2.9: By Problem 1.2.8, it is enough to show that f has lateral limitsat all points. We have, for any x E IR,
—00 <sup(f(y)) = f(x—) f(x+) = inf(f(y)) <ooy<x y<x
since f is an increasing function.
Solution to 1.2.10: Fix s > 0. For each x E [0, 1], let be as in the hypothesisand = (x x +&). The open intervals (Ii) cover [0, 1] so, by compactnessand the Heine—Borel Theorem [Rud8l, p. 30], we can choose a finite subcover
[0,1] C UI,2
Let M = + s}. If x E [0, 1] then f(x) <M and f is bounded fromabove.
Let N be the least upper bound of f on [0, 1]. Then there is a sequence ofpoints such that tends to N from below. Since [0, 1] is compact,by the Bolzano—Weierstrass Theorem [Rud87, p. 40], [MH93, p. 153], has
a convergent subsequence, so (by passing to a subsequence) we may assume thatconverges to some p [0, 1]. By the upper semicontinuity of f and the
convergence of we have, for n sufficiently large, < f(p) + s andN + s. Combining these, we get f(p) N <1(p) + 2e. Since thisholdsforalle > 0,f(p) = N.
Solution to 1.2.11: Suppose f : JR —÷ JR is continuous, maps open sets to opensets but is not monotonic. Without loss of generality assume there are three realnumbers a <b <c such that f(a) <f(b) > f(c). By Weierstrass Theorem[MH93, p. 189], f has a maximum, M, in [a, c], which cannot occur at a or b.Then f((a, c)) cannot be open, since it contains M but does not contain M + efor any positive s. We conclude then that f must be monotonic.
Solution to 1.2.12: The inequality given implies that f is one-to-one, so f isstrictly monotone and maps open intervals onto open intervals, so f(JR) is open.
Let Zn = be a sequence in f(IR) converging to z JR. Then Zn IS Cauchy,and, by the stated inequality, so is x = limxn. By continuity we havef(x) = = Jim = z so f(JR) is also closed. Thus, f(IR) = JR.
Solution to 1.2.13: 1. For s > 0 let
L= max(If(x)I+1) and o<8<minf—f-.11j.x€[O,1J 2L
We have
f(x)dx x If(x)Idx LS —
2
andpl—6
J x"f(x)dx j (1 —0 0 -
so(I
urn I x'2f(x)dx =0.n-+ooJO
2. We will show that
flurn n xhl(f(x) — f(1))dx =0.
n-+oo
Fors>Oletöbesuchthatlf(x)—f(1)I <e/2ifx EEl —&1].Wehave
nf xldxand, letting L = SUPXE[O11 11(x) — 1(1)1,
1—6 '1 —n x'2(f(x)—f(1))dx /
io o n+1
and the result follows.Now it suffices to notice that
p1 p1 p1n j x"f(x)dx=n I x'2(f(x)—f(1))dx+n I f(l)fdx.
JO JO Jo
Solution to 1.2.14: Suppose that f is not continuous. Then there exist e > 0,x E [0, 1], and a sequence tending to x such that f(x) — ?for all n. Consider the sequence in Gf. Since the unit square iscompact, by Bolzano—Weierstrass Theorem [Rud87, p. 40], [MH93, p. 153], thissequence has a convergent subsequence; using this subsequence, we may assumethat ((xn, converges to some point (y, z). Then we must have the se-quence converging to y; so, by the uniqueness of limits, x = y. Since Gf isclosed, we must have z = f(x). Hence, converges to f(x), contradictingour assumption.
The converse of the implication is also true, see the Solution to Problem 2.1.2.
Solution to 1.2.15: For each y E [0, 1], consider the function gy(x) = f(x, y).Then g(x) = sup gy(x). The family (gy) is equicontinuous because f is uni-formly continuous. It suffices then to show that the pointwise supremum of anequicontinuous family of functions is continuous. Let e > 0, xo E [0, 1]. There is
such thatgy0(xo) g(xo) <gy0(xo) + e.
Let S be positive such that if Jr — SI < 8, then — gy(s)I <e for all y, and
Ixo — xi <8. For some we have that
(g(xi) <gyj(Xi)+S.
Further, by equicontinuity of (gy). we have the two inequalitiesIgy0(xo) — gy0(xI)J < 8 and — gy1(xi)I < c. By combining them weget
gy0(xo) <g(xi)+e <gy1(Xi)+28and
gy1(xi) <gy1(xo) + s <g(xo) + e +2e.
These two inequalities imply (xi) — gy0(xo)I <2s. This, combined with thefirst two inequalities, shows that — g(xi)I <3c. Since this holds for all cand and all x1 close to g is continuous.
Solution to 1.2.16: We need only show that (F) is closed whenever F isclosed. Let F be a closed subset of R. Let be a convergent sequence inf'(F) with limit xo. We need only show that xo is in f'(F). For n > 0let = The sequence (yn)?° is in F and it is bounded (since the con-vergent sequence (xn)r is). Passing to a subsequence, we can assume the se-quence converges, say to yo, which lies in F because F is closed. The setK = {yo, yi, . . .} is then compact, so f'(K) is closed. Hence f'(K) con-tains its limit points. But the sequence (xn)r lies in f1 (K) and converges to xo.Therefore xo is in I (K), and so also in (F), as desired.
1.3 Sequences, Series, and Products
Solutionto 1.3.1: A? (A? + ( we have
A1 = lim ( lim urnn-+oo n-.+oo n-+oo
showing that the limit equals A1.
Solutiontol.3.2:Letp1 = 1,p2 = (2/l)2,p3 = (n/(n_l))P?.Then
P1P2"Pn — flfl—
n n!and since —+ e, we have = e as well (using the fact that
= 1).
Solution 2. As the exponential is a continuous function, L = exp( urn where
= logn — !(log 1 + Iog2 + . • + logn).
Since
log 1 + log2 + + log(n — 1) j logx dx = n logn — n + 1,
we have
(1—1/n) logn—logn+l—1/n = 1—(l+logn)/n —÷ I as n -+00.
On the other hand,
fnlog 1 + log 2+ • + logn logxdx = n log n — n + 1,
JI
so
1/n.
Hence,1—(1
soLo -÷ landL=exp(1)=e.
Solution to 1.3.3: Obviously, 1 for all n; so, if the limit exists, it is 1, andwe can pass to the limit in the recurrence relation to get
3+2xc,o3+x00
in other words, + —3 =0. So is the positive solution of this quadraticequation, that is, = +
To prove that the limit exists, we use the recurrence relation to get
(3
Hence, I — Iteration gives
JXn+I xoJ =
The series — of positive terms, is dominated by the convergentand so converges. We have (xn÷i — = — xi
and we are done.
Solution 2. To prove the existence of the limit it is enough to notice that if g isdefined by
3+2xg(x)= 3+x
we have
for
and apply the Fixed Point Theorem [Rud87, p. 220].
Solution to 13.4: We prove, by induction, that 0 < 1/2 for n 1. First,
0 < 1/3 < 1/2. Suppose for some n 1, that 0 < < 1/2. Then2/5 <Xn+I = 1/(2 + < 1/2. This completes the induction.
Let f(x) = 1/(2 + x). The equation f(x) = x has a unique solution in theinterval 0 <x <1/2 given by p = — 1. Moreover, lf'(x)I = 1/(2 + x)2 <1/4 for 0 <x <1/2. Thus, for n 1,
— —
by the Mean Value Theorem.Iterating, we obtain
12—p1
1
— p1
Hence the sequence tends to the limit — 1.
Solution 2. Define f(x) = 1/(2 + x) for 0 x 1. Then f maps the closed
interval [0, 1] to [1/3, 1/2] C [0, 1]. Also If'(x)I = (2 +x)2for x E
[0, 1]. Therefore f : [0, 1] —÷ [0, 1] is a contraction map with Lipschitz constant1/4 < 1, so f has a unique fixed pointy, and the sequence = definedaboveconvergestoy.Wehavey = 1/(2+y)ory2+2y = 1,whencey
(1).
Solution to 1.3.5: By the given relationXn+i —Xfl = Therefore,by the InductionPrinciple [MH93, p. 7], we have = (a— i)n_1 (xi —xo),showing that the sequence is Cauchy and then converges. Hence,
— = — Xk_1) = — xO)—
Taking limits, we get(l—a)xo+xi
hm =n-+Oo 2—a
Solution 2. The recurrence relation can be expressed in matrix form as
= A ( where A(a I —a
\ / \Xn_1J \1 0
Thus,
\XnJ \XOJ
A calculation shows that the eigenvalues of A are 1 and a —1, with correspondingeigenvectors vi = (1, 1)' and = (a — 1, 1)'. A further calculation shows that
f(l—a)xo+xi\1 1=1 1v2.\XOJ 2—a / \2—aj
Hence,
(xn÷i) == (1
vi +(a —
1 we have urn (a — = 0, and we can conclude thatn-+oo
(1—a)xo+xin-÷oo 2—a
Solution to 13.6: The given relation can be written in matrix form as =
where A = The required periodicity holds if and only if
A is of A
A necessary condition for A" = (i?) is that the eigenvalues of
A be kth roots of unity, which implies that c = cos j = 0, 1,..., "c has the preceding form and 0 < j (i.e., —I <c < 1), then the eigenvalues
of A are distinct (i.e., A is diagonalizable), and the equality = (i?) holds.If c = 1 or —1, then the eigenvalues of A are not distinct, and A has the JordanCanonical Form [HK61, p. 247]
or respectively, in which case A" Hence, the desired
periodicity holds if and only if c = cos where j is an integer, and o < jk/2.
Solution to 13.7: If = E R, we have = a + xi,; so
-4a2
and we must have a 1/4.Conversely, assume 0 < a 1/4. As = we conclude, by
the Induction Principle [MH93, p. 7], that the given sequence is nondecreasing.Also, 2111
Xn+1 = + =
if < 1/2, which shows that the sequence is bounded. It follows that the se-quence converges when 0 <a 1/4.
Solution to 1.3.8: First we show, by induction, that the sequence is bounded.
For that, choose M large so that max lxi — M, for all n.
= M
showing that is bounded. Now to compute the limit write
Xj!j —xc)Xn+1
2
and taking the urn sup, we have
2
showing that urn SUPXn x. In the same way we obtain urn x, showingthat = x.
Solution to 1.3.9: Let = It will be shown that -+ 1. The sequence(y,,) satisfies the recurrence relation
if 1
Yn=(Yn_i+Yn—1
We have 1 for all n because for any positive b
Hence, for every n,
if 1\Yn—1Yn(Yn—1
yn—iJ
so the sequence (Yn) is nonincreasing. As it is also bounded below, it converges.Let -÷ y. The recurrence relation gives
1/ 1\Y=flY+—),
i.e., y = l/y, from which y = 1 follows, since y 1.
Solution to 1.3.10: Clearly, 0 = — ... for all n.Thus,
—x7),
and therefore
xj 11(1 —4) = xi exp _4))
Since log(1 —4) = 0(4) as k —÷ oo, the sum converges to a finite value L asn -÷ 00 and we get
exp(L) > 0.n—+oo
Solution to 13.11: 1. We have
so is bounded by 1/2 and, by the Induction Piinciple [MH93, p. 7], nonde-creasing. Let A be its limit. Then
and, as the sequence takes only positive values,
2
2. It is clear, from the expression for f above, that
for
and1 3
for
therefore, the sequence diverges for such initial values.On the other hand, if Jx — 1/21 < 1, we get
so, for these initial values, we get
1< =o(1).
Solution to 1.3.12: Suppose that urn = a <co. a 1 since the the
sequence is increasing. We have
___
-In- In
Taking the limit as n tends to infinity, we get (since a 0)
ora2—a—l=0.
This quadratic equation has one positive root,
4° 2
We show now that the sequence (fn+I is a Cauchy sequence. Applying thedefinition of the 's, we get
fn+i In -1n1 -
Since is an increasing sequence,
fn—i(fn—1
or+fn-lfn-2
By substituting this in and simplifying, we get
fn+i_In In-I 2 fn-i fn-2
By the Induction Principle [MH93, p. 7], we get
fn+1 fn<1f3f2Ia-i 12 fi
Therefore, by the Triangle Inequality [MH87, p. 20], for all m > n,
rn—Ifm-I-I fn+I 13 12
fm In 12 fi k=n2
Since the series converges, the right-hand side tends to 0 as m and n tendto infinity. Hence, the sequence (fn+i is a Cauchy sequence, and we aredone.
Solution to 1.3.13: We have
1 1 1 1
n+1 2n k=1l+n n
which is a Riemann sum for (1 + x)1dx corresponding to the partition of theinterval [0, 1] in n subintervals of equal length. Therefore, we get
11 1
we get
<e< (1+kul)k
therefore, we have
urnn—,.oo
k=n+1
I I k \k
i)
and the result follows.
Solution 3. We have
1 1 1
and the result now follows from the Maclaurin expansion [PMJ85, p. 127] oflog(1 +x).
Solution to 13.14: We have
1
+n+l n+2 nk
ci
Jo
Solution 2. Using the inequalities
1 dx=log2.1+x
log2 = logf 2n(I-I\k=n+I k=n-4-1
k=nl-1 k=n-f-1
fk+_1\kk )
12n+ 1\=log( 1;\n+1 I
=iog(
1 +...+—n+1 2,z
1 1 1 1 1 1
1 1
1 1=-I + +...+-k
The right side is the lower Riemann sum of the integral / — dx over the parti-J1X
tion of [1, k} into (k —1 )n intervals each of length Because the lower Riemannsum is less than the integral, whose value is log k, we get — <logk.
To get the other inequality note that the change in on the interval [1, 1+ of
the partition is 1 —= 1
, and it is less than that on every other interval
1 1 k—iJ < =1x nn+1 n+1
since each term in the Riemann sum contributes an error of less than —.n n-i-i
and there are (k — 1)n terms. So we can take C = k—I.One can deduce a more precise error bound as follows. For each subinterval of
the partition, translate the region between the curve y = and the approximatingrectangle horizontally so as to put it above the interval [1, 1 + The translated
regions are then nonoverlapping. As they lie within a rectangle of width andheight 1, the sum of their areas, which bounds the difference log k — —
is less than We can thus take C = 1. In fact, one easily sees that the translatedregions fill up less than half of the preceding rectangle, so C = works.
Solution to 1.3.15: Let n > 0. For m n, we have
Xm
where
Taking the lim sup, with respect to m, we have
limsupxm —m-÷oo
The series converges, so urn = 0. Considering the urn inf with respectn—).00
ton, we get
limsupxm — limsupxm.m 4
The reverse inequality also holds, so exists.
Solution to 1.3.16: Fix 8> 0, and choose such that <8 for all n no.Then
ano+I kane + Sn0 +8
+ k8 + + (1 + k)8
+ (k + k2)S + Sno+2 + (1 + k + k2)8
and, by the Induction Principle [MH93, p. 7],
ano÷m <k"1 + (1 + k + • . • + km—1 )8 +1 — k
Letting m 00, we find that
8
1—k
Since S is arbitrary, we have urn sup,,...÷00 0, and thus (since > 0 for all n)urn
fl—+ 00
Solution to 1.3.17:11 is unbounded, then, without loss of generality, it has nofinite upper bound. Take x,,1 = xj and, for each k E N, Xnk such thatXnk > max(k, Xnk_l }. This is clearly an increasing subsequence of
If is bounded, it has a convergent subsequence: urn y,, = say. containsa subsequence converging to or one converging to Suppose (Zn) is asubsequence of (Yn) converging to Let Z1 and, for each k 1, let
<Znk_1. This is a monotone subsequence of (Xn).
Solution to 1.3.18: Suppose that there are x> 1,6 > 0 such that I bm/bn —xI sfor all 1 n <m. Since = 1, for all k sufficiently large there existsanintegernk > ksuchthatbm/bk <xifm <nkandbm/bk >xifm >nk.Inparticular, for each k,
bflk+1
bflk bflk X
As tends to infinity as k does, the left-hand side should tend to 0 as k tends toinfinity, a contradiction.
Solution to 1.3.19: 1. Using the Ratio Test [Rud87, p. 66], we have
— n ! (4n) ! (2n + 2)(2n + 1) (2n) ! (3n + 3) (3n + 2)(3n + 1) (3n)!(2n+2)!(3n+3)! — (2n)!(3n)!(n + 1)n !(4n + 4)(4n + 3)(4n + 2)(4n + 1)(4n)!(n+I)!(4n+4)!
— (2n + 2)(2n + 1)(3n + 3)(3n + 2)(3n + 1)
— (n + 1)(4n + 4)(4n +3) (4n + 2)(4n + 1)
27
so the series converges.
2. Comparing with the series E 1/(n logn), which can be seen to diverge using
the Integral Test [Rud87, p. 139],
lognurn =lim
1=00
l/nlogn n /n
we conclude that the given series diverges.
Solution to 1.3.20: 1. Assume that <oo. As = an+1 +— 2&Janan÷i, we have
+ 1) = +n=2
<00.
2. Since + an+i) = 2 + E —we require a
sequence = > 0, such that <00 butE — = 00.
One such example is
1 n isodd
I.if n
Solution to 1.3.21: As
nt'(logn)'lim =IaI
(n + 1)b(logn + 1)' a?R
the series converges absolutely for lal < 1 and diverges for al> 1.
• a=1.
(i) b> 1. Let b = 1 + 2e; we have
1 fl\nI+2€(logn)C
= ° (n -+ oo)
and, as the series converges, the given series convergesabsolutely for b> 1.
(ii) b = 1. The series converges (absolutely) only if c> 1 and diverges ifc ( 1, by the Integral Test [Rud87, p. 1391.
(iii) b < 1. Comparing with the harmonic series, we conclude that theseries diverges.
• a = —1. By Leibniz Criterion [Rud87, p. 71], the series converges exactly
when1
urn =0nb(log
whichisequivalenttob>Oorb=O, c>0.
Solution to 1.3.22: Note that
+1'2 (n-+oo)j
that is,
urnn-÷oo 1/nx+V2 —
so the given series and00
1/2n=1
converge or diverge together. They converge when x> 1/2.
Solution to 1.3.23: If a 0, the general term does not go to zero, so the seriesdiverges. If a > 0, we have, using the Maclaurin series [PMJ85, p. 127] for sinx,
1 1 1— — sin — = + o(n3) (n -+ 00)n n 6n'
and, therefore,
(1— . iy = 6an3a
+ O(fl—3a) (n -÷ oo).
Thus, the series converges if and only if 3a> 1, that is, a> 1/3.
Solution to 13.24: For n = 1,2,... the number of terms in A that are less thanis — 1, so we have
= I> = <00.
aEAaEA
Solution to 1.3.25: Let S be the sum of the given series. Let No =0. By conver-gence, for each k > 0 there exists an Nk > Nk_1 such that
N*+1
For Nk + 1 n Nk+j let = We have = 00. As the terms are allpositive, we may the sum and get
00
=n=1 k=ONk+I
Ic=O Nk+1
Solution 2. The convergence of the given series shows that there is an increasing
sequence of positive integers (Nk) with <I / k3 for each k. Let
fi ifcn=s if
Then —+ oo, and
Co N1—i 00 co
+ k akn=i n=1 k=1 n=Nk
Solution 3. Let converge to S. For positive n let = and Ro =0,so we have = — Rn_i for all n. Consider the sequence defined by
= + As -+ S we get -+ oo.We have
00 00
>cnan =
Solution to 1.3.26: Using the formula sin 2x = 2 sin x cos x and the InductionPrinciple [MH93, p. 7], starting with sin = 1, we see that
it it it I
cos cos COS2" 2" 1 sin
So we haveI 2
= ; — oo)
since sin x x (x -÷ 0).
1.4 Differential Calculus
Solution to 1.4.1: Lemma 1: If (xe) is an infinite sequence in the finite interval[a, b], then it has a convergent subsequence.
Consider the sequence Yk = SUP(Xn I n k). By the least upper bound property,we know that Yk exists and is in [a, b] for all k. By the definition of supremum, itis clear that the Yk'S form a nonincreasing sequence. Let y be the infimum of thissequence. From the definition of infimum, we know that the Yk'5 converge to y.Again, by the definition of supremum, we know that we can find arbitrarilyclose to each so we can choose a subsequence of the original sequence whichconverges to y.
Lemma 2: A continuous function f on [a, b] is bounded.Suppose f is not bounded. Then, for each n, there is a point E [a, b] such
that > n. By passing to a subsequence, we may assume that thex E [a, b]. (This is possible by Lemma I.) Then, by the
continuity of f at x, we must have that If(x) — < 1 for n sufficientlylarge, or <If(x)I + 1, contradicting our choice of the
Lemma 3: A continuous function f on [a, b] achieves its extrema.It will suffice to show that f attains its maximum, the other case is proved in
exactly the same way. Let M = sup f and suppose f never attains this value.Define g(x) = M — f(x). Then g(x) > 0 on [a, b], so l/g is continuous. There-fore, by Lemma 2, hg is bounded by, say, N. Hence, M — f(x) > 1/N, orf(x) <M — 1/N, contradicting the definition of M.
Lemma 4: if a differentiable function f on (a, b) has a relative extremum at apoint c E (a, b), then f'(c) =0.
Define the function g by
(x) = ff(4=f(c)
for x c0 for x=c
and suppose g(c) > 0. By continuity, we can find an interval J around c such thatg(x) > 0 if x E J. Therefore, f(x) — f(c) and x — c always have the same signin J, so f(x) <c if x <c and f(x)> f(c) if x > c. This contradicts the factthat f has a relative extremum at c. A similar argument shows that the assumptionthat g(c) <0 yields a contradiction, so we must have thatg(c) =0.
Lemma S (Rolle's Theorem [MH93, p. 2001): Let f be continuous on [a, b]and differentiable on (a, b) with f(a) = f(b). There is a point c E (a, b) suchthat f'(c) = 0.
Suppose f'(c) 0 for all c E (a, b). By Lemma 3, f attains its extrema on[a, b}, but by Lemma 4 it cannot do so in the interior since otherwise the derivativeat that point would be zero. Hence, it attains its maximum and minimum at theendpoints. Since f(a) = f(b), it follows that f is constant, and so f'(c) = 0 forall c E (a, b), a contradiction.
Lemma 6 (Mean Value Theorem [Rud87, p. 108]): If f is a continuousfunction on [a, b], differentiable on (a, b), then there is c E (a, b) such thatf(b) — f(a) = f'(c)(b — a).
Define the function h(x) = f(x)(b — a) — x(f(b) — f(a)). h is continuous on[a, b}, differentiable on (a, b), and h(a) = h(b). By Lemma 5, there is c E (a, b)
such that h'(c) =0. Differentiating the expression for h yields the desired result.There is a point c such that f(b) — f(a) = f'(c)(b — a), but, by assumption,
the right-band side is 0 for all c. Hence, f(b) = f(a).
Solution to 1.4.2: 1. We have exp(f(x)) = (1 + 1/x)x, which is an increasingfunction. As the exponential is also increasing, so is f.2. We have
tim f(x) = urnlog(x + 1) — logx
= tim1/(x + 1) — 1/x
=x-÷O x-÷O 1/x x—÷O —l/x
On the other hand,
so lim f(x) = 1.
Solution to 1.4.3: Let 0 <s <t. Then by the Mean Value Theorem, there exist uin the interval (0, s) such that f(s) = f(s) — f(O) = sf'(u) and v in the interval(s, t) such that f(t) — f(s) = (t — s)f'(v). Then
f(s)= f'(u) f'(v) = f(t)
If(s)
This inequality reduces to f(s)/s f(t)/t. Hence g is an increasing function on
the interval (0,00). Now g(O) = f'(O) = f(x)/x = g(x). Fixx > 0. For 0 < t <x, g(t) g(x). Taking limits as t —÷ 0 we conclude that
g(0) g(x). Thus g is increasing on [0, oo).
For x > 0, g(x) is the slope of the chord of the graph of f joining the points(0, f(0)) and (x, f(x)).
Solution 2. Since f(0) = 0, f(x) = f'(t)dt and using the fact that f'(x) isincreasing, we get
fx
f(x)= J
f'(t)dt < I f'(x)dt = xf'(x),0 Jo
xf'(x)—f(x)so g (x) = is positive and g an increasing function of x.
Solution to 1.4.4: 1. The function f(x) given by
f(x) = x2 sin —x
has a derivative that is not continuous at zero:
f'(x)= {
2x sin — cos forfor x=O
2. Consider the function g given by g(x) = f(x) —2x. We then have g'(O) <0 <g"(l). Therefore, g(x) <g(0) for x close toO, and g(x) <g(1) forx close to 1.Then the minimum of g in [0, 1] occurs at an interior point c E (0, 1), at whichwe must have g'(c) =0, which gives f'(c) 2.
Solution to 1.4.5: Suppose y assumes a positive maximum at Then > 0,= 0, and 0, contradicting the differential equation. Hence, the
maximum of y is 0. Similarly, y cannot assume a negative minimum, so y isidentically 0.
Solution to 1.4.6: 1. Suppose u has a local maximum at xo with u(xo) > 0. Thenu"(xo) 0, but = eX0u(xO) > 0 and we have a contradiction. So u cannothave a positive local maximum. Similarly, if u has a local minimum at xo, thenu"(xo) 0, so we must have u(xo) 0 and u cannot have a negative localminimum.
2. Suppose u(O) u(1) = 0. If u(xo) 0 for some E (0, 1), then, as Uis continuous, u attains a positive local maximum or a negative local minimum,which contradicts Part 1.
Solution to 1.4.7: From the given inequality we get
0 — Ke_Kty(t) = (e_Kty(t)) (r 0).
Integrating from 0 to t we find that, for t 0,
e_Kty(t)— y(O) 0,
from which the desired inequality follows.
Solution 2. Let z(t) logy(t). Then z'(t) = y'(t)/y(t) K. Fix t > 0. By theMean Value Theorem, there exist u in the interval (0, t) such that z(t) — z(O) =z'(u)t Kt. Hence z(t) Kt + logy(O). Since the exponential function isstrictly increasing, we obtain y(t) This inequality also holds fort = 0.
Solution to 1.4.8: Without loss of generality assume xo = 0. As f is continuousand f(0) = 0, we have f'(x) > f(x) > 0 in some interval [0, e). Suppose thatf(x) is not positive for all positive values of x. Let c = inf{x > 01 f(x) 0).Since f is continuous and positive in a neighborhood of the origin, we havec > 0 and f(c) = 0. By Rolle's Theorem, [MH93, p. 200] there is a pointd with 0 < d < c and f'(d) = 0. However, by the definition of c, we havef'(d) > f(d) > 0, a contradiction.
Solution 2. Let g(x) = Then g'(x) = — f(x)) > 0. As g isan increasing function, we have g(x) = e_Xf(x) > = 0 for x > xo, andthe conclusion follows.
Solution to 1.4.9: Suppose that f(x) 0 for some positive value of x. Thena = inf{x > 0
If(x) 0) is positive. Since f is continuous, f(a) = 0. Let
b E (0, a), then f(b) > 0. By the Mean Value Theorem [Rud87, p. 108], thereexists c E (b, a) such that f'(c) = (f(a) — f(b))/(a — b) <0. Applying thesame theorem to f' on the interval (0, c) we get a number d E (0, c) such thatf"(d) = (f'(c) — f'(O))/c <0. Since 0 <d <a, f(d) > 0, contradicting theassumption that f"(x) 1(x) for all x E (0, a).
2Solutiontol41O:Letf :IR —÷ ]Rbedefinedbyf(x)=a?_l_x-_f_-.We have 2
lim f(x) = —oo and lim f(x) = oo
so f has at least one real root, say. We have
forall xER;
therefore, by Problem 1.4.8, f has no other root.
Solution to 1.4.11: Let g : [0, 1] ]R be defined by g(x) = e2'f(x). We have
g'(x) = e_2X(f(x) — 2f(x)) 0
so g is a decreasing function. As g(0) =0 and g is nonnegative, we get g O,sothe same is true for f.
Solution to 1.4.12: Consider the function g defined on [0, 1] by g(x) = e"f(x).We have
g"(x) = ? (f"(x) + 2f'(x) + 1(x)) 0
so g is concave upward; that is, the point (x, g(x)) must lie below the chordjoining (0, g(0)) and (1, g(1)) = (1,0) for x (0, 1). Then g(x) 0 and theconclusion follows.
Solution to 1.4.13: As 601 and 602 satisfy the given differential equations, we have(t) = v1 (çoi (t)) and = v2 (602(t)). Since (ta) = ço2(to), it follows from
our hypotheses that Hence, there exists a point so> such that(t) for t so. Suppose there existed a point <t <b such that
q'j (t)> Let ti so be the infimum of all such points t. By continuity, wemust have that 60i (ti) = Hence, repeating the above argument, we see thatthere must be a point Si > ti such that 601(t) if <t <SI, contradictingour definition of ti.
Solution to 1.4.14: Let f(x) = x2 sin and g(x) = x. Then we havetim f(x) = lim g(x) = Oand
limf(x) = tim x =0.
x-'O g(x) x-+O X
1 1
As g'(x) = 1 and f'(x) = 2x sin — — cos —, we have
urnf'(x) = tim 2x —
x-+O g'(x) x-+O x x
which does not exists.
Solution to 1.4.15: The given conditions imply that f'(O) = 0, so the question is
whether f'(x)/x must exist. The answer is no and one counterexampleis f(x) = x3 sin A calculation gives
f'(x) I I= 3x sin — — cos —.
x x x
Solution to 14.16: By Rolle's Theorem [MH93, p. 200], f'(xi) = 0 for some
xj E (0, 1). Then, since f'(O) = 0, f"(x2) = 0 for some X2 E (0, xi). Repeated
applications of Rolle's Theorem give 1(n) = 0 for some E (0, andtherefore, f(fl+I)(X) = 0 for some x (0, C (0, 1).
Solution to 1.4.17: Since 1' 0 and f 0, we deduce that f is monotone andurn f(t) = exists. Hence, for every 6 > 0, we have
t-*00
f(t+6)—f(t)tim =0.
(-+00 6
On the other handf(t +8) — f(t) = f'(t) +
6 -+ 0 we get tim f'(t) =0.t-,00
Solution to 1.4.18: Let x > 0 and 6 > 0. Since f is positive and log is continuous,
• ff(x+Sx)\"8 . ff(x+Sx)\"8loghm i = lirniogi
o-+O\ f(x) / o-+O \ f(x)
— timlogf(x+Sx)—logf(x)
_o-+o S
— timx (log f(x + Sx) — log f(x))
Sx
= x (log f(x))'xf'(x)f(x)
and the result follows, by exponentiating both sides.
Solution to 1.4.19: It is enough to show that
limf(h) — f(0)
and timf(h) — f(0)
h h
both exist and are equal. By L'Hôpital's Rule [Rud87, p. 109]
timf(h) — f(0) = lim
f'(h)= tim f'(h).
h I
The other lateral limit can be treated similarly.
Solution to 1.4.20: We have
pt(x) (1 +t2)x3 —3t3x+t4
p,'(x) = 3(1 +t2)x2 —
p'(x) = 6(1 + t2)x
• t < 0. In this case, > 0 and p,(x) < 0 for x sufficiently negative,and Pt(x) > 0 for x sufficiently positive. Hence, by the Intermediate ValueTheorem [Rud87, p. 93], Pt has exactly one root, of multiplicity 1, sincethe derivative is positive.
• t = 0. Now p,(x) = x3, which has a single zero of multiplicity 3.
• t > 0.Wehave
and p'(x) for positive x. So
is a local minimum, and
is a local maximum of p,.
We will study the values of Pt at these critical points. As pz(O) > 0 and<0, the relative maximum must be positive.
We have
(i) 0< t > 0,sop,hasonesingleroot.
(ii) <Oandp,hasthreeroots.
(iii) t > > Oandpt hasoneroot.
Solution to 1.4.21: Let
f(x)—f(a) —f'(a)
so that Jim h(x) = 0 and f(x) = f(a) + (f'(a) + h(x)) (x — a). Then
f(yn) — — f'(a)(yn — + — a) — — a)
—
=0
Pt
Pt
Pt
say. We get
h(x) =
so that
f'(a) (Yn a+
(a _Xfl)— — —
+=o(1) (n—+oo).
Solution to 1.4.22: By changing variables, itis enough to show that f(l) f(O).Without loss of generality assume f(O) =0. Consider the function g defined by
g(x)=f(x)—f(1)x.
As g is continuous, it attains a maximum at some E [0, 1]. We can assume< 1, because g(1) = g(0) =0. As g(x) for <x < 1, we have
urn supg(x) — = —f(1) + 1(x) —
As the rightmost term is nonnegative, we have f(1) 0, as desired.
Solution to 1.4.24: We have
1 = f(z)( 1) (i
Multiplying this out, we get = 1 and
0.
From this, it can easily be seen by the Induction Principle [MH93, p. 7] that allthe 's are rational.
Solution to 1.4.25: The function f given by
f(x) = J for 0 <x <3/20 for
is such a function.
1 3/2
This is based on the example of a nonconstant function having derivatives ofall orders, vanishing for negative x:
J for x>010 for
Solution to 1.4.26: Let g(x) = sinZ x; it follows from the Taylor series of sinxthat the first nonzero derivative of g at 0 is g(1z)(O) = n!. For any positive numberX, setc = a/(n!A'1); then f(x) = satisfies = cAtzn! = a.
There is a real number M such that M for x E [0, 2ir] andk = 0, 1, . . . n — 1. Since g and its derivatives are periodic, with period 2,r,
(k)(x) I M for all real x and k <n. Therefore (x) I = Ic IAk (IC)
(x) IIcIXICM = CAIC_1z for all real x, where C — Ja I/n!. Choosing A max(1, C/e)ensures that If(k)(x)I e for all x E R and k = 0, 1, .. . n — 1.
Solution to 1.4.28: We claim there is an e > 0 such that f(t) 0 for allt E (0, e). Suppose, on the contrary, that there is a sequence —÷ 0 such that
=0. Considering the real function to each subintervalwe find a sequence -+ 0, E [Xn+1, such that = 0 for all n, butsince f'(t) = C, this would imply C = 0. In the same fashion, usingthe imaginary part of f(x), we see that C =0, which is a contradiction.
Since f(t) is nonzero on a small interval starting at 0, the composition with theC°°—function absolute value
I
will give a C1—function g(t) = If(t)I, on a small neighborhood of zero.
Solution to 1.4.29: By Taylor's Theorem [Rud87, p. 1101, there is a constant Csuch that
If(x) — f(0) — f'(O)xI Cx2
when lxi <1. Since f'(O) = 0, we actually have f(x) — f(0) — f'(O)l Cx2and, consequently, by the triangle inequality, f(x) f(0) + Cx2 when lxi <I.We conclude:
If (x, y) lies on or below the graph of f and Ixl < 1, then
Now consider the disc D centered at (0, f(0) + b) with radius b, where0 < b < 1 will be chosen at the end. Clearly, (0, f(0)) is on the boundary ofD.Ontheotherhand,if(x,y) E D, then lxi <b <land
Iy—f(0)—bl <1b2_x2
y> — x2 = f(0)-Fb—bJl — x2/b2 f(0)-Fb—b(1 —x2/2b2)
since — x 1 — x/2 when 0 x 1. Thus,
y> f(0) + x2/2b.
If 1/2b c, then it follows that (x, y) must be above the graph of f. So we aredone if we take b = min(1/2, 1/2c).
1.5 Integral Calculus
Solution to 1.5.1: Let a = xo <xi < ... b be any partition of [a, b].Since f' is Riemann integrable, on each interval [Xj_!, x,] it has a finite supre-mum M and infimum m-.
By the Mean Value Theorem [Rud87, p. 108],
>mi(xj >2(f(Xj) —
The middle sum simplifies to f(b) — f(a), which therefore lies between the lowerand upper sums for the partition. As these sums both converge to the integral of
b
f' on [a, b], we must have f f'(x)dx = f(b) — f(a).
Solution to 1.5.2: Assume f does not vanish identically. Then, for some 6 > 0,the set {X E (0, 1) f(X) > 8) is nonempty and, as f is continuous, open;therefore it contains a closed interval I of length L > 0. Let g be the function thatequals 6 on I and 0 off I. Then g is Riemann integrable with g(x) dx = 8L.Since f g we have
flI I g(x)dX=SL>0,Jo Jo
as desired.
Solution 2. The properties of the Riemann integral we use are
• If f, g are (Riemann) integrable on [a, b] and f(x) g(x) for all X thenff(x)dx
• If f is integrable on [a, b} and a <c <b then f is integrable on [a, c] andon[c,b},andff(x)dx
Suppose that f(c) > 0 for some c. We consider 0 < c < 1. Since f is con-tinuous at c, there is an interval [a, b] containing c in which f(X) > f(c)/2.Then
b
f f(x)dx f(c)(b —
and hence
ff(x)dx=ff(x)dx+ff(x)dx+ff(x)dx>o
a contradiction. A similar argument applies when c is an endpoint of [0, 1].
Solution to 1.5.3: Since I is continuous, it attains its minimum and maximum atxo and yo, respectively, in [0, 1]. So we have
r1f (xo) J x2 dx x2f(x) dx x2 dx,0 JO J0
or
f(xo) 3 x2f(x)dx f(yo).JO
Therefore, by the Intermediate Value Theorem [Rud87, p. 93], there is a pointE[O, 1] with
f1
f(4)=3 I x2f(x)dx.JO
Solution to 1.5.4: Since the discontinuities are only of the first type (the limitexists), they do not have any accumulation point (for a detailed proof of this, seethe Solution to Problem 1.2.8), and form a finite set. Let d1 <d2 <be the set of discontinuities of f. Then f is continuous in every interval [x, y]with <x <y using the Solution to Problem 1.5.5 (on both endpointsof the interval), f is integrable on each interval of the type so f isintegrable on [a, b].
Solution to 1.5.5: 1. Let (f(x)( M for x E [0, 1]. If (ba) is a decreasingvanishing sequence, then fb' Ill M is a bounded, increasing sequence, so itmust converge. We conclude that If I is Riemann integrable over [0, 1], and so is
1.2. The function f(x) = 1/x is integrable over any interval [b, 1] for positive b,but is not integrable over [0, 1].
Solution to 13.6: Using change of variables,
p00
I If(x)Idx= I (Jf(x)I+If(—x)I)dx.J—oo JO
If, forx> 1, we had x(If(x)J + f(—x)I)> 1, we would have the convergenceof the integral f10° So, there is such that xi(If(xi)I + If(—xi)I) 1.
If, for all x > max(2, xi), we have x(If(x)l + > 1/2 we similarlyconclude the convergence of J(2X1)(1/x) dx, thus, there is > xi with
X2(lf(X2)l + Jf(—x2)I) 1/2. Recursively we can then define a sequence
such that > max(n + 1, and + If(—xn)l) 1/n. It followsthat —* ooand
urn = urn xnlf(—xn)I =0.n-+00 n-+00
Solution to 13.7: 1. Letting t = x + s, we get
f(x) = ex2/2 / =00
0ds.
Since s > 0, < 1, so for all positivex; then
(000<f(x)< I
Joe_SXds
2. Let 0 <X2. For s > 0, e5Z1 _52/2> so
00
f(xi)= IJo
00
ds = f(x2)
Solution 2.
1. The function f is clearly positive. Using integration by parts, we get
2. We have
1
I dtJx
=x
dt>0,
e_t2/2dt.
t2
Solution to 1.5.8: Integrating by parts and noting that q.' vanishes at I and 2, weget
/2=
21 j'2
ço(x) ——IiA
00
x
Therefore,
p 00 1e_t2/2dt
= ix 7(_e12/2)' dt
=x
00
x
f(x)= ! _ec2 '2/00 e_t212 1
xdt<—.
x
1
f'(x)=
—f00
1+—x
so is an increasing function.
applying integration by parts a second time and using the fact that ço' also vanishesat the endpoints, we get
=
Taking absolute values gives
Since ço E C2, the integral on the right-hand side is finite, and we are done.
Solution to 1.5.9: Suppose that f is such a function. Cauchy—Schwarz Inequality[MH93, p. 69] gives
cia= I xf(x)dx
Jo
1 11/2
(j x21(x)dxf
a chain of equalities. For equality to hold in the Cauchy—SchwarzInequality, we must have = for some constant k, so
0, which contradicts
piI f(x)dx=1.
Jo
Thus, no such function f can exist.
Solution 2. Multiplying the given identities by a2, —2a, and 1, respectively, weget
I f(x)(a — x)2dx = 0Jo
but the integral above is clearly positive for every positive continuous function, so
no such function can exist.
Solution to 1.5.10: Dividing the integral in n pieces, we have
f(j/n)— f =
— j(j+I)/nf(x)dx)
n—i
I )f(j/n) — f(x)I dx.j=O Jj/n
Therefore,
f'(c) =f(x) — f(j/n)
x—j/n
Solution to 15.11: Suppose not. Then, for some ö > 0, there is a sequence of realnumbers, (xe), such that oo and If(xn)I 5. Without loss of generality,we can assume 5.
Lets> Overify
then
for Ix—yI<s,
f(x)dx = 00
The result follows from the following claims.Claim 1: liminff(x) 0.
X-+ 00
For every x E (f/n, (j + 1)/n), applying the Mean Value Theorem [Rud87, p.108], there is c E (f/n, x) with
As the derivative of f is uniformly bounded by M, this gives us the inequality
— f(j/n)I — f/n).
f(j/n)— [Iz-1 n JO
I i=°
n—i
j=O
M(x — j/n)dx
n.— I
j=O
j22n2 — 2n2
3
M
2n
fxn+S
>1A
f00contradicting the convergence of I
JO
Solution to 13.12: Let
f(x)dx.
fXg(x)=f(x)-I- I f(t)dt.
JO
If not, there are e, > 0 such that f(x) > e for x > xo. Then, we have
g(x)=f(x)+ f(t)dt+ f(t)dtJO Jx0
'1xo
f(t)dt+s(x—xo).Jo
This is a contradiction since the right side tends to 00 with x.Claim2: limsupf(x) 0.
X—) 00This follows from Claim 1 applied to the function —f.
Claim 3: urn sup f(x) 0.X—) 00
Assume not. Then, for some e > 0 there is a sequence X2, ... tending to oosuch that > e for all n. By Claim 1, the function f assumes values e/2for arbitrarily large values of its argument. Thus, after possibly deleting finitelymany of the we can find another sequence ... tending to oo such that
< for all n and e/2 for all n. Let Zn be the largest number in[Yn, Xn] where f takes the value s/2 (it exists by the Intermediate Value Theorem[Rud87, p. 93]). Then
fXn
8(Xn) — 8(Zn) = f(Xn) —+ J
f(t)dt
s>e—--+! —dt
2 2e
which contradicts the existence of lim g(x).X—) 00
Claim 4: lim inff(x) 0.X—) 00
Apply Claim 3 to the function —f.
Solution to 1.5.13: Suppose that for some e > 0, there is a sequence Xn 00
with Xnf(Xn) s. Then, as f is monotone decreasing, we have f(x) e/x forx large enough, which contradicts the convergence of f00° f(x)dx, and the resultfollows.
Solution to 1.5.14: Let 0 < e < 1. As
f(x)dx<oo,JO
thereisanN > Osuchthatforn> N,(00
Jf(x)dx<s.
n
Therefore, for n large enough, that is, such that ne> N, we have
MW xI f(x)dx = (—') f(x)dx + J (—)
f(x)dxJO 'fl' JO 'fl' nE
<eI f(x)dx+! f(x)dxJo Jne
1nfl8
<sJ f(x)dx+s
<e(ff(x)dx + i).
Since this inequality holds for all c > 0 and for all n sufficiently large, it followsthat
1urn — I xf(x)dx=0.
fl Jo
Solution to 1.5.15: Using the Maclaurin expansion [PMJ85, p. 127] of sinx, weget
SiflXlflx (2n+1).
The series above is alternating for every value of x, so we have
IC 2n+1 2k-i-2
(2k+3)!
Taking k =2, we have
1/2 / x2\ 1/2 x4
1_f
which gives an approximate value of7l/144 with an error bounded by 0.00013.
Solution to 1.5.16: Let
dxi(t)= f 114+logt.io (x4+t4)'
It suffices to show that for t > 0, the function 1(t) is bounded below and mono-tonically increasing. For x, t 0, we have (x + t)4 x4 + t4, so
dx fI+tdu1(t)
J + log t = j — + log t = log(1 + r) 0.ox+f Jt U
We now show that I'(t) 0 for t > 0. We have
dx dx1(t) = I
1 4 + +logt,Jo t + i) /
Ji t ((x/t)4 + 1)1/4
letting y = x/t, we get
so
Itt) £' (y4 +1)1/4 + f(y4 +1)1/4
+ log t,
1'(t)= —1
t2(1/14+1)uI'4 '
Solution to 1.5.17: Integrate by parts to get
f(x)2dx = 100 (- x) f (x)2dx =—f x 2f(x)f'(x)dx.
The boundary terms vanish because xf(x)2 = 0 at x = 0 and 00. By the Cauchy—Schwarz Inequality [MH93, p. 69],
1
I xf(x)f'(x)dx (Jo
Solution to 1.5.18: Consider the figure
b
/Y=f(x)
K
The left side of the desired inequality is the sum of the areas of the two shadedregions. Those regions together contain a rectangle of sides a, and b, from whichthe inequality follows. The condition for equality is b = f(a), the condition thatthe two regions fill the rectangle.
100
Jo
x2f(x)2dx f'(x)2dx.
/,
a
Solution 2. Without loss of generality, assume f(a) b. We have
pa pa
ab= I f(x)dx+ I (b—f(x))dx.Jo JO
The second integral is
limn n
k=O
ForO 1,
a
Substituting in the limit above,
lim
Multiplying out each term in the sum and rearranging them, and noting thatf(O) = g(O) =0, we get
— I (v)) +ab —af(a).
Since g is continuous, this equals
p1(a)I g(y)dy+a(b—f(a)).
Jo
As g(y) a for y E (f(a), b), we have
pba (b — f(a)) I g(y) dy.
Jf(a)
This gives the desired inequality. Also, we see that equality holds if f(a) = b.
Solution to 1.5.19: Given e> 0, choose R so that If(x)Idx <e/4. ThenIf(x) cos xyldx <e/4 for all y. So
Ig(z) — = I f(x) (cosxz — cosxy)dxlxi
+ f(x) (cos xz — cos xy) dxJ lxI )R
+ f lf(x)Ilcosxz — cosxyldx.2
The latter integral approaches 0 as z •—+ y by uniform convergence of cos(xz) tocos(xy) on the compact interval —R x R. Hence, for iz — yJ sufficientlysmall,
lg(z)—g(y)l <
and g is continuous.
Solution to 13.20: Let M be an upper bound of and let K = Ig(x)Idx.Fix s > 0. Since f(x) = 0, there exists R1 > 0 such that lf(x)l <sfor lxi > R1. As K is finite, there is R2 > 0 such that Jjxt>R2 Ig(x)Idx <s. Forlxi > R1 + R2 we have then
plh(x)I I lf(x — y)Iig(y)tdy + I f(x — y)Ifg(y)Idy
J—R2 JIyI>R2
fI Jg(y)Idy+M I jg(y)ldy
J—R2 JIyI>R2
s(K +M).
As s is arbitrary, the desired conclusion follows.
Solution to 13.21: We will do the proof of the sine integral only. For n 0, let
I sinx2dx.J
We show that the series Sn converges and use this to show that the integralconverges.
By the choice of the domains of integration, the alternate in sign. Also,
setting u = x2, we get
(n+1)ir sin u
= f du
ç(n+1)ir>1 du
Jnir
p(n+2)ir sinu=1J(n-i-1)ir
=
Finally, the tend to 0:
(n+1)ir 1
f
and the right-hand side gets arbitrarily small as n tends to infinity. Therefore, byLeibniz Criterion [Rud87, p. 71], the series converges. Let a > 0 and n besuch that a <4/(n + l)ir. Then
a 4J(i4-1)ir =0
sinx2dx_>sk=j sinx2dx_>2Sk.
The second term tends to zero as n tends to infinity. By estimates almost identicalto those above,
f%f(fl+1)lt• 2 I(n+1)7r—a21
Ja 2a 24fiuiI
so the first term does as well. Therefore, we have
00
I sinx2dx=>2Sn<oo.JO n=O
k
Solution to 13.22: Let p(x) = a polynomial. We havej=0
So the result holds for polynomials. Now let f be a continuous function ands> 0. By the Stone—Weierstrass Approximation Theorem [MH93, p. 284], thereis a polynomial p with hf — <s. So
(n+1) IJO
x'7p(x)dx — f(1)
j'18+ (n+l) I fp(x)dx—f(1)
JO
s + Ip(I) — 1(1)1
<2e.
Since e is arbitrary, the desired limit holds.
Solution to 15.23: Jogx is integrable near zero, and near infinity it is domi-nated by so the given integral exists finitely. Making the change of variablesx = alt, it becomes
+
dx=°o logx loga dt — 1 P00 logt
Jo x2+a2 a o 1+t2loga arctantl—J
a 10
— irloga2a
If we treat J in a similar way, we get J = —J, so J =0 and the given integralequals
irloga2a
Solution 2. We split the integral in two and use the substitution x = a2/y.
P°° logxdx
=
f Iogx iogx
Jo x2+a2 o
dx
a logx 0 log(a2/y)
(a2
= x2 + a2dx + f
a logx a 2 log a — logydyx2+a2dX+fo
pa 2loga
Jo
a a10— irloga— 2a
Solution 3. Using residues, and arguing as in the solution to Problem 5.11.28, welogz
z2+a2'
( f(z)dz + [ f(z)dz = 2iriRes (f, ia) = (log a +
On the negative real axis, we have
Jo jo log(—x) + in
f(z)dz = x2+a2
=J dx+°° logx
J°°in
0 x2+a2 0
x2+a2 2a
therefore,logx 2T2i Jr I ut
21J0 2a 2
and (00 logx Jrlogai dx=
J0 2a
Solution to 13.24: As sinx 1, to show that I converges, it is enough to showthat I > —oo. By the symmetry of sin x around ir/2, we have
pitI = log(sin x) dx + log(sin x) dx
Jn/2pir/2
=2 I log(sinx)dxJo
I log(2x/ir)dxJo
> —00.
The first inequality holds since on [0, ir/2], sinx 2x/ir; see Problem 1.1.25.Letting x = 2u, we get
I = 2 I log(sin 2u) duJo
/ pir/2 pir/2 pir/2= 2 ( I log2 du + I log(sin u) du + I log(cos u) du
\Jo Jo Jo
The first integral equals (ir/2) log 2. As cos u = sin(ir/2 — u), the last integral is
flr/2 pir/2 Prlog (sin(ir/2 — u)) du = log(sin u) du = log(sin u) du.
JO JO J,r/2
The above equation becomes I = it log2 + 21, so I = —it log 2.
Solution to 1.5.25: The integrand is continuous at x =0, so that no trouble arisesthere. For r > 0 let
I(r)= I dx.
For n a positive integer,
I(nir) =
whereIsinxl dx.
J(k—I)ir
From the n-periodicity of I.sinxl it follows that > ak+1 for all k, whileak < 1)jr 0. Hence urn I(nir) exists finitely by the alternating
n-+ooseries test. For r > 0 let ii,. be the smallest integer such that r <nit. Then
fl\I(r)—J(nrir)=O(—J
so limI(r) exists finitely, which means f dx converges. On the other
hand,
CO • 00 if IsinxI IsinxI—c dxJkir+
dx
1
3
= 300.
An alternative proof of the convergence of the integral uses integration by parts:
çr sinx çr1
I dx=— I —d(cosx)Ji Ji
cosx T1 f' cosx
+—Ix3'
cosr 1=cosl— dx.x3/2
The right side converges as r —+ oo because f x312dx <00.
1.6 Sequences of Functions
Solution to 1.6.1: Let B be the set of function that are the pointwise limit ofcontinuous functions defined on [0, 1]. The characteristic functions of intervals,Xi' are in B. Notice also that as f is monotone, the inverse image of an interval isan interval, and that linear combinations of elements of B are in B. Without lossof generality, assume f(0) = 0 and f(1) = 1. For n E N, let the functions bedefined by
n—2
=+
Lemma: Let c B with max andx€[O,IJ
n=I< 00. Then
B. As — Jg2k÷1 — f — fI 2k-I and2k—1
we get
— g2k) = f — B
sof—g2+g2 = f€B.Proof of the Lemma: For each n let be the pointwise limit of } C B such
that on [0, 1]. Consider the functions = Given e > 0,takem such that <s/3. Then the sum Ihn(x)I <e/3 and
kok(x)I <6/Forx [0, 1],takeK so that
SJhn(x) —
3m
For k> K we then have
for n=1,...,m.
n=1 k=1 n=m+1 n=m+1
<S
B.
-I--- IIIr7T(7
I
From the construction we can easily see that maxI
considerXEEO,I) n
now the following result:
Solution to 1.6.2: Let a <b be real numbers and s > 0. Take n large enough so
and
Then, using the Mean Value Theorem [Rud87, p. 108],
lg(a)—g(b)I
where a <b. As the inequality holds for any s > 0,
Ig(a) — g(b)I lb — al
and the continuity of g follows.
Solution 2. Let N > 0. We will show that g is continuous in [—N, N]. We have,for any n E N, by the Mean Value Theorem,
— = — y)J 2N forx, y E [—N, N].
So the sequence {f,,} is bounded. The relation
— = — y)l ( Ix — yl
shows that it is also equicontinuous. By the Arzelà—Ascoli Theorem [MH93, p.273] we may assume that } converges uniformly on [—N, N]. The function g,being the uniform limit of continuous functions, is continuous as well. As N isarbitrary, we are done.
Solution to 1.63: Let e > 0. By uniform continuity, for some ö > 0 we have
let = k/N, and divide [0, 1] into the intervals 1 k N. Since
f pointwise, by taking the maximum over the finite set {4k} we know thatthere exists M> 0 such that if n M, then <s for 0 k N.Each of the is nondecreasing, so we have, forx E
<
or
— <2€.
Therefore,
— f(x)l Ifn(x) — + — f(x)l <3€.
Since this bound does not depend on x, the convergence is uniform.
Solution to 1.6.4: Assume the contrary. Then, for some real x and positive s,given any 5 > 0, we can find y E (x —8, x +8) such that If(x) — 8.
For each positive integer m let Ym (x — x + with If(x) — f(ym)1 8.
We know that fn(ym) = f(Ym), therefore, given s > 0, we can find, for
each m, an integer such that
(Yin) — f(Ym)I <s/2,
andchoosethemsoni <fl2 <.Consider the sequence (xv) defined by = Yin for flm—I < fl Then
—÷ x, but f(x) since, for each m,
1mm — f(x)I = Ifflm (Ym) — f(x)I11(x) — f(ym)I — If(Ym) — fnrn (Ym)I
>8—8/2 = sf2.
Solution to 1.6.6: 1. For k E N, consider the continuous functions given by
I ifgk(x)=
{
0 if
Define fo 0, and, fork > 0, fk(x) = gk(t)dt. We have fk Efk(O) =0, and fk'(x) = gk(x) -÷ 0= for all x E However,
cxurn fk(x) = urn gk(Odt = gk(t)dt = I # fo(x)
n-+oo Jo
2. -÷ f0'(x) uniformly on IL
Solution to 1.6.7: As f is a homeomorphism of [0,1] onto itself, we may assumewithout loss of generality (by replacing f by 1 — f) that f is strictly increas-ing, with f(0) = 0 and f(1) = 1. We first treat the case where f' is a con-tinuous function. By the Stone—Weierstrass Approximation Theorem [MH93, p.284], there is a sequence of polynomials which converge to f uniformly.Since f' > 0, we may assume (by adding a small constant) that each of the ispositive. Further, since the Ps's converge uniformly,
pI p1
J -+ Jf'(t)dt = f(l) = I.
0 0
Defining byAPIç'=J0
we can replace each by so we may assume that
= 1.
Now consider the polynomials
txj
Jo
= 0, = 1, and = > 0 for all x and n. Hence, eachis a homeomorphism of the unit interval onto itself, and by their definition,
the 's converge to f uniformly.It is enough now to show that any increasing homeomorphism of the unit inter-
val onto itself can be uniformly approximated by C' homeomorphisms. Let r > 0and
f forfr(X)—10for x=0
Acalculationshowsthatfr isC1 on[0, 1],fr(l) = 1,f,'(O) =0, and f,'(l) =r.For r, s > 0, let
fr(X) for xE[0,1]X
— —f3(—x) for x E [—1,0].
Each is a C'—function such that grs(O) = 0, = 1, == s, and = r. By scaling and translating, we can find a C1
function on any interval such that its values and the values of its derivative at bothendpoints are any given positive values desired.
We can now approximate any continuous homeomorphism f as follows:Given s > 0, choose n > 0 such that if x — yI < 1/n, for these valueslf(x) — < s. (This is possible since [0, 11 is compact, so f is uniformlycontinuous there.) Partition [0, 1] into 2n intervals of equal length. On the inter-vals [2k/2n, (2k + 1)/2n], 0 k n — 1, approximate f by the line segmentjoining f(2k/2n) and f((2k + 1)/2n). On the other intervals, join the line seg-ments by suitable functions as defined above to make the approximating functionC'. Since f is an increasing function, this approximating function will always liewithin s of it.
Solution to 1.68: 1. Let S8 = (x E [0, 1] I f(x) M — e) for positives. f(x)M—s if and only if for each n, (x) M—e, so = (EM —e, oo)).So each is closed. By definition of supremum, each set S6 is nonempty. Also, ifSj are finitely many positive numbers, = Smjnej 0. As [0, 1] is compact,the intersection of all sets S6 is nonempty. Let t belong to this intersection. ThenM f(t) M — s for arbitrary s> 0, so f(t) = M.2. Take = min(nx, 1 — x}.
Solution to 1.6.9: Fix e > 0. For each n, let = {x I In (x) <e}. Then
• is open, since is continuous.
• C since fn+1.
• [0, 1] = since fn(x) —± 0, for each X.
Since [0, 1] is compact, a finite number of the cover [0, 1], and by the secondcondition above, there is N such that Gn = [0, 1] for all n N. By the definition
N, we have0 <E foralix E [0, 1].Thisprovesthatthe sequence converges uniformly to 0 on [0, 1].
Solution to 1.6.10: If fn (x) -÷ f(x) pointwise on E, then converges to funiformly on E if and only if supE I (x) — 1(x) I tends to 0 as n tends to 00. We
note that if Ac B then SUPA f(x)I — f(x)I.
k =0. In this case I (x) I 1 / n so I -÷ 0 as n oo. Henceconverges uniformly on R, and also on every bounded subset of R.
x n—x2k 1. fn(X) =
-÷ Opointwise, and the derivativeis = (2+)2'so attains its maximum at x = where
sup If1d ==
so that sup I-+ 0 as n -+ oo, hence the convergence is uniform on R
and also on bounded sets.
k =2. (x) = and sup un I = 1, thus the convergence is not uniform onx2 +n
R. However, since the function is even and increasing on the supremuma2
on [—a, a] is sup I = fn(a)a2 +
so that sup I —÷ 0 as n -÷ 00
and the convergence is uniform on bounded subsets.
k 3. Again (x) -÷ 0 pointwise, but sup I In I= oo, thus the convergence is not
uniform on ]R. In the same way as above, IJ
is even, and increasing onR÷, this can be seen directly or by computing the derivative
xk_I((k_2)x2+kn)(x2+n)2
so sup I In (a) —+ 0 as n —÷ 00. Thus the convergence is uniform onbounded sets.
Hence the convergence is uniform on R when k = 0 and 1 only, and uniformon bounded sets for all k.
Solution to 1.6.11: Let Pk be the set of polynomials of degree k, and let L =d(f', l'k) = inf{IIf — Ph I
P Pk}, where the norm is that of C(o,j). Then, thereexists a sequence (Pa) contained in such that Ill — PnII —+ L. Then, for someM > 0, IIPnII IIPn — III + 11111 M, for all n E N. Letxj, ...,Xk E [0, 1] be
k distinct points. Thus, M, for all i = 1, .. ., k and n E N.The sequence (xl)) is bounded, so, passing to a subsequence if necessary,we may assume that it converges. We can repeat this argument and suppose that
for any i = 1, ... , k, for some constants y,. Let P(x) =ycoj(x) where
= (x — xi). • (x — X1_1)(X Xj+1) • • (x — Xk)
• —xk)
i.e., P is Lagrange's interpolation polynomial. Clearly, there are positive con-
stants L1 such that L, for all i = 1, ..., k, and all x E [0, 1]. There-
fore, for all x E [0, 1], we have I (x) — P(x)I (xi) — I
(x)I
L1 — -÷ 0. This means that for this polynomial P in l'k. wehave P; but ii — f II -÷ L, whence L = lIP — f II, i.e., d(f, Pk) =hf — pt,, p E Pk, therefore P is a polynomial of degree k, which minimizesd(f,l'k).
Solution to 1.6.12: Let ao, ... ,aD be D + 1 distinct points in [0, 1]. The polyno-mials fin' defined by
fm(X)=fl for m=0,...,D1=0
i fm (am) = 1. Any polynomial of degree, at most,D can be written
P(x)
since the right-hand side is a polynomial of degree, at most, D which agrees with
P in D + 1 points.Let M be an upper bound of lfm(X)i for x E [0, 1], m = 0, ..., D. Given
e > Olet N EN be such that IPn(am)l (D+l)M forn N. Then we have
lPn(am)IIfm(x)I <s
therefore the convergence is uniform.
Solution to 1.6.14: Suppose that -÷ f uniformly. Then f is continuous, andf(0) = 1. So thereise > Owithf(x)> 1/2 for lxi <e.Jfjis large enough, we have, by uniform convergence,
foralix,
Foronesuchj,andx=2flj
1 1 1
a contradiction.
Solution to 1.6.15: 1. In an obvious way we see that d(f, f) = 0 and d(f, g)d(g, f), so we only need to verify that d(f, g) > 0 for f g and the TriangleInequality [MH87, p. 20].
If f g then If — gJ is positive in a small interval and so is the iuntegrandand the integral is nonzero. Now the function a i-+ is increasing in [0, oo).
c a+b a b <a+b1+c 1+a+b 1+a+b 1+a+b 1+a 1+b
which implies the Triangle Inequality.
2. For natural n define fn by
In2x,fn(x)_l 1/x,
It is easy to verify that the sequence is Cauchy, since for any n, m we have,
I (x) — fJo 1+lfm(X)—fn(x)l
X
pmax(1/m, I/njldx
Jo= max{1/m, 1/n).
Suppose that (C[O,I], d) is complete and take f as the limit of the sequence
{f,j. If f(a) 1/a for some a E (0, 1], then, by continuity, there exists > 0such that l1/x — f(x)I s forx E (a —s, a]. Therefore,
pa edx
a—s 1+E
for sufficiently large n. But the right hand side is a positive constant independentof n, which contradicts the convergence -÷ f. Thus f(a) = 1/a for alla E (0, 1]. This contradicts the continuity of f on [0, 1]. We conclude then that(C[o,l], d) is not complete.
Solution to 1.6.16: (a). Let : [0, 1] -÷ R be defined by [0, 1] iscompact, fl II = 1, but the sequence is not equicontinuous.(b). Let = [0, 1], and = n. This sequence is clearly equicontinuous, iscompact, but no subsequence of can converge.
2
I
1
(b) (c)
(c). Consider now : R R, = X(n_1,n+1](X)COS((X — n)ir/2), whereX(a,b] is the characteristic function for the interval [a, b}. II 1 and the se-quence is equicontinuous, however no subsequence can converge.
Solution to 1.6.17: For each n, let be the function that equals In at the pointsk/n, k = 0, 1, .. . , n, and that on each interval [(k — 1)/n, k/n] interpolates itsendpoint values linearly. By the assumption on In, the slope of on each ofthe preceding intervals is at most 1, so is a Lipschitz function with Lipschitzconstant at most 1. Hence the sequence is uniformly equicontinuous, soby the Arzelà—Ascoli Theorem [MH93, p. 273], it has a uniformly convergentsubsequence Let g be the limit function. Fix a point x in [0, 1]. Foreach j, let Xj be a point in [0,1] of the form k/ni (k = 0, ... , Iij) such that
( — <2/ni. We have
Ig(x) — (x)I Ig(x) — gnj (x)J + (x) — + Ifn, (Xj) — fflj (x)I
2 2Ig(x) + — +
flj hf
As j -÷ Co. the first summand on the right tends to 0 uniformly, showing that-+ g uniformly.
Solution to 1.6.18: By the Arzelà—Ascoli Theorem [MH93, p. 273], it will sufficeto prove that the sequence is equicontinuous and uniformly bounded.Equicontinuity. For 0 x <y 1 and any n,
— f f =
The function F(x) = is continuous on [0, 1], hence uniformly continuous.Therefore, given s > 0, there is a 5 > 0 such that — F(x)I <s whenever
I I
I
(a)fn
n
x and y are in [0, 1] and ly — xl < ö. By the inequality above, we then have— <e for all n when — xI <ö, establishing the equicontinuity
of the sequence.Uniform boundedness. Since (x)dx = 0, the function cannot be alwayspositive or always negative, so there is a point on [0, 1] such that = 0.
Then, by the estimate found above, for all x:
Solution to 1.6.19: We claim that a subset A of M is compact if A is closed,
bounded and f 1' I f A) is equicontinuous. If A satisfies all conditions andis a subsequence in A then and are bounded and equicontinuous andby the Theorem of Arzelà—Ascoli [MH93, p. 273], there is a subsequence
such that { and { }are uniformly convergent and therefore, sequences of
Cauchy. Since M is complete and A is closed, j converges to f E A in M,and A is compact.
On the other hand, if A is compact, consider the spaces:
Ü={(f,f')If EM) A={(f,f')IIEA)
X is compact in M, and so are each of the projections, and, by Arzelà—AscoliTheorem, { f' J f E A) is equicontinuous.
Solution to 1.6.20: We consider three cases.
• (an) has a vanishing subsequence. Then, the corresponding subsequence of} converges to x + cos x, a continuous function.
• (an) has a subsequence with limit a 0. Then, the corresponding subse-quence of } converges to
— sin(ax) + cos(x + a),
which is continuous.
• I 00. In this case, sin —÷ 0, and cos(x + depends onLet (ba) be the sequence defined by (mod
E [0, 2,r]. As [0, 2,r] is compact, (ba) has a convergent subsequence, tob say. Then the corresponding subsequence of converges to cos(x +b),a continuous function.
Solution to 1.6.21: The answer is no. Consider the sequence of functions: [0, 1] -÷ R whose graphs are given by the straight lines through the points
(0, 0), (l/2n, n), (I/n, 0) to (1, 0).
f3
ci1
The sequence approximates the zero-function pointwise, but j (x)dx = —
foralin. Jo 2
Solution to 1.6.22: We have
so
gn(x) = + + = for some (0, 1)
for x E [0, 1].
= — — ( 1 for x [0, 1].
— Ix — for x, y E [0, 1].
Also,
Therefore,
The sequence is then equicontinuous and uniformly bounded; so, by theArzelà—Ascoli's Theorem [MH93, p. 273], it has a uniformly convergent subse-quence.
Solution to 1.6.23: Fix e > 0. Since K is continuous on a compact set, it isuniformly continuous. So, there is a 8 > 0 such that IK(xi, Yi) — K(x2, Y2)I <cwhenever — x2)2 + (Yl — Y2)2 <8. Let f and & be as above, and suppose
12
11
1/3 1/2 I
and x2 are in [0,1] and satisfy lxJ — <8. Then
jilf(xi) — f(x2)J = 1 g(y) (K(xi, y) — K(x2, y)) dy
Jo
1
1
I 1.sdy=s.Jo
As the estimate holds for all f in F, the family F is equicontinuous.
Solution to 1.6.25: 1. By the Cauchy—Schwarz Inequality [MH93, p. 69], we have
2. Since + y is a continuous function of x and y on the compact unit square,it is uniformly continuous there. Hence, given any s > 0, there exists S > 0 suchthat
<8
whenever lxi — x21 + lyi — Y21 < ö. In particular, + y — .Jx2 + )'I <ewhenever lxi — X21 <8, therefore,
=('xi
Jo
whenever Ixi — x21 <8. Since the same value of S works for all values of n simul-taneously, the family is equicontinuous. Using the uniform bound establishedabove the conclusion follows from the Arzelà—Ascoli's Theorem.
Solution to 1.6.26: We will first show that is a Cauchy sequence in sup-norm.
Using the Cauchy—Schwarz Inequality [MH93, p. 69], we have
sup (x) — supxe[O,I] xE[O,1]
lfn(y) — fm(y)I2dy
1 I I
'I
—J
IK(x, — fm (y))dy0
hence,
I
Iffl(y) — fm(y)I2dy
I I
Since K is continuous, it is integrable, and taking M = SUPX yE[O 1] IK(x, y)I. wehave
I i.i— — fm(y)I2dy -+ 0
JO
showing that the sequence is a Cauchy sequence in the sup-norm; as C[0, 11is complete on this norm, the sequence converges uniformly.
Solution to 1.6.27: We'll use the Stone—Weierstrass Approximation Theorem[MH93, p. 284] in the space C[OI) equipped with the sup norm
Ill — g II = sup(If (x) — g (x) I I x E [0, 1]). Let f E C[011] and let (pa) be asequence of polynomials converging to f in the sup norm. Using our hypothesiswith k = 0 we conclude that, for some positive M, we have I M forall n. Given s > 0 there exists an integer k(s) such that Ill — pk(e)II for
k k(s). As Pk(s)c°n converges, it is a Cauchy sequence, therefore there is aninteger n(s) such that
p1 p1 6I Pk(e)c0n / <—
for m, n n(s),
f f f'Pm icon f+ f Pk(e)c°n
— f0
If_Pk8)conl+—+f0 30
therefore the sequence Jbeing Cauchy, converges.
0
Solution to 16.28: As
1°°
1
and
by the Weierstrass M-test [Rud87, p. 148], the given series converges uniformly
on JR to a continuous function f. We have, since the convergence is uniform,
i ,T 00>e dxn2
CO T IXflX 00e
—T
n2
we have, again by the Weierstrass M-test, that
00sin
2XTn=In n
converges uniformly in T. Therefore, we get
hm—I2T J-T
00
f(x)dx = urnsin
n=1
00
=V' urn =/_dT.÷oon=1
Solution to 1.6.29: Let a > 1. It suffices to show that is defined and hascontinuous derivatives for x a. The series > converges for such x. As
it follows from the Weierstrass M-test [Rud87, p. 148] that the seriesconverges uniformly, so is a continuous function. To see that it has continuousdeiivatives of all orders, we formally differentiate the series k times, geuing
00 k
>.:(—logn)n=2
It is enough to show that this series converges uniformly on k. Since
(logn)k
by the Weierstrass M-test, it will suffice to show that the series
00 k> (logn)
n=2
converges. But— 11
01"no—8
(n —÷ oo),
for any positive e. As
converges for a — c> 1, we are done.
Solution to 1.630: Fix an interval [a, b] and s> 0. Since f is continuous, it isuniformly continuous on the interval [a, b + 1], then there exists an N > 0 suchthat if n N and x — yl <1/n we have Jf(x) — <e. We will show that
(x) converges uniformly to
J f(y)dyx
for all x in the given interval. Fix x and n N. We have
x k=O x+k/n
By the Mean Value Theorem for Integrals [MH93, p. 457], for each k there isa, E (x+k/n,x +(k+ 1)/n)suchthat
I f(y)dy=f(ak)/n.Jx+k/n
Substituting this in the above, expanding using the definition of and usinguniform continuity, we get
x+i 1n—i
f f(y)dy — — — +k/n)IX k=O
Since this holds for any x, we are done.
Solution to 1.6.31: Let a > 0, then for ml > 4a, the bound on f gives, forX E [—a, a],
As the series
)Mn
converges, by the Weierstrass M-test [Rud87, p. 148], the series
f(x+n)tnl>4ce
converges uniformly. So the series for F(x) converges uniformly on [—a, a] and
F is continuous there. As a is arbitrary, F is continuous on IL
We have
F(x+1)—F(x)= urn >2(f(x+1+n)—f(x+n))—a
= urn (f(x+1+a)—f(x—a))00
=0
the last equality holding by our assumption on f.If G is continuous and periodic with period 1, then, since the series for F
converges uniformly,
f F(x)G(x)dx=
f(x
In each integral on the right-hand side, let y = x + n, and get, sinceG(y —n)= G(y),
00 pn+1 p00f(y)G(y)dy
= Jf(y)G(y)dy.
—00 —00
Solution to 1.632: Given f and s, let h : [0, 1] R be defined byh(x) = f By the Stone—Weierstrass Approximation Theorem [MH93, p.284], there is a polynomial P such that IP(x) — h(x)I <sf2 for x E [0, 1], fromwhich it follows that
p (x4) — = Jp (x4) — h (x4)I <s/2 for x E [0, 1].
lIP = > _oakxk,takeCo, ..., EQ — CkI <s/2.Thenwe have
— — + — <s.
1.7 Fourier Series
Solution to 1.7.1: 1. We have, forn E N,
1
— I f(x)cosnxdx=0
because the integrand is an odd function. Also
f(x)sinnx dx — f f(x)sinnx dx
2 1 x cos nx ir cos nx+1n 0 Jo fl
=
so the Fourier series of f is
00 n±1sInnx.
2. If the series converged uniformly the function f would be continuous, whichis not the case.3. As f and f' are sectionally continuous, we have
— f(x—) + f(x+) J f(x) if x (2n + 1)irn — 2 —l 0 if x=(2n+1)ir
n=I
where n E Z.
Solution to 1.7.2: 1. Since f(x) is an odd function, the integrals
— f(x)cosnxdxITJ_,r
vanish for n E N. The Fourier series has only terms in sin nx given by
1
I x3sinnxdx.
2. As f and f' are sectionally continuous, we have
00
b— f(x—) + f(x+) — J f(x) if x (2n + 1)n
fl Sifl flX— 2 — 0 if x = (2n + 1 )ir
n= I
where n E Z.3. Using Parseval's Identity [MH93, p. 577]
+ + = ! 12(x) dx
and the fact that all = 0,
26
Solution to 1.73: The answer is no; f(x) = 1 satisfies the above equation and isnot identically zero.
Solution to 1.7.4: As f is we have
f(n)= j 2i
dx= f f(x + dx.
Letting y = x + we get
-fn =
J
f is also 1-periodic, we have
f(n) = f dy =
Iforn
Solution to 1.7.5: Suppose that such an f exists. As the power series for e-7'converges uniformly, we have, for n > 0,
/(n)= f f f(x)x" dx = —27rni.
This equality contradicts the Riemann—Lebesgue Lemma [MH93, p. 628], whichsays that Jim f(n) = 0, so no such a function can exist.
n-÷00
Solution 2. From the assumptions the function x2f(x) is orthogonal to all polyno-mials, so by the Stone—Weierstrass Approximation Theorem [MH93, p. 284] it isidentically zero, and so it is f(x), which is a contradiction with the first integral.
Solution to 1.7.6: The Fourier series of f converges to f because f" exists. Letthe Fourier series of f be
00
+ cos nx + sin nx).n=1
As f" = g — kf is continuous, we obtain its Fourier series by termwise differen-tiating the series for f, and get
00
+ cos nx + sinnx).
So we have
aoao=, 2' 2for
k k—n k—n
Solution to 1.7.7: Consider the Fourier series of f,
f(x) = + cos nx + sin nx).
We have ao = 0, and, by Parseval's Identity [MH93, p. 577],
f2irf2(x)dx = jr + yr + = j2"(fF(x))2dx.
Solution to 1.7.8: The Riemann—Lebesgue Lemma [MH93, p. 6281 states thatthe result is valid for all functions g (x) of the type
cos kirx and sin
using linearity of the integral the result extends to all finite trigonometric polyno-mials
We will now use the fact that the set of trigonometric polynomials is dense in thespace of continuous functions with the sup norm (Stone—Weierstrass Approxima-tion Theorem [MH93, p. 2841) to extend it to all continuous functions. Given anye > 0, there exists a Pe (x) as above such that
Ig(x) — <c
then
f f(x)g(nx)dx— f f(x)dx f g(x)dx
—
(urn f f(x)p6(nx)dx — 101 f(x)dx jf(x)(g(nx) —
+f(x)dx —
lim J If(x)I Ig(nx) — p6(nx)I dx+
j ff(x)j dx j Ig(x) — dx
I f(x)IdxJo
1.8 Convex Functions
Solution to 1.8.1: Let M = maxxe[o,1] If(x)t. Consider a sequence such
that xo = 1, and 0 satisfies for 0 <xDefine g: [0, 1] -÷ [0, 1] by g(0) 0 and, using the fact -÷ 0,
M M
forO <x +(1 E [0,1], n =0,1,.... Wehaveg fand
g(xn) — M/2'1 = g(xn_i) — g(xn)
— Xn+1 — — Xn+1 — Xn+1 Xn—1 —
so g is concave.
Solution to 1.8.2: For x y and t E [0, 1], let
c=(logf(y)—logf(x))/(x — y).
By hypothesis, we have
+ (1 — t)y) + (1 —
so
f(tx + (1 — t)y) + (1 —
t f(x) + (1 — t)(f(x))t
= f(x)t
f is convex.
Solution to 1.8.3: Consider an interval [a, bJ and suppose that the maximum offdoes not occur at one of its endpoints. Then, by Weierstrass Theorem [MH93, p.189], there is c E (a, b) maximizing f. By the continuity of f, there are intervalsA=[a,ao]andB=[bo,b]in[a,b]withf(x)<f(c)ifxliesjnAorB. Bythe Mean Value Inequality for Integrals [MH93, p. 457], we have
11' If Iff(c) — I f(y)dy + — I f(y)dy + — I f(y)dy2h JA 2h J[aoboJ 2h JBao—a bo—ao b—bo
2hf(c)
+ 2hf(c) + 2h
f(c)
=f(c).
This contradiction shows that f must attain its maximum at a or b.If L(x) is any linear function, a straightforward calculation shows that L is
convex and satisfies the mean value inequalizy above, and that both of these in-equalities are, in fact, equalities. Now let L be given by
L(x)(x — a)f(b) — (x — b)f(a)
b—a
and consider G(x) = f(x) — L(x). By the linearity of the integral, since f and Lsatisfy the Mean Value Inequality, G does as well. Therefore, G takes its max-imum value at a or b. A calculation shows that G(a) = G(b) = 0. There-fore, we must have that f(x) L(x) for all x [a, b}. For any t [0, 1],(1 — t)a + tb E [a, b}. Substituting this into the inequality gives that is convex.
2
Multivariable Calculus
2.1 Limits and Continuity
Solution to 2.1.1: Let x e > 0, and let B denote the open ball withcenter f(x) and radius s. For n = 1, 2,..., let be the closed ball with cen-ter x and radius 1/n. By (ii) we have = (f(x)). By (1) the sets(R'2 — B) fl are compact for n = 1,2,.... They form a decreasing se-quence, and their intersection is empty, by the preceding equality. Hence, there isan no such that — B) fl = 0, which means that — f(x)I <swhenever — xl <1/no. So f is continuous at x.
Solution to 2.1.2: Let (x, y) G(g). Then y g(x), and there exist disjointneighborhoods V of y and V' of g(x) in By hypothesis, we can find a neigh-borhood U of x such that g(U) C VI. Then U x V is a neighborhood of (x, y)disjoint from G(g), and this proves that G(g) is closed.
The converse is false. Take n = 1, and let g : R —÷ R be defined byg(x) = 1/x, for x 0; g(0) = 0. Then the graph of g is closed in R xand g is discontinuous at 0.
Solution 2. Let p : x R'3 -÷ R?Z defined by p(x, y) = y — f(x). Then p iscontinuous and G1 = p1(0) is closed being the inverse image of a closed set.
Solution to 2.1.3: If U W2 let x be in the boundary of U. Therefore there is asequence of elements of U converging to x. is a Cauchy sequence,so it is convergent, to y say. Let y = h(z). Then z x since z E U and x U.
Then,h (xv) -+ h(z) and 74 z,
which contradicts the fact that h is a homeomorphism.
Solution to 2.1.4: We show that f is continuous at (0,0); for the general case,use a change of variables. By adding a constant, if necessary, we may assumef(0, 0) = 0. Suppose f is not continuous at the origin. Then, for any c> 0, thereis a sequence tending to the origin with If(Xn, s for each n.Since f is continuous in the first variable, there exists aS > 0 such that if lxi <ô,then If(x, 0)J < s/2. Applying this to our sequence, we see that there is anN > 0, such that if n N then <5, so If(xn, 0)1 < s/2. However, foreach such n, y) is continuous in the second variable, so by the Interme-diate Value Theorem [Rud87, p. 93], there exists 0 < < such that
= ne/(n + 1). Since the tend to 0 as n tends to infinity; thedo so as well. Hence, the set E = (n N)U((0, 0)) is compact. Then byour hypothesis, the set f(E) is compact. But f(E) = (ns/(n +1) I n N) U (0),and s is a limit point of this set which is not contained in it, a contradiction. Hence,f is continuous at the origin and we are done.
Solution to 2.1.5: Continuity implies f(0) = 0, so if any Xk is 0, then so areall subsequent ones, and the desired conclusion holds. Assume therefore, thatXk 0 for all k. The sequence (IIxkII) is then a decreasing sequence of positivenumbers, so it has a nonnegative limit, say c. Suppose c > 0. The sequence(xk), being bounded, has a convergent subsequence, say (xk1). with limit a. Thenhail = JIxk1 II = c. Hence, hlf(a)I1 <c. But, by the continuity of f,
f(a) = •lim I (xkj) = •lim xk,+1,J-+oo J—*oo
and c for all j, so we have a contradiction, and the desired conclusionfollows.
Solution to 2.1.6: 1. If (ei, ... , denotes the standard basis of andthe norm of each base vector, then any vector v in the unit sphere can be writtenas
where
so N(v) IN(vi) + •+ n max(N(e1), . . . showing thatN is bounded on the unit sphere.
2. Let B be the supremum of N in the unit sphere, then for any vector x y
y=x+lIy—xfly—x
—xli
so N(y) N(x) + — xIIN that is\IIy—xfIj
N(y)—N(x)
changing the places of x and y we then get N(x) — N(y) Bifx — yli. that is
IN(x) — N(y)( — yff
and N is continuous.3. Let A > 0 be the minimum of the continuous function N on the compact
unit sphere, where B > 0 is already the maximum, then
N(v) = ItvIIN
so taking the maximum and minimum over all such v,
A be a closed subset of Rm, and suppose b IR'1 \ f (A).For r > 0 let Br be the closed ball in W with center b and radius r. Since B.is compact and f is proper, I 1(B,.) is compact. As A is closed, A 11 f1(Br)is compact. Since flr>0(A fl f1(Br)) = A fl =0, A fl = 0for some s > 0, so f(A) is disjoint from the neighborhood of b, so f(A) isclosed.2. Suppose B is a compact subset of W'. Let AR = fx E JRtm I lix II R}.Since f is closed, f(AR) is closed, so f(AR) fl B is compact. As f is 1-1,flR>o(f(AR) fl B) = 0. Therefore f(As) fl B = 0 for some S > 0,so f'(B)is contained in the compact set 1W' \ Since f is continuous, f'(B) is closed.Therefore (B) is compact.
2.2 Differential Calculus
Solution to 2.2.1: We maximize the function f(x, y) = (x2 + in thefirst quadrant, x 0 and y 0. The function attains a maximum there because itis nonnegative and tends to 0 as (x, y) tends to infinity. We have
— = (2x — x — y — = (2y — x — y )eax
The critical points of f are thus the points (x, y) that satisfy
2x—x2—y2=0=2y—x2—y2.
These equalities imply x = y and 2x2 — 2x = 0. Hence, the only critical pointin the open quadrant is (1, 1).
On the x—axis, we have f(x, 0) = and = (x2 — so
= 0 only for x = 0 and x = 2. The point (2,0)is thus another candidate
for the point at which f attains its maximum. By the same reasoning, the point(0,2) is another such candidate. The points (2,0) and (0,2) are the only candidates
on the boundary of the quadrant.We have
f(1, 1) = 2e2, f(2, 0) = 4e2 = f(0, 2).
Hence, the maximum value of f in the quadrant is 4e2, that is,
(x2 + y2)e_X_)1 <4e2
X +)'
Solution 2. Letting u = t + 1 in the inequality et t + 1, which holds for anyreal t, we get u. For u = x/2 + y/2 we obtain
x+y x2+y2As
2 4, squaring we get
2 2 2x+y—2> (x + y \ > x + y
e4
Solution 3. On each level set x + y = k the maximum for the left-hand side occursat the extremes, where one of the variables is equal to 0. To see this observe thatx2 + y2 is the square of the distance of (x, y) to the origin, which is maximal atthe extremes of the segment x + y = k in the first quadrant. So we only need toprove that
but this can be easily seem by the same argument as in the last solution, usubstituting u = x/2 and squaring both sides. Alternatively, we can also
differentiate the function f(x) = e_xx2 to see that the maximum for positivevalues of x happens at 2 and then obtain the inequality again.
Solution 4. On each quart-of-circle 0 of radius r the maximum of theleft-hand side is r2/4 and the value of the right-hand side is
O+sin O)—2
which is a one variable function assuming the minimum at the extremes 0 andir/2, where the value is Using the same argument as before we complete theproof.
Solution to 2.2.2: Let the arc length of the circle between the points of contactof the (i — l)th and ith sides be 2çoj. The area of the corresponding part of thehexagon is tan çoj, the total area is A = i tan çoj. This is the function to beminimized on (0, with the restriction çoj = it. Since A is a convexfunction on (0, C R6, any critical point is an absolute minimum.
We use the method of Lagrange Multipliers [MH93, p. 414]. Lagrange's equa-tions for a critical point are sec2 çoj = X, which are satisfied by çoj ir/6, corre-sponding to a regular hexagon. The minimal value of A is
Solution 2. With the same notation as above, the function tan is convex in theinterval (0, so, by Jensen's Inequality [Str8l, p. 201], the minimum forthe area will happen when all angles are the same, that is, when the hexagon isregular.
Solution to 2.23: 1. If(x, 2.Jx2 + y2, and when (x, y) approaches theorigin the limit is 0, and so is the limit of f(x, y).2. Computing the the directional derivative of f in the direction of (x, y) withy
f(hx,hy) '2 2lim =lim vx +y =0h
and in the direction of the x-axis the limit is trivialy zero, because the function isidentically zero.3. From the previous calculation, we know that, if f were differentiable, thederivative would be zero, and then the quotient
f(x,y)0
as (x, y) -+ (0,0), but is false. To see it, approach the origin following the curvey x2/Jr, the limit of the quotient is then 1 — cos =2.
dl— J' x113cos(y/x) if
if x=O
x=O
which are continuous on JR2 \ {(O, y) I y E IR). Thus, f is differentiable there. Atany point (0, y), we have
f(h, k) — f(0, y)lI(h, k)lI
= O(Ihl"3) = o(l) (h 0)
= ciIxII2
where we used Cauchy—Schwarz's inequality.Choose e positive such that R(x) Ill lix < I — for lix < e, for these
values of x, we have
IIF(x)iI IlAxil + IIR(x)li + (1 — llxH = lix II
hence F maps the ball centered at 0 with radius s into itself.
Solution to 2.2.4: The function f is differentiable at the point z = (xo, yo) E Uif there is a linear transformation f'(z) E L (R2, RI) such that
urnIf(z + h) — 1(z) — f'(z)hl
—h-÷O Iihii —
Continuity of the partial derivatives is a sufficient condition for differentiability.A calculation gives
) — J (4/3)x1/3 sin y/x — yx213 cos y/x ifif
so f is differentiable at these points also.
Solution to 2.2.5: Let A = (ask) be the Jacobian matrix of F at the origin.According to the definition of differentiability we have F(x) = Ax + R(x) withurn ii R(x) ii / lix Ii = 0. Moreover,x-+0
2
j=1
j=I
Solution to 2.2.6: The Jacobian determinant of f is given by
Jf(x, y) = det(
÷ ÷ ) = —2p'(x + y)p'(x — y)
The derivative Df(x, y) is invertible exactly when Jf(x, y) 0, thus if and onlyif p'(x + y)p'(x — y) 0.
Since p has positive degree, its derivative p' is not identically zero, so it hasonly finitely many zeros, say A1, ... , The set (if = 0) is the union of finitelymany lines, thelinesx+y = Ak,k = 1,...,n,andthelinesx—y = Ak,k = 1, ... , n. This is a closed nowhere-dense set, so (if 0)is an open denseset.
Solution to 2.2.7: We have
Vh(x, y) = (g'(x)g(y), g'(y)g(x)).
For the desired condition to hold, Vh(x, y) must be a scalar multiple of (x, y), atleast for (x, y) (0, 0). Thus,
g'(x)g(y) = g'(y)g(x)x y
assuming xy 0. This can be written as
g'(x) g'(y)xg(x) = yg(y)'
which implies that = A = constant. The preceding differential equation
for g has the general solution
g(x) =
B a constant, positive because g is positive by assumption. Every such gobviously has the required property.
Solution to 2.2.8: Since f is continuous and RPZ is connected, f(RPz) is connected.We will prove that f(R'1) is both open and closed in This will imply that
= because 0.Let y = f(x) E f(IR'1). As the rank is n, by the Inverse Function
Theorem [Rud87, p. 221], there are open neighborhoods and of x and ysuch that : —÷ is a diffeomorphism; therefore, is an open neigh-borhood of y, and is open.
Let be a sequence in converging to y Ri?, = say.
The set K = In E N) U (y) is compact; therefore, is also com-pact. But
Jn N) c therefore, it contains a convergent subse-
quence, say with —* x Ri?. Since f is continuous, —÷ f(x).But lim = y; therefore, f(x) = y, and f(R') is closed.
f-400
Solution to 2.2.9: We have R2 = f(S) u (R2 \ f(s)) where S is the set ofsingularities of f. It suffices to show that f maps R = R2 \ (f(S)) onto
1R2 \ f(S). f(S) is finite, so R2 \ f(S) is connected. As f(S) is closed, R is open.It suffices to show that f(R) is open and closed in R2 \ f(S).
As RflS = 0, by the Inverse Function Theorem [Rud87, p. 221], f is invertible
in a neighborhood of each point of R. Hence, locally, f is an open map, and as
the union of open sets is open, f(R) is open.Let be a limit point of f(R) in R2 \ f(S), and be a sequence in f(R)
converging to 4. The set is bounded, therefore, by hypothesis, is
bounded. Let (xv) be a sequence such that = This sequence is bounded,so it must have a limit point, x, say. As f is continuous, we have f(x) = Since
g f(S), x f1(f(S)), so x E R. Therefore, 4 E f(R), so f(R) is closed inR2\f(S).
Solution to 2.2.10: Consider the scalar field G : -÷ IR given byG(y) = II Vf(y)112. We have
DG(y)—
By the hypothesis, we have G(x) =0, G is C1, and G'(x) 0. Therefore, by theInverse Function Theorem [Rud87, p. 221], G is locally a diffeomorphism onto aneighborhood of 0 in IL In particular, it is injective in some neighborhood of x,so it has no other zeros there.
Solution to 2.2.11: 1. Since f is C2, we can expand f in a Taylor series [MH93,p. 359] around a and obtain
f(a + h) = f(a) + f'(a) h +
f"(a) = h'Hh = (h, Hh), H = and the big Oh notation
means that for h in some neighborhood of 0, h E Vo, we have IO(1h13)I <K1h13for some K > 0.
The hypothesis that a is a critical point implies that f'(a).h =0 for all h E R'3,and the hypothesis that H is positive definite implies that (h, Hh) 0 for allh E 1W1 and zero only at h = 0. Therefore, all the eigenvalues of H are positiveand there exists some c > 0 such that (h, Hh) cIhl2 (namely c is the minimumof all the eigenvalues, which are all real because H is symmetric, by the SchwarzTheorem [HK6I, p. 340]). Let Wa = (a + h I h E Vo, IhI <c/k). We have
f(a + h) — f(a) = (h, Hh) + O(1h13) + O(1h13)— K1h13 = 1h12(c — KIhI) > 0
which shows that f(a + h)> f(a) for h 0 and h Wa, and, therefore, a is alocal minimum.
2. Assume f has two critical points, P1 and P2. Since the Hessian matrix is posi-tive definite, pj and are local minima. Let tpi + (1 — t)p2, t E be the linecontaining P1 and Consider the real function g given byg(t) = + (1 — t)p2). g has local minima at t = 0 and at t = 1. Therefore,g has a local maximum at some 10 (0, 1). We have g"(to) 0, but
g"(to) = f"(tpi + (1 — to)p2)(pl — P2)2 = (P1 — H(pi — P2))
and our assumptions on H imply — — P2)) > 0, a contradiction.
Solution to 2.2.12: As the Laplacian of f is positive, the Hessian of f has positivetrace everywhere. However, since f E C3, for f to have a relative maximum itsHessian must have negative eigenvalues and so its trace must be negative.
Solution to 2.2.13: With respect to polar coordinates r, 9, we can write u asu(r, 9) = We need the expression for the Laplacian in polar coordinates,which one can derive on the basis of the easily obtained identities
8r x 8r y 89 sin9 8y cos9
8xr 8yr 8x r ' r
After some computations one finds that
82 82 82 18 182or r8r r 89
Hence
0 = A(r"g(O)) = d(d — 1)r"_2g(9) + += r'1_2 (d2g(O) + g"(9)).
Thus g must be a solution of the ordinary differential equation g" + d2g = 0.
This equation has the general solution
g (9) = a cos dO + b sin dO.
Since g must be 2ir-periodic, d is an integer.
Solution to 2.2.15: The derivative of T is given by
which is always nonsingular since det(DT) = 2v — 2u is never 0. By the InverseFunction Theorem [Rud87, p. 221], this means that T is locally one-to-one.
2. Considering the function f(u, v) = u+v restricted to = y, we conclude
that u + v therefore, the range of T is
1(x,Y)IY>O,
Let (x, y) E ran ge(T). u +v = x is the equation of a straight line with slope —1 in
the u, v-plane which intersects the circle u2 + v2 = y centered at the origin withradius These two lines intersect exactly at one point in U, so T is globallyinjective.
U
Solution to 2.2.16: Letting fi and f2 be the components of f, we have
=
Assume t > 0 and use the Mean Value Theorem [Rud87, p. 108] to rewrite theright side as
2t +
where0 <tand0 <t.Ast \0,thecontinuityoff' gives
+ > 0.
Hence, there is an e > 0 such that > 0 for 0 < t <s. and the desiredconclusion follows.
Solution to 2.2.17: For X E E, we have
OA—X112= (1 —x)2+y2+z2+(2—t)2=y2+z2+1 _2x+x2+(2_t)2
—2x+21x1(2—t)1.
FFI
u+v=x
We can choose the sign, so ±yz —lxi: =0 because det X = 0. As 4lxl — 2x 0,we have hA — 1 with equality when 4lxi — 2x = 0, lxi = 2 — t, y =and detX = xt — yz =0, which give S =Solution 2. We can use the Method of Lagrange Multipliers [MH93, p. 4141 tofind the minimum of the function
d(x,y,z,t) =(x — l)2+y2 +Z2+(t _2)2
over the surface defined by E = der'(O). Setting up the equations
x—1 =Aty=—Azz=—Ay
t—2 =Ax
substituting the third equation into the second we see that if y 0 then A = ±1,and substituting back into the first and last equations we get a contradiction, soindeed y = 0, and then so is z = 0. Now solving the first and last equationstogether we find
2A+ Ix=
A+2t
I —
To have determinant zero then either A = —1/2 or A = —2, and the two matricesare 10 o\ 11 0
and
and we can easily see that is the closest one.
Solution to 2.2.18: Let d denote the Euclidean metric in ]R3. We first prove that
there exist p E S and q E T such that d(p, q) = d(S, T), where
d(s,T)=inf(d(s,t)lt T)
and
d(S,T) =inf{d(s,t)is E S,tE TI
= inf{d(s, T) I s E S).
The function S —÷ R defined by x i-+ d(x, T) is continuous and attains itsinfimum on the compact set 5, i.e., there exists p E S such that d(p, T) =d(S, T). The function ç : T -÷ R given by x t-+ d(p, x) is continuous, and
infr = infr' ço where T' is the compact set
T'=(x
Hence there exists q E T' C T such that d(p, q) = d(p, T) — d(S, T).If (x, y, z) E S then z 9, whereas if (x, y, z) E T then z 1. Therefore
p q. Now S, T are level sets of
gi(x,y,z) =2x2+(y— l)2+(z— 10)2
g2(x,y,z) =z(x2+y2+ 1)
respectively.Define, on x R3,
F(x,y,z,u,v,w)=(x —u)2 +(y— v)2+(z— w)2
fi(x,y,z,u,v,w) =gi(x,y,z)f2(X, y, z, u, v, w) = g2(u, v, w).
Then S x T = E 1R3 x iiV hi = 1, = 1). We showed abovethat there exists (p, q) E S x T which minimizes F on S x T. For ,j) E S x T,the two vectors Vfi ij) = (v), 0), = (0, Vg2(?7)) are linearlyindependent.
By Lagrange's Multiplier Theorem [M1193, p. 414], there exist A, E R such
thatVF(p, q) = AVf1 (p, q) + q).
NowVF(p, q) = (2(p — q), —2(p —
and
XVf1(p, q) + jiVf2(p, q) = A 0) + (0, Vg2(q))
=
It follows from the equations above that = p — q =None of p — q, Vgi (p) and Vg2(q) are zero. Thus the vector p — q is parallel toboth (p) and Vg2(q), from which the result follows.
Solution to 2.2.19: Each element of P2 has the form ax2 + bx + c for (a, b, c) ER3, so we can identify P2 with 1R3 and J becomes a scalar field on R3:
J(a, b, c) f(ax2 + bx + c)2 dx= ç + + + ç + bc + c2.
To Q corresponds the set {(a, b, c) I a+b+c = 1). If J achieves a minimum valueon Q, then, by the Method of Lagrange Multipliers [MH93, p. 414], we know thatthereisaconstantAwithVJ = XVg,whereg(a,b,c)= a+b+c— l.Wehave
f2a b 2ca 21, 2a
and Vg = (1, 1, 1). These and the constraint equation g(a, b, c) = 0 form thesystem
/2/5 1/2 2/3 —i\ a\ /011/2 2/3 1 bl fO12/3 1 2 c110
1 1 0/ A)which has the unique solution A = 2/9, (a, b, c) = (10/3, —8/3, 1/3). There-fore, if J attains a minimum, it must do so at this point. To see that J does attaina minimum, parameterize the plane Q with the xy coordinates and consider thequadratic surface with a linear z term defined by z = J(x, y, 1 — x — y) in JR3.The surface is the graph of the map J : P2 -÷ JR. Rotating around the z—axis willeliminate the xy cross-terms in the equation, reducing it to the standard equationof either an elliptic paraboloid or a hyperbolic paraboloid. However, J is alwaysnonnegative, so the surface must be an elliptic paraboloid and, as such, has a min-imum.
Solution to 2.2.20: First observe that the group of orthogonal matrices is transitivein R", that is, given any two vectors u and v there is a orthogonal matrix A, suchthat Au = v. Now using the Cauchy—Schwarz Inequality [MH93, p. 69] we seethat the maximum and minimum of the inner product of any two vectors of thesame size is ±IIv 112 and they are both achieved when the vectors are the sameand have opposite directions, respectively. So the maximum of the function f isachieved on the matrices that leave vo invariant and the minimum on the ones thatreverse it.
Solution to 2.2.21: Let el, ... , be the unit vectors in JR'2. Let a, b E W be suchthat b = a + h some and that the line segment
[a, b] = (a + 10 i h) lies in W. The function 1 i-÷ f(a + tei) hasderivative =0, by hypothesis. By the mean value theorem, f(b) = f(a).
In order to prove that f is constant, it is therefore sufficient to note that any twopoints in W can be joined by a sequence of line segments in W each of which is
parallel to some axis of R'2.
Solution to 2.2.22: Let
g(x, and f(x, y) =X
Theny2—x2 81
8y (x2+y2)2 9x
Let y be the curve in .4 given by
(x, y) = (2 cost, 2 sin t), 0 1 2ir.
Then
I' f—2sint 2costI g(x, y) dx + f(x, y) dy = j (—, (—2 sin 0 + (2 cos 1)) di
Jy Jo \
2,r
0
=2ir.
dt
Suppose there is a function h such that (g, f) = Vh = Then, since y isclosed, by the fundamental theorem of vector calculus, we obtain
fgdx+fdY=0.
which contradicts the calculation above.
Solution to 2.2.24: Let (x, t) E R2. By the Mean Value Theorem, [Rud87, p.108] and the hypothesis, we have, for some ij) in the segment connecting(x,y)to(x+y,0),
(x +t,0))
= ii), ii)) (—t, t)
sof(x,t)=f(x+t,0)> 0.=
ij) — 17))
Solution to 2.225: Given two points x and y E one can build a polygonalpath from x to y with n segments all parallel to the axis (adjusting one coordinateat a time). Applying the Mean Value Theorem [Rud87, p. 108] to each of thesegments of the path, we have
y
x
If(xi, . . Yi+1, ... ... Ys Yi+I, ..., yn)I
and then
11(x)
—f(xi, . . . , Yi' Yi+i, • • •
—YiI.
Now applying the Cauchy—Schwarz Inequality [M}193, p. 69] to the vectors(1,l,...,1) andx—y,weget
f(x) — 1 —
IIx—yfl.
Solution to 2.2.26: Let ((xl, .. . , be a sequence in converging to 0 eThis sequence is Cauchy, so there is an N > 0 such that if k, I > N, then foreach of the coordinates we have IXik — xj, J <s/2n M. Then we draw a polygonalpath, as in the Solution to Problem 2.2.25, from (xlk, . . . , x,,j) to (xl,, . . . ,
parallel to the axes.If this path does not goes through the origin, then as before
and if the origin is in one of the segments of the polygonal path, we can perturb ita bit, by traversing in the same direction but s/4M away from the origin. On thisaltered path
If(xlk,
in both case the sequence (f(xi, ... , is Cauchy and, thus, it converges.
Given any other sequence ((Yt. ..., Yn))k. an identical argument shows that. .. , — ..., tends to 0, so all sequences must converge
to the same value, which can be defined as the continuous extension of I to theorigin. For n = 1, consider the function f(x) = 1 if x < 0 and f(x) = 0 ifx>0.
Solution to 2.2.27: 1. The answer is no, and a counterexample is the function
xyf(x,y)= for(x, y) (0,0)
Ui)2=limt 2+v2°t—+O U
g(x,y)urn =lim
(x.y)—*(O.O) II(x,x=y
f is differentiable everywhere, but cannot be extended continuously to the origin,because it is constant equal to k/(l + k2) on each line y = kx passing throughthe origin.2. The answer is again no, with the counterexample a variant of the previous one,the function
xy2g(x,y)= and g(0,0)=0
g is now continuous everywhere, but not differentiable because the directionalderivative does not depend linearly on the vector. Let (u, v) (0,0). We have
• g((0, 0) + t (u, v)) — g(0, 0)urn
t
So the directional derivatives at the origin exist in all directions. If g were differ-entiable at (0, C)), as all the directional derivatives vanish there, we would haveDg(0, 0) = 0 (the zero linear map). Then, by definition of differentiability, wewould have
which is absurd, since
g(x, y) = ((x, y) -+ (0,0))
Solution to 2.2.28: A simple counterexample is
L
if y=Oif
and not even continuity at the origin and C1 on the rest of the plane is enoughto guarantee differentiability, as shown in the counterexample of Problem 2.2.27,Part 2.
Solution to 2.2.29: Let s > 0. By the hypothesis, there is S such thatIIDf(w)II <s if lIwil <8. For lIxil <8, by the Mean Value Theorem [Rud87, p.108], applied to the line segment joining 0 and x, we have
111(x) — 1(0)11 IlDf lix — Oil <dxli for some 0 <1,
which implies differentiability at the origin.
Solution to 2.2.31: Define f : R2 -÷ IR by
p(x,y)f(x,y)= I
J (0,0)udx+vdy,
the line integral taken along the straight line segment from (0,0) to (x, y). Fixa = (at, a2) E JR2 and t 0. Using Green's Theorem [Rud87, p. 253], onthe triangle with vertices (0,0), (ai, a2) and (ai + t, CZ2), the parametrizationx = aI + r, y = a2, 0 r t, and the mean value theorem for integrals, weobtain
f(ai + t, a2) — I (ar, a2)t
_!— t
udx+vdy
Both examples are from [Lim82].
I "I u(al+r,a2)dttJo
= u(a + Ot, a2) for some 0 <0 < I.
afTaking the limit when t -÷ 0; we obtain — = u and similarly, --- = V.
ax
2. Let U = 1R2 \ fØ} if such a function f: U —>. R existed then, for any closedcurve C in U,
fcax ôy
However, for the unit circle with the parametrization x cos t, y = sin t,
r —y
___
I dx+ dy=2ir.Jcx2+y2 x2+y2
which is a contradiction.
Solution to 2.232: The answer is no; to see it, consider the functionf(x,y,z) = 1 —x2 —y2 —z2.
Solution to 2.233: Fix a pointx R'. By the Chain Rule [Rud87, p. 214],
D(g o f)(x) = ((Dg) (f(x))) ((Df)(x)) =0.
The transformation (Dg) (f(x)) : R" —* R is nonzero because g has no criticalpoints. The preceding equality therefore implies that the transformation(Df)(x) : 1W' is noninvertible, so its determinant vanishes. That deter-minant is the Jacobian determinant of f at x.
Solution to 2.234: 1. Let F : 1R2 —÷ ]R be defined by F(x, t) = f(x) — tg(x).Then F is a smooth scalar field with F(0, 0) = 0 and
0) f'(O) — Og'(O) 0.ax
Therefore, by the Implicit Function Theorem [Rud87, p. 224], there exists a pos-itive 5 such that, for Iti <5, x is a smooth function of t, withx(0) = 0.2. Differentiating both sides of f(x(t)) = tg(x(t)) with respect tot, we have, forIti <8,
x'(t) = g(t)f'(t)
As x(0) =0, the desired expansion of x(t) is
f,(0)
Solution to 2.2.35: Subtracting the second and third equations from the first weget u4 — 3u = 0, so that u = 0 or u =
Ifu = 0, then
x—y+2z=02x+2y—3z=0
whose solution is given by x = —z/4, y = 7z/4.Ifu = then
3x+y—z=
x = —(z/4) — y = 7z/4.The solution set of the system of equations is a disjoint union of two parallel
lines in R4:
/17(x,y,z,u)=j:'
1,0
Thus, for any E > 0, the system can be solved for (x, y, z) as a function ofz E [—E, E}, withx(0) = y(O) = u(0) =0. Namely, let A, B be disjoint sets withA U B = [—E, E] \ {0), and define
I —z/4 zEA 10 z€Ax(z) = { —(z/4) — z E B y(z) = 7z/4 u(z) = { z E B
(0 z=0 [0 z=0The function y (z) is unique, continuous, and differentiable. Neither functions
x(z) and u(z) are unique. The function x(z) is continuous, and differentiable, ifand only if B = 0. The function u(z) is continuous, and differentiable, if andonly if B = 0.
2. We have seen above that the system has a solution if and only if u = 0 oru = So if 0 < u the system has no solutions. Thus, the system cannotbe solved for (x, y, z) as a function of u E [—S,S], for any S > 0.
Solution to 2.236: Parametrizing the circle by the angle 0, the function becomes
fa,b(O)aSifl2O+bcOSO
and the derivative is
= 2asinOcosO —bsin9
= sin8(2a cosO — b)
So fa,b always has at least two critical points at 0 = ±ir/2 and maybe twoadditional ones when cosO = b/2a. For this to happen Ib/2a1 1, that is, the
(a, b) region is the region in between the lines b = ±2a, containing the x-axis,including the lines itself and excluding the origin.
Solution 2. Using the method of Lagrange Multipliers [MH93, p. 414] for laband g(x, y) = x2 + y2, we end up with the system of equations:
b=Ax2ay=Ay
x2+y2=1.From this we can easily see that the pair (x, y) = (±1,0) is always a criticalpoint, and if we suppose y 0 then A = 2a, which implies that the pair of points
— b2
2a
might be two additional critical points. They will occur only when a 0 and4a2 — b2 0, that is 21a1, that is, the region in between the lines b = ±2a,containing the x-axis, including the lines itself and excluding the origin.
Solution to 2.237: Consider the function G : —÷ R2 given by
G(x,y)=
Since Vu and Vv are linearly dependent and Vu is never 0, G' has rank 1 every-where. Therefore, by the Rank Theorem [Bar76, p. 391], given a point E R2,there is a neighborhood V of an open set W C R2, a diffeomorphismh: W —÷ V, anda C1—functiong = : R —÷ R2 such that G(h(x, y)) =g(x) on W. So (x) is never 0. Therefore, by the Inverse Function Theorem[Rud87, p. 221], is locally invertible. By shrinking the set W (and so theset V), we may assume that it is invertible. Therefore, gj' (u (h(x, y))) = xor gz o (u (h(x, y))) = v (h(x, y)) for all (x, y) E W. Since h is a diffeomor-phism of W onto V, it follows that (u(x, y)) = v(x, y) for all (x, y) E V.F = o satisfies the required condition.
Solution to 2.2.38: The conclusion is trivial if f is constant, so we assume f isnot a constant. There is (xo, Yo) E R2 such that Df(xo, Yo) 0. After perform-ing a rotation of the coordinates, if necessary, we assume
f(xo, yo), and consider the function F : R2 given by
F(x,y)=(f(xy),v).The Jacobian of F is nonzero at (xo, so, by the Inverse Function Theorem[Rud87, p. 221], the function F has a local inverse, G, defined in a neighbor-hood of (a, yo). Thus, F (G(a, jo)) = (a, y) for all y in some closed interval I
containing y be any one-to-one map of [0, 1] onto 1. The function
g(t) = 0 (a, y(t)) (t E [0, 1])
has the desired properties.
Solution to 2.239: Consider F : R2 given by
F(x, y) = (f(x), —y +xf(x)).
A calculation gives that the Jacobian of F at (xO, is —f'(xo) 0. So, bythe Inverse Function Theorem [Rud87, p. 221], F is invertible in a neighborhoodof (xo, yo). Similarly, f has a local inverse, g, close to xo. In a sufficiently smallneighborhood of (xo, Yo) we can then solve for each component of explicitlyand get
g(u) =g(f(x)) =xand
y = —v + xf(x) = —v + g(u)f(g(u)) = —v + ug(u).
Solution to 2.2.40: f is clearly a C°° function, so
f(X + hY) — f(X) = (X + hY)(X + hY) — X2
hXY + hYX + h2Y2
thereforef'(X).Y = xY+YX
Solution to 2.2.41: 1. Let F : -÷ be defined by
F(x,y,z,w)=(x2+yz,y(x+w),z(x+w),zy+w2).
This map is associated with the given map because
(x y x2+yz y(x+w)z w ) — z(x + w) zy + w2
The Jacobian of F at (1,0, 0, 1) is therefore, F is locally invertible near thatpoint.2. We have F(1,s,e, —1) = (1,0,0, 1) for any e, so F is not invertible near(1,0,0, —1).
Solution to 2.2.42: Identify the matrix X = with the element of (x, y, z, w)
in the usual way. Let F be defined by F(X) = X2 + Xt. We have
f2x+1 z y 0
I y x+w 1 yDF(X)(x,y,z,w)=I 1
0 z y 2w+1
so 1000DF(X)(0,0,0,0)=
0001is invertible; therefore, by the Inverse Function Theorem [Rud87, p. 221], thereis such an s.
Global unicity fails for X = since X2 + = 0 = + 0.
Solution to 2.2.43: Since F(0) = 0 and F is clearly a C°°—function, the In-verse Function Theorem [Rud87, p.22 1] will yield the result if we can prove thatDF(0) is invertible. We have
F(X+hY)— F(X)=X+hY +(X +hY)2 —x —X2 =hY+hXY +hYX +h2Y2
therefore,DF(X)Y = Y +XY+YX.
In particular, DF(0) is the identity operator which is invertible.
Solution to 2.2.44: Define the function F : -+ by F(X) = X4F is clearly a C°°—function and F(I) = I, so the Inverse Function Theorem[Rud87, p. 221] will yield the result once we prove that DF(1) is invertible, butthe computation of the derivative, as above, is
DF(X) • Y = X3Y + X2YX + XYX2 + YX3.
In particular, DF(I) = 4! which is invertible, showing that F is a cliffeomor-phism in a neighbourhood of the Identity matrix.
Solution to 2.2.45: 1. Using the method of Laplace Expansions [HK6 1, p. 179]we can see that finding the determinant involves only sums and multiplicationsof the entries of a matrix, therefore, it is a C°°—function.2. For i, j = 1, . . -, n, let Xjj denote the (i, J)th entry of X, and let Xü denote thecofactor of Xjj, so that
det(X) = (j 1,... , n).
Since = 0 for each i f 'k it follows from the preceding expression that
d det=
Thus, X is a critical point of det if and only if = 0 for every i and j or, whatis equivalent, if and only if the rank of X does not exceed n —2.
Solution to 2.2.46: u(t) is differentiable as the composition of differentiable func-tions, and the derivative can be computed using the chain rule.
u'(t) = tr(F2 (t)F'(r) + F(r)F'(t)F(t) + F'(r)F2(r))= tr(F2 (t)F'(t)) + tr(F(t)F'(r)F(t)) + tr(Fh1(t)F2(t))
= 3tr(F2(r)F'(t)).
Solution to 2.247: Let be the 1—column matrix
Ai—
)(auanj
so A is the matrix whose column j is that is, A = (Ai, ..., Since det isan n—linear function of (R'1y' into R, the derivative is given by
..., An).dt
j=i
Let A(i, j) denote the cofactor of that is,
I . . . • • 1
A(i, j) = (_1)i+J al_il •.• ai_lj_l ai_1j÷i ... al_In• .. ai÷lj÷1 . . .
an! •.. anj_1
Using Laplace's Expansion Method to evaluate the determinant [HK61, p. 179]
dt
of each component of the derivative, by developing the column we get
=
andn n n n
det(A) = j) =j=i i=1 j=I 1=1
where the last equality follows from the fact that the inverse matrix is given by
b13 = det(A) . A(j, i). Therefore, we have
log(det(A)) = det(A)det(A)
= det(A)det(A)
j=1 1=1
=j=I i=1
Solution to 2.2.48: 1. The measure preserving condition is that the derivative, f',has absolute value I at every point. Since the function is continuous it must beeither f' = —1 or f' = I on the entire domain. Therefore, f(x) = —x + c orf(x) = x + c where c is a constant. Thus, f cannot map R into IR+.2. Take f(x, y) = er).
2.3 Integral Calculus
Solution to 2.3.1: Let f : R3 -÷ R, f(x, y, z) = (ax, by, cz). The volume givenis the image, under f, of the unit ball of B. As the Jacobian of f is abceverywhere, we have
vol (f(B))= fff dxdydz
= fff abc dxdydz =
Solution to 2.3.2: Using polar coordinates, we have
22ir I
IL e_X dxdy= f f=__f I= j(e1 —1)
=ir(e1 —1).
Solution to 2.3.3: We write
I = j°°e_3Y212 e 2+2xy_y2/2 dx) dy
= f (f dy
making the substitution t = (x — dt = on the inner integral, weget
= I
f00
e_t2dt) dy
= f e3)?212 dy
now making the substitution: s = ds = we obtain
=
Solution 2. As the integrand is positive, the iterated integral equals the double
integral:
I= f d(x, y).
Let u = x + y, v —x + y; this map is a differentiable bijection R2 R2.Moreover
x2 + (y — x)2 + y2 = + 3v2) and 2.
By the Change of Variable Formula for multiple integrals [MH93, p. 505],
f d(x, y)= f e 2+3v2) d(x, 3')
R2 R2 8(u,v)
=2R2
=!2JR R
=
Solution to 23.4: Let f(x, y) = x3 —3xy2. Derivating twice (see Problem 5.9.1),wecanseethau\f=0,sofisharmonic.LetBi =((x,y) I(x+I)2+y2
1).Wehave
andjbjl ,'3/p2,r
JJ f(x, y) dxdy= J
+ dO ) rB1 O\O /
where = (—1,0) is the center of As f is harmonic, the inner integral isequal to = —2ir and the integral of f over is —9ir. An identicalcalculation shows that the integral over 132 is it, so the integral of f over R. is—lOir.
Solution 2. The integrand is harmonic, being the real part of the holomorphicfunction z3. Its integral over any disc is therefore the value at the center of thedisc times the area of the disc. Hence, if we call Di and V2 the external andinternal discs that make up the boundary of R, respectively, we have
ff(x3 — 3xy2) dxdy= ff (x3 — 3xy2) dxdy
— ff (x3 — 3xy2) dxdy
=—lOir.
Solution to 2.3.5: Consider the annular region A between the circles of radius rand R, then by the Green's Theorem
= f eX dx + SOx dy)
= j7
= 'IA_eXSoyy dy A dx + + A dy
= 'IA + excoxx + dx A dy 0
showing that the integral does not depend on the radius r. Now, parametrizing thecircle of radius r
[ eX dx + cox dy) =Jcr= 12,r
rsinO)sjn9 + cox(r cosO, r sinO)cosO)rdO
but when r -÷ 0 the integrand converges uniformly to
sinO • cosO IsinO-i- •cosO=—
2ir 2ir 2ir
so the integral approaches I when r -÷ 0 and that is the value of the integral.
Solution to 2.3.6: Using the change of variables
x = singocosOy = sin gosin8Z=C0S97
0<8 <2ir0<go<ir
we have
and
dA =singodOdgo
ff(x2 +y+z)dA =,r 2ir
00 (sin2 cocos2 8 + sin go sinG + cosgo)singodGdgo.
Breaking the integral in three terms, we get
ff sin go cos go dO dgo = 2ir.
2g
f f sin2 go sin 8 dO dgo =
Therefore,
= (fgsin3 go
=1 —(3singp—sin3go)dgp
=
= (_3(;2) + ir
fj(x2+Y+z)dA =
Solution 2. The outward unit normal to S is ii = (x, y, z). Consider the vectorfield (x, 1, 1). Then ii = x2 + y + z, and = 1, Using theDivergence Theorem,
ff(x2+Y+z)dA
jig
I sin2godgo=0Jo
sin2 go dco)f sinG dO = 0
pg p2g
J J sin3 go cos2 8 dO dgo002,r
0cos2 8 dO
1 +cos2O2
= 111 divFdVJ J Jx2+y2+z2
=fff
Solution to 23.7: 1. We have
j ka/ax a/ay a/az = —2zj + (3y — 1)k.
x2+y—4 3xy 2xz+z2
2.Letfl={(x,y,z)EIR3Ix2+y2-f-z2= 16,given by V = ((x, y, 0) E 1EV 1x2 +y2 16). fl and V have the same boundary,so, by Stokes' Theorem [Rud87, p. 253]
f (VxF).dS=f F.Jl=f7-t an av
= J(VF x F) dS= ff (_2z7 + (3y — 1)1) dxdy
= If (3y — 1)dxdy = —l6ir.JJv
Solution to 23.8: Let C be a smooth closed path in R3 which does not containthe origin, and let L be any polygonal line from the origin to infinity that does notintersect C. V = R3 \ L is simply connected; so to show that
IJc
it suffices to show that V x F = 0. Let r = (x, y, z) and F = (P, Q, R). Wehave
F(r) = (g (IIrIDx, g (OrII) y, g (liril) z).By the Chain Rule [Rud87, p. 214],
— = g' (IJrII)aUrII
z + g' (IIrII) — g' (lIrII)811r11
— g (IIrII)
= g' (liril) — g' (liril)YZ
11'Il 11'•lI
=0.
Similarly,aP aRaQ ap0az azax
and we are done.
Solution to 23.9: 1. From the identity
V. = (Vf). fV. 1= Vf
and Gauss Theorem [Rud87, p. 2721, we obtain
fff (Vf) . J dx dy dz= fff v. (fI) dx
=1/=0.
2. Apply Part I with f(x, y, z) = x.
Solution to 23.10: By Gauss Theorem [Rud87, p. 2721,
IL div (u grad u)dxdy = j(u grad u) ds,
where ii is the unit outward normal, and ds is the differential of arc length. Theright-hand side vanishes because u = 0 on The left side equals the left sideof (*) because
a /au\ a/audiv (u grad u) =
i—) +/au\2 /au\2 /a2u a2u
=1—I +1—I +u( +\axj \8x2 ay
= Igradul2+Au2.
Solution to 23.11:
+:2;
substitutinga2\
u tU +
1 = V (uVu) — II Vu 112
the equality becomes:
jjV.(uVu)_JJIIVu(12_jjAu2=0
Now applying Green's Theorem [Rud87, p. 253], to the first term of the equation,
we get:
fuDnu =o
and since = au on 8V, we get
afu2_ffsince u is not identically zero inside the disk, the last integral is positive. Now ifu were constant on V. the second equation = au on 8D would force u —0,so Vu is not identically zero on V and the integral Ii Vu 112 is also positive,
showing that a lay u2 ? Oand consequently that A <0.
Solution to 2.3.12: By the Change of Variable Formula for multiple integrals{MH93, p. 505],
VOlf(Qr(XO))= f JJ(x)Idx.
Qr(Xo)
Hence, If Mr is the maximum and mr the minimum of I J (x) I for x E Qr (xo), wehave
mr ( r3 vol f (Qr(X0)) Mr.
By the continuity of J, we have mr —4 IJ(xo)1 and Mr -÷ IJ(xo)i as r -÷ 0,from which the desired equality follows.
To establish the inequality, we note that the same reasoning gives
(*) IJ(xo)i = limVOlf(Br(x0))
r-+O
where Br (xO) denotes the ball of radius r and center xo. Let
K = urn sup111(x) — f(xo)II
lIx—xoII
Then, given s > 0, there is an r8 > 0 such that IIf(x)— f(xo)Ii (K +s)llx —xoIlfor lix — xoli r6. The latter means that, for r r8,
f (B,(X0)) C (f(xo))
so thatVOlf(Br(XO)) 4(K +s)3r3
4 3 4 =(K-fs)3.
In view of the equation (*), this gives 1J(xo)i ( (K + s)3. Since s is arbitrary,we get IJ(xo)l K3, the desired inequality.
3
Differential Equations
3.1 First Order Equations
Solution to 3.1.1: For (xo, Yo) to be the midpoint of L(x0, yo). the y intercept ofL must be 2yo and the x intercept must be Hence, the slope of the tangentline is —yo/xo. Let the curve have the equation y = f(x). We get the differentialequation
or1 f'(x) ldyxf(x)ydx
By separation of variables, we get
logy —logx +C.
Hence,Df(x)=y=—
for some constant D. Solving for the initial condition f(3) = 2, we getf(x) = 6/x.
Solution to 3.1.2: We have g'(x)/g(x) = 2, so g(x) = where K is aconstant. The initial condition g(O) = a gives g(x) = ae2X.
=(1 —n)x+c.
1
x(t) = ((1 —
Solution to 3.1.6: Separating the variables, we obtain
dy
dxfdyJy—3
log(y —3)
y—3
Solution to 3.1.3: Suppose y is such a function. Then
y'(x) =
or
=dxyfl
d —dx
Moreover, c = > 0. We thus have
Iy(x) — 1/(n—1)(c—(n—1)x)
This function solves the initial value problem y' = y", y(O) = inthe interval [0, and, by the Picard's Theorem [San79, p. 8], it is the onlysolution. Since the function tends to 00 as x —÷ there is no function meetingthe original requirements.
Solution to 3.1.4: 1. The zero function x(t) 0 is a solution for all a > 0.2. If a > 1, then f(x) = is lipschitzian in a neighborhood of zero and byPicard's Theorem this is the only solution.
If 0 <a <1, then, as in the solution to Problem 3.1.3, we get
=(l —a)t+c,
and the initial condition implies c = 0. Therefore
is a solution defined on IL
=x2(y—3)
fx2dx
—
= Cex313
y=3+Cex3 /3
From initial conditions, C = —2. The unique maximal solution is given by
y = 3 — 2eX3/3.
Solution to 3.1.7: Picard's Theorem [San79, p. 8] applies because the func-tion x2 — x6 is Lipschitzian on finite subintervais of the x—axis. Thus, two dis-tinct solution curves are non intersecting. The constant functions x(t) 0 andx(t) 1 are solutions. Hence, if the solution x(t) satisfies x(0) > 1, thenx(t) > 1 forallt,andifitsatisfIes0 <x(0) < 1,thenO < x(t) < 1 forallt.
Since
x2 —x6 =x2(1 —x4) = (1 —x)x2(1 +x +x2 +x3),
wehaved 2 2 3(*) 1(x—1)=—(x—1)x (1+x+x +x).
We see from this (or directly from the original equation) that if x(0) > 1, thenx — 1 decreases as t increases, and if 0 <x(O) < 1, then 1 — x decreases as xincreases.
Case 1: x (0)> 1. In this case, (*) implies
(since x(t) > 1 for all t), so that
1) x(0) — 1, that is, x(t) — 1 e_'(x(O) — 1), from whichthe desired conclusion follows.
Case 2:0 <x(0) <1. In this case, (*) implies
—x) _x(O)2(1 —x)
(since x (t) x (0) for all t), so that
—x)) 0.
Therefore, — x(t)) 1 — x(O), that is, 1 — x(t) — x(O)),
and the desired conclusion follows.
Solution 2. The constant solutions are given by the roots of x2 — =0, and are
x(t) 0, 1, —1. The phase portrait for this equation is
—1 0 1
so 1 is a stable singularity and the only one in the positive axis, so it is the limitof any orbit starting on the positive real axis.
Solution to 3.1.8: From the equation, we get x' = 0 if x3 —x =0, so the constantsolutions are x —1, x 0, and x 1.
2. Considering the sign of x', we get the phase portrait
• S S
—1 0 1
so 0 is a stable singularity, and 1 a unstable one. There are no other singularitiesin [0, 1], so the limit of the orbit of the solution x (t) that verifies x (0) = 1/2 is 0.
xSolution 2. The function is lipschitzian on the finite subintervals of the
1 + exx-axis, so by Picard's Theorem the solutions by each point are unique. For anyo <x < 1, dx/dt <0 and since x(t) 0 is one of the solutions of the equation,the solution by x(0) = 1/2 is strictly decreasing and in between 0 and 1/2 for allt > 0. Now for these values of x (0 <x < 1/2) we can easily estimate
1 1—x2 1-> >-2 1+ex 4
dx 1—x2so — = —x can be estimated asdt
dx 1 1for
andd :14 :41'x dx\x)__el
Hence e'4x <x(0), that is, x from which the conclusion follow.
Solution to 3.1.9: The given equation satisfies the hypotheses of Picard's Theo-rem [San79, p. 8], so a solution x(t) exists in a neighborhood of the origin. Sincex'(O) = 231 +85 cosl7 0, by the Inverse Function Theorem [Rud87, p. 221],x is locally invertible. Its inverse satisfies the initial value problem:
dt 1— t(77)=0.dx 3x+85cosx
So Ix It(x)
= 177 +in some neighborhood of 77. Let a be the first zero of 3x +85 cos x to the left of77, since there are no zeroes to the right of it, t(x) above is defined in the entireinterval (a, oo), our maximal solution. The series expansion around a is
3x +85 cosx = (3— 85 sina)(x — a) + O((x — a)2) (x a).
We cannot have 3 — 85 sin a = 0, since this, together with 3a + 85 cos a = 0imply 9 + 9a2 =0. Therefore
• 3x+85cosxurn
x-+cr x—a
It is clear that• 3x+85cosx
lirn =3.x—+oo x—a
Thus,lim t(x) = —oo and lim t(x) = 00.x-4a
We may take the inverse of t : (a, oo) —÷ R and get a function x(t) that solvesour initial value problem and is defined in all of R.
Solution 2. Because the function 3x +85 cosx is continuous in Ix — bi K and— aj T, for any positive K and T, the differential equation has a unique
solution with x(a) = b defined for
/ KIt — al mm
T = 1/4 and K = 3lb1 + 85, the solution is defined for It — al ( 1/4,since a and b are arbitrary, the solution can then be continued in patches of size1/2 to the entire real line.
Solution to 3.1.11: Suppose y(t) > 0 for t (tO, ti), y(to) = 0. Integrating theequation
Yl
we get the solution y(t) = (t + c)2/4 where c is a constant. Each such solutioncan be extended to a differentiable function on R:
to ify(t)—j (t—to)2/4 if
We must have t0 0 for y to satisfy the given initial condition. y 0 is also asolution.
Solution to 3.1.12: The two functions y 0 and y 2 are obviously solutions tothe differential equation, and the first one satisfies the given initial condition. Wewill show that this is the only one. Suppose there is a solution y(x) that satisfiesthe initial condition and it is not identically zero, then there is a point xo wherey(xo) 0 and let (a, b) denote a maximal interval where y(x) 0 containing
Ate least one of the two extremes of the interval exists, let's call it a, the othermight be infinite and for that extreme y(a) 0.
Inside this interval the separable differential equation can be written as
dy =dx
and with the substitution w = x — 1 the first integral becomes
fdw
so integrating both sides we get
which is the same as
y — 1 + i/y (y — 2) = kex for some k> 0
but taking the limit when x —÷ a, y is zero at the limit and we have a contradictionbecause the right side is positive. So no such functions exist, except for y(x) 0.
Solution to 3.1.13: The given equation is equivalent to = x. Integrating
bothsideswegetxy= — +k,soy= +—andkhastobe0forthefunction2 . 2 x
to be differentiable in the given interval.
Solution 2. The equation is equivalent to
dx x
which is homogeneous in the sense of = f(x, y) and f(tx, ty) = f(x, y),so the substitution y = xv(x) converts it into a separable equation, in this case
dvv+x—=1-v(x)dx
or, after the separation of variables,
dv dx1—2v — x
integrating both sides, we get,
mu —2v1
—2=lnIxI+C
lnul—2vu=ln1
k2x2
IIi
= k2x2
solving both cases and renaming the constant we get y = + We obtain
the only one that is differentiable in the given domain making kX= 0, that is,
Solution to 3.1.15: 1. Let u be defined by u(x) = exp(—3x2/2)y(x). The givenequation becomes
+ (3xY(x) +Y(x))
u(x)— 1+e3x2u(x)2= f(x, u).
f is clearly C1, so it satisfies the Lipschitz condition on any compact convexdomain. The initial value problemdu
dxl+e3x2u'then has a unique solution for any n E N.2. f 0 is the unique solution of the initial value problem associated withthe condition u(0) = 0. Therefore, cannot have any zero, so (x) > 0 forx E [0, 1]. For u(x) = exp(—3x2/2)fn (x), we have
U' 1
2u 1+e3xu2
so
n n
therefore,13 15-e2n n
Owhenn -÷ Co.
Solution to 3.1.16: If y(t) 0 for some t, then y'(t) 1, so y(t) is growingfaster than z(t) = t for all t where y(t) 0. Hence, there is a with y(to) > 0.
For y > 0, y' > 0, so for t y is positive. Further, for y > 0, >so y' > Now consider the equation z' = e5z. Solving this by separationof variables, we get z(t) = log(t/5 + C)/5, and for some choice of C, we haveZ(tO) = y(to). For all t to, y' > z', so z(t) y(t) for t > to. Since z(t) tendsto infinity as t does, so does y.
Solution to 3.1.17: Since = ).eAt, the function y(t) = is a solutionif and only if A = in other words, if and only if A is a zero of the entirefunction 1(z) = z + et0z.
Assume 0 <to <ir/2. We must show that f has no zeros in the closed half-plane 0. If 0 and IzI> 1, then
If(z)I IzI — > 1 — 1 — 1 = 0,
so 1(z) 0. 0 and 1, then
IRf(z) = + = +
cos 0 0 0 f has no zeros for
Solution 2. Suppose with A = a + ib is a solution. We must prove a < 0.Assume a 0. Then, from f(A) =0, we get
a + ib = = (cos bt0 — i sin bt0).
Hencea —e° cos bt, b = e° sin bt0.
Since a 0, the second equality implies that Ibi 1. Then cos bt0 > 0 since0 <to whence a <0 by the first equality, a contradiction.
Solution to 3.1.18: Multiplying the first equation by the integrating factorexp q(t)dt), we get
(1(x) exp(fX
=
The general solution is therefore,
f(x) = Cexp(fX
where C is a constant. The hypothesis is that q(t)dt = +00. Even ifthe corresponding property may fail for p. For example, for p —1
and q 1, the general solutions are respectively C exp(—x) and C exp(x).
Solution to 3.1.19: Consider the equation
0 = F(x,y,z) = siny)z3 +(excosy)z
F(O, 0,0) = 0, and all the partial derivatives of F are continuous, with
aF /— = (3z2(eXsiny)+?cosy) = 1.(0,0,0) / (0,0,0)
By the Implicit Function Theorem [Rud87, p. 224], there is a real valued functionf with continuous partial derivatives, such that, F (x, y, f(x, y)) = 0 in a neigh-borhood of (0, 0, 0). Locally, then, the given differential equation is equivalent toy' = f(x, y). Since f satisfies the hypotheses of Picard's Theorem [San79, p. 8],there is a unique solution y in a neighborhood of 0 with y(O) =0.
Solution to 3.1.20: Since f and g are positive, the solutions of both problemsare monotonically increasing. The first differential equation can be rewritten asdx/f(x) = dt, so its solution is given by x = h1 (t), where the function h isdefined by
h(x)=Jo
Because the solution is defined for all t, we must have
(-co
Jo Jo
Since g it follows that
1 00, I
Jo Jo
Using a similar reasoning we can see that the solution of the second equation isgiven by x H'(t), where
H(x)=J
The conditions = oo, and f0 = —oo guarantee that H maps ]R
onto R, hence that is defined on all of R. Thus, the solution of the secondequation is defined on IL
Solution to 3.1.21: Let y be a solution of the given differential equation. If y'never vanishes, then y' has constant sign, so y is monotone.
Suppose that y'(xi) = 0 for some Then the constant function yi (x) = y(xi)is a solution of = 0. Consider the function z, z(x) = yi for all x. Then the
differential equation y' = f(y) with initial condition y(xj) =y is by Picard's Theorem [Sanl9, p.
8], y = z, so y is constant.
3.2 Second Order Equations
Solution to 3.2.1: By Picard's Theorem [Sanl9, p. 8] there is, at most, one real
valued function f on [0,00) such that f(0) = 1, f'(O) = 0 and
f"(x) = (x2 — 1)f(x). Since the function satisfies these conditions, we
must have f(x) = We then have
urn f(x) = urn e_x2/2x-+oo
Solution to 3.2.2: The characteristic polynomial of the given differential equationis (r — 1)2 so the general solution is
+fltet.
The initial conditions give a = 1, and fi =0, so the solution is y(t) = e.
Solution to 3.2.3: The characteristic polynomial of the associated homogeneousequation is
r2—2r+1 =(r—1)2
so the general solution of the homogeneous equation
d2x dx-2—+x =0dt2 di
IsAet + Btet (A, B E R).
(cos t)/2 is easily found to be a particular solution of the original equation, so thegeneral solution is
cos tAet+Btet+
The initial conditions give A = and B = so the solution is
1—tet +cost).
Solution to 3.2.4: The characteristic polynomial of the given equation is
5r2+ lOr+6
which has roots —1 ± it'lL so the general solution is given by
x(t) = cie' cos + sin
where cj and C2 are constants. We can assume Cl 0 or C2 0. Using calculus,we can see that u2(I + u = ±1. Then f attainsa maximum of 1/2 iffx attains one of the values ±1. We have = 0.Suppose cj 0. Then, if k is a large enough integer, we have
>1
so, by the Intermediate Value Theorem [Rud87, p. 93], x attains one of the values±1. If c2 0, a similar argument gives the same conclusion.
Solution to 3.2.5: We first solve the homogeneous equation x" + 8x' +25 = 0.The general solution is = clelT1t + where cj and c2 are con-stants and = —4 ± 3i, k = 1,2, are the roots of the characteristic equationr2+8r+25=0.
All the solutions of the differential equation x" + 8x' + 25x = 2 cos t are ofthe form x(t) = xO(t) + s(t), where s(t) is any particular solution. We solve foran s(t) by the Method of Undetermined Coefficients [BD65, p. 1151. Considers(t) A cos t + B sin t. Differentiating this expression twice, we get
2 cos t s" + 8s' + 25s = (24A + 8B) cos t + (24B — 8A) sin t.
Solving the two linear equations gives A = 3/40 and B = 1/40. Therefore,the desired solution x(t) is given by + s(t), where and c2 are cho-sen to give the correct initial conditions. tends to 0 as t tends to infin-ity; therefore, to finish the problem, we need to find constants a and 8 withacos(t—8) = Acos t+B sint. Wehavea cos(t—8) = a cost cos8+a sint sin8,so the problem reduces to solving a cos 8 = 3/40 and a sin 8 = 1/40. These equa-tions imply tan8 = 1/3 or 8 = arctan(1/3). Hence, by elementary trigonometry,cos 8 = so a =
Solution to 3.2.6: 1. The differential equation is equivalent to y' = z andhave
II(zi, IYII) — (z2, 1Y21)II = — z2)2 + (1Y21 — IyiI)2
— Z2)2 + (Y2 — Yl)2
= II(zi, yl) — (z2, Y2)ll
so the Lipschitz condition is verified and our initial value problem has, by P1-card's Theorem [5an79, p. 8], a unique solution. If y is such a solution, define the
function z by z(x) = y(—x). We have z"(x) = y"(—x) = —Iy(—x)I = —Iz(x)I,
z(0) = y(O) = 1 and z'(O) = —y'(O) = 0, so z = y and y is even.
2. We have
y'(x) = I y"(t)dt = — I < 0Jo Jo
so y is a decreasing function; therefore, it has, at most, one positive zero. If y is
positive on by continuity, y is positive in some interval of the form (—s, oo)
for some 8> 0. Together with y(O) = 1, y'(O) = 0 gives y(x) = cosx, which is
absurd. We conclude then that y has exactly one positive zero.
Solution to 3.2.7: Substituting y(x) = gives the quadratic equationa(a — 1) + 1 =0. The two roots are
so the general solution is
y(x) = + B4jicos
The boundary condition y(1) =0 implies A =0 and then the boundary conditiony(L) = 0 can be satisfied for nonzero B only if
sin =0.
Equivalently,
L =
where n is any positive integer.
Solution to 3.2.8: The general solution of the corresponding homogeneous equa-tion is AeYX + Be" where r, s are solutions of the characteristic equationx2+2p+xl =0,i.e.,
—r,s—
2
Since the given equation has a constant solution, y 3, it will admit solu-tions with infinitely many critical points exactly when r, s have a nonvanishingimaginary part, because in this case the homogeneous equation will admit solu-tions which are product of exponentials and sines or cosines. This happens when
< l,pER.
Solution to 3.2.9: Multiplying the first equation by a(t)/p(r) where a is a differ-entiable function, we get
a(t)x"(t) +a(t)q(t),()
+ = 0.p(t) p(t)
Expanding the second given equation, we get
a(t)x"(t) + a'(t)x'(t) + b(t)x(t) = 0.
For the two equations to be equivalent, we must have a'(:) = a(t)q(t)/p(:).Solving this by separation of variables, we get
= (ftq(x) dx).p(x)
Letting b(t) = a(t)r(t)/p(t), we are done.
Solution to 3.2.10: The function x' (1 )x (t) —x' (O)x (t +1) satisfies the differentialequation and vanishes along with its first derivative at t = 0. By Picard's The-orem [San79, p. 8] this function vanishes identically. Assuming x is not the zerofunction, we have x'(O) 5& 0 (again, by Picard's Theorem), so x(t + 1) = cx(:),where c = x'(l)/x'(O). It follows that the zero set of x is invariant under transla-tion by one unit, which implies the desired conclusion.
Solution to 3.2.11: The characteristic equation is 12_ 2c1 +1 = 0, which has theroots c ± — 1.
Case 1: c$> 1. Let w = — 1. Then the general solution is
x(t) = é(Acosh wt + B
The condition x(O) = 0 implies A 0, and then the condition x(2nk) = 0
implies B = 0, that is, x(t) 0. There are no nontrivial solutions in this case.Case 2: c = 1. The general solution is
x(t) = Ac + Btet.
The condition x(0) = 0 implies A = 0, and then the condition x(2.irk) = 0implies B =0. There are no nontrivial solutions in this case.Case 3: c = —1. Similar reasoning shows that there are no nontrivial solutions inthis case.Case 4: —1 <c < 1. Let co = 4/i — c2. The general solution is then
x(t) = edt(A cos cot + B sin cot).
The condition x (0) = 0 implies A =0. If B 0, the condition x (2irk) = 0 then
implies 2irkco = irn (n E Z), that is, co = n/2k, and
2
C = 1—co
The right side is nonnegative and less than 1 only for 0 < ml 2k. The required
values of c are thus
n=1,2,...,2k.
Solution to 3.2.12: By the basic Existence—Uniqueness Theorem, the solution set
is a two-dimensional vector space, V. Fix a real numbers, and define the functions
and q.ç by p9(1) = p(t +s), = q(t +s). Let f be in V. By the translation
invariance of V, f is a solution of the equation
d2x dxdt2
=
Hence it is also a solution of the equation
dx (*)
If p — p.c does not vanish identically, there is an interval on which it is nowherezero. But on such an interval the solutions of (*) form a one-dimensional vectorspace (by the Basic Uniqueness Theorem). Hence p — p3 0, from which onesees that also q — 0. Therefore p and q are constant.
3.3 Higher Order Equations
Solution to 3.3.1: 1. Since the differential equation is linear homogeneous withconstant coefficients we can build its characteristic equation, which is,
A3+A2—2=0.
By inspection we can easily see that 1 is one of the roots, factoring as
(A— l)(A2+2A+2)=0
so the roots are 1 and —l ± i, and the solution to the differential equation has theform
f(x) = ci? + cosx + sinx
so the space of solutions has dimension 3.2. E0 is the two-dimensional subspace defined by = 0, that is, the functiong(t) = c2e cost + c3e sin t, and after the substitution for the initial conditionwe see that C2 = 0 and C3 =2, so
g(t) = 2e1 sint.
Solution to 3.3.2: The four complex fourth roots of —1 are± 1 ±_i•
Therefore the
functions and satisfy the differential equa-tion =
For u(t) = we have u(0) = lim u(x) = urn u'(x) = 0.x—*oo x—).oo
As u'(O) = y(t) = also satisfies y'(O) = 1.
Solution to 3.3.4: 1. The characteristic polynomial of the equation is
7 x8—1x
x—1
which has roots —1, ±i, and (±1 ± For each such root Zk = Uk + ivk,k = 1, .. ., 7, we have the corresponding solution
= (cos Vkt + i sin vkt)
and these form a basis for the space of complex solutions. To get a basis for thereal solutions, we take the real and imaginary parts, getting the basis
xi(t) = e_t, x2(t) = cost, x3(t) Sint, X4(t) = cos
x5(t) = sin x6(t) = cos x7(t) = Sin
2. A solution tends to 0 at 00 if it is a linear combination of solutions in the basiswith the same property. Hence, the functions xi, x6, and form a basis for thespace of solutions tending to 0 at oo.
Solution to 3.3.5: The set of complex solutions of the equation forms a com-plex vector space which is invariant under differentiation. Hence, the functionscos t and cos 2t are also solutions, and, therefore, so are e±u cos t ± i sin tand e' = cos 2t ± i sin 2t. It follows that the characteristic polynomial of theequation has at least the four roots ±i, ±2i, so it is divisible by the polynomial(A2 + l)(A2 + 4). The differential equation is therefore, at least of order 4. Thesmallest possible order is, in fact, 4, because the given functions are both solutionsof the equation
1d2=0,
that is,d4x d2xdt4+5 dt2
+4x=O.
The preceding reasoning applies for both real and complex coefficients.
Solution to 3.3.6: Solving the characteristic equation r3 —1 =0, we find that thegeneral solution to y" — y = 0 is given by
—/2 —x/2y(x) = cieX + X C05 + c3e sin
with Cl, C2, and C3 E JR. y(x) = 0 when Cl = 0. But the solution above
with =0, is the general solution of the differential equation with characteristic
polynomial (r3 — 1)/(r — 1) = r2 + r + 1, that is,
y"+y'+y =0.So (0)+y' (0)+y(0) —0,and we can
3.4 Systems of Differential Equations
Solution to 3.4.2: Rewritting the system as a liner homogeneous system of equa-tions we have
(x'(t)\ — (2 —1\ (x(t)— 0 )
so the solutions are of the form
(x(t)\ — A:
By the Cayley—Hamilton Theorem [HK61, p. 194], the computation of the ex-ponential of a matrix reduces to a finite sum of terms, in this case two,
=aiAt+aolwhere the are functions of t depending on the matrix A. Furthermore, ifr(A) = aiA + ao, then for each eigenvalue A1 of A, r(A1) = and for each onewith multiplicity two
e" =dA
which is a way to compute the values of ao and aI, and then the exponentialMore generally, if eAt can be computed with the polynomial
eAt + . ''+aiAt +aolthen if we consider the polynomial
rCA) •••+aiA+aofor each eigenvalue A1 of At
=r(A,)and furthermore, if the multiplicity of A1 is k > 1, the each of the equations isalso valid:
fordA
For more details on this and other interesting ways to compute the exponential ofa matrix see [MVL78], specially Section 5, and the references cited there.
The characteristic polynomial of At is
XA: (A) = A2 — (2t)A +
so the elgenvalues are Aj =12 = t, and using the above formulas for the compu-tation of a', we get
r(A) =ail+aodr(A)
dA
and we end up with the system
e =tai+ao
=
Therefore the solutions are ai = et and ao = e (1 — t) so the exponential is
—t
t 1—t
The solutions are then
(x(t)\ ,(l+t —t
that is,
x(t) =e(k1 +k2t)
y(t) = e'(ki — k2 + k2t).
Solution 2. Alternatively, we can follow the previous solution closely, but computethe matrix etA by decomposing A = S + N, where S is complex diagonalizableand N nilpotent, for details we refer the reader to {HS74, Chap. 6]. This decom-position changes the computation of the exponential into a finite sum, in our case,
since A has only one eigenvalue with multiplicity two
1)obviously N2 =0, and the exponential can now be computed as
etA = etSetN = e'(I +tN)= et(1 +t
and the solutions follow as before.
Solution 3. Computing the Jordan Canonical Form [HK61, p. 247] of the systemwe have
(0 1'\j'2 —l\(l i\_(i 1
i
which we will write T1AT = JA, so with the change of coordinates
\yJ \WJ
that is,fx\ (1 i\(z
the system becomes, according to the form JA,
z'(t) z(t) + w(t)w'(t) = w(t).
These equations can be readily solved in the reverse order, and we obtain,
w(t) = clet z(t) = cltet + c2et
and the solutions x(t) and y(t) can now be written as:
x(t)=et (cit +ci +c2)y(t) = e' (cjt + c2)
Solution 4. Derivating the first equation and substituting the second we end upwith a second order homogeneous differential equation
x"(t) — 2x'(t) + x(t) =0
whose characteristic equation )i. — 2A + 1 =0 has one double root A = 1, so thegeneral solution will be
x(t) = cIe + c2te
substituting back in the first equation we get
y(t) = (Cl _c2)et +c2te'.
Solution to 3.4.3: We have
d 2 2 dx dy+y)=2xT+2y_
= 2x(—x + y) + 2y(log(20 + x))= —2x2+2xy —2y2+2ylog(20+x).
As, for any positive e,
log(20 + x) = o(x8) (x -+ +oo)
and
—2x2 +2xy — 2y2 ( —2(x2 + y2 — Ixyl)( —2(x —
we conclude that
for (x, y) II large enough so the distance of (x, y) to the origin is bounded.
Solution to 3.4.4: 1. Using poiar coordinates, x = r cos 9 and y r sin 9, we get
dr xdx ydy———+——=r(1 —r2)dt rdt rdtd9xdy ydxdt r2dt r2dt — 1
solving these, we get
etr= and O=—t+c2+
where cj, c2 are constants.For (xo, = (0,0), we have = — 0; therefore, x = y 0. For
(xO, yo) (0, 0) let = ro cos and yo = ro sin We have
TOand C290
so
x(t) —Clet
cos(9o — t), y(t) = — t).\f1+Cle2t
2. We havedc'
lim r = lim = 1.t-÷oo
+ C]
Solution to 3.4.5: We have
dfX\ /0 1 0\/x—1 ' 1=12 0 0 ydt\Zl \0 0 3J\z
Let A denote the 3 x 3 matrix above. Its characteristic poiynomial is
(x
therefore is a simple eigenvalue of A. Using an eigenvector v E R3 associ-
ated with this eigenvalue, we get a solution v which tends to 00 as t -4 —00and to the origin as t -÷ +00.
Solution to 3.4.6: We have
=
= (x'(t),x(t)) + (x(t),x(t)')= (Ax(t), x(t)) + (x(t), Ax(t))= ((A + A*) x(t), x(t))
so it suffices to prove that A ÷ A* is positive definite. We have
/2 2 —2
A+A*=( 2 8 22 16
and by the Solution to Problem 7.8.2 it is enough to check that the determinant ofthe principal minors are positive, which is a simple calculation.
Solution to 3.4.7: We have
so, for small positive t, (x(t), y(t)) lies in the right half-plane. For 0 <x < 1, we
havedx 12. 12 1
Thus, the function x(t) is increasing with slope at least 1/2. Therefore, by thetime t = 2, the curve (x(t), y(t)) will cross the linex = 1.
Solution to 3.4.8: 1. Let the function g be defined by g(t) = f(x(t), x'(t)). Wehave
(*) g'(t) = —2(x'(t)2 + x(t)4)
so g is a decreasing function.2. It is enough to show that g(t) = 0.
Since g is a positive decreasing function, the limit exists and satisfiesg(t) = c 0. If c > 0, then, for some s > 0, T E R we have
x'(t)2 + x(t)4 > s for t T. Then, by (*), we have
g(T)— C = +x(t)4)dt > fedtwhich is absurd. We must then have c =0, as desired.
Solution to 3.49: 1. We have
aF aFar 9Fdy 3= + = y(x + x5) — y(ay + x3 + x5) = --ay2.
Thus, a F/at 0, which implies that F decreases along any solution (x(t), y(t))of the differential equation.2. Let e > 0. Since F is continuous, there exists a 5 > 0 such thatF(x, y) < 62/2 if 1I(x, y) vary over all pointssuch that II(x, y)fl = s, elementary calculus shows that F(x, y) 62/2.
Let the initial conditions x(0) and y(O) be such that II(x(0), y(O))II < S.By Picard's Theorem [San79, p. 8], there exist unique solutions x(t) and y(t)to the differential equation satisfying these initial conditions. Since F(x, y) de-creases along solutions, we must have that F(x(t), y(t)) < for all t > 0
in the domain of the solution. Now suppose that for some t> 0 in this domain,II(x(t), y(t)) 62/2, a contradiction. There-fore, II(x(t), y(t))II < s for all t > 0 in the domain of the solution. But thisbound is independent oft, so the Extension Theorem [San79, p. 131] shows thatthis solution exists on all positive t.
Solution to 3.4.12: 1. We have
dH lddi'
= x'(t)) + (grad V(x(t)), x'(t))
= (x"(t), x'(t)) + (grad V(x(t)), x'(t))
=
f, (x(t)) x(t)
+
since f, = Therefore H(t) is constant on (a, b).2. Fix tO E (a, b). Since H is constant, H(t) = for all t E (a, b). We have
H(t0) = + V(x(t)) + M
and then0 2(H(to) — M)
showing that x'(t) is bounded. Noting that
f12Ix(ti) — x(t2)J
= Jx'(t)dt — M) fri — t21
we conclude that x is bounded and has limits at the endpoints of the intervalbecause it is Lipschitz. Using the continuity of f we get
urn x'(t) = urn f, (x(t)) = f, (lim x(t))t-+b
so limt...÷bx"(t) exists, fori = 1, .. . , n. The corresponding facts fort —+ a can betreated similarly. Therefore the closure of x((a, b)) is compact and x' is boundedbecause its image is contained in the image of a compact under the continuousextension of f,.
Using the maximum of IIx"(t)II on (a, b) we see thatx' is aLipschitz function,and that the limit x'(t) exists, completing the proof.
Solution to 3.4.14: Let ... , be a basis for V. Since V is closed underdifferentiation, we have the system of equations
Let This system has solutions of the form (x) = where theC1 's are constants and the A, 's are the (complex) eigenvalues of the matrix A. Bythe properties of the exponential function, we immediately have that f, (x + a) =Cf, (x) for some constant C depending on a, 50 (x + a) E V. Since the f, 'Sform a basis of V, V is closed under translation.
Solution to 3.4.15: Let V be such a space, and let D denote the operator of dif-ferentiation acting on V. If x is the characteristic polynomial of D, the functionsin V are solutions of the linear differential equation x (D)f = 0. The set of so-lutions of that equation is a three dimensional vector space, since x has degree 3,so it equals V.Case 1. x has three distinct roots A2, A3. The functions eA1, form abasis for the set of solutions of x(D)f = 0 and so for V.Case 2. x has a root of multiplicity 2 and a root A2 of multiplicity 1. Thefunctions te?'11, eX2t form a basis for V.Case 3. x has a single root of multiplicity 3. In this case the functions tea,
form a basis for V.
Solution to 3.4.16: We will show that for 1 k n + 1, fk(t) =where the 's are constants depending on k. From this, it follows that each 1k (r)approaches 0 as t tends to infinity.
Solving the second equation, we get
fn+1 (t) =
for some E R, which has the desired form. Assume that, for some k, theformula holds for fk+I. Differentiating it and substituting it into the first equationgives
n+ I
j=k+1
This is a first order linear differential equation which we can solve. Letting= (k + 1 + we get
fk= ds+C)e_kt
where C is a constant. Changing the order of summation and evaluating, we get
n+11k = =
j=k
where the 's are some real constants, and we are done.
Solution to 3.4.17: We solve the case n = 1 in two different ways. First method:Let B be the indefinite integral of A vanishing at 0. One can then integrate the
equation = Ax with the help of the integrating factor namely
_Bdx dx di...O=e ——e A—=—(e 8x),dt dt dt\ /
giving x(t) = eB(t)x(O). Since A(t) 13, we have B(t) fit for t > 0, so
)x(t)I =
as desired.Second method (n = 1): Consider the derivative of e_Ptx:
= — fix) = (A —13) x.
By Picard's Theorem [San79, p. 8], x either has a constant sign or is identically0. Hence, is nonincreasing when x is positive and nondecreasing when x isnegative, which gives the desired conclusion.
n> 1. We have for a solutionx(t),
= x) = 2(Ax, x) 213 lix 112
whichreducesthecasen> 1 tothecasen 1.
Solution to 3.4.18: 1. We have
IIX(t)112 = X(t)) = 2X(t)•dX(t)
= WX(t) = 2W'X(t) X(t)
= —2WX(t) . X(t) = —2X(t)- WX(t)
= IIX(t)112.
from which it follows that X (t) 112 = 0, hence that IlK (t) Ills constant.
2. We have
_(X(t).v)=dX(t).v=WX(t).V__X(t).WtV=_X(t).WV=0.dt dt
3. It will suffice to show that the null space of W is nontrivial. For if v is a nonzero
vector in that null space, then
1IX(t) — vu2 = IIX(t)112 + 11v112 — 2X(t)
which is constant by Part 1 and Part 2, implying that X(t) lies on the intersectionof two spheres, one with center 0 and one with center v.
The nontriviality of the null space of W follows from the antisymmetry of W:
det W = det Wt = det(—W) = det W = — det W.
Hence, det W =0,50 W is singular.
Solution to 3.4.19: Consider the function u defined by u(t) = IJx(t)112. We have,
using Rayleigh's Theorem [ND88, p. 418],
u'(t) = 2(x(t), x'(t)) = 2(x(t), P(r)x(t)) —2(x(t), x(t)) = —-2u(t)
which implies that u(t) u(O) exp(—2t) for t > 0, so lim u(t) =0, and theresult follows.
Solution to 3.4.20: Expanding the matrix differential equation, we get the familyof differential equations
dfr
dt =fj—1J—i,
where f&, 0 if i or j equals 0. Solving these, we get
XQ=I
+where the 's are constants.
Solution 2. We will use a power series. Assume X(t) = The givenn=O
equation gives
n=1
= >JAth2CflB
which can be written as
>2t12 ((n +
1
n+1so we have
=
Since A4 = B4 = 0, the solution reduces to a polynomial of degree at most 3:
X(t) Co +tAC0B + +
where Co = X (0) is the initial value of X.
4Metric Spaces
4.1 Topology of
Solution to 4.1.1: Suppose x is a point of S \ C. Then there is an open intervalcontaining x whose intersection with S is finite or countable and which thus isdisjoint from C. By the density of Q in R, there is such an interval with rationalendpoints. There are countably many open intervals in R with rational endpoints.Hence S \ C is a finite or countable union of finite or countable sets. Therefore,S \ C is finite or countable.
Solution to 4.1.2: 1. For n E Z let A AA is uncountable, at least one of these sets, Ak say, is uncountable. Then
Ak contains a bounded sequence with distinct terms, which has a convergent sub-sequence. The limit of this subsequence is an accumulation point of A.2. Let M be the set of accumulation points of A. Suppose M is at most countable.As M is closed, its complement IR \ M is open. Then we can represent it as acountable union of closed sets, R \ M = If M is at most countable, then Amust have uncountable many elements in R \ M, therefore in some Cm. By part1., A fl Cm has an accumulation point. Cm is closed, so this accumulation point isin Cm. This contradicts the fact that all accumulation points of A are in M.Solution 2. Suppose there are no accumulation points, then for each point r Rthere is a open interval 1r around r with only finitely many points of A. Reducingthe interval a bit further we may assume it contains at most one point of the set A(the possibility being r itself). This open set form an open covering of the compactset [n, n + 1], so it has a finite subcover and as such A only has finite number of
elements in the interval [n, n + 1], hence at most a countable number of elements,
a contradiction, so A must be uncountable.
Solution to 4.1.3: For each i, the given condition guarantees that the sequence
(x(i, is Cauchy, hence convergent. Let a, denote its limit. We have, by the
same condition,11 1
giving p(a,, ak) ( rnaxf}, Hence the sequence is Cauchy, therefore
convergent. Let a denote its limit.Each sequence (x(i, is Cauchy, hence convergent, say to and the
sequence is Cauchy, hence convergent to b.Again, by the same condition as before,
11p(x(i, i), x(i, j)) max(-, -).
In the limit as j -+ oo this gives p(x(i, i), a') —• Hence urn x(i, i) = a. By1 i—+OO
the same reasoning, urn x(z, z) = b.
Solution to 4.1.4: The closures form a nested sequence of compact sets whosediameters tend to 0. The intersection of the closures therefore consists of a singlepoint. The question is whether that point belongs to every
It is asserted that the centroid xo = + B + C) of the vertices A, B, C isin every It is in To, being a convex combination of the vertices A, B, C withnonzero coefficients, in fact, = Ja, $, y > 0, = 1).
Moreover the centroid of the vertices of Ti is
lfA+B B+C C+A\2
+2
+2
)X0.The obvious induction argument shows that xo is the centroid of the vertices of
for every n. Hence fl0 = {xo}.
Solution to 4.1.5: Suppose there is no such e. Then there exists a sequence inK such that none of the balls is contained in any of the balls Since Kis compact, this sequence has a limit point, by the Bolzano—Weierstrass Theorem[Rud87, p. 40], [MH93, p. 153], x E K. Then, since the B3's are an open cover ofK, there is a j and an e > 0 such that C Let 1/N <e/2, and choosen > N such that Ix — <e/2. Then C C B3, contradictingour choice of 's. Hence, the desired s must exist.
Solution 2. Suppose the conclusion is false. Then, for each positive integer n, thereare two and in K such that — < 1/n, yet no B3 Contains both
and Since K is compact, the sequence has a convergent subsequence,say, with limit p E K. Then, obviously, Ynk -+ p. There is a that contains
p. Since B3 is open and p = urn Xnk urn Ynk' both Xnk and Ynk must be in B3 fork sufficiently large, in contradiction to the way the points and were chosen.
Solution 3. By compactness, we can choose a finite subcover of K,[Rud87, p. 301. For x E K, define
f(x) =max(dist(x,IR!z\Bj)Ix E
Then f(x) > 0 for each x E K, because each is open and there are onlyfinitely many of them. Since K is compact and f is continuous and strictly pos-itive on K, f has a positive minimum c > 0. By definition of f, every s—ballcentered at a point of K is contained in some B3.
Solution 4. For each positive integer k let Ek = fx E M I B(x, 1/k) Cfor some a I). For each x E Ek we clearly have B(x, 1/k) C so Ek isopen. The sets Ek form an open cover of C, and are nested. Since C is compact,C C EK for some K, and we are done.
Solution to 4.1.6: Suppose is an open set of real numbers forn N, such thatQ = Then each set R \ is nowhere dense, since it is a closed set whichcontains only irrational numbers. We then have
U Un U(q)nEN qEQ
but R is not a countable union of nowhere dense sets, by Baire's Category Theo-rem [MH93, p. 175]. So Q cannot be a countable intersection of open sets.
Solution to 4.1.7: Suppose x, y X. Without loss of generality, assume x <y.Let z be such that x < z < y (for instance, z irrational verifying the doubleinequality). Then
(—oo,z)flX, (z,oo)flXis a disconnection of X. We conclude then that X can have only one element.
Solution to 4.1.8: The Cantor set [Rud87, p. 41] is an example of a closed sethaving uncountably many connected components.
Let A be an open set and suppose Ca, a F are its connected components.Each Ca is an open set, so it contains a rational number. As the components aredisjoint, we have an injection of F in Q , so F is, at most, countable.
Solution to 4.1.9: We show S is bounded and closed.To establish boundedness, choose no such that fl f — I on K for n no.
The function f is continuous, being the uniform limit of continuous functions,so f(K) is bounded, hence f(K) U fn(K) is bounded. Each is
bounded. Therefore so is (K), whence S is bounded.To prove that S is closed, let (yj} be a convergent sequence in S with limit y.
It will suffice to show that y is in S. if infinitely many terms of the sequence lie
in f(K), or in for some n, then y belongs to the same set, and hence toS. (Reason: f(K) and all the (K) are compact.) In the contrary case we mayassume, after passing to a subsequence, that Yj is in fmj (K), where <
say Yj = (Xj). Since K is compact we may assume, passing to a further
subsequence, that the sequence (Xf) converges, say to x. Then
— f(x)II ( flY — fmj(Xj)II + Ilfmj(Xj) — f(x)II
and all three summands on the right tend to 0 as j -+ oo. Hence y = f(x), so y
is in S.
Solution to 4.1.10: Suppose we have
[0,1] =lEN
where the [a1, b']'s are non empty pairwise disjoint intervals. Let X be the set ofthe corresponding endpoints:
X = (al, a2, ...) U fbi, b2, •
We will show that X is a perfect set, so it cannot be countable.
The complement of X in [0, 1] is a union ofopen intervals, so it is open, and Xis closed. By the assumption, there must be elements of X in (a1 — a) for each
e > 0, and each i E N, and similarly for the b, 's. Each element of X is then anaccumulation point, and X is perfect.
Solution to 4.1.11: 1. Let X = {x) and be a sequence in Y such thatIx — <d(X, Y) + 1/n. As is bounded, passing to a subsequence, wemay assume that it converges, to y, say. As Y is closed, y E Y and, by the conti-nuity of the norm, lx — yl = d(X, Y).2. Let be a sequence in X such that Y) <d(X, Y) + 1/n. As X iscompact, by the Bolzano—Weierstrass Theorem [Rud8l, p. 40], [MH93, p. 153],we may assume, passing to a subsequence, that converges, to x, say. We thenhave d(X, Y) = d((x), Y) and the result follows from Part 1.3.TakeX=f(x,1/x)Jx >0)andY=f(x,0)lx >0)inR2.
Solution to 4.1.12: Suppose that S contains no limit points. Then, for each x E S.there is a > 0 such that fl S fx). Let = The balls Bex(x) aredisjoint, so we can choose a distinct point from each one with rational coordinates.Since the collection of points in RiZ with rational coordinates is countable, the setS must be countable, a contradiction. Hence, S must contain one of its limit points.
Solution to 4.1.13: Let y be a limit point of Y and a sequence in Y con-verging to y. Without loss of generality, we may suppose that — yI <r. Bythe definition of Y, there is a sequence in X with Ixn — I r. Therefore,
,xfl — — + — <2r, so the sequence is bounded. Hence, ithas alimitpointx E X. By passing to subsequences and (yn), if necessary,we may assume that = x. Let e > 0. For n large, we have
1xY1 IX—XnI+LXn —Ynt+IYn--YI
and
Since s is arbitrary, lx — = r. Hence, y E Y and Y is closed.
Solution to 4.1.14: Let A be an infinite closed subset of For k = 1,..., letBk be the family of open balls in whose centers have rational coordinates andwhose radii are 1/k. Each family Bk is countable. For each ball B E Bk suchthat B fl A 0, choose a point in B fl A, and let Ak be the set of chosen points.Each Ak is a countable subset of A, so the set = UAk is a countable subsetof A. Since A is closed, the inclusion A A
a a lies in some ball B E Bk, and this ball B mustthen contain a point of Ak, hence of Thus, some point of lies within adistance_of 2/k of a. Since k is arbitrary, it follows that a E and, thus, that
Solution to 4.1.15: For k = 1, 2, .. ., let Bk be the closed ball in R" with center ato and radius k. Each compact of is contained in some Bk. As each Bk is com-pact, it is covered by finitely many 's, by the Heine—Borel Theorem, [Rud87, p.30]. Let Jk be the smallest index such that Bk is covered by U1, ..., Ujk. DefineV3 by setting = UJ \ Bk for 3k + 1 j (if the indices Jk are all equalfrom some point on, set V3 = 0 for j larger than their ultimate value.) The setsV1, V2, ... have the required property.
Solution to 4.1.16: If K is not bounded, then the function x lix ills notbounded on K. if K is bounded but not compact, then it is not closed, by theHeine—Borel Theorem, [Rud87, p. 40], [MH93, p. 155]; therefore, there exists
K. In thiscase, the functionx i-* lix isnotboundedon K.
Solution to 4.1.17: 1. Suppose not. Then there is a positive number S and a sub-sequence of (Xe), such that
As A is compact, by the Bolzano--Weierstrass Theorem [Rud87, p. 40], [MH93,
p. 153], has a convergent subsequence, which, by hypothesis, converges tox, contradicting the inequality.2. Let A = R and consider
i if i isodd1/i if i iseven.
All the convergent subsequences converge to zero, but (xe) diverges.
Solution to 4.1.18: Let s > 0. As f is uniformly continuous on X, there is aS > 0 such that
If(x)—f(y)l <s
forix—yl <8andanyMiAssume Ix — 8. As the function f is bounded, there is an M2 > 0 with
11(x) — for all x and y.Let = M2/ö. We have
11(x) — f(y)I MiIx — MiIx — yI + e
forallx, X.
Solution to 4.1.19: Let5= =AUB
where A and B are open. The origin belongs to A or to B. Without loss of gener-
ality, assume 0 E A. For every a, we have
Sa=(SuflA)U(ScxflB)
so, as is connected and 0 E Sa fl A, we get 5a fl B =0. Therefore,
SflB=U(SaflB)=O
and S is connected.
Solution to 4.1.20: Proof of 1 2: Suppose that G(f) is disconnected. LetA, B be disjoint non-empty open subsets of G(f) with A U B = G(f). LetAc, = (x IR'1 I (x, f(x)) E A), Bo = (x E I (x, f(x)) B). If x E Ao then(x, f(x)) E A. Since A is open in G(f) there exist neighborhoods U of x, andV of f(x), with (U x V) fl G(f) C A. Since f is continuous at x, we can finda neighbothood U' of x such that f(U') C V. Then U fl U' is a neighborhoodofx,andforanyz E UflU',(z,f(z)) €(U x V)flG(f) C A, andhence Ac,contains U fl U'. This proves that Ac, is open in Similarly, Bc, is an open set.Clearly Ac, and Bo are non-empty, Ac, fl Bc, = 0 and Ac, U Bo = This is acontradiction since R'1 is connected.
Counterexample to 2 1: Take n = 1, and let S1 = ((x, sin(1/x)) Ix <0),S2 = {(x, sin(1/x)) I x > 0). Si is the image of the continuous map (—oo, 0) —+RxR, x i-+ (x, sin(1/x)) and therefore itis aconnected subset of]RxR. Similarly
is a connected subset of R x R. Since (0,0) belongs to the closure in IR x IRof both and S2. the sets Si U ((0,0)) and S2 U ((0,0)) are connected. Sincethese sets have a point in common, their union (S1 U ((0,0))) U (S2 U ((0,0)))is connected. This union is the graph of the function f : R -+ IR defined byf(x) = sin(1/x), x 0, f(0) =0, which is not continuous at x =0.
Solution to 4.1.21: Let S be a countable dense subset of U, for example, the setof points in U with rational coordinates. Let I be the function that equals 0 offS and, at any point x in 5, equals the distance from x to 1W2\U. Then f has therequired property.
Solution to 4.1.22: The assertion is true. Let S be such a set and a a point in S.Define the set Sa = (b E S I a and b are connected by a path in S). 5a is open(because S is locally path connected) as well as its complement in 5, so these twosets make up a partition of 5, which is connected, therefore, the partition is trivialand 5a is the whole S.
Solution to 4.1.24: As X is compact, there is a constant B > 0 such that B I
for all matrices A = (ajj) in X. This bound, together with the equation Ax = Axgive us the bound IAI
In2B, so S is bounded.
Let be a sequence in S converging to For each A, there is an elementM1 of X such that is an eigenvalue of M,. The sequence (M,) has a convergentsubsequence, by compacity, so we may assume it converges to M E X. The set ofsingular matrices in is closed, since it is the inverse image of [0} under thecontinuous map det : -+ R, therefore — A1) is singular. Asthis limit equals M — Al we conclude that A E S, so S is also closed, therefore itis compact.
Solution to 4.1.25: 1. P2 is the quotient of the sphere S2 by the equivalencerelation that identifies two antipode points x and —x. If ir : -+ P2 is thenatural projection which associates each point x to its equivalence classjr(x) = [x, —x) P2. the natural topology is the quotient topology; that is,A C P2 is open if and only if jr'(A) C 52 is open. With this topology, theprojection 7r is a continuous function and P2 = ir(S2) is compact, being theimage of a compact by a continuous function.
Another topology frequently referred to as the usual topology of P2 is the onedefined by the metric
d(x, y) = min(Ix — yl, Ix + yl).
It is a straightforward verification that the function d above satisfies all axioms ofa metric. We will show now that it defines the same topology as the one above, onthe space that we will call (P2, d).
The application 7T : S2 -+ P2 with the metric as above satisfies the inequality
ir(y)) x —
so d is continuous. This defines a function W on the quotient which is
(P2,d)7/the identity and then continuous. Since P2 is compact and (P2, d) Hausdorff,is a homeomorphism and the two topologies in P2 are equivalent.
Now S0(3) is the group of orthogonal transformations of R3 with determinant1, so every matrix in this set of satisfies
therefore, = 1, for i = 1,2,3, implying that S0(3) is bounded. Con-sider now the transformation
f is continuous and S0(3) = f'(J, 1), that is, the inverse image of a closed set,then itself a closed set, showing that S0(3) is compact. Another way to see thisis to observe that the function
X-÷ ,/tr(rX)is a norm on the space of matrices and that for matrices in the or-thogonal group tr (XtX) = n, so 0(n) and, consequentially, S0(n) are compact.2. To see the homeomorphism between P2 and Q, first define the applicationço : P2 —÷ Q given by the following construction: For each line x through theorigin, take c°x : R3 -+ R3 as the rotation of 1800 around the x axis. This is welldefined and continuous. To see that it is surjective, notice that every orthogonalmatrix in dim 3 is equivalent to one of the form
/1 0 00 cosO sinO0 —sinO cosO
and with the additional condition of symmetry 0 = ir, which is a rotation of 180°around an axis. For more details see the Solution to Problem 7.4.24. Since iscontinuous and injective on a compact, it is an homeomorphism.
Solution to 4.1.27: Convergence in is entrywise convergence. In otherwords, the sequence (Ak) in converges to the matrix A if and only if, foreach i and j, the (i, entry of Ak converges to the (i, entry of A. It fol-lows that the operator of multiplication in is continuous; in other words,if —÷ A and Bk -÷ B, then AkBk —÷ AB. Now suppose (Ak) is a sequenceof nilpotent matrices in and assume Ak —÷ A. Then -÷ by thecontinuity of multiplication. But = 0 for each k since Ak is nilpotent. Hence,
= 0, that is, A is nilpotent. As a subset of a metric space is closed exactlywhen it contains all its limit points, the conclusion follows.
Solution to 4.1.28: For real 1 let T1 ft E IRI Sfl (—oo, t) is countable). If
is sequence in with limit t E R, then
S n (—oo, t) C fl (—oo, t1)).
As a subset of a countable union of countable sets is countable we conclude thatt E T1, so T1 is closed. If S fl (—oo, n) were countable for every integer n, then Swould be countable, a contradiction. Therefore Ti JR.
Similarly, T,. = {t E JR I S fl (r, oo) is countable) is closed and TR JR. Wehave T1 fl Tr = 0 S1flCC jf E T1 fl Tr then S c (S fl (—oo, z)) U (S fl (r, 00)) U{t) would be countable. If T1 U Tr = JR. then we would have JR as a union ofdisjoint proper subsets, contradicting the connectivity of JR. Hence there existst EJR\TUTr.ForsUChatSfl(—O0,t)andSflfr,00)&eunCoufltable,
4.2 General Theory
Solution to 4.2.1: We'll prove that p(g(y), (y)) —÷ 0 uniformly as n -+ 00.Since a uniformly continuous map preserves uniform convergence, and g is uni-formly continuous, it will suffice to show that ci(y, 0 uniformly asn —+ 00. But
o(y, =
so the desired conclusion follows from the hypothesis that f unifonnly asfl -+ 00.
Solution to 4.2.2: As the En's are non empty we can form a sequence sat-isfying E for all positive integers n. Given s > 0, there is an integer n(s)such that diam <s for n n(s). Thus, Xm) <e for n, m n(s). Thissequence is Cauchy, so it converges, to x, say. As is closed and contains thesequence we must have x E E
Solution to 4.2.3: Suppose that p is in the closure of A. Define the functionf(x) = —d(x, p) on X. By assumption, the restriction f attains a maximumon A. But since p is in the closure of A, f is nonpositive while attaining valuesarbitrarily close to 0,so the maximum value can only be 0; hence p E A. Thus Ais a closed subset of a compact space, so A is compact.
Solution to 4.2.4: Let U be an open cover of C. Then there is a set Uo in U thatcontains xo. Since = xo, there is an no such that is in Uo for alln > no. For each n no there is a set U that contains The subfamily
(Uo, U1,..., is then a finite subcover of C, proving, by the Heine—BorelTheorem [Rud87, p. 30], that C is compact.
Solution to 4.23: Let X be a compact metric space. For each n N, consider
a cover of X by balls with radius 1/n, 13(1/n) = {Ba(Xa, 1/n) E X). As X
is compact, a finite subcollection of 13(1/n), 13'(l/n), covers X, by the Heine—
Bore! Theorem [Rud87, p. 30]. Let A be the set consisting of the centers of theballs in 13'(l/n), n E N. A isa countable union of finite sets, so it is countable. It
is also clearly dense in X.
Solution to 4.2.6: Suppose x f(X). As f(X) is closed, there exists a positivenumber s such that d(x, f(X)) e.
As X is compact, using the Bolzano—Weierstrass Theorem [Rud87, p. 40],[MH93, p. 153], the sequence of iterates (f's (x)) has a convergent subsequence,
(x)), say. For i <j, we have
d (x), (x)) = d (x, fflifli (x)) S
which contradicts the fact that every convergent sequence in X is a Cauchy se-quence, and the conclusion follows.
Solution to 4.2.7: For x E C we clearly have f(x) = 0. Conversely, if f(x) =0,then there is a sequence in C with d(x, —÷ 0. As C is closed, we havex€C.
Given x, z E M and y E C, we have, by the Tiiangle Inequality [MH87, p.20],
d(x, y) d(x, z) +d(z, y).
Taking the infimum of both sides over y E C, we get
f(x)
or
f(x) d(x,
z)
and f is continuous.
Solution to 4.2.8: hf II 0 for all f in C1!3 is clear. If f 0, it is obvious thatIll H =0. Conversely, suppose that hf II = 0. Then, for all x 0, we have
0< If(x) — 1(0)1
— Ix —0111111 =0.
Since f(0) = 0, this implies 1(x) = 0 for all x. Let f, g E C"3 and e > 0.There exists x y such that
Ill +gii÷
< jf(x)—f(y)IIx—yI'/3 +
Ix—yI'/3IIfhI+hIghI+e.
Since e was arbitrary, the Triangle Inequality [MH87, p. 20] holds.
The property lid II = Icilif II forf E C1'3 andc ER is clear.Let {f,j be a Cauchy sequence in C"3. By the definition of the norm, for all
XE [0, 1]andanys > OthereisanN > Osuchthatifn,m > N,wehave
— fm)(X) (In — fm)(0)1 Ix —
orfm(X)I 8.
Hence, the sequence } is uniformly Cauchy. A similar calculation shows thatfunctions in C"3 are continuous. Since the space of continuous functions on [0, 1]is complete with respect to uniform convergence, there exists a continuous func-tion f such that the f uniformly. Suppose f C1'3. Then, forany M > 0, there exist X y such that
11(X) — >M.Ix—yI'/3
SoLfn(x)—fn(y)I
MIx — yl'/3
+Ix — yI'13
+x — >
Since the f uniformly, for fixed x and y we can make the first
and third terms as small as desired. Hence, fJ fl > M for all M and n sufficiently
large, contradicting the fact that E C1/3 and that, since the 's are Cauchy,their norms are uniformly bounded.
Suppose now that the sequence does not converge to f in C1!3. Then thereis an e > 0 such that — Ill > e for infinitely many n's. But then there exist
withIIn(X) — f(x)I —
Ix—yI"3+
Ix—y11/3>8
for those n's. But, as we have uniform convergence, we can make the left-handside as small as desired for fixed x and y, a contradiction.
Solution to 4.2.9: We first show that is equicontinuous. Fix e > 0. By theequicontinuity of the sequence }, there is a S > 0 such that
lIn(X) — <8
for all n, whenever d(x, y) < e. Fix n and fix x and y with d(x, y) < S. Let
j, k n be such that gn(x) = and = Ik(y). Then
gn(x) — = — Ik(Y) = — f(y) —<8.
By the same reasoning, —gn (x) <s, from which follows the equicontinuity
of the sequence
The sequence is clearly uniformly bounded, so, by the Arzelà-Ascoli The-orem, it has a uniformly convergent subsequence with limit g, say. Sincethe sequence (ga) is nondecreasing, for n > we have
g uniformly.
Solution to 4.2.11: Since f(K) c f(K)a n, the
fl is nonempty and compact (the latter because it is a closed subsetof the compact set Ku). Also, fl c f'((y}) fl Hence, by theNested Set Property [MH93, p. 157], the set
ñ (r' (fy)) n K
is nonempty; that is, y E f(K).
Solution to 4.2.12: 1. The completeness of Xj implies the completeness of X2.In fact, assume Xi is complete, and let be a Cauchy sequence in X2. Theconditions on f imply that it is one-to-one, so each can be written uniquelyas with in X1. Then dj(Xm, d2(ym, yn), implying that isa Cauchy sequence, hence convergent, say to x. Since f is continuous, we thenhave lim = f(x), proving that X2 is complete.2. The completeness of X2 does not imply the completeness of Xi. For an exam-ple, take X1 = (—i, X2 = IR, and f(x) = tanx. Since f'(x) = sec2x 1
on Xi, the condition Ix — 11(x) — f(.y)l holds.
4.3 Fixed Point Theorem
Solution to 43.1: The map is the image, by a contraction, of a complete met-tic space (California!). The result is a consequence of the Fixed Point Theorem[Rud87, p. 220].
Solution to 43.2: Let g(x) = (1 + x)1. We have
g'(x)= (1 +x)2
therefore,
(1<1 for x >
Then, by the Fixed Point Theorem [Rud87, p. 220], the sequence given by
xn+1 =
converges to the unique fixed point of g in [xo, 00). Solving g(x) = x in thatdomain gives us the limit
2
Solution to 4.3.3: Let S = (x E [0, oo) I x — f(x) 0). S is not empty because0 E S; also, every element of S is less than 100, so S has a supremum, xo, say.For any s > 0, there exists an element of S. x, with
so
and we conclude, since e is arbitrary, that xo f(xo).Supposef(xo)—xo =8> 0.Then,forsomex E S,wehavex <xo+8;
therefore,x <f(xo)—xo+x+8
and we getx (xo<f(xo)—xo+x
from which follows, since f is an increasing function,
f(xo) — xo + x f(xo) f — xo + x)
but then x E S, which contradicts the definition of Wemust then have f(xo) =
Solution to 4.3.4: Consider F: K -+ IR defined by F(x) = d (x, ço(x)). çô, beinga contraction, is continuous, and so is F. Since K is compact and nonempty, Fattains its minimum E at a point m E K, d(m, ço(m)) = From the minimalityof s, it follows that d (ço(m), ço (ço(m))) = F (ço(m)) ? s = d (m, ç9(m)). Thecontractiveness assumption implies that m = ço(m).
Suppose n E K also satisfies n =çp(n). Then d (ço(n), ço(m)) = d(n, m),which, by the contractiveness assumption, implies n = m.
Solution to 433: As the unit square is compact, max jK(x, = M < 1. Con-sider the map T : C[O,I] —* C[o,l] defined by
2T(f)(x)=eX — I K(x,y)f(y)dy.Jo
We have
plIT(f)(x) —
= JK(x, y)(g(y) — f(y))dy
0
MIg(y) — f(y)Idy
max Ig(y)—f(y)I
By the Contraction Mapping Principle [MH93, p. 275], T has a unique fixed point,
h E We have
(1 2h(x) + I K(x, y)h(y)dy = e'
Jo
Any such a solution is a fixed point of T, so it must equal h.
Solution to 43.6: Consider the map T : C[o,l] —÷ defined by
T(f) = g(x) + f f(x — t)e_t2 dt.
Given f, h E C[o,l], we have
IIT(f) — sup f If(x — t) — h(x — t)(e_t2 dtXE[0,I] 0
Ill — sup dtXE[OI] 0
p12
= hf — j dt < Ill —JO
so T is a contraction. Since C[o,I] is a complete metric space, by the ContractionMapping Principle [MH93, p. 275] there is f E C[0,J] such that T(f) = f, asdesired.
Solution to 4.3.7: Define the operator T on C[o,i] by
T(f)(x) = sinx + f dy.
Let f, g E C[o,lJ. We have
IIT(f) — T(g)hI sup 1101 If(y)g(y)I dY}
supflf(x) — f
where 0 <X < 1 is a constant. Hence, T is a strict contraction. Therefore, by theContraction Mapping Principle [MH93, p. 275], there is a unique f E C[O, 1) withT(f)=f.Solution to 4.3.8: Since M is a complete metric space and S2 is a strict contrac-tion, by the Contraction Mapping Principle [MH93, p. 2751 there is a unique pointx E M such that S2(x) = x. Let S(x) = y. Then S2(y) = S3(x) = S(x) = y.Hence, y is a fixed point of so x = y. Any fixed point of S is a fixed point
so S has a unique fixed point.
5
Complex Analysis
5.1 Complex Numbers
Solution to 5.1.1: We have
for k€Z;
therefore,
14+i _=
= (2k,r
+2kJr)
(k Z).
Solution to 5.1.2: We have i1 = and log i = log lii + i arg i =+ 2kir), k E Z. So the values of i k E Z}.
Solution to 5.1.3: Let A, B, C be the points a, b, c in the Argand diagram. Wesuppose that a, b, c are noncollinear.
To prove necessity, suppose ABC is equilateral. To be definite suppose ABChas counterclockwise orientation. Then c — a = (b —
a — b = (c — Therefore (c — a)(c — b) = (a — b)(b — a) which isequivalent to the condition a2 + b2 + c2 = bc + ca + ab.
Conversely, suppose a2 + b2 + c2 = bc + ca + ab. Then (c — a)(c — b)(a —b)(b--a),and
c—a a—b= =re say.b—a c—b
HenceAC BA
AB = BC=r
andLBAC = LCBA = ±9 (mod 2,r).
From the above equations we get AC = BC. Therefore ABC is an equilateraltriangle.
Now consider the case that a, b, c are collinear. The condition a2 + b2 + C2 =bc + ca + ab is equivalent to (c — a)(c — b) = (a — b)(b — a). The latter conditionis clearly invariant under translation and rotation about the origin. Thus we mayassume that a, b, c are real and that a = 0. The latter condition then reduces toC2 + b2 = bc, which is satisfied only by the real numbers c = b = 0, as canbe seen by using the quadratic formula to solve for c. Thus in this case the givencondition is equivalent to a, b, c forming a degenerate equilateral triangle.
Solution to 5.1.4: Multiplying by a unimodular constant, if necessary, we canassume c = 1. Then + = 0. So a = b. Their real part must be nega-tive, since otherwise the real parts of a, b, and c would sum to a positive num-ber. Therefore, there is 6 such that a = cosO + i sinO and b cos9 — i sin6,cos9 =—1/2.ThenO
Solution 2. Since b+c = —a, we get 1 = IIb+c112 = thus = —1,
hence IIb — = 2— =3. The same calculation gives ha — b112 = 3 andha—cl2 =3.
Solution to 5.1.5: 1. We have
=x—1
= +... + 1
forx 1, so = n.2. Let
2,ri(k— 1)
pk=e " fork=1,...,n= 1,wehave
J](z — Pk) =
Letting z = 1, and using Part 1, we get the desired result.
Solution to 5.1.6: Let q = so 2cosz = q + q1, and 2cosnz = qfl + q_•fl•
The problem reduces to find the polynomial such that TN(q + = +Now
+ {if n is even,
Now assuming that we know T, for j <n and proceeding by induction,
_j)/2) (Tj (x))
— fif n is even,
0<j<si 0 otherwise.n—jeven
is the sought polynomial.
Solution to 5.1.7: Consider the complex plane divided into four quadrants by thelines = and let Aj be the set of indices j such that Zj lies in the jth
quadrant. The union of the four sets 4, is (1, 2, ... , n), so there is an i such thatA = 4, satisfies
IzJI.jEA j=I
Since multiplying all of the Zj by a unimodular constant will not affect this sum,we may assume that A is the quadrant in the right half-plane, where > 0 and
lz.jI So we have
Combining this with the previous inequality, we get the desired result.
Solution to 5.1.8: The functions 1, ... , are orthonormal on [0,1].Hence,
2
dx= 1.k=1
Since the integrand is continuous and nonnegative, it must be 1 at some point.
Solution 2. Since = 0 for k 0, we have
1
=t0
= + q_" +0<) <ii
n—f even
((n _j)/2) (qJ +qJ)+
otherwise.
Zjj€A
n
k=1
Now argue as above.
Solution to 5.1.9: We have
eb_ea
for all complex numbers a and b, where the integral is taken over any path con-necting them. Suppose that a and b lie in the left half-plane. Then we can take apath also in the same half-plane, and for any z on this line, JezI 1. Therefore,integrating along this line, we get
blezlldzl lb—al.
Solution to 5.1.10: The boundary of N(A, r) consists of a finite set of circulararcs C,, each centered at a point a' in A. The sectors Si with base C, and vertexa1 are disjoint, and their total area is Lr/2, where L is the length of the boundary.Since everything lies in a disc of radius 2, the total area is at most 4n, so L8,T/r.
Solution to 5.1.11: Without loss of generality, suppose that
IaiI IaiI <Iai÷iI ... lakl
that is, exactly k — 1 of the a's with maximum modulus (1 may be zero.)We will first show that lak I = Jaj I is an upper bound for the expression,
and then prove that a subsequence gets arbitrarily close to this value. We have
I/n
(k = Jakl
the limit on the right exists and is lakl = supJ
1/n
limsupn j
Now dividing the whole expression by we get
= (EL)" = +j=1 j=l j=l+l
since the last k — I terms all have absolute value 1.
It suffices to show that
(*) limsup (EL)" + = k —1
sincek
1/n
a consequence of the the fact that orbits of irrational rotation on the circle
are dense in the unit circle. To see that, discard the first term because
its limit exists and equals zero, being a finite sum of terms that converge to zero,and distribute the rest in two sums, one containing all rational angles (pj/q,), andanother one containing all irrational angles (Sj). Without loss of generality we areleft to prove that
I
limsup 1.n j=k'+l
If the sequence contains only rational angles choose P — qj, twice theproduct of the denominators of the rational angles. Then the sequence n P wheren E N will land all angles at zero and the summation is equal to k —1.
Now if there is at least one irrational angle among them the set of points
/ . . \, ... , e , em +1 , , n E N
in the tori S1 x S1 x .. . x S1 is infinite (the last coordinates will never repeat) andso has an accumulation point, that is, for any s > 0 there are two iterates m> nsuch that
and then
... , . . .
is s-close to (1,..., 1), and we are done.
5.2 Series and Sequences of Functions
Solution to 5.2.1: Multiplying the first N + 1 factors we get
1 — z1° 1 — z'0° I — z'00° 1 — 1 —
1 — z 1 — z10 1 — I — — 1 — z
so the product converges to 1/(1 — z) as N -+ 00.
Solution to 5.2.2: From the recurrence relation, we see that the coefficients
grow, at most, at an exponential rate, so the series has a positive radius of conver-gence. Let f be the function it represents in its disc of convergence, and consider
the polynomial p(z) = 3 + 4z — z2. We have
p(z)f(z) = (3 + 4z — z2) an?
= 3ao + + 4ao)z + + —
=3+z.
So 3+zf(z)=
The radius of convergence of the series is the distance from 0 to the closest sin-gularity of f, which is the closest root of p. The roots of p are 2 ± Hence,the radius of convergence is —2.
Solution to 5.23: Let
1(z) =
be the Maclaurin series of f. Substituting the equation we get:
ao=0, a2=0,
and by induction for k 1, we get,
k1
k
a3k = LI 3j (3j — 1)' a3k+1 = 3f(3f + 1) a3k÷2 = 0.
So dividing the sum up in two components and analyzing the convergence of each
piece we see that a3k÷3 = 0, so the series > a3kz3tc has infinite radius ofconvergence. The series relative to the indexes 3k1 + 1 can be treated in a similarway. We conclude then that the series for f has infinite radius of convergence,which proves the unicity of the solution to the given problem.
Solution to 5.2.4: Let f(z) = exp (z/(z — 2)). The series can then be rewrittenas so, by the standard theory of power series, it converges ifand only if If(z)I 1. The preceding inequality holds when 0, so theproblem reduces to that of finding the region sent into the closed left half-plane
by the linear fractional map z i-+ The inverse of the preceding map is themap g defined by g(z) = Since g(O) = 0 and g(oo) = 2, the image of theimaginary axis under g is a circle passing through the points 0 and 2. As g sendsthe real axis onto itself, that circle must be orthogonal to the real axis, so it is thecircle z — ii = 1. Thus, g sends the open left half-plane either to the interior orto the exterior of that circle. Since g (— 1) = 1, the first possibility occurs. We canconclude that If(z)I I if and only if Jz — lJ 1 and z 2, which is the regionof convergence of the original series.
2
Solution to 5.2.5: The radius of convergence, R, of this power series is given by
For Izi < I, we have
I— =R n.-+oo
=/00 F
(1—z)2
By the Identity Theorem [MH87, p. 397],
1
1(z) =(1 — z)2
where the right-hand side is analytic. Since this happens everywhere except atz = 1, the power series expansion of f centered at —2 will have a radius ofconvergence equal to the distance between —2 and I. Hence, R =3.
Solution to 5.2.6: As
11 —x2+x4 —x6 +... = I +x2
which has singularities at ±i, the radius of convergence of
— 3)fl
0
is the distance from 3 to ±i, 13 j iJ = We then have
/ I10
Solution to 5.2.7: As urn 'i = 1, we haven.-+oo
= urn sup = urn supR ,2—*oo
so E anz'1 and n2anz7z have the same radius of convergence, and the conclu-sion follows.
Solution to 5.2.8: We'll show that the series converges for Izi = 1, z 1, whichimplies that the radius of convergence is at least 1, therefore the series is conver-
< 1.
Let (where p is a positive integer) and be defined by
n+p
b,z', b,z'.i=n+1
We have,
n+p—1
(z — = + (l'j — +i=n+i
therefore,
n+p—1
Iz — + — +i=n+1
Letting Izi = 1, z 1 in this inequality, we get,
i / n-i-p—I \—
+ (b1 — + bn+p) =i=n+I /
Since this holds for any positive p, we may let p —+ 00 and obtain,
IRn(z)I = b1z' =o(1) (n —* oo),
as we wanted.
Nothing can be paid about convergence at z = 1. The series diverges atz = 1, while converges at the same point. 1
Solution to 5.2.9: The series converges for all z.n—O
1/n = (I/fl)2
Izi Izt
By Cauchy's Root Test, the series converges for IzI> 1, and diverges forIzI <1. Hence the series + fl2/ZiZ) converges for IzI> 1, and divergesfor Izi < 1; the series is undefined for z =0.
Let IzI = 1. The sequence z'7n! tends to zero and the sequence tends to00. Therefore the sequence (ZZ /n! + n2/zhz) tends to oo, which implies that theseries +n2/zt2) is divergent for IzI = 1. Thus the given series convergesexactly when IzI> 1.
Solution to 5.2.10: Fix z ±i. Then
1/n1=
(1 + II + z2 I Ii + z2(
By Cauchy's Root Test the given series converges if Ii + z21 > 1, that is, forpoints exterior to the lemniscate II + z2 I 1.
Cauchy's test also shows that the given series is divergent for 11+ z21 <1. Theseries clearly converges for z = 0. ff11 + z21 = 1, z 0 then Izi/Il + = Iziwhich does not tend to 0 as n -÷ oo. Thus the series diverges for points z 0 onthe lemniscate.
Thus the series is convergent precisely for points exterior to the lemniscate,together with the point z = 0 on the lemniscate, is undefined at z = ±i, anddiverges at all other points.
Solution to 5.2.11: Let R denote the radius of convergence of this power series.
2R = urn sup = n = =
The series and all term by term derivatives converge absolutely on Izi < I anddiverge for Izi > I. Let Izi = 1. For k ? 0 the kth derivative of the power seriesis
To see that this converges absolutely, note that
001 1
n(n 1) . . . (n — k +n=k n=k
Since, for n sufficiently large, log n —k > 2, and 1/n2 converges, by the Com-parison Test [Rud87, p. 601 it follows that the power series converges absolutely
on the circle Izi = 1.
Solution to 5.2.12: We have1/n
=
1 .1/n
=—hmsup —R n!
R
=0
so h is entire.Let 0 <r < R. Then 1/R < 1/r, so there is an N > 0 such that
for n > N. Further, there exists a constant M > 2 such that Ian I for1 n N. Therefore, for all z,
Ih(z)I M mn!=
Solution to 5.2.13: We have
n=k+1 n=k+1
Therefore
IcnIIz'21k=O k=O n=k+I
oo n—i oo
=>: =n=1 k=O n=1
From the basic theory of power series, the series z'2 and its formal deriva-tive, the series i have the same radius of convergence. Therefore theseries also has radius of convergence R, so it converges absolutelyfor lzI <R, and the desired conclusion follows.
Solution to 5.2.14: Let the residue of f at 1 be K. We have
an Zn= 1
+ Z1Z with I bn 11/n> 1.
Therefore,
= + bn)z'2
and K + As <oc, we have limbo =0 and = K.
Solution to 5.2.15: The rational function
1— 2
f(z)= Z
l—z
has poles at all nonreal twelfth roots of unity (the singularities at = 1 areremovable). Thus, the radius of convergence is the distance from 1 to the nearestsingularity:
= = — 1)2 + sin2(ir/6) =12—
Solution to 5.2.16: Suppose that (f,j is a sequence of functions analytic in anopen set G, and that converges to the function f uniformly on G. Then f
E G.ThereisR > C G.Letybea triangular contour in R). Then (z) dz = 0 for each n, by Cauchy'sTheorem [MH87, p. 152]. As f uniformly on the image of y, we deducethat f 1(z) dz = 0. By Morera's Theorem,[MH87, p. 173] f is analytic in R5. This proves the result.
The analogous result for real analytic functions is false. Consider the function1: (—1, 1) —÷ JR defined by f(x) = lxi. As f is continuous, it is the uniformlimit of polynomials, by Stone—Weierstrass Approximation Theorem [MH93, p.284]. However, f is not even continuous in all of its domain.
Solution to 5.2.17: By the Hurwitz Theorem [MH87, p. 423], each zero of g isthe limit of a sequence of zeros of the 's, which are all real, so the limit will bereal as well.
Solution to 5.2.18: Let 6k = Then, clearly, IskI pf(k)(Ø)1 forall k. Since f is an entire function, its Maclaurin series [MH87, p. 234] convergesabsolutely for all z. Therefore, by the Comparison Test [Rud87, p. 60], the series
converges for all z and defines an entire function g(z). Let R > 0 and e > 0. ForIzi R, we have
— g(z)l — SkJR"
— €kIR'C + i!k=N+I
taking N sufficiently large, the second term is less than e/2 (since the power seriesfor f converges absolutely and uniformly on the disc Izi R). Let n be so large
that J <s/2M for 1 k N, where
N
M = >2Rk.k=O
Thus, for such n, we have Jgn(z) — g(z)J <s. Since this bound is independent ofz, the convergence is uniform.
Solution to 5.2.19: We have
fz(2—z)\log( )=log(2_z)+log(1 1
)1—z /
=log2+log(l —1
In the unit disc, the principal branch of log is represented by the series
which one can obtain by termwise integration of the geometric series=
and
I Z
log
00'c—' Z
n=1
(IzI <2),
(Izi > 1),
fz(2—z)\1—z
Zn
2'1nn=—00 n=1
for
5.3 Conformal Mappings
Solution to 53.1: We will show that the given transformations also map straightlines into circles or straight lines.
z i-+ z + b and z i-÷ kz clearly map circles and straight lines into circles andstraight lines.
LetS=(zjIz—aI=r),ax0+jyo,andf(z)=1/z=wu+jvTheequation for S is
=r2or
2 2—lalWV) W W
• TSr = that is, when S contains the origin, we get
I
or
This is equivalent to1
uxo—vyo=
which represents a straight line.
• If r IaI, we obtain
a —1WW—I lw—I lw=\1a12_r21 1a12_r2
Letting
laP—r2we get
2
2 r2\IaI —r
which represents the circle centered at with radius r/(Ia j2 — r2).
If S is a straight line, then, for some real constants a, b, and c, we have, forZX+iy ES,
ax+by—c.
Letting a = a — ib, we get
or
and it follows, as above, that f(S) is a straight line or a circle.
Finally, letaz+b
f(z)=—+dIf C = 0 f is linear, so it is the sum of two functions that map circles and linesinto circles and lines, so f itself has that mapping property. If c 0, we have
az+bl( ad—bccz+d
so 1(z) = 13 (12 (fI (z))) where
1 a ad—bcf'(z) = cz + d, f2(z) = —, 13(z) =
—z,
each of which has the desired property, and so does f.
Solution to 5.3.2: 1. The locus Iz+I1 = Jz —fl is the perpendicular bisector of theline segment joining —1 to 1, that is, the imaginary axis. The set Iz +11 <Iz —is then the set of points z closer to —I than to 1, that is, the left half-plane <0.Hence, IJtZ <0 if f : z -÷ w = maps (z I Wtz <0)
onto f w lwl < 1), and is conformal as f' 0. The inverse map is easily seentobewi—÷ z=2. The map 1(z) = iz has the specified properties.3. The map z w = z3 is conformal (when z 0), and sends { z 0< argonto (w 0 < arg w < f-). The inverse is the map w i-÷ z = =
(log Iwl + i arg w)) where arg w is chosen in 0 <arg w <
Solutionto5.33:LetA=(zlIzI <1, lz—l/4I> l/4),B=fzlr < IzI <I).Let 1(z) = (z — a)/(az — 1) be a linear fractional transformation mapping Aonto B, where —1 <a < 1. We have
f(fzllz—I/41= l/4})=(zIIzI=r}so
{f(0),f(l/2)) = (—r,r)
and
0=r—r=f(0)+f(I/2)a+ 1/2—aa/2— 1
which implies a = 2— Therefore, r = 11(0)1 = 2 —Suppose now that g is a linear fractional transformation mapping
C = (z {s < Izi < 1) onto A. Then is a straight line through the on-gin, because the real line is orthogonal to the circles (z
I lz — 1/41 = 1/4) and(z I IzI = 1). Multiplyingby a unimodularconstant, we may assume = IR.Then f o g(C) = A and f o g(R) = IR. Replacing, if necessary, g(z) by g(s/z),we may suppose f o g((z I Izi I)) = {z I Izi 1), so
z—awith IaI<lflI=1.az—I
Using the relation 0 = f(s) + f(—s), we get cx = 0, so f o g(z) = f3z and
Solution to 5.3.4: Suppose f is such a function. Let g : A —÷ B be definedby g(z) = f(z)2/z. Then, as on C1 U C4, the absolute value of ,g is 1, then gis a constant, c, say. Therefore, f(z) = which is not continuous on A. Weconclude that no such function can exist.
Solution to 5.3.5: The map z i-+ iz maps the given region conformally ontoA > 0J.Themap
1+wwF-÷
1—w
maps A onto the first quadrant, Q. The square function takes Q onto (z I > 0).Finally,
takes > 0) onto ID). Combining these, we get for the requested map:
(1 + iz)2 — i(1 — iz)2
(1 +iz)2 +i(1 —iz)2
Solution to 5.3.6: The map çoi (z) = 2z — 1 maps conformally the semidisc
1 1
onto the upper half of the unit disc. The map
1+zco2(z)=
1—z
maps the unit disc confonnally onto the right half-plane. Letting z = re'0, itbecomes
1 + rë0 — 1 — r2 + 2ir sin 01 — — 1 + re10 12
Since sinG > 0 for 0 < 0 < co2 maps the upper half of D onto the upper-right quadrant. The map = maps the upper-right quadrant conformallyonto the upper half-plane. The composition of qi, c02, and is the desired map,
namely the function z i-÷
Solution to 5.3.7: Let R be the given domain. The map
I
z
transforms R onto the vertical strip S = E C <1). Now
maps S onto the horizontal strip T = Di E C I0 <it). Finally
maps T onto the upper half-plane U = {p E C I> 0). Putting everything
together we get the conformal mapI I
z F-+
Solution 2. Let C1, C2 denote the circles Iz — ii = 1, Iz — Irespectively.
Consider the map z t-+ = l/z. Ci contains the points 0, 1 + i, 2 which are sentto 00, — respectively. Thus C1 is sent to the line = and the inside of
Ci is sent to (consider the image of 1). C2 contains the points 0, + 1
which are sent to 00, 1 — i, 1; thus C2 is sent to the line = 1 and the outside ofC2 is sent to < I (consider the image of 2). Thus the lune is sent to the strip
< 1. We follow with the map i-÷ w = which sends thisstrip to the upper half-plane. Composing, we obtain the map
f 2,ri
which is a conformal map from (z E C I Iz — ii 1) fl( z E C I Iz — }
onto the upper half-plane.
Solution to 5.3.8: Suppose fis such a map. f is bounded, so the singularity at theorigin is removable, p = 1(z). Since f is continuous, p is in the closureofA.
Suppose that p is on the boundary of A. Then f (G) = A U (p), which is notan open set, contradicting the Open Mapping Theorem, [MH87, p. 436].
Let p E A and a E G be such that f(a) = p. Take disjoint open neighborhoodsU of 0 and V of a. By the Open Mapping Theorem, f(U) and f(V) are open setscontaining p. Then f(U)fl 1(V) is a nonempty open set. Take x E f(U)fl f(V),x p. Then x = 1(z) for some nonzero z E U and x = f(w) for some w E V.Then z and w are distinct elements of G with 1(z) 1(w), contradicting theinjectivity of f.
Solution to 53.9: Let w be a primitive nih root of unity. The transformations1/Ij(Z) = w3z, j 0,... ,n — 1, formacyclic groupofordern and fixthepoints
= x oi,1r3 j = 0, ...,n— 1, thusformacyclicgroupof order n that fix the points 1 and —1. We have explicitly
— — (wi—1)z+cti+1 —
5.4 Functions on the Unit Disc
Solution to 5.4.1: The function f is continuous on Sa, and = Sa U (0) iscompact, so it will be enough to show that f has a continuous extension towill be shown that (z) -÷ 0 as z —÷0 from within We have
10 1 —e10 \ f — cos 8 + i sin 8 1— cos 8I (r e ) I = exp ) = exp
r )= exp
Inside we have cos 8 cos a, so
f—coscñIzi )
when z —* 0, as desired.
Solution to 5.4.2: Let b/a = re113 and consider the function g defined byg(z) = f (z)a_'e"912. We have
g = += (1 + r) cos (8 — p8/2) + i (1 — r) sin (8 — p6/2)
so the image of the unit circle under g is the ellipse in standard position withaxes 1 + r and Ji — rJ. As 1(z) = a exp(i16/2)g(z), f maps the unit circle ontothe ellipse of axes a + r) and Ia(1 — r)(, rotated from the standard position
Solution to 5.43: We have
L
=
dO= j dO
= 2yr If'(°)I
by the Mean Value Property [MH87, p. 185].
Solution to 5.4.4: As the Jacobian of the transformation is f'(z)12, we have
A = j If'(z) 12 dx dy
f'(z) can be found by term by term differentiation:
f'(z) =
so
If'(z))2 =j,k=1
We then have
A= Jf j dxdy.
j,k=1
Letting z = re'0, we get
A
=fkcJ?kff d9dr.
Sincep2ir
IJo
for n 0, we have
A = 27r>Jfl2ICnl2f dr =
Solution to 5.4.5: We have, for z, w E D,
f(w)=f(z) 1ff
so f in injective. Then the area of its image is given by
ID If'(z)12 dxdy= f (i + + 1z12)
= f(i ± 2x + x2 + dxdy
=f f (1 + 2r cos 9 + r2)
Solution to 5.4.6: Fix E 11) and let h be defined by h(z) = 1(z) — f(zo).As h has only isolated zeros in D, we can find an increasing sequence p,-+ 1 withh(z) 0 for IzI = p,1 i = 1,.... Let g be the function given byg(z) = al (z — zo). For = pj, we have
Ig(z) — h(z)I = >Janz" —
max Iz — zol (w in the segment [z. zo])
IaiIIz—zol
=Ig(z)I.By Rouché's Theorem [MH87, p. 421], h has a unique zero in the disc{z I IzI <nj), so f assumes the value f(zo) only once there. Letting p1 —÷ 1, weget that f is injective in ID).
Solution to 5.4.7: Let * be a linear fractional transformation which maps D ontoitself with = Then the conjugateg = ifr1 of o ijr is analytic in the unitdisc f, and has two fixed points, namely 0= and w =
As g(0) = 0 we can define the analytic function h by h(z) = g(z)/z. For each0 <s < 1, on the circle = 1 — e, we have Ih(z)I = Is(z)/IzI 1/(1 — e), sothe maximum modulus principle implies Ih(z)I 1/(1 — s) on IzI 1 — s. As ecan be arbitrarily small, we have I for any z E D.
We know that h(w) = 1, so h attains its maximum inside 1]). By the MaximumModulus Principle h must be constant, that is, h is identically 1, and in this caseg(z) = f(z) = z.
Solution to 5.4.8: Let f Xk, z II), and y be the circle around z with radiusr = (1 — IzI)/2. y lies inside the unit disc, so, by Cauchy's Integral Formula forderivatives [MH87, p. 169], we have
1 1' If(w)I C 2CIf (z)I
J,, Iz — w12 r(1 — IzI)k = (1— IzI)k+1'
where C is a constant, so f' Xk÷I.Let f' Xk+1 with f(0) = 0 (the general case follows easily from this).
Letting z = re10, we have
jf(z)I j If'(w)I Idwi
= jrlf'(tei°)l dt
fo(1 _r)k+I dt
(1 _r)k
Hence, I Xk.
Solution to 5.4.9: Let 1(z) = for z E D = {z I Izi < 1].
As f(z) is analytic, so is g : -÷ C defined by g(z) = 1(z) — We have,for realz ED,
g(z) =
___
== f(z)—f(z)=0
therefore, g(z) 0 on D, so the coefficients are all real.
Put zo = and let a* be the nonzero coefficient of smallest index. Wehave
f(tzo) — ao k c' n n—k+1—Zn+ /" a'
tZO E D for t E [0, 1) and the left-hand side expression is a real number for all t,
limf(tzo) — = zko E R
aktk
which implies k =0, by the irrationality of thus f is a constant.
Solution to 5.4.10: Let be the Maclaurin series for f. Thenf'(z) = The Cauchy Inequalities [MH87, p. 170] give
I f f'(z)dz
2K1 JgzI=r Z
,2jr/ f'(re'0JO
0<r<1.
Lettingr—+
p 1'1
If(x)Idx I dxJfO,I) JO
1dx
Jo
Solution to 5.4.11: As Ih(0)I = 5, by the Maximum Modulus Principle [MH87p. 185], h is constant in the unit disc. Therefore, h'(0) = 0.
M
Solution to 5.4.12: For r = log 2,
z—rI — rz
maps the closed unit disc onto itself with T(0) = —r. Then g(z) = f(T(z)) isanalytic on the closed unit disc and g(0) = 0. For z on the unit circle we have(g(z)I I. Therefore h(z) = e_T(z)g(z) is analytic in the closed unit disc,h(0) = 0, and Ih(z)I I for all z with = 1. By Schwartz Lemma, [MH87,p. 190},Ih(z)I Izlforallzwithlzl l,hencelf(T(z))Iclosed unit disc. If w T(z) then z = —T(—w), giving If(z)I IezT(_z)I.Setting z = r = log 2, we get
r 2r 41og2=1+(log2)
This bound is attained by the function e1 +zlog2
Solution to 5.4.13:.11 +z
co(z) = z
—
maps the unit disc to the upper half-plane with = i. Thus, 10 çø maps the unitdisc into itself fixing 0. By the Schwarz Lemma [MH87, p. 190], Jf op(Z)J Izi.Solving go(z) = 2i, we get z = 1/3. Hence, If(2i)( 1/3. Letting f = wesee that this bound is sharp.
Solution to 5.4.14: The function
1+zço(z) = — z
maps the unit disc, ID), onto the right half-plane, with ço(O) 1. Therefore, thefunction f c ço maps 11) into itself, with f o ço(O) = 0. By the Schwarz Lemma{MH87, p. 190], we have 1(1 o ço)' (0)1 1, which gives 1
and If'(l)I 1/2. A calculation shows that equality happens for f =
Solution to 5.4.15: Suppose f has infinitely many zeros in D. If they have a clus-ter point in D, then f 0 and the result is trivial. Otherwise, since(z C 1) is compact, there is a sequence of zeros converging to a point inthe boundary of 11), and the conclusion follows.
Assume now that f has only finitely many zeros in ID), Wi, ..., Wm. Then fcan be written as
1(z) = (z — (z —
where g is analytic and never zero on II). Applying the Maximum Modulus Princi-ple [MH87, p. 185], we get that 1/g attains a maximum in the disc Izi (1—1/n)
at a point Zn with = 1 — 1/n (n 2). Then Ig(z2)I .... The
product (z — wi)a1 • (z — is clearly bounded, and so is
Solution to 5.4.16: 1. We can assume I has only finitely many zeros. (Otherwise,assuming f 0, its zero sequence has the required property, since the zeros ofa nonconstant analytic function in an open connected set can cluster only on theboundary of the set.) That done, we can, after replacing f by its quotient witha suitable polynomial, assume f has no zeros. Then 1/f is analytic in the disc.For n = 1, 2, ..., let be the maximum of 1/1(z) 1 for
I z = 1 — By the
Maximum Modulus Principle [MH87, p. 185], Mj for all n. Hence, foreach n, there is a such that IanI = 1 — and If(an)1 = 1/Ma 1/Mi =11(0)1. Then is a bounded sequence of complex numbers and so has aconvergent subsequence, which gives the desired conclusion.2. Let (Zn) be a sequence with the properties given in Part 1. Subtracting a constantfrom f, if needed, we can assume urn = 0. We can suppose also that
> > 0 for all n. For each n, let be the maximum of If(z)I forIzi = IZnI. The numbers are positive (since f is nonconstant) and increasewith n (by the Maximum Modulus Principle [MH87, p. 185]). Since -÷ 0,there is an such that < M1 for n no. For such n, the restriction ofIf I to the circle Izi IZnl is a continuous function that takes values both largerthan M1 and smaller than M1. By the Intermediate Value Theorem [Rud87, p.93], there is for each n a point such that = Iz,d and = M1.Then, for the desired sequence (wa), we can take any subsequence of (ba) alongwhich f converges. (There will be such a subsequence by the boundedness of thesequence
Solution to 5.4.17: Suppose f(a) = a E 11), f(b) = b E IL), and a b. Let: D -+ II) be the automorphism of the unit disc that maps 0 to a, that is,
= (a — z)/(1 — az). Then the function g = a f o maps 11) intoitself with g (0) = 0 and g '(b)) = 1(b). Since ço is one-to-one and ab, (b) 0. Hence, by the Schwarz Lemma [MH87, p. 190], there exists aunimodular constant A such that g(z) = Az, and letting z = (b), we see that
= 1; that is, g is the identity map and so is f.
Solution to 5.4.18: Using Cauchy's Integral Formula [MH87, p. 167], for0<r< l,wehave
1 f If(w)IIdwi
1ldwI=
1 —.
2ir JIwI_—r (1 — (1 —
Letting r = n/(n + 1), we get (n + 1)(1 + 1/n)I? <(n + 1)e.
Solution to 5.4.19: By the Schwarz Lemma [MH87, p. 190],
If(z)I Izl.
Iff(z)=alz+a2z2+...,letgbedefinedby
= 1(z) +f(—z) = + a4z3 + a6z5 +
g is analytic in D, and since
If(z)I + If(—z)IIg(z)I 1
g maps 11) into D. Hence, by the Schwarz Lemma,
Ig(z)t Izi
or11(z) + f(—z)I
Now suppose equality held for zo E D. We would have Ig(zo)I= Izol so, by theSchwarz Lemma,
g(z) = Az
for some unimodular A, or
1(z) + f(—z) = 2Az2.
Plugging this back into the power series for g(z), we get = A and a4 = ==0. Hence,
1(z) = Az2 + h(z)
where h(z) is odd. We have
1 Jf(z)I = IAz2 + h(z)I
and1 ) If(—z)I = IAz2 + h(—z)I = IAz2 — h(z)I.
Therefore,
(Az2 + h(z))(Az2 + h(z)) 1
(Az2 — h(z))(Az2 — h(z)) 1.
Expanding and adding, we get
1z14+Ih(z)12 1
Jh(z)12 1—1z14
which, by the Maximum Modulus Principle [MH87, p. 185], implies
Solution to 5.4.20: Schwarz's Lemma [MH87, p. 190] implies that the functionfi(z) = f(z)/z satisfies hi (z)1 1. The linear fractional map z '-÷ sends
the unit disc onto itself. Applying Schwarz's Lemma to the function
12(z) = we conclude that the function 13(z) = fi(z)/satisfies I
1. Similarly, the map z sends the unit disc onto
itself, and Schwarz's Lemma applied to the function 14(z) = f3(z)/
implies that the function (z) = 13 (a,) satisfies 115 (z) I 1. All
together, then,
z—r z+r z—r z+r1—rz 1+rz 1—rz i+rz
which is the desired inequality.
Solution to 5.4.21: Let be the automorphism of the unit disc given by
Z — ZO
we have— 1—fzol2
(1
Now consider the composition
g(z) = 010 = 010
then g(0) = 0 and as composition of maps of the unit disc into itself, we canapply the Schwarz Lemma [MH87, p. 190] to obtain Ig'(O)I 1. Computingg'(O) using the chain rule, we have
= 1— If(—zo)12(1— 1z012J 1
so we can conclude that
1—lf(z)12 1If(z)k 2 2l—IzI 1—lzI
The first inequality is known as Picks' Lemma and is the main ingredient in theproof that an analytic map of the disc that preserves the hyperbolic distance be-tween any two points, preserves all distances, for more detail see [Car6O, Vol. 2,§290] or [Kra9O, p. 16].
Solution 2. Using the same notation as above,
1 Ifocozo(w)I1(10 qzo)'(O)I — I 2 I'!CUI
21t JIoiI=r 1(01
that is2 1 f2JZdO 1
If(—zo)I l—IzoI ;
which holds for any IzoI <r = IwI <1, so the conclusion follows.
Solution to 5.4.22: Fix z E 11) and let y be the circle centered at z parametrizedby
w = z +
which is completely contained inside the unit disc. Using Cauchy's Integral For-mula for derivatives [MH87, p. 169], we have
I f f(w)f(z)= .1 dw.2iri (w—z)
taking absolute values and applying the inequality we get
[ C2IdwI.2irj, I1—wIIw—zI
From the parametrization we cann see that Idwi = IzildO, and obtain
l—zI 2d0.
IT j0 (1 — IwD(1 — JzI)
Now I — IwI is minimum at the point of y farthest from the origin, at which wehave 1 — IwI = (1 — IzI)/2, therefore,
/ C f2n 2(1 — IzI) 4CI (z)I<—, d6=
n Jo (I — IzI)(I — IzI) (I — IzI)
5.5 Growth Conditions
Solution to 5.5.1: Consider the function g(z) = it is entire with Ig(z)I =Liouville's Theorem implies that g is constant, and since it is
obviously a nonzero contant, f maps the connected set C into the discrete set ofthe logarithms, hence f itself is constant.
Solution to 5.5.2: Let g(z) = 1(z) — f(O). Then g(O) = 0, so g(z)/z has aremovable singularity at 0 and extends to an entire function. g(z)/z tends to 0as IzI tends to infinity since f(z)/z does. Let e > 0. There is an R > 0 suchthat Ig(z)/zI <e for Izi R. By the Maximum Modulus Principle [MH87, p.185], ig(z)/zI <e for all z. Since s is arbitrary, g(z)/z is identically 0. Hence,g(z) = 0 for all z and f is constant.
Solution to 5.5.3: Let f = p/q where p, q are polynomials with q zero freein the upper half plane. For positive R consider the curve FR = [—R, R] U CR,
where CR is the upper semicircle centered at the origin, with radius R. By theMaximum Modulus Principle [MH87, p. 185], we have
sup{If(z)I I0) = lim sup{If(z)I I z
R-÷oo
Let's call this value M. If deg q > deg p. we have If(z)I = 0 so the
result is clear. If deg q p. then If(z)I = M and the result follows
also.
Solution 2. Case 1. f has a pole at oo. Then 1(z) = c'o,so both the supremaequal oo.
Case 2. f does not have a pole at oo. Then f has a finite limit at oo, so I extends
continuously to the closure of H = {z E C ) > 0) in the extended plane. Since
that closure is compact, Ill attains a maximum there. By the Maximum ModulusTheorem,
Icannot attain a local maximum in H unless f is constant. Thus, the
maximum of If I on the extended closure of H is attained at a point of the realaxis or at 00. In either case the desired equality holds.
Solution to 5.5.4: The function g(z) 1(z)! sin z is analytic in the open con-nected set G = C \ ,rZ. Write g = u + iv. By hypothesis, u2 + v2 = 1 in G.Taking partial derivatives, we have
uux+vvx=0, uuy+vvy=0.
Using the Cauchy—Riemann equations we get
— VUy = 0, UUy +VUx = 0.
G. It follows that = 0 = Vy on G. By the Mean Value Theorem, both u andv are constant on every line segment lying in G parallel to the real or imaginaryaxis. G is step connected: that is, given any two points Zi in G there is a pathin G from zo to Zi, consisting of line segments parallel to the axes. Thus u and vare constant in G, whence g = C in G, for some constant C, with Id = = 1.That is, 1(z) = C sin z in G. By continuity this equality holds in C.
Solution to 5.5.5: If g 0, the result is trivially true. Otherwise, the zeros of gare isolated points. If/SI is bounded by 1 in C , so all the singularities of f/g areremovable, and f/g can be extended to an entire function. Liouville's Theorem[MH87, p. 170] now guarantees that f/g must be a constant.
Solution to 5.5.6: Since p and q are of the same degree, the ratio p/q has anonzero finite limit at oo. The function 1(z) = p(l/z)/q(1/z) thus has a remov-able singularity at 0, as does 1/f. Moreover f and 1/f are both analytic in aneighborhood of the disk IzI I (since p and q are without zeros in Izi 1) andf and I/f have unit modulus on Izi = I. By the Maximum Modulus Principle[MH87, p. 185], If(z)I ( 1 and I1/f(z)I 1 for IzI I. So If(z)I = 1 and for
Izi 1, and another application of the Maximum Modulus Principle shows thatf = constant, the desired conclusion.
Solution to 5.5.7: Let g be defined on C by g(z) = f(z)g is an entire
function such that, on the circle of radius R,Z
<Jg(z)I-...
R R
where the inequality for z = —R follows by continuity. Using the MaximumModulus Principle [MH87, p. 185], we get, for Izi ( R,
R + fl2 +11(0)1Ig(z)I
R
Taking the limit R -+ 00 we get g 0 so f is a constant function. As11(1)1 I log II we conclude that f vanishes identically.
Solution 2. Since f is entire, its Maclaurin series converges everywhere,1(z) = Zn. Using Cauchy's Integral Formula for derivatives [MH87,p. 169], we have
— ' 1— 2iri ICR
where CR is the circle of radius R centered at the origin, positively oriented. Usingthe given inequality we get, for n 1,
I IogR logR—2irR = =o(R) (R —÷ oo).2ir R
Therefore f equals a constant, ao. As in the previous solution, we must haveao=0.
Solution to 5.5.8: Let h(z) = 1(z) — kg(z). Then h is entire and 0. Wethen have
1 for all z C
therefore, by Liouville's Theorem [MH87, p. 170], eh is constant, and so is h.
Solution to 5.5.9: Suppose the maximum of over the compact set K is attainedat a point in the interior of K. Let r be the distance of zo from the complementof K. The circle Iz — zol = r is then in K, so cp(z) co(zo) on that circle, giving
r2ir
— I ço(zo + re'0 )dO.
Hence equality holds by the property. Thus
I (q(zo) — w(zo + re'0))dO = 0.Jo
The integrand here is nonnegative, so, being continuous, it must vanish identi-
cally. Thus ço(z) = co(zo) for lz — zol = r. Since r is the distance from zo tothe complement of K, the circle lz — zol r contains at least one point of theboundary of K. Hence attains its maximum over K on the boundary of K.
Solution to 53.10: 1. Using Cauchy's Integral Formula for derivatives [MH87,
p. 169], we get
f ldzI2ir IzI=R Z
k!k If2irR + IzI=R
—
=o(1) (R-÷oo)f(k)(0) =0 fork 1, and f reduces to a constant, f(0).
2. Using the same method as above for k 3, we get
(0)1 f Idz I
IzI=R Z
k!2itRk+ IzI=R
k!(avIi+b)= Rk
=o(1) (R-÷oo)
f(k)(0) = 0 for k 3 and f reduces to a polynomial of degree, at most, 2,f(0) + f'(O)z + f"(0)z2/2.
Solution to 5.5.11: For r > 0, let z = re'9 in Cauchy's Integral Formula forderivatives [MH87, p. 169] to get
f(n)(Ø) = fl• ,dO.2iri j0
Combining this with the inequality given yields
<2,r
For n > 5, letting r tend to infinity, we get =0. If n 5, letting r tend to0 gives the same result. Hence, the coefficients of the Maclaurin series [MH87,p. 234] off areall 0,sof
Solution to 5.5.12: If such a function f exists then g = 1/f is also analytic onC \ (0), and satisfies Ig(z)I Asg is bounded on (z 10 < Izi < 1), g has a
removable singularity at 0, and extends as an analytic function over the complexplane. Fix z, choose R> Izi, and let CR be the circle with center 0 and radius R.Then
g'(z) = 1
fg(w)
2dw2iri CR(WZ)
soI 2—*0asR-÷oo.
2ir (R — IzI)
Thus, g' = 0 everywhere, so g (and, hence, f) is constant. But this contradictsthe hypothesis If(z)I for small z, so no such function exists.
Solution to 5.5.13: By Liouville's Theorem [MH87, p. 170], it will be enough toprove that f is bounded. For 1/2, we have If(z)I Let zo be a pointsuch that < 1/2. Let S be the square with vertices ± 1 ± i, orientedcounterclockwise.
S
zo
Then zo is in the interior of S. so Cauchy's Integral Formula [MHS7, p. 167]gives 1(1(z)
f(zo) = . 1 dz.2iri js z — zo
The absolute value of the integrand is, at most, The contribution to theintegral from each vertical edge is thus, at most, 4 in absolute value. The contribu-
tion from each horizontal edge is, at most, 2 1x1"2dx = 8 in absolute value.Hence,
If(zo)I I2ir
proving that f is bounded.
Solution to 5.514: Since f is nonconstant its zero at the origin is isolated. Hencethere is a circle C with center at the origin on and within which f does not vanish,
except at 0. Let e be the minimum of Jf I on C. Then the set {z I If(z)I < s)intersects the interior of C but is disjoint from C. Since it is connected it mustbe contained in the interior of C. It follows that the origin is the only zero of fand that f is bounded away from 0 near oo. The latter conclusion implies that
00 is not an essential singularity of f (Casorati—Weierstrass Theorem [MH87, p.256)), so oo must be a pole of f (it is not a removable singularity, by Liouville'sTheorem [MH87, p. 170]). Therefore f is a polynomial. Since its only zero is at
the origin, it is a constant times a positive power of z.
5.6 Analytic and Meromorphic Functions
Solution to 5.6.1: By the Cauchy—Riemann equations [MH87, p. 72],
and Uy=Vx.
Thus, au + by = c implies
and, therefore,
In matrix form, this reads
(a
Since the matrix has nonzero determinant a2 + b2, the homogeneous system hasonly the zero solution. Hence, = Uy = 0. By the Cauchy—Riemann equations,v, = = 0. Since D is connected, f is constant.
Solution to 5.6.2: 1(z) = is a counterexample. Define it by making a cut onthe negative real axis and choosing an associated branch of the logarithm:
f(z)=110 z=0.
f is analytic in the right half-plane and so on the disc Iz —1 I <1. Since tendsto 0 as z tends to 0, f is continuous on the disc Iz — ( 1. However, f cannotbe analytic on any open disc of radius larger than 1. For if it were, f would beanalytic at 0, so
f'(O) = lim 1(z)z-+O z
would exist and be finite, which is absurd.
Solution to 5.6.3: 1. 1(z) = z2 is entire and satisfies
1(1/n) = f(—1/n) =
2. By the Identity Theorem [MF187, p. 397], in a disc centered at the origin, gwould have to be z3 and —z3, which is not possible; therefore, no such function gcan exist.
Solution to 5.6.4: Let be the power series of such an f at the origin.
ThenCO = f(O) = (f(0))k =
SO cç is either 0 or a (k 1):h root of unity. Assume co 0 but that f is notconstant. j be the smallest positive integer such that Cf 0. Then f(z1') =co + cjzik+ higher order terms, so (f(Z))k CO + higher order terms,which gives a contradiction. Hence, if f(0) 0, then f is constant and equal toa (k — 1 root of unity.
Assume f(O) = 0 but f is not identically 0. Let j, as above, be the smallestpositive integer such that Cj 0. We then have f (z) = z3 g(z), where g is entireand g(0) = The function z z3 satisfies the condition imposed on f, SO galso satisfies that condition. By the preceding argument, g is constant, equal to a(k — 1 )th root of unity.
Then f is either the zero function or 1(z) = where c is a (k — I root ofunity and j is a nonnegative integer.
Solution to 5.6.5: We first prove existence. Multiplying the rational function Rby a suitably high power of F, we can reduce the general case to the case where Ris a polynomial. Assuming R is a polynomial, we can use the division algorithmto write R = ao + R1 F where ao and R1 are polynomials and deg ao < d.Necessarily, deg Ri <deg R. If deg R1 d we can use the division algorithmagain to write R1 = aj + R2 F where deg ai < d and deg R2 < deg R1. Theprocedure can be iterated and with each iteration the degree of the polynomialmultiplying F is reduced. After finitely many iterations we arrive at the desiredrepresentation.
To prove uniqueness it will suffice to show that if
= 0 (*)
then every ak is 0. Assume not. After multiplying the expression by a suitablepower of F, we can reduce to the case where m = 0 and ao 0. Then the
ak Fk 0, and (*) implies that F divides ao, a contradiction becausedegao <d=degF.
Solution to 5.6.6: Let f be such a function. For n = 2,3,... the singularityat 1/n is removable because f is bounded in a punctured neighborhood of 1/n.
Therefore f extends holomorphically to 11) \ (0). Then the origin becomes anisolated singularity of the extended f, so, by the same reasoning, f extends holo-morphically to 11).
Solutionto5.6.7:NoLetU =C \(z E IRkon U with f(l) = 1, and let a = —l + i. Then the Taylor series for f at a hasradius of convergence since there is a holomorphic branch of defined on
D = (z IIz — a I <,h) that agrees with f in a neighborhood of a = —1 + i.
But f cannot extend to a holomorphic function on U U D, because —l E D,
and the limit of 1(z) as z approaches —1 along the unit circle does not exist:
urn f(eit) = urn cIt12 = i and urn f(e') = urn —i
Solution to 5.6.8: 1. Write f = u + iv. Then the Cauchy—Riemann equationshold in U. By hypothesis, v = 0 and hence = Uy = 0. The open connectedset U is step-path connected in that given Z2 in U there is a path consisting ofline segments parallel to the axes contained in U from to Z2. The Mean ValueTheorem applied to u then shows that u(zi) = u(z2), and thus f = u + iv isconstant.2. Write A = (z) (z) I z E W} C IR, and let h : C -+ R; w i-+Then A = h(g(W)). Fix x = h(g(z)) A. The hypothesis implies that g isnonconstant. By the Open Mapping Theorem (MH87, p. 436] g(z) is an interiorpoint of g(W) and thus g(W) contains some disc centered at g(z) with radiusr > 0. It is clear that h(g(W)) then contains the open interval (x — r, x + r),proving that A is an open set.
Solution to 5.6.9: Suppose f(C) is not dense. Then, for some w E C and s> 0,we have 11(z) — wI s for all z E C. The function l/(f(z) — w) is then entireand bounded in modulus by 1/s, so, by Liouville's Theorem [MH87, p. 170], isa constant, and so is 1.
Solution to 5.6.10: We may assume that 0 L, otherwise we use f + c insteadof f, where c is a constant. We may even assume that L is the imaginary axis,otherwise we use af in the place of f, where tx is a constant of modulus 1. Since1(C) is connected, it is contained in one of the halfplanes 91z > 0, < 0.We may assume that 1(C) is contained in the left halfplane (otherwise we use—f instead of f). Then g is an entire bounded function, where g(z) = ByLiouville's Theorem, g is constant, g(z) = k, say. Therefore, f(C) is containedin the solution set of ez = k, which is discrete. As f(C) is connected we concludethat it must contain only a point, so f is constant as well.Solution 2. By Picard's Theorem the image of an nonconstant entire functionomits at most one complex number, [Ah179, Sec. 8.3] and [Car63b, Sec 4.4], sothe result follows. See the previous problem.
Solution to 5.6.11:1. 1ff is entire and > 0, then 1/(1 +1) is bounded andentire, hence constant by Liouville's Theorem [M1187, p. 170].
2. Assume F(z) + F(Z)* is positive definite for each z. Since the diagonal entriesof a positive definite matrix are positive, it follows that and f22 have positivereal parts. Hence fj i and f22 are constant by Part 1. Also, the determinant of apositive definite matrix is positive, so
111122 —1112 o.
The function 112 + is thus bounded. It follows that the real part of the entirefunction 112+121 is bounded, hence 112+121 is constant, by Part 1. Similarly, thereal part of the entire function i(f12 — 121) is bounded, so 112 — 121 is constant,by Part 1, therefore 112 and 121 are constant.
Solution to 5.6.12: Let f(x + iy) = u(x, y) + iv(x, y), where u(x, y) = exs(y)
and v(x, y) = ?t(y). From the Cauchy—Riemann equations [MH87, p. 72], weget = so s(y) = t'(y). Similarly, s'(y) = —1(y). This equationhas the unique solution s(y) = cosy satisfying the initial conditions s(O) =I ands' (0) = —t (0) =0, which, in turn, implies that t(y) = —s'(y) = sin y.
Solution to 5.6.13: 1" + f is analytic on 11) and vanishes on X = (I/n I n 0),so it vanishes identically. Using the Maclaurin expansion [MH87, p. 234] ofwe get
k — —f(k+2)(Ø)
L_i k!Z L..# k!
So we havef(0) = f"(O) = = (_1)kf(2k)(Ø) =
andf'(O) = —f"(O) = ... = (_1)kf(2k+1)(0) =
Therefore,
(_1)k1(z) = f(0) + 1)
,Z +
= f(0) cos z + f'(O) sin z.
Conversely, any linear combination of cos z and sin z satisfies the given equation,
so these are all such functions.
Solution to 5.6.14: Let a be the intersection of the lines. Replacing 1(z) byf(z + a) — f(a), we may reduce to the case a = f(a) = 0. Then 1(z) =
+ o(l)) as z 0, for some nonzero c E C and n ? 1. In particular,
1(z) is nonvanishing in some punctured neighborhood of 0. If the lines are in the
directions of e'dr and then for real r, and are real, and so is
li — urntrz (I + o(I)) =
— + o(1))
This forces ni(cx — /3) E iriZ, so —/3)/7r is rational, as desired.
Solution to 5.6.15: It is enough to show that for any z E the derivative in the
sense of 1R2 has an associated matrix
Df(z)=(
satisfying a = d and c = —b. As Df(z)(1, 0) 1. Df(z)(0, 1), we have c = —kb
and d = ka for some k. As f preserves orientation, det Df(z) > 0, so k > 0.If a2 + b2 =0, then Df(z) =0 and there is nothing to show.Assume a2 + b2 0. As, for (x, y) 0, Df(z)(x, y) I Df(z)(—y, x), we
have 0 = (k2 — 1)(a2 + b2)xy. Therefore, k = I and the result follows.
Solution to 5.6.16: We have f = f3/f2 = g/h. It is clear that f3 and f2 havethe same zero set. If zo is a common zero, there are analytic functions gi and hiwhich are not zero at zo such that 13(z) = (z — zoYhi (z) and
f2(z) = (z — But (f3)2 = (f2)3, (z — =(z — Rearranging, we get = (z — Neitherh1 nor gi are zero at zo, so the left side is analytic and nonzero at Hence, wemust have 3j — 2k = 0, so k> j. Therefore, zo has a higher multiplicity as azero of f3 than it does as a zero of f2• Thus, the function has a removablesingularity at zo. Since this holds for every zero, f = f3/f2 can be extended toan analytic function on 11).
Solution to 5.6.17: Let g = f2 and let the common domain of f and g be G. Wewill show that g(G) contains no path with winding number 1 about 0. Supposethat for some path y: [0, 1] —÷ G, g(y) had winding number 1 about 0. Sinceg(y) is compact, there is a finite cover of it by n open, overlapping balls, noneof which contain 0. In the first ball, define the function h1 (z) = wherethe branch of the square root is chosen so that hi(y(0)) = f(y(0)). In eachsuccessive ball, define the function hk (z) = with the branch chosen so thathk is an analytic continuation of hk_1. This implies that if y(t) is in the domainhk, we must have hkQ'(t)) = f(y(t)). However, since these analytic extensionswrap around the origin, = hiQ'(O)), which contradicts thecontinuity of f.
Therefore, since g(G) contains no path with winding number 1 about the ori-gin, there exists a branch of the square root on it which is analytic and such thatf(z) = for all z in G. Thus, f is an analytic function.
Solution to 5.6.18: 1. Let w be an primitive root of unity. Then the function0 k n — I are all roots of 1' analytic and distinct.
rootoff.Fixzo E G,s> Osuchthatf(z) Ofor— zol <S. h/g is continuous in — zol <s and since h/g has
its range among the n points 1, w,..., Since it is continuous, it must beconstant. Therefore, h = for some k in this little neighborhood, so h
2. The function f : [0, 1] —÷ R defined by f(x) = (x _1/2)2 has four continuoussquare roots, 12, 13, and given by
fi(x)=x—1/2 f2(x)=—fI(x)
Ix—1/2 forf3(x)
= 1/2 — x for 1/2 x i= —f3(x).
Solution to 5.6.19: 1. Since 1(z) 0 for all z E D, is a holomorphic
(. 4'I(\function on D. For z E D set h(z) = I
' ' / dz, where [0, z] stands for theJ[O,z] 1(z)
onented segment from the origin to z. Using the independence of path, which is
a consequence of Cauchy's Theorem [MH87, p. 152], we get h'(z) = ' ' / for1(z)
all z 11), so 1(z) has derivative (f'(z) — h'(z)f(z)) = 0, therefore= f(0) for all z E 11). Since f(0) 0 there exists c C such that
e" = f(0). Set g(z) = h(z) + C; then = 1(z) for all z2. No. Consider D = C \ (0) and 1(z) = z. Suppose z = e8(Z) with g holomor-
fuzphic on D. Then g (z) = l/z, so, for any closed curve in D, — would vanish,Jczwhich is absurd since I = 2iri if C is the unit circle.
JcZSolution to 5.6.20: The function g(z) = I is analytic in the same region asf, and f — g = 0 on (1, oo). Since the zero set of f — g has limit points in theregion IzI > 1, the Identity Theorem [MH87, p. 397] implies that f — g 0.Hence, 1(z) f(i). In particular, forx in (—co, —1), f(x) = f(x).Solution 2. Let be the Laurent expansion [MH87, p. 246] of f aboutoo. It will suffice to show that is real for all n. The series con-verges everywhere the original series does (since its terms are dominated in ab-solute value by those of the original series); let g(z) = For x in(1,oo),
As above, the Identity Theorem [MH87, p. 397] implies g = f, so each is real,as desired.
Solution to 5.6.21: The function g(z) = f is analytic and coincides with fon the real axis; therefore, it equals 1. The line in question is its own reflectionwith respect to the real axis. Since it also passes through the origin, it must be oneof the axes.
Solution to 5.6.22: The function is also entire, and it agrees with f on thereal axis. By the Identity Theorem [MH87, p. 397], it equals f everywhere. Inparticular, then, f(x — in) = f(x + in) for all real x. On the line = —in
the function f(z + 27ri) — 1(z) vanishes identically, so it is identically 0, by theIdentity Theorem.
Solution to 5.6.23: By the Schwarz Reflection Principle for circles [BN82, p.85], we have
______
f(z) =
Forx real, we getf(x) = f(l/x) = f(x)
so f(x) is real.
Solution to 5.6.24: 1. Let zi, Z2, ..., be the zeros of p, ennumerated withmultiplicities, so that
p(z)= c(z
where c is a constant. Then
p'(z) 1 1 1
= + +...+p(z) z—zi z—z2 z—zn
and, for x real,
p'(z) —
p(z) Ix — zi I Ix — Z21 Ix —
Part 1 is now obvious.= Xj +Yj, so that
—fdx
Ix —
=
—00 (x
=1 —Iarctan—Idxj0dx\ yjJ
Hence,p'(x)
J dx=irn=irdegp.p(x)
Solution to 5.6.25: Let D be an open disc with D C G. It will suffice to showthat there is an n such that has infinitely many zeros in D. For then, the zerosof will have a limit point in G, forcing to vanish identically in G by theIdentity Theorem [MH87, p. 397], and it follows that I is a polynomial of degree,atmost,n — 1.
By hypothesis, D is the union of the sets = (z E D I (z) = 0) forn 1,2 Since D is uncountable, at least one is, in fact, uncountable(because a countable union of finite sets is, at most, countable).
Solution to 5.6.26: Let u = and v = For x real, we have
f'(x) = =
where the first equality holds because v = 0 on the real axis and the second onefollows from the Cauchy—Riemann equations [MH87, p. 72]. Since v is positivein the upper half-plane, 0 on the real axis. It remains to show that f' doesnot vanish on the real axis.
It suffices to show that f'(O) 0. In the contrary case, since f is nonconstant,we have
1(z) = (1 + O(z)) (z —+ 0)
where c 0 is real and k 2. For small z, the argument of the factor 1 + 0(z)lies between and say, whereas on any half-circle in the upper half-planecentered at 0, the factor assumes all possible arguments. On a sufficientlysmall such half-circle, therefore, the product will assume arguments between irand Zir, contrary to the assumption that > 0 for > 0. This provesf'(O) 0.
Solution to 5.6.27: 1. The function f is not constant, because if f took theconstant value c, then ((c)) would equal U, a noncompact set. Since f isholomorphic and nonconstant, it is an open map, and f(U) is open. Since Vis connected, it only remains to show that f(U) is closed relative to V. Leta V fl f(U). There is a sequence (Wa) in f(U) such that a = Foreach n, there is a point Zn in U with = The set K = (a, Wi, W2,...) isa compact subset of V. so (K) is also compact. Since the sequence (Zn) liesin a compact subset of U, it has a subsequence, (Znk), converging to a point I' ofU. Then f(b) = lim f(Zflk) = lim Wnk = a, proving that a is in f(U) and hencethat f(U) is closed relative to V.2.TakeU=V=C andf(z)=IzI.
Solution to 5.6.28: By the Inverse Function Theorem [Rud87, p. 221], it willsuffice to prove that Jh(0) 0, where Jh denotes the Jacobian of h:
/Jh=detl
\By the Cauchy—Riemann equations [MH87, p. 72],
du dv du — dv — dq ——
— dx dx — dy — dx
Hence,
II ax ' ax ax ax
Jh=detl' ax ax ax ax
(0u\2 (0p\2 (0v\2 (0q'\2- + vax)= — 1g12.
Since Ig'(O)I < If'(O)I, it follows that (Jh)(O) 0, as desired.
Solution to 5.6.29: f does not have a removable singularity at oo. If f had anessential singularity at infinity, for any w E C there would exist a sequence
—+ oo with lim f has a pole at infinity and is a polyno-mial.
Solution to 5.630: Clearly, entire functions of the form 1(Z) = az + b, a, b E Ca 0, are one-to-one maps of C onto C. We will show that these are all suchmaps by considering the kind of singularity such a map f has at oo. If it has aremovable singularity, then it is a bounded entire function, and, by Liouville'sTheorem [MH87, p. 170], a constant.
If it has an essential singularity, then, by the Casorati—Weierstrass Theorem[MH87, p. 256], it gets arbitrarily close to any complex number in any neighbor-hood of 00. But if we look at, say, f(0), we know that for some and 8, the discIZI <6 is mapped onto 11(0) — ZI <e by f. Hence, f is not injective.
Therefore, f has a pole at oo, so is a polynomial. But all polynomials of degree2 or more have more than one root, so are not injective.
Solution to 5.631: 1. If w is a period of f, an easy induction argument showsthat all integer multiples of w are periods of f. It is also clear that any linearcombination of periods of f, with integer coefficients, is a period of f.2. If f had infinitely many periods in a bounded region, by the Identity Theorem[MH87, p. 397], f would be constant.
Solution to 5.632: For 0 < r < ro, by the formula for Laurent coefficients[MH87, p. 246], we have
L=r If(z)I IdZI
,, J lf(re'°)IdOo
If n < —2, as r gets arbitrarily close to zero, this upper bound gets arbitrarilysmall. Hence, for n <—2, = 0.
Solution to 5.633: We think of the rational function f as a continuous map of C,
the Riemann sphere, into itself. The roots of the denominator of f arewhose real parts are positive. Hence f is holomorphic on the closed left half-plane. Let M = sup (If(z)I I 1Hz < 0). By the Maximum Modulus Principle[MH87, p. 185], there is no point z in the open left half-plane where If(z)I = M.Take a sequence in the open left half-plane such that -+ M.Passing to a subsequence if necessary, we can assume, by the compactness oft,that the sequence converges in C, to zo, say. Then, as f is continuous, we haveIf(zo)I = M; therefore zo is not in the open left half-plane. Hence, either zo = 00or zo is on the imaginary axis. We have f(oo) = 1. On the imaginary axis
If(iy)I = 1 + — ay2=
1 —;y—ay
Since numerator and denominator are complex conjugates of each other. HenceM = If(zo)I — 1, as desired.
Solution 2. The three facts below are verified by straightforward algebra.
•
u2+v2—1(u+1)2+v2
1
1 —z+azw—l z
w+1 = 1+az2
• Ifz=x+iy,then
( z \ — x(1 + ax2 + ay2)
Il+az212
Thus,whena <0.
Now let 1Hz <0. From the last statement, we have 1H <0 and from the
first two it follows that 11(z) I < 1.
00
Solution to 5.634: 1. The Maclaurin series for cos z is , and it(2n)!
n—O
converges uniformly on compact sets. Hence, for fixed z,
00
f(t)coszt=> (2n)!
with the series converging uniformly on [0,1]. We can therefore, interchange theorder of integration and summation to get
00
in other words, h has the power series representation
Zn
Since h is given by a convergent power series, it is analytic.2. Suppose h is the zero function. Then, by Part 1, f0' = 0 for
n = 0, 1,2 Hence, if p is any polynomial, then p(t2)f(t)dt = 0. Bythe Stone—Weierstrass Approximation Theorem [MH93, p. 284], there is a se-quence fPk} ofpolynomials such that pk(t) —* f (Ii) uniformly on [0,1]. Thenpk(t2) -÷ f(t) uniformly on [0,11, so
I f(t2)dt = Jim I pk(t2)f(t)dt = 0k—ooj0
Solution to 5.6.35: Let z E C . We have
g(z)= f
f in t, so we can change theorder of summation and get
g(z)=00
n=O
Ii )f(t)t'1dt =
00
n=O
1
I f(t)t"dt.n! j0
We have1
I If(t)Idtn!so the radius of convergence of the series of g is oo.
Solution 2. Let zo E C. We have
g(z) — g(zo)
=etZ0
dt.Z—zo 0 ZZ0
I
n=O
h(z) with(_1)12 P1
= /(2n)! jçj
implying that f 0.
where
From the power series expansion et Z0)
=
t"(z — zoY',one gets
etZ — etZ0= tetZ0 + O(z — zo)
z — zo
uniformly on 0 t 1, when z Thus, as z —÷ zo, the integrand in theintegral above converges uniformly on [0, 1] to tf and one can pass to thelimit under the integral sign to get
urng(z) — g(zo) = 11 tf(t)etZodt,
Z—ZO 0
proving that g is differentiable at zo.
Solution 3. The integrand in the integral defining g is a continuous function of thepair of variables (t, z) [0, 1] x C, implying that g is continuous. If R C C isa rectangle, then by Fubini's Theorem [MH93, p. 500] and the analyticity of etz
with respect to z,
JRg(z)dz
= fJ f(t)etzdt dz= ff f(t)etzdzdt = 0.
By Morera's Theorem [MH87, p. 173], g is analytic.
Solution to 5.6.36: 1. Let f and g have the Maclaurin expansions [MH87, p.234]
1(z) = Eanz", g(z) =
By Cauchy's Integral Formula [MH87, p. 167] and the uniform convergence ofthe series for g, we have
27ri 'Cr1f(w)g (!)dw
= 27ri 1Cr
=1Cr
= >anbn?.k=O
As
n—3.oo n—).oo n-400
the radius of convergence of h is at least 1.
2. If we take f(z) sin z and g(z) = cos z, we have = 0 forn = 1,2,..., so h 0.
Solution to 5.6.37: 1. The function f is defined on the open set = C \ [0, 1].
Suppose D(a, R) C For every z E D(a, R) and t [0, 1],
z—a !z—aI <1.t—a R
Thus,F(t)(z—aY' — F(t) 1 — F(t)
t—aj
The convergence of the series, for a fixed z D(a, R), is uniform in t E [0, 1].This follows from Weierstrass's M-test and the inequality
< IIFII (Iz—aI\"R R I
Thus we have the power series expansion, valid for z D(a, R),
1(z)= f dt
= f dt=
—
whereI" F(t)
dt (n=0,1,2,...).j0 (t—a)"
This proves that f is analytic in c�.
2. For IzI> 1 and 0 t 1, we have,
F(t) — F(t) I — F(t) °°
z z z"n=O
where the convergence, for a fixed z in Izi > 1, is uniform for t [0, 1] sinceIt/zi 1/IzI <1.
We thus have the Laurent expansion, valid in IzI> 1,
f(z) = tPzF(t)dt =
where 'I(n=1,2,...).
JoThe above is a transform F C[0, 1] i—f f E We have seen above thatthe Laurent coefficients of f are the moments dt (n 0) of F. Now
the Laurent coefficients are determined by f. To show that F is determined byf it is sufficient to show that the moments of F determine F. This in turn is aconsequence of the following result.
lfgo EC[O, =OforallnIt suffices to consider real-valued go. By hypothesis, J0' p(t)go(t) d: = 0 for
all polynomials p. The Weierstrass Approximation Theorem asserts that there isa sequence of polynomials Pk which converges to go uniformly on (0, 1]. Thus
JO' Pk(t)co(t)dt -÷ ask -÷ co, and so f0' go2(t)dt = 0. This impliesthatgo=Oon[O, 1].
Solution to 5.6.38: First we will show that Jr p(z)f(z)dz = 0 for every polyno-mial p.To see this consider the polynomial p(z) and p(z) +1 apply the conditionand subtract them, we find that
[(2p(z) + 1)f(z)dz = 0.
and since every polynomial can be written in the form 2p(z) + 1, we concludethat
f p(z)f(z)dz = 0,r
for every polynomial p.Suppose that f has a pole of order n at a E C. Then (z — a
nonzero residue at a, therefore,
IJrfor a sufficiently small loop around a. Thus, f cannot have any poles and must beentire.
5.7 Cauchy's Theorem
Solution to 5.7.1: By Cauchy's Integral Formula (MH87, p. 167], we have
1' z
1
2ir
I e'dO=2ir.Jo
Solution 2. We can also compute it using residues. Parametrize the unit circle byy(t) = elt for 0 t Then
eZ p2ir es" • ç2ir
I I y'(t)dt= I —7-ie1dt=i I ee dt.J,z Jo y(t) e'
Since ez/z is a quotient of analytic functions, it is itself analytic everywhereexcept at z =0, hence the only singularity lying inside y is the origin, so
ç2ir lfeZ fez \ -
I eedt=_, —dz=2jrRes(—,Op=2JrlImz—=27r.JO 1JyZ \Z / z—+O z
Solution to 5.7.2: By Cauchy's Integral Formula for derivatives [MH87, p. 169],we have
1z p2ir
= . I dz = — I ee'°_bodOdz 2irz Jpzi=i Z
therefore,p2ir
I = 2ir.Jo
Solution to 5.7.3: We'll apply Cauchy's Integral Formula [MH87, p. 167] tof(z) = 1/(1 — which is holomorphic on a neighborhood of Izi 1. Wehave
( IdzJdO
Jizi=i Iz — a12 le'9 — a12
1
(e' — —
etO
= (e18 — a)(1 —dO
_1f 1
— Jizi=i (z — a)(1 —dz
2irI — IaP
Solution 2. The function f(z) = l/(?iz — 1)(z — a) has a simple pole inside thecircle IzI = 1 at z = a, with residue I/(1a12 — 1). Therefore, we have, by theResidue Theorem [MH87, p. 280],
I IdzI _.f dz1 —1 I
Iz — a12 Jiz,=i (?iz — l)(z — a)
=2nRes( a
2,r1_1a12
Solution to 5.7.4: The integrand equals
1
— b)2 (ae'° — b)2 = (a — belO)2
Thus, I can be written as a complex integral,
= (az — b)2(a — bz)2dz
L=1 (az — b)2(zdz.
By Cauchy's Theorem for derivatives, we have
11d( z a2-i-b2— b2 dz (az — b)2) = — a2)3
Solution to 5.7.5: Let p(z) = anz'2 +••• + ao. If p has no zeros then I/p isentire. As p(z) = 00, i/p is bounded. By Liouville's Theorem [MH87,p. 170] i/p is constant, and so is p.
Solution 2. Let p(z) = anz'1 + + ao, n 1. If p has no zeros, then i/p isentire. As p(z) = 00, the Maximum Modulus Principle [MH87, p. 185]gives
1 . 1 . 1max = urn max = urn max = 0z€C Ip(z)I R-÷oo Ip(z)I R+oo JzI=R Ip(z)I
which is a contradiction.
Solution 3. For p(z) = anz" +••• + ao with n 1 let the functions f and gbe given by f(z) = g(z) = p(z) — 1(z). For R > 1 consider the circlecentered at the origin with radius R, CR. For z E CR we have
jf(z)I = and (IaoI + •+
Therefore, on CR,
IgI<IfI if R>Ian
so, by Rouché's Theorem [MH87, p. 421], f + g = p has n zeros in(zEClizi <R).Solution 4. Let P(z) be a nonconstant polynomial. We may assume P(z) is realfor real z, otherwise we consider P(z)P(z). Suppose that P is never zero. SinceP(z) does not either vanish or change sign for real z, we have
do(*)
P(2cosO)
But
dO if dz
Jo P(2cosO) ZJIzI=1
_1 fQ(z)
where Q(z) = zP(z + z') is a polynomial. For z 0, Q(z) 0; in addition, ifis the leading coefficient of P, we have Q(0) = 0. Since Q(z) is never
zero, the last integrand is analytic and, hence, the integral is zero, by Cauchy'sTheorem [MH87, p. 152], contradicting (*).This solution is an adaptation of [Boa64J.
SolutionS. Let p(z) = + •• + ao, n 1. We know that Ip(z)Ioo, thus the preimage, by p, of any bounded set is bounded. Let w be in theclosure of p(C). There exists a sequence C p(C) with limo hEn W. Theset n E N) is bounded so, by the previous observation, so is its preimage,
((Wn n N)) = X. X contains a convergent sequence, Zn zo, say. Bycontinuity we have p(zo) = w, so w E p(C). We proved then that p(C) is closed.As any analytic function is open we have that p(C) is closed and open. As onlythe empty set and C itself are closed and open we get that p(C) = C and p isonto. In fact, all we need here, in order to show that p is open, is that p : R2 -÷ R2has isolated singularities, which guides us into another proof.
Solution 6. By the Solution to Problem 2.2.9 the map p : —÷ JR2 is ontoguaranteeing a point where p(x, y) = (0,0).
Solution 7. Let p(Z) = anz't + • • + n 1. Consider the polynomial q givenby q(z) = +•• Assume p has no zeros. As the conjugate of any root ofq is a root of p. q is also zero free. Then the function I /pq is entire. By Cauchy'sTheorem [MH87, p. 152], we have
Jr p(z)q(z)=0
where F is the segment from —R to R in the horizontal axis together with thehalf-circle C = {z E C
I IzI = R, > 0). But we have
f dz f dz 1R dzJr p(z)q(z) — Jc p(z)q(z) + J—R Jp(z)J2
dz
—R Ip(z)1
this gives, for large R,çR dz
J—R Ip(z)12=0
which is absurd since the integrand is a continuous positive function.Solution 8. Let p(z) = + + n 1. For R large enough, as
p(z) oo, IpJ has a minimum in (z E C I Izi <R), at Zo, say. Supposep(zo) 0. Expanding p around zo we get
p(z) = p(zo) + >Jbj(z — bk 0.
Let w be a k-root of —p(ZO)/bk. We get, for e > 0,
p(zo + we) = p(zo) + bkwkd +j=k+1
= — ek) +j=k+1
therefore, for e small enough, we have Izo + wel <R and
Ip(zo + we)I Ip(zo)IIl —
j=k+1
= (p(zo)I — ek
< Ip(zo)I
which contradicts the definition of zo. We conclude then that p(zo) =0.
Solution 9. Let p(z) = + • + ao, n 1. We have
fp' \ . p'(z)Res —, = — lim z —n.j JzI-+oo p(z)
As the singularities of occur at the zeros of its denominator, the conclusionfollows.
Solution to 5.7.6: By Morera's Theorem [MH87, p. 173], it suffices to show that
I f(z)dz=0Jy
for all rectangles y in C. Since f is analytic on fz I 0}, which is simplyconnected, it is enough to consider rectangles which contain part of the real axisin their interiors.
Let y be such a rectangle and I be the segment of R in its interior. For e> 0small enough, draw line segments li and 12 parallel to the real axis at distance eabove and below it, forming contours Yl and Y2.
Since is continuous, its integral depends continuously on the path. So, as etends to 0,
(*) f 1(z) dz+
f 1(z) dz=
1(z) dz> f 1(z) dz,
since the integrals along li and 12 have opposite orientation, in the limit, theycancel each other. By Cauchy's Theorem [MH87, p. 152], the left side of (*) isalways 0, so
jf(z)dz=0.
j=k+I
Solution to 5.7.7: Since p(z) has lower degree than q(z), 1(z) = O(1/z) as
z oo, so the integrand is O(1/t2) as t oo, and the integral converges.For R > Izol, let yi be the straight-line path from —R to R, and let Y2 be thecounterclockwise semicircle y2(t) = Relt, t [0, ir]. Let y be the closed contourconsisting of followed by Y2. The function 1(z) is holomorphic on and insidey. By Cauchy's formula,
f(zo) = f 1(z)dz
= 11R 1(z)
dz+ 1
f 1(z)dz.
2iri z—zo 2irz —RZ'ZO
We will obtain the desired formula by taking the limit of both sides as R 00.The first term on the right tends to
1 (+°° f(t).1 dt as R—oo.t—zo
The second term on the right tends to 0 as R —÷ 00, since the integrand isO(1/R2) while the length of Y2 is 0(R).
Solution to 5.7.8: We have, using the fact that the exponential is -periodic,
if z"'Jf(z)I2dz= if dz2iri 2irz IzIR
2iri f > a1z' dzIzl=R i=O i=O
= 1j2ir
dOz,j =0
2ir=
i— f aj dO
11'21r
= — dO
=
Solution to 5.7.9: Let be the power series for f. Since the seriesconverges uniformly on each circle Izi = r, we have
j27rIf(r = J f(rS°)dO.
= fn=Om=0 0
=
Thus, for each n we have 27r Icn 12 ( If n > k we let r -÷ 00 to get=O.ffn <kweletr —÷ =Owhenevern
the desired conclusion follows.
Solution 2. Assume f is not the zero function, and let be its lowest ordernonzero power series coefficient at the origin, that is,
1(z) = + +••then z —k 0, 50
1,'2ir
lim I If(re'°)12d0 =r
On the other hand,
1p2,r
lim sup I If (re'8 ) I 2d0 A,r-+0 r
so stays bounded as r -+ 0, implying that n k. Hence the functiong(z) = is entire.
Fix zo in C. For r> Izol, Cauchy's formula gives
2 1 1' g(z)2g(zo) = . dz,
2irz JIzI=r z — zo
sof27r
tg(zo)I I Ig(re )I dO2it(r — Izol) io
1'2jr
= I If(re'°)12d82ir(r — Izol) Jo
rA A>—.2ir(r—IzoI) 22r
It follows that g is bounded, hence constant, as desired.
Solution to 5.7.10: By Cauchy's Theorem [MH87, p. 1521,
1• [2,ri JIzI=r Z
for r> 1. Parameterizing the domain of integration by z = re'0, we find
1f27r
f(re'0)2rie'0dO
f(O)2.o re
Simplifying and taking real parts gives
— v(re10)2) dO = 2ir (u(o)2 — v(O)2) o.
Solution to 5.7.11: We have
-'
dzm
for all z. Dividing both sides by IzIk and taking the limit as Izi tends to infinity,we see that dmf/dzm has a pole at infinity of order at most, k so dmf/dzm isa polynomial of degree, at most, k. Letting n = m + k + 1, we must have thatd'1 = 0 and that n is the best possible such bound.
Solution to 5.7.12: 1. We have
f(z) = (z — (z—
where g is an analytic function with no zeros in c�. So
f'(z) flk g'(z)= +...+ +f(z) z — z — z* g(z)
Since g is never 0 in g'/g is analytic there, and, by Cauchy's Theorem [MH87,p. 152], its integral around y is 0. Therefore,
= k
2. We havezf'(z) — z
+ g'(z)1(z) z—zi g(z)
so1 (zf'(z) dz=zi.2in Jy 1(z)
Solution to 5.7.13: Suppose f(zi) f(z2) and let y be the segment connectingthese two points. We have 0 = f'(z)dz. Hence,
f(f'(z) — f'(zo))dz = —f'(zo)(z2 — zi).
Taking absolute values, we get
jf'(zo)I 1Z2Z11 f If'(z)—f'(zo)I <f If'(zo)I IdzI = If'(zo)I 1Z2—ZI I,
an absurd. We conclude, then, that f is injective.
Solution to 5.7.14: It suffices to show that there exists an integer n such thatthe image of under h(z) = f(z)/z'2 contains no curves with positive windingnumber about 0; because it implies the existence of an analytic branch of thelogarithm in h Each closed curve in h is the image of a closed curve inso it is enough to show that the images of simple closed curves in have windingnumber 0 about the origin. Consider two classes of simple closed curves in
• r1, the curves with 0 in their interiors, and
• r2, the curves with 0 in their exteriors.
Since f has no zeros in it is clear that if y E r'2, then (0) = 0. Fromthe shape of it follows that all the curves in r1 are homotopic. Let n be thewinding number about 0 of f(y) for y E r'1. Since h has no zeros in we musthave =Ofory E T2.Fixy E F1;then
Indhy(0) = f dz = 2iri —
dz = Indf(y)(O) —
and we are done.
Solution to 5.7.15: The analyticity of f can be proved with the aid of Mor-era's Theorem [MH87, p. 173]. If y is a rectangle contained with its interior inC\[0, 1], then
f f(z)dz = j (f dz) dt = 0
(by Cauchy's Theorem [MH87, p. 152]). Morera's Theorem thus implies that fis analytic. Alternatively, one can argue directly: for zo in (C \[0, 1]
urn1(z) — f(zo) = urn I dt =
f12
dt,z — zo z-+z,o Jo (t — ZO)(t — z) o (t — zO)
where the passage to the limit in the integral is justified by the uniform conver-gence of the integrands.
To find the Laurent expansion [MH87, p. 246] about oo we assume IzI> 1 and
write
f(z)=_!f1 too 00
di'.zJo n=O
The series converges uniformly on [0,1], so we can integrate term by term to get
1(z) =- (L'
=-
Solution to 5.7.16: Let c> 0. It suffices to show that there is a constant M suchthat
If(zI)—f(z2)I —z21 forall Z1,Z2 E > C, JZi Z21 <C).
Fix two such points and let y be the circle of radius c whose center is the mid-point of the segment joining them. y lies in the right half-plane, so, by Cauchy'sIntegral Formula [MH87, p. 167], we have
—
2ir
NJz1 — f2,r J,,
where N is the supremum of Ill in the right half-plane. On y, — fori=1,2,so 2
If(zi) — f(z2)I — Z21.
5.8 Zeros and Singularities
Solution to 5.8.1: F is a map from to the space of monic polynomials ofdegree 3, that takes the roots of a monic cubic polynomial to its coefficients,
because if a, f3, andy are the zeros of z3 — Az2 + Bz — C, we have
A=a+fl+y, aj5+ay+j3y=B, afly=C.
Thus, by the Fundamental Theorem of Algebra (for several different proofs seethe Solution to Problem 5.7.5), it is clear that F is onto. F(1, 1,0) = F(l, 0, 1),so F is not injective, in fact, F(u, v, w) = F(v, w, u) = F(w, u, v).
Solution to 5.8.2: Using Rouché's Theorem [MH87, p. 421], it is easy to concludethat p(z) has two zeros inside the circle Izi = 3/4.
Solution 2. The constant term of p is 1, so the product of its roots is 1, in absolutevalue. They either all have absolute value 1, orat least one lies inside Izi < 1. Theformer is not possible, since the degree of p is odd, it has at least one real root,and a calculation shows that neither 1 nor —us a root. So p has a root in the unitdisc.
Solution to 5.83: For Izi = 1, we have
I—z31 = 1 > lf(z)I
so, by Rouché's Theorem [MH87, p. 421], f(z)—z3 and z3 have the same numberof zeros in the unit disc.
Solution to 5.8.4: Let fi and 12 be defined by (z) = 3z100 and f2(z) = _ez.On the unit circle, we have
Ifi(z)l = 3> 1 = 1f2(z)I.
By Rouché's Theorem [MH87, p. 421], we know that f and 1' have the samenumber of zeros in the unit disc, namely 100.
Let be a zero of f. Then
= — = — = (100— 0
so all the zeros of f are simple.
Solution to 5.8.5: 1. Let 1(z) = 4z2 and g(z) = 2z5 + 1. For Izi = 1, we have
By Rouché's Theorem [MH87, p. 421], f and p = f + g have the same numberof roots in 14 < 1. Since f has two roots in IzI <1, so does p.2. There is at least one real root, since p has odd degree. The derivative isp'(z) = 10z4 + 8z, so p' has two real zeros, namely at 0 and More-
over, on the real axis, p' is positive on (—oo, and (0, oo), and negativeon 0). Thus, p is increasing on the first two intervals and decreasingon the last one. Since p(O) = 1 > 0, also p has no root
in oo) and exactly one in (—oo, (The real root is actually in(—2, —1), since p(—l) > 0 and p(—2) <0.)
Solution to 5.8.6: Let p(z) = 3z9 + 8z6 + z5 + 2z3 + 1. For Izi = 2, we have
Ip(z) — 3z91 = 18z6 + z5 + 2z3 + ii
+ 1z15 + 21z13 + 1
= 561 <1536 = 13z91
so, by Rouché's Theorem [MH87, p. 4211, p has nine roots in <9. For Izi = 1,we have
Ip(z) — 8z61 = 13z9 + + 2z3 + ii31z19 + 1z15 + 21z13 + 1
=7 <8=18z61
and we conclude that p has six roots in Izl < 1. Combining these results, we getthat p has three roots in 1 <IzI <2.
Solution to 5.8.7: For z in the unit circle, we have
15z21=5 1z5+z3+21
so, by Rouché's Theorem [MH87, p. 421], p(z) has two zeros in the unit disc. ForIzI =2,
so p(z) has five zeros in (z I IzI <2). We conclude then that p(z) has three zerosin 1< lzI< 2
Solution to 5.8.8: Let p(z) = — 4z3 — 11. For z in the unit circle, we have
Ip(z) — ill = 1z7 11
so, by Rouché's Theorem [MH87, p. 421], the given polynomial has no zeros inthe unit disc. For IzI =2,
p(z) — = + 43 < 128 =
so there are seven zeros inside the disc (z I IzI <2) and they are all between thetwo given circles.
Solution to 5.8.9: Since p(O) = —1 and p(x) -÷ +00 as x -÷ oo on the positivereal axis, p has at least one root on (0, oo). On the negative real axis p is negative,in particular nonzero.
We have p'(z) = 3sz2 — 2z, which has roots at 0 and 2/3s. The roots of p'lie in the convex hull of the roots of p. by the Gauss—Lucas Theorem [LR7O, p.
94] On the imaginary axis, p has a nonzero imaginary part, except at the origin.Therefore p has no roots on the imaginary axis. Consequently p must have atleast one root in <0. Such a root is nonreal (since p 0 on (—oo, 0)), andnonreal roots of p occur in conjugate pairs. Therefore p has two roots in <0.Solution 2. From the expression above for p' one sees that p is decreasing on(0, 2/3s) and increasing on (2/3s, oo). Therefore p has exactly one root on (0, oo).Let the root be A. Since p is nonzero on (—00,0], the other two roots form a con-jugate pair, say and j1.. Since the linear term in p vanishes, we have
which can be written as
A11L12
implying that <0.
Solution to 5.8.10: Rescale by setting z = s"5w. Then we need to show thatexactly five roots of the rescaled polynomial
p8(w)= w7 +w2+6,
with ô = s2/5 —+ 0 as s —+ 0, converge to the unit circle as s —÷ 0. We havepo(w) = w2(w7+1). Sincetworootsofpo are atw = Oand the otherfive are onthe unit circle, the result follows from the continuity of the roots of a polynomialas functions of the coefficients, see Problem 5.8.30.
Solution 2. Let q(z) = z2 + 1,so
Ip(z) — q(z)I = sIzI7 = r7s215
on the circle Izi = rs115. Also,
Iq(z)I = 1z2 + ii > 2/5—
on Izi = rs115 . Since r < 1, r7 < r2, and r7e215 <r2s2/5 — 1 for s suffi-ciently small. Then Ip(z)—q(z)I <Iq(z)I on Izi = and by Rouché's The-orem [MH87, p. 421], p and q have the same number of zeros inside Izi rs115,namely two. By the Fundamental Theorem of Algebra (for several different proofssee the Solution to Problem 5.7.5), the other five roots must lie in Izi > r81'5.
Now take q(z) = ez7, so
Ip(z) — q(z)I = I + ii R2s2'5 + 1
on Izi = Rs115, whereJq(z)i = R7s215.
SinceR> 1,wehaveR7 >
R2e215 + 1 <R7s215
for e sufficiently small. Thus, Ip(z) — q(z)I < Iq(z)I on Izl = Re115 , so p andq have the same number of zeros inside = namely seven. This leavesprecisely five roots between the two circles.
Solution to 5.8.11: The determinant of A(z) is 8z4 + 6z2 + 1. For z in the unit
circle, we have
so, by Rouché's Theorem [MH87, p. 421], det A(z) has four zeros in the unit disc.
Also,
(det A(z)) = z(32z3 + 12)dz
with roots
0,
which are not zeros of det A(z). Thus, all the four zeros are simple, so they aredistinct.
Solution to 5.8.12: Lety : [a, b] —÷ C be a continuous function, with 0where we denote the image of i' by y*• There is a continuous branch of arg y on[a, b]. If 0, are branches of arg y on [a, b], then = 0(b) — 0(a) =We may therefore define w(y, 0) = (0(b) — 0(a))/2ir, where 9 : [a, b] —+ R isany branch of arg y. Note we are not assuming that y is closed. If y lies in someopen half-plane, then IcoO', 0)1 < If y is closed then w(y, 0) is an integer.If y(t) = where Y2 are curves not containing 0, then coQ', 0) =
O)+ø(y2, 0); inparticular, if y?z(t) y(tY' thenco(yJz, 0) = nco(y, 0) forn 1.Ify = yi +"+Yk isajoinofcurvesnotcontaining0,thenco(y,0) =
0).
Choose R sufficiently large, so that the zeros of f(z) = z4 + 3z2 + z + 1 lieinside Izl = R. Let y = yj + Y2 + where [iR, 0] (line segment fromiR toO on imaginary axis), Y2 = [0, R] and y3(t) = Reit, 0 t Denote
Forz = iy E y, f(z) = f(iy) = y4 —3y2 + 1 +iy O.AsO =y <R, Fi lies in the upper half-plane and therefore Ico(Fi,
For z = x E 1(z) = .x4 + 3x2 + x + 1 0, this being a strictly increasingfunction in x 0. F2 is a subset of the positive real axis. Thus &)(F2, 0) = 0.
We have f(z) = z4(1 + g(z)), where g(z) = (3z2 + z + 1)/z4. Now g(z) -+ 0as z —+ oo. For large R, Ig(Re')I < 1, and I +g(Rett) lies in the disc centered at1 radius 1; that is I + ,g(Re") lies in the right half-plane, and thus 0)1where 0(t) = I + g(y3(t)). Thus co(r'3, 0) = wQ4, 0) + w(o, 0) = 4w(V3, 0) +w(o, 0) = I + cv(o', 0).
Thus 0) — = Ico(Fi, 0) + 0) + 0)1 + = 1. As F isclosed, w(F, 0) is an integer, and so co(f', 0) = 1. The Argument principle assertsthat co(l', 0) equals the number of zeros of f inside y. As R is arbitrarily large,we deduce that f has one zero in the first quadrant. The conjugate of this root
is the only other zero in the right half-plane. Thus f has two zeros in the righthalf-plane.
Solution to 5.8.13: Let . .. , z zero of zi = 1,...n.Wehave
1
P
Using the fact that i/a = and conjugating we get
11T112 =
which is clearly impossible if 91z 0.This result can be generalized to give the Gauss—Lucas Theorem [LR70, p. 94]:
The zeros of p' lie in the convex hull of the zeros of p. If zi, ... , Zn are the zerosof p, andz is azero of p', z Zi, i = 1, ...,n. We have, similartotheabove,
=0
which is impossible if z is not in the convex hull of Zi, ..., Zn.
Solution to 5.8.14: Let a be a real zero of f. Then at a the function f restrictedto R has a minimum, therefore f'(a) = 0. Hence the order of the zero of f at a isat least 2. The function g(z) = f(z)/(z — a)2 thus satisfies the same hypothesesas f. If g(a) = 0 we can repeat the preceding argument. After finitely manyrepetitions we reach the desired conclusion.
Solutionto5.8.15: Wemay assumer =degpandxi <x2 <xk be the roots of p, with multiplicities ml, . . . , m,, respectively. If any mjexceeds 1, then p — rp' has a root at of multiplicity — 1 (giving a total ofn — k roots all together). We have
p(x) = c(x — Xl)mI . - — Xk)mk.
The logarithmic derivative p'/p is given by
k
p(x)
Its range on the interval (Xj, (j = 1, ... , k — 1) is all of IR, since it iscontinuous there and
p'(x) . p'(x)lim = +00, lim = -00.
p(x) p(x)
Hence, there is a pointx (Xi, Xj+I) where p'(x)/p(x) = 1/r; in other words,
where p — rp' has a root. Thus, p — rp' has at least k — I real roots other than the
n — k that are roots of p. Hence, p — rp' has at least n — 1 real roots all together,and the nonreal ones come in conjugate pairs. Hence, it has only real roots.
Solution to 5.8.16: Let R > A+1 and considerthe contour CR = FRU[—Rj, Ri],
Let the functions f and g be defined by f(z) = z + A, g(z) = _ez. On FR, wehave
and on [—Ri, Ri],
If(z)I Izi —A>1 ? Ig(z)I
If(z)I A> 1 = Ig(z)I.
By Rouché's Theorem [MH87, p. 421], fx and f have the same number of zerosinside the contour CR, so has exactly one zero there. As this conclusion is validfor every R > A + 1, we conclude that fx has one zero in the left half-plane. Asf is real on the real axis and f(x)f(0) <0 for x small enough, we get that thezero of fx is real.
Solution 2. We find the number of zeros of in the left half-plane by consider-ing a Nyquist diagram [Boa87, p. 106] relative to the rectangle with corners iy,—x+iy, —x--iy, and —iy,x, y > A. This wiligive thechange in (1/2ir) arg fx(z).Then we let x, y —÷ 00.
On the right side of the rectangle, as t ranges from —y to y, thenfx (it) = it + A — cos t — i sin t has a positive real part, and its imaginary partchanges sign from negative to positive. On the top of the rectangle, as s ranges
r
—iR
from 0 to —x, fA(s +iy) s +iy +A —e cos y sin y has positive imaginarypart, and its real part changes sign from positive to negative.
Similar reasoning shows that on the left side of the rectangle, <0 and $3fxchanges sign from positive to negative. On the bottom of the rectangle, <0and changes sign from negative to positive. Hence, fj. is never 0 on thisrectangle and the image of the rectangle winds around the origin exactly once. Bythe Argument Principle [MH87, p. 419], fx has exactly one zero in the interiorof this rectangle. Letting x and y tend to infinity, we see that has exactly onezero in the left half-plane. As fx is real on the real axis and fx(x)fx(0) <0 forxsmall enough, we get that the zero of fj. is real.
Solution to 5.8.17: For Izi = 1, we have
I> = I —
so, by Rouché's Theorem [MH87, p. 421], the given equation has one solution inthe unit disc. Let f(z) = z real, f increases from f(O) = 0 tof(l) = > 1, by the Intermediate Value Theorem [Rud87, p. 93], f(4) — 1
for some E (0, 1).
Solution to 5.8.18: By the Gauss—Lucas Theorem [LR7O, p. 94] (see Solution to5.8.13), if p(z) is a polynomial, then all of the roots of p'(z) lie in the convex hullof the roots of p(z). Let z = 11w. The given equation becomes, after multiplyingby w", w" + w"' + a = 0. The derivative of the polynomial on the left-handside is nw"1 +(n — which has roots 0 and —(n — 1)/n 1/2. For thesetwo roots to lie in the convex hull of the roots of w" + + a, the latter musthave at least one root in Iwl 1/2, which implies that az" + z + 1 has at leastone rootin Izi 2.
Solution 2. The product of the roots of p(z) = az'3 + z + 1 is its constant term,namely 1, so all p's roots are unimodular or at least one is in Izi < 1.
Solution to 5.8.19: Let FR be the closed curve in the first quadrant consisting ofthe interval [0, R], the interval [0, iR], and the circular arc centered at 0 joiningthe points R and iR. It suffices to show that, for R arbitrarily large, the functiong(z) = z4 + z3 + 1 has exactly one zero in the interior of FR. We'll compare gwith 1(z) = z4 + 1. The function f has one simple zero in the interior of rR,at z = By Rouché's Theorem [MH87, p. 421], it remains to verify thatIf(z) — g(z)I <If(z)I for z in FR. We have
• If(x) — g(x)I = 1x13 <x4 + 1 = ff(x)f forx in R (since 1x13 1 forxin [—1, 1] and 1x13 <x4 forx in JR \ [—1, 1])
• If(iy) — g(iy)I = 1y13 <y4 + I If(iy)I for y in JR
• If(z) — g(z)I = 1z13 = R3 — 1 If(z)I for Izi = R (with R 2, SO
R4—1 >2R3—R3=R3).
Solution to 5.8.20: Suppose that f is zero free in the disc Izi < 1, and hence also
in the disc Izi 1. Then f has no zero in an open set containing the latter disc, so
the function g = 1/f is holomorphic there. By the Maximum Modulus Principle
[MH87, p. 1851 and our hypotheses, we have
! < Ig(0)Im m
a contradiction.Solution 2. For Izi 1 Let g(z) = —f(0). Then on Izi = 1,
Ig(z)l = 11(0)1 <m < If(z)I.
By Rouché's theorem [MH87, p. 421], f and f + g have the same number ofzeros in the unit disc. Since f + g = f — f(0) has at least one zero, the resultfollows.
Solution to 5.8.21: Let p(z) denote the polynomial and suppose p(zo) = 0 for
some zo 11). Then zo is also a root of (z — 1)p(z). We then have
0 = + (a 1 — f. .. + (an — an—i )zO —
Since all the aj's are positive and Izol < 1, we have, by the Triangle Inequality[MH87, p. 20],
= + (al — +• '. + —
a contradiction.
Solution to 5.8.22: By Descartes' Rule of Signs [Caj69, p. 7], [Coh95, vol. 1,pag. 172], the polynomial p(z) has zero or two positive real roots. Asp(O) = 3 and p(l) = —2, by the Intermediate Value Theorem {Rud87, p. 93],p(z) has one and so, two, positive real roots. Replacing z by —z, and again ap-plying Descartes' Rule of Signs, we see that p(—z) has one positive real root, sop(z) has one negative real root. Applying Rouché's Theorem [MH87, p. 421] tothe functions f = p and g = 6z on the unit circle, we see that p has exactly onezero in the unit disc, which is positive as seen above. Hence, the real roots aredistinct. (The same conclusion would follow from noticing that p and p' have nocommon roots.) The imaginary roots are conjugate, so they are distinct as well.
Solution 2. Graphing the polynomial y = — 6x + 3 (for real x), we can seethe result easily. First, y' = 5x4 —6 and the only two real roots are x =and none of them are multiple. Now looking at the limits when x —* —00 andx -+ 00, we can conclude that the graph looks like
So there are three distinct real roots. There cannot be a forth, otherwise y'would have a third root. The other two roots are then complex and not real; since
they are conjugate, they are distinct, making for five distinct roots, three of themreal.
y
x
Solution to 5.8.23: Let z be a zero of the given polynomial with Izi = r. If r 1,
then z lies in the given disc. For r> 1, we have
= —
By the Cauchy—Schwarz Inequality [MH93, p. 69}, we get
12>J
— IThe second sum is a finite geometric series which sums to
2• Combining
these, we have r — 1
r212 <
2_iMultiplying both sides by
rwe get the result wanted.
Solution to 5.8.24: The product of all zeros of P(x) (with multiplicities) equals± I, so, if there are no roots inside the unit circle, then there are no roots outsidethe unit circle either. Hence, all roots are on the unit circle. From P (0) = —1 <0and P(x) = +00, it follows that P(x) has a real zero in the interval(0, oo). Since it lies on the unit circle, it must be 1, so P(1) = 0.
Solution to 5.8.25: Let R > 0. Consider the semicircle with diameter [—Ri, Ri]containing R and its diameter. We will apply the Argument Principle [MH87, p.419] to the given function on this curve.
Suppose n is even. Then
+ +462 = y2fl + 462 —
is always in the first quadrant for y <0, so the change in the argument when wemove from 0 to —Ri is close to zero, for R large. On the semicircle,
which is close to for R large. So the argument changes by when we gofrom —Ri to Ri. From Ri to 0,
y2fl +462
is always in the fourth quadrant, so the change in the argument is close to zero,for R large. The total change is then 2jrn, so there are n roots with positive realpart.
Now, suppose that n is odd. We have
+ +462 = +462+
so when we go from the origin to —Ri, for R large, the argument change is closeto —ir. The variation on the semicircle is again about 2,rn. The change when ygoes from R to 0 in the argument of
+462+
is about —n. Therefore, the number of zeros with, positive real part is now(—pr + — ir)/2ir = n — 1.
Solution to 5.8.26: Let p > 0 and consider the functions
z28n(Z) = fn(l/z) = I +z+ — +..+ —.
2!
Since 0 for all n, has a zero in Izi p if and only if has azero in Izi p. gn(z) is a partial sum of the power series for ez. Since this seriesconverges locally uniformly and {z I Izi p} is compact, for any e > 0 there isN > 0 such that if n ? N, then — ezI <s for all z in this disc. eZ attains itsminimum m > Gin this disc. Taking N0 such that ifn ) No, — e9 <m/2for all z in the disc, we get that gn(z) is never zero for Izi ( p. Therefore,has no zeros outside this disc.
Solution to 5.8.27: If z = x + iy then eZ = eXetY, so w = exp z maps thehorizontal line = A onto the ray w = rep', r > 0. Thus if 0 <
then ez lies in the first quadrant. As is also positive, we have ez + z 0.If < < 0 then eZ lies in the fourth quadrant; whereas z lies in thelower half-plane. Again eZ + z 0. Consider z = x real. It is easy to see that
= —x has one root. Thus 1(z) eZ + z has precisely one zero in the strip—ir/2 <ir/2.
We now consider the zeros of g(z) in the same strip. As arg(—l/z) = IT — arg zwe deduce that, (writing z = x + iy), if x > 0, 0 < y < then —l/zlies in the second quadrant, whereas ez lies in the first quadrant; and if x > 0,—ir/2 <y <0 then —l/z lies in the third quadrant, whereas eZ lies in the fourthquadrant; that is, eZ —liz. When z = x > 0 then zez + 1 > 0. Thereforezez + 1 0 for x > 0, < y <ir/2.
We'll use the notation of the solution to Problem 5.8.12. Consider the rect-angular contour y, which is the join of the line segments yi = [—iir/2, 0],Y2 = [0, 1'3 = [ijr/2, —R + = [—R +
i —i Write h(z) = Zez, Fj = h o We will study each pieceseparately.
r'1: We see that h(—i,r/2) = —ir/2, h(0) = 0. If 0 <y <ir/2 then h(—iy) == and —it <argh(iy) = + y) <—it/2. Thus r1 is a
curve from — to 0 which, apart from the end points, lies in the third quadrant.
r2: We find that h(in/2) = h(iy) = = ForO < y <,r/2,wehave <argh(iy) =apart from the end points, lies in the second quadrant.
F3:Ifz = > Othenh(z) =quadrant. Thus F3 is a curve from to = h(—R + = + iR)),which apart from the initial point, lies in the third quadrant. —
r'4: We see that = h(—R — ii.) = + iR) = If z = —R+iy,<y < then Jh(z)I = I — R + (R + thus h(z) lies
in the unit disc, for large R. For such an R, F4 is a curve in the unit disc fromto
F5: If z = —x — i x > 0 then h (z) = — + which is in the secondquadrant. Thus F5 is a curve from to — which lies, apart from the final point,in the second quadrant.
The curve I + F1 (t) lies in the tower half-plane; by choosing a branch ofarg(1 + F1 (t)), we see that co (F1, —1) = 1/2. Likewise co (F2, —1) = 1/2. Ifa <a <ir/2,thenco(F5, —1)= =w(F3, —1).The curve 1 +F4(t) lies in the disc D(1; 1), and thus in the right half-plane. Hencew(F4, —1) = a/it. Adding, we have co(F, —1) = 2. By the argument principle,h(z) = Zez takes the value —l twice inside y. As R is arbitrarily large, we con-clude that g(z) = zez + 1 has two zeros in the specified strip. So f has one zero,whereas g has two zeros in the strip.
Solution to 5.8.28: The function sin z satisfies the identity sin(z + = — sin z,and vanishes at the points nit, n Z, and only at those points. For m a posi-tive integer, let Rm denote the closed rectangle with vertices (m — + is and(m + ±ie. The function sin z has no zeros on the boundary of Rm, so its abso-lute value has a positive lower bound, say ö, there. (The number 8 is independentof m because of the identity sin(z-I-ir) = —sinz.)Suppose(m — — al> 1Then, forz in Rm, we have
<8lz—aI lzl—a
implying thatIz—aI < I on the boundary of Rm. By Rouché's Theorem
[MH87, p. 421] then, the functions sinz and f(z) = sin z + have the samenumber of zeros in the interior of Rm. Since sin z has one zero there, so does 1(z).As the condition on m holds for all sufficiently large m, the desired conclusionfollows.
Solution to 5.8.30: We will prove that the simple zeros of a polynomial dependcontinuously on the coefficients of the polynomial, around a simple root.
Consider
For . .. , E C's, let F be the polynomial given by
F(z) = + + + , • + + +
We will show thatforsomee > 0, <efori = 0,... ,n— 1 impliesthat Fhas zeros inside the circle Ck centered at Zk with radius rk.
Let be the polynomial given by
On Ck, we have
Mk=
(rk +
and
Ip(z)I JJ — ZkJ = >0.
Taking e < 8k/Mk, we get < Jp(z)I on Ck; therefore, by Rouché's The-orem [MH87, p. 421], F has the same number of zeros in Ck as p. As in thisdomain p has a single zero with multiplicity we are done.
Solution to 5.8.31: From
f(k)(z) = > n(n — 1). • (n — k +
we conclude that( Jf(k)fr)J
for 0 <r < 1, and 0 <0 2ir. Suppose f can be analytically continued in aneighborhood of z = 1; then its power series expansion around z = 1/2,
has a radius of convergence R> 1/2. Let be the Taylor coefficients [MH87,p. 233] of the power series expansion off around the point (1/2)e10. By the aboveinequality, Therefore, the power series around the point (1/2)e'0 hasa radius of convergence of at least R. So f can be analytically continued in aneighborhood of every point of the unit circle. However, the Maclaurin series[MH87, p. 234] of f has radius of convergence 1, which implies that at least onepoint on the unit circle is a singularity of f, a contradiction.
Solution to 5.8.32: Without loss of generality, assume r = 1.
Suppose f is analytic at z = 1. Then f has a power expansion centered at 1with positive radius of convergence. Therefore, f has a power series expansioncentered at z 1/2 with radius 1/2 + s for some positive s.
= (k
(i)k_fl
=n'(k — n)'
n=O
=
which is absurd because we assumed the radius of convergence of to be 1.This contradiction shows that f cannot be analytic at z = 1.
Solution to 5.8.33: As
2 ,,—2 —' Z —
(tanz) —z '=(tanz)
the Maclaurin expansion [MH87, p. 234] of the numerator has no terms of degreeup to 3, whereas the expansion of the denominator starts with z4, therefore, thelimit is finite.
As
(z-+O)
wehave,forl <x <1+e,
f(x) =(n!
k! (i\k_fla,
(k—n)!
/ 1\fl
t\2)
we have
2 22 2 z —z
(tanz) — = A A (z -÷ 0)z' + o(z')
so the limit at 0 is —2/3.
Solution to 5.834: We use the fact that an entire function is a polynomial exactlywhen it has limit 00 at 00. (If the function tends to 00 at 00 then it has a pole at00. In this case, its Laurent series about 00, which is its power series, has onlyfinitely many nonzero terms with positive exponent. The other direction is clear.)
Suppose g is not a polynomial. Then (g(zn)) is bounded for some sequence(Zn) tending to 00. Therefore is bounded, against the hypothesis. Henceg is a polynomial.
Suppose f is not a polynomial. Then is bounded for some sequencetending to oo. As g is onto, there is, for each n E N, E C such that
= As -÷ oo we have —÷ 00, which is contrary to the hypothesissince stays bounded. Hence f is a polynomial.
Solution to 5.835: When we write f as a single fraction, with denominator(Z — Zj) we see that the number of zeros of f in C is given by the de-
gree of the numerator of f, since the Zj 's are no zeros of f. The order of the zeroat infinity of f(1/t) gives the difference between the degree of the denominatorand the degree of the numerator. We have
n nI Qjt 22 +)= = >
p=m
Thus f(1/t) has a zero of order m + 1 at t = 0. Since the denominator of 1(z)hasdegreen,thenumeratorhasdegreen—m— 1,so fhasn—m—l zerosinC.
We cannot have m n because the Vandermonde determinant, see the Solu-lion to Problem 7.2.11 or [HK61, p. 125]
1 •••
1 Z2 4 4'
:i Zn ..:
is not zero, so the linear system = 0 has only the trivial solution fork = 0, . . . , n — 1 . i=1
5.9 Harmonic Functions
Solution to 5.9.1: Derivating twice, we can see that
82u 82u
8x2
so Au =0. The function f is then given by (see [Car63b, pp. 126-127])
fz z' =2\2 2i'
so, up to a constant, v(x, y) = 3x2y — y3 + k.
Solution 2. Using the Cauchy—Riemann equations [MH87, p. 72], we see that
ay
and integrating with respect to y we obtain
v(x,y)=3x2y—y3+t/'(x)av au
and using the Cauchy—Riemann equations again, this time — = —— we seethat *'(x) = 0, so v(x, y) = 3x2y — y3 + k. ax Fly
Solution to 5.9.3: 1. Let f = u + iv. Then v is identically 0 on the unit circle.By the Maximum Modulus Principle [MH87, p. 185] for harmonic functions, vis identically zero on ID). By the Cauchy—Riemann equations [MH87, p. 72], thepartial derivatives of u vanish; hence, u is constant also and so is f.2. Consider
•z+1f(z)=iz—1
f is analytic everywhere in C except at 1. We have
e'0 + 1 e'812 +e'012f (1°) = — = —= cot ER.
Solution to 5.9.4: We have u = 91z3, zlog with —it < argz < Jr. is analytic in the slit plane C \ (—oo,O] and
= Hence, u = is harmonic in the same domain.
= z3.
Solution to 5.9.2: Since u is the real part of an analytic function, it is harmonicin the unit disc D. By Green's Theorem [Rud87, p. 253],
f Flu FluI —dx——dy=—J9D8Y Flx D
dxdy =0.
Solution to 5.9.5: Let v be the harmonic conjugate of u. Then, f = u + iv is anentire function, Consider h = e1. Since u ? 0, IhI 1, h is a bounded entirefunction and, by Liouville's Theorem [MH87, p. 170], a constant. Therefore, u isconstant as well.
Solution to 5.9.6: Let v be a harmonic conjugate of ii, and let f = Thenf is an entire function and, for IzI> 1, we have
If(z)I = = ebizia
Let n be a positive integer such that n a. Then the function has anisolated singularity at oo and, by the preceding inequality, is bounded in a neigh-borhood of 00. Hence, 00 is a removable singularity of and, thus, is, atworst, a pole of f. That means f is an entire function with, at worst, a pole at00, and so f is a polynomial. Since nonconstant polynomials are surjective and fomits the value 0, f must be constant, and so is u, as desired.
Solution to 5.9.7: The linear fractional transformation w = (1 + z)/(1 — z)maps the unit disc to the halfplane > 0, with the upper and lower boundarysemicircles mapped to the halflines and iR_, respectively. A branch of log w,defined onC \R_,has
J jr/2,L —ir/2, WER_,
so a solution is given by f defined by
2 1+zf(z)= log1'
5.10 Residue Theory
Solution to 5.10.1: Since the r1 's are distinct, f has a simple pole at each of thesepoints. If Ai, A2, ..., are the residues of f at each of these points, then
A1 A2g(z) = 1(z)
— z — ri — z — T2 — -. z —
is entire. Clearly, g tends to zero as z tends to infinity, so, by the MaximumModulus Principle [MH87, p. 185], g must be identically zero and we are done.
Solution to 5.10.2: We have
a_i Res (cot 7rz, —1) + Res (cot irz, 0) + Res (cot Jrz, 1) =
For n <—1, the coefficients are given by
1 /' cotirz=JIzI=3/2
dz
= _i) +Res i)cotirz cotirz
= urn (z + 1) + hm(z — 1)
= ((_1)_fl_1 1
z
Solution to 5.10.3: The function 1(z) = z/(z — 1)(z — 2)(z — 3) has simplepolesatz E A = (1,2,3},withresiduesRes(f, 1) = 1/2,Res(f,2) = —2,andRes (1,3) = 3/2. For f to have a primitive in U, it is necessary and sufficientthat f(z)dz = 0 for all (piecewise C1) closed paths y in U.
Lety be a closed path in U, with image Then UC = fizi 4) is connectedand lies in one connected component of C \Therefore the winding numbersare the same, w(y, 1) = w(y, 2) = w(y, 3).
By the Residue Theorem, [MH87, p. 280],
f f(z)dz = 2iri z)w(y,
zEA
=0.
Therefore f has a primitive in U. On the other hand, if
g(z)=(z—1)(z—2)(z—3)
then Res (g, 1) = 1/2, Res(g, 2) = —4 and Res(g, 3) = 9/2. If y is the circleIzI =5, then
and g has no primitive in U.
Solution to 5.10.4: If the roots of I are not distinct, then some satisfiesf(xo)= f'(xo)=O.Butf'(x) = 1+x+".+
m!
and = 0. However, 0 is clearly not a root of f. Hence, the roots of f aredistinct and nonzero.
For 0 k n —2, consider the integral
f zk
= iCrwhere Cr is a circle of radius r centered at the origin such that all the roots of Ilie inside it. By Cauchy's Theorem [MH87, p. 152], Ik is independent of r, and asr -+ 00, the integral tends to 0. Hence, 'k =0. By the Residue Theorem [MH87,
p. 280],
Since 2 m — k n, we get the desired result.
Solution to 5.10.5: Let the disc centered at the origin with radius r contain all thezeros of Q. Let CR be a circle centered at the origin with radius R > r. Then, bythe Deformation Theorem [MH87, p. 148],
f P(z) f P(z)1 dz=i dzJc Q(z) JCR Q(z)
where C is any closed curve outside Izi = r. As
P(z) /=O(IzI j (IzI-+oo)Q(z) /
we have
fP(z)
dz = o (IzI_2) 2irR = o(1) (R -+cR Q(z)
and the result follows.
Solution to 5.10.6: Letting z = iG, we have
cosO = + z')and dO = dz/iz, so that
p2jr1 ' d
I e2C c0s0d9 = — I2irJ0 z
where y is the unit circle. Next,
1 = 1 f2ir i z 2ir i \ Z I Z
1
'—'2lriJ},n! z'z zn=O
Now,
1= +—+...+--—+ +....n! z (n+1)!
Thus, the residue at zero is — and
It n
I dz=—2jrz n!
henceI [ dz
fl!Zrz t
and the result follows.
Solution to 5.10.7: As the singulaijties of the integrand, call it f, all lie inside theunit disc, we have [MH87, p. 286],
I —Res (1(z), oo) = Res G) , o).
We have I (!) = (1±2z)which has residue at 0. Therefore I =
Solution 2. Let f(z) = Then f is meromorphic with a double pole atz = 0 and a simple pole at z = 1/2. Both singularities are inside the contour of
integration. By the residue theorem we have
I =Res (f(z),0)+Res
We have, near the origin,
1(z) =+ = =
so the residue at the origin is —12.
To find the residue at we write f as
(z + 2)2f(z)=
2 12z (z —
Res=
Therefore, I = —12 + = 0.5.Solution 3. Since f has no singularities outside the unit disc, we have, by Cauchy'sTheorem [MH87, p. 152],
foranyr> 1. Thus
11=—I2irj0 dO (r>1).1)
It is easy to see that the integrand approaches uniformly when r -+ 00, so
Solution to 5.10.8: We have
I
2dz=
z (z — 2) (z_2)dz+fCa
eZ 1
We will use the Residue Theorem [MH87, p. 280], to compute the integral, andfor the first integral:
Res(12,0)=0
1 3
= 2z2 4z
and
Res (1(z), 0) =
and get
(z + 2)2
2z2
25
2
I— 2iri Jfzj=r
f(z)dz
(re1° + 2)2 re'8
(2re'° —
Let f(z) =
1
ezThe following expansions hold:
z2(z—2)
1 z2 z3
z—2
1 1 1 1
z2(z—2) 2z2 4z 8
eZ
z2(z—2)
/ z2=
2
z3
3!
thus
I
4z
1
8
5
8
Also,
Res (f(z), 2) =z.-+2 Z
Therefore, we have, for a > 2,
z2(z—2)dz = 2jri (Res
(1
2'2) + Res(f, 0) + Res(f, 2))
andfora <2,
fz2 + dz = 2iri Res (f(z), 0) =
—___
CaZ(Z2) 2
r2Solution to 5.10.9: For Izi = r we have = —, so we can rewrite I(a, r) as
z
r 1 c zI(a,r)=I 2dz= I 2dzJIzl=r a— Jlzl=r az—rif z 2dza JIzJr Z —
If lal <r the integrand is analytic in and on the circle Izi = r, so I(a, r) = 0by Cauchy's Theorem [MH87, p. 152]. (Use the second-last expression for thecase a = 0.)
If lal > r then Cauchy's Integral Formula [MH87, p. 167] gives
23ri I r2\ 2irir22a a
Solution to 5.10.10: Let f(z) = — 1 = (1/2)
z2k. f is analytic forIz I > 1. therefore, k—O
fVz2_ldz=
We have
1/2 )(l)kZ2k
= (1/2)
Res (1(z), 00) = Res , o) =
We then obtain
1c2_1dz=
Solution to 5.10.11: The function sin z = (eIZ — e1Z)/(2i) vanishes if and onlyif = 1, which happens if and only if z is an integer multiple of it. Since2 <ir <4, the zeros of sin 4z within the unit circle are 0, ±ir/4.
The residue of 1/ sin4z at z = 0 is 1/4, because the numerator is nonvanishing,while the denominator sin4z has a simple zero and its derivative at z = 0 is 4.Since sin(4(z + ir/4)) — sin 4z, the residue of 1 / sin 4z at z = ±ir/4 is —1/4.By the Residue Theorem [MH87, p. 2801, I = 1/4+(—1/4)+ (—1/4) = —1/4.
Solution to 5.10.12: The integrand has poles at the integer multiples of it, allsimple poles except for the one at the origin. By the Residue Theorem [MH87, p.2801,
In=2Yti>2ReS(sin zk=-n
Since the integrand is an even function, its Laurent series at the origin containsonly even powers of z, implying that the residue at the origin is 0. Hence
I \ 1/z3 l/k3ir3Resi 'k,ri= = =
sin z / z) cos kit k3yt3
4i3itkl k
Solution to 5.10.13: Using the change of variables z = !, we obtain
1 fdz if —dwlf dw2iri Ic sink = IC' w2sinw — 2iti Ic" w2sinw
where the circles C' and C" have radius 5 and are oriented negatively and posi-tively, respectively. By the Residue Theorem [MH87, p. 280], we have
1f—dw 11 11I =Resi ,itp+Resi 22iti Jc w2 sinw sinw / \w sinw
+Res(1• ,o).
\W sinw
We have
/ I
w w2 sin w
w w2
1 1
= w3 1 — (w2/3! — w4/5! +
= (i + (w2,3! — w4/5! +...)
= + (w2,3!.)2
+...)
Then, 1( dw 1 2
2iri Ic w2sinw = 6 ir2
Sohition to 5.10.15: We assume C has counterclockwise orientation. First,Jo = = fc = 2iri. Now consider k 1. We have, for some constantsAo, . . . ,
1 A0 A1 Akf(z)= =—+ +•••+
z z—1 z—kThen
k k
1 =>2Am fl(z—n).
Equating coefficients gives=
Am. For n =0, 1, .. , k
LHence
=2Jri>An 0.
On the other hand,
Jk = 2iri Res ((z — 1)... (z — k)
=(—1)1'k!2iri.
Solution to 5.10.16: + 1)2 has a double pole at ±i/2. By the ResidueTheorem [MH87, p. 280], the value of this integral is 2iri times the sum of theresidues of + 1)2 at these two points. We have
= = = 1 + 2ir(z — i/2) + 2ir2(z — i/2)2 +•hence,
and so
+ 1 = —2ir(z — i/2) — 2ir2(z — i/2)2
(e22r2 + 1)2 = 4ir2(z — i/2)2 + 8ir3(z — i/2)3 +"•1
The residue at — is
Using the fact that = an identical calculation shows that the otherresidue is also —1/27r, so the integral equals —2i.
Solution to 5.10.17: A standard application of the Rouché's Theorem [MH87,p. 421] shows that all the roots of the denominator lie in the open unit disc. Bythe Deformation Theorem [MH87, p. 148], therefore, the integral will not changeif we replace the given contour by the circle centered at the origin with radiusR> 1. Using polar coordinates on this circle, the integral becomes
Re1 1W iRe'0 dO
1 =
j 1'2w
2ir Jo 12— 9R3e310 + 2R6e610 — + R—12
which has the limit as R -+ 00. The value of the given integral is then,
Solution to 5.10.18: We make the change of variables u = z — 1. The integralbecomes
J(+1I=2(2u +
Using the power series for the exponential function,
00
e en!(z —
n=O
ddz
(z — i/2)2
+dl 1
z=i/2
=+ — i/2) + O((z —
—8ir3+O(z—i/2) I
= + 0 (z — i/2))2 I
1
1f2lr
2iri Jo
dO
we get
The residue of this function at zero, which lies inside Ju + ii = 2, is 2e, so the
integral is 4eiri.
Solution to 5.10.19: Denote the integrand by f. By the Residue Theorem [MH87,
p. 280], 1 is equal to the sum of the residues off at —1/2 and 1/3, which lie inthe interior of C. 1 is also the negative of the sum of the residues in the exteriorof C, namely at 2 and oo. We have
Res(f,2)=z-+2 5
As urn 1(z) = 0,z-+00
Res (f, oo) = — lim zf(z) =0.00
SoI=
Solution to 5.10.20: The numerator in the integrand is times the derivative ofthe denominator. Hence, I equals times the number of zeros of the denominator
inside C; that is, I =
Solution 2. For r> 1, let Cr be the circle Izi = r, oriented counterclockwise. ByCauchy's Theorem [MH87, p. 152], and using the parameterization z = rd0,
1
— I rlzeutzO
As r —+ 00, the integrand converges uniformly to giving I =
Solution to 5.10.21: The integrand has two singularities inside C, a pole of orderI at the origin and a pole of order 2 at —1/2. Hence,
Ic z(2z+ 1)2dz = 2iri (Res
1)2+ Res _!))
The residues can be evaluated by standard methods:
/ Z
Rest ,o'I= e
\z(2z + 1)2 / z(2z + 1)2
I ez 1\ ldfez\Res + 1)2'
— = z=-4
Hence,
1 fez eZ\ 3e
2
I' eZ 1 3
Solution to 5.10.22: Let 1(z) denote the integrand. It has a pole of order 2 at aand a pole of order 3 at b, and no other singularities. The contour r' has windingnumber 1 about a and —1 about b. By the Residue Theorem [MH87, p. 280],
We have
Hence
Res(f, a) = —
1d2
—3
= (a—b)4
3
= (b—a)4
1=— (a — b)4) =
12 = —1
Solution to 5.10.23: The function f has poles of order 1 at z = ±1 and a pole oforder 2 at z = 0. These are the only singularities of f. The winding numbers ofy around —1, 0, 1 are 1, 2, —1, respectively. By the Residue Theorem [MH87,
p. 280],
1• If(z)dzReS(f,1)+2RC5(f,0)_ReS(f,1).2ii'i Jy
1 1+i
—1
I = 2iri(Res(f, a) — Res(f, b)).
Since 1 and —1 are simple poles, we have
_eZ I —eRes(f,1)= lim(z—1)f(z)= 2(Z+1)Ii 2Z
_eZI
Res(f—1)= jim (z+I)f(z) 2z
To find the residue at 0, we use power series:
eZ z
z2(1—z2)(l+z2+z4+...)= +z+
I
It follows that Res(f, 0) = 1. Hence,
1 (f(z)dz 2 + cosh 1.
Solution to 5.10.24: The function f(z) = (eZ — 1)/z2(z — 1) has a pole of order
2atz=0,andasimplepOleatz 1.Wehave
d ez_1I=—1,
z=oez —1
Res(f,I)=lim =e—l,z—+1 Z2
co(C,0) = —2,
co(C,1)=2.
By the Residue Theorem [MH87, p. 280], we have then
IeZ_1
dz = 2iri (Res (f, 0) co(C, 0) + Res (f, 1) w(C, 1))c z2(z— 1)
= 4jrie.
Solution to 5.10.25: The roots of 1 — 2z cos 0 + z2 = 0 are z = cos 9 ±i/i — Using the Residue Theorem, [MH87, p. 280], we get
1J 1 — 2zcosO + = (z — — e'0)
—Res(Z
— (z — —
/ Z —iO"+Res— e'0)(z — )
z —
e_u1ZO
= e'0 — e'0 + e'9 — e'0
—
— e'8sin nO
sinO
Solution to 5.10.26: Substituting z = we have
dO e'0d0I(a)=1 . =21Jo a + J0 2ae'9 + e20 + 1
2f dz— i Jizi=i Z2 + 2az + 1•
The roots of the polynomial in the denominator are —a + Ia2 — 1 and—a — — 1, of which oiily the former is within the unit circle. By the ResidueTheorem [MH87, p. 280],
I(a)=4JrRes( 1
z2+2az+ 1
Since the function in question has a single pole at z = —a + -Ia2 — 1, the residueequals
1 — 1
2z + 2a — — I
2irgiving 1(a) =
Consider the function F defined for 0 [—1, 1] by
dO
F is analytic on its domain. Combining with the previous results,we have that the function
—
is analytic and vanishes for > 1; therefore, it must be identically zero. Fromthis, we obtain that
dO — 2ir
Jo
in the domain of F.
Solution to 5.10.27: Let f be the function defined by
f(z) = (z—r)(rz —1)
We have
ff(z)dz=fForIrl <1,
Res(f(z),r)1
andf27r dO — Zir
Jo I — 2r cos 0 + r2 — 1 — r2
For rI> 1, we get
Res = —
anddO — 2ir
Jo I — 2r cos 0 + r2 — r2 — 1
Solution 2. Suppose that r E R and 0 < r < 1. Let uo be defined on the unitcircle by uo(z) = 1. The solution of the corresponding Dirichiet Problem [MH93,
p. 600], that is, the harmonic function on D, u, that agrees with uo on is givenby Poisson's Formula [M1187, p. 195]:
I—r2 f27T dOu (r)
= 2ir Jo I — 2r cos 0 + r2
but u 1 is clearly a solution of the same problem. Therefore, by unicity, we get
dO 2ir
Jo I_2rcosO+r21_r2If r> 1, consider a similar Dirichiet problem with uo(z) = 1 for Izi = r. We get
2_i dO
J0 1—2rcos9+r2so
(271 dO —
2r cos 0 + r2 — r2 — 1
If r <0, a similar argument applied to u (—r), u (— 1) leads to the results above,noting that cos(O — 7r) = — cos 0.
fir=1R1
e410dO
0 1 ± +e'0 )24
fir=491j
o 6 + +dO
z3
IzI=1
z2dz
IzI=1 z2+6z+1
(Res(z2+6z+1'
=—l2ir17)
—
1 e6'°+l5429d0
=—1 ,2ir e8'0 +e2'0
dO2 2e40 — Se210 + 2
1 •f z3-i-1dz.
2 —5z+2
We evaluate this integral using the Residue Theorem [MH87, p. 280]. As theintegrand has a simple pole at z = 1/2 inside the unit circle and no others, wehave
Solution to 5.10.28: Evaluating the integral using the Residue Theorem [MH87,p. 280], we have
Jo
cos4O e4'0dO=?Iij dO1+cos2O Jo 1+cos29
17+—7r.
Solution to 5.10.29: As 2 cos2 x = cos 2x + 1, we have
[27r cos6O+1dO
2Jo 5—4cos2O
3ir
S
Solution to 5.1030: The integrand is holomorphic except for a pole of order threeat the origin. By the Residue Theorem, we have
I = 2iriRes
Nearz =Owe have
cos3 z—
(1 — z2 + 0(z4)) (1 — + 0(z4))
=
showing that the residue in question is Hence I = —3iri.
Solution to 5.1031: Consider the contour in the figure
We have
ye
— 1dz,
(z— 1)2
1 —dO,
1 —cos9
vanishes (since
2,r—e1 — cos nO 1 — cos nOdO = tim IdO tim
Jo 1 — cosO 1 — cosOisinnO
the last equality because the integral of f(8) =1—cosOf(2,r —8) = —1(0)). Next,
1—
f1—z'2 dz
Je
dO— 1 — (z + 1 /z)/2 iz'1—cosO r
where is the almost-circle (e10 e 0 — e), traversed counterclockwise.The latter integral equals 2[
1
which, when e —÷ 0, tends to
\
so the integral in the statement of the problem has the value 2mn.
Solution to 5.10.32: Let z = e'0. Then
SindO = f
z — z1 iz
If—dz
1 JIzi=1 Z" — 1
= f (i + z2 + • +1 IzI=1
k=O 1z11
If n is even, only even powers of z occur, and each term vanishes. For odd n, wehave
sinn8 1 f dz- —=2,r.
sin 0 i z
Solution to 5.1033: We have
j'K(a) + (a) = I dO
a —cosO
=—21 . dOe2-'6 — 2ae'8 + 1
= 2iz2 — 2az +
dz.
Let 1(z) denote this last integrand. Its denominator has two zeros, a ± — 1,
of which a — v'a2 — 1 is inside the unit circle. The residue is given by
(a_V'a2_1)'t
Therefore, by the Residue Theorem [MH87, p. 280],
2Jr(a_v1a2_2 —1
Since the right-hand side is real, this must be the value of and = 0.
5.11 Integrals Along the Real Axis
Solution to 5.11.1: We have
(001
(00 (x—i)m(x + j)m+l (x — j)fl+l dx.
Ifm =n,weget
1(°° dx 1—I =—arctanx =1.
ir
Ifm <n, as x2 + 1 = (x — i)(x + i), we have
1 f°° 1
(x — j)n_flZ
Since the numerator has degree 2 less than the denominator, the integral convergesabsolutely. We evaluate it using residue theory. Let CR [—R, R] U FR be thecontour
We evaluate the integral over CR. For R > 0 sufficiently large, the integrand hasa pole at x = i inside the contour. Calculating the residue, we get
,in—m+J I r ,
Idx
x 1(x — j)n—m+1
—
By the Residue Theorem [MH87, p. 280], the integral over CR is 0 for all suchR. Letting R tend to infinity, we see that the integral over the semicircle tendsto 0 since the numerator has degree 2 less than the denominator. So
1 i
it J_oo x2 + 1 (x — j)n-m dx =0.
Solution to 5.11.2: Consider the following contour around the pole of the function
1 _euIaIzf(z) =
_____
On the larger arc z = R(cos + i sine), so
1 — I + = o (R_2) (R -÷ oo).
Then
f°°1—cosax 1 / •
f2
2
Solution to 5.113: Let (z) = eutz/(z + i)2, and consider first the case t > 0.Then (z)l is bounded in the upper half-plane by Iz + For R > 1 letCR = FR U [—R, R], where rR is the semicircle centered at the origin joining Rand —R, oriented counterclockwise. The function f is holomorphic on CR andits interior, so, by Cauchy's Theorem, we have
0= f f1(z)dz
= f f,(x)dx + f
The absolute value of the second summand on the right is at most irR/R2, sinceLfz(z) R -+ 00 we obtain 1(t) = 0(f 0).
Suppose now t <0. Then lit is bounded in the lower half-plane by Iz + i12.Let CR be the reflection of CR with respect to the real axis, oriented clockwise.By the Residue Theorem, we have
f ft(z)dz = —2iriRes (fi(z),
iR
it (z)dz.
A calculation shows that the residue equals ite', so
f f,(z)dz =CR
As R oo, the contribution to the last integral from the semicircle tends to 0since (f, I (R — 1)2 on the semicircle, giving 1(t) = 2ntet (t <0).
Solution to 5.11.4: For t = 0 the integral is elementary:
çR dx R
I dx= —0 as R—÷oo,J_R(X+1)3 2(x+02 —R
hence F(0) =0.
For t <0 the function (z) = is bounded in the upper half-plane, and(z+i)3
in fact is O(1z13) there. We integrate around the contour FR consisting of theinterval [—R, R] on the real axis and the semicircle in the upper half-plane withcenter 0 and radius R (R > 1), oriented counterclockwise. The integral is 0 byCauchy's Theorem [MH87, p. 152]. The length of the semicircle is nR and theintegrand is O(R3) on it, so the contribution to the integral due to the semicircletends to 0 as R -+ oo. The contribution due to the interval [—R, R] tends to F(t),so we conclude that F(t) = 0 for t <0.
For t > 0 we integrate ft around the reflection of FR with respect to thereal axis (oriented clockwise). The integrand has one singularity in the interior of
a pole of order 3 at z = —i. The residue there is
1 d2 _t2et= 2
By the Residue Theorem [MH87, p. 280],
f (z)dz = —2jri' =
By the same reasoning as above, the preceding integral tendsto F(z) as R —* 00.Hence F(t) = for t > 0.
Solution to 5.11.5: To see that the integral exists, notice that
sin2x sin2x / I \hm2 =1,
2x-÷OX X
As the integrand is an even function, we have
r°° sin2 x r°° sin2 x
J dx=2J dx._oox 0 X
Letf(z)=ForO <8
I —• Then f is analytic except for a simple pole at 0.
<R, consider the contour C8IR = FR U U [—R, —s] U [e, R]
iR
by Cauchy's Theorem [MH87, p. 152]. We have, by the Residue Theorem,
o=fCE.R
f(z)dz= IJ ÷1 ÷1÷1[—R,—e) Ye [eR]
IiIf(z)I=
z2
therefore100 l_e21x
f—00
Since sin2(x) = (i — e2ix), we have
00
0
1 f°o,
XL 4 J—00
(00 sin2xI
JO
1 — e21X-) dx,
X'.
Solution 2. Assuming Dirichiet's Integral
f°°sinx yrdX=—
J0 X 2
whose evaluation can be found in [Boa87, p. 861 and [MH87, p. 313], we can use
integration by parts:
00
0
-2 i 00S1flX -2dx=——sinX —
X 0 0
1.—— sin2x dx
X
It is easy to see that on rR we have
2
dx=iriRes (f,0) =,ri(—2i) =2,r.
soJr
2
=o+f000s1;Y dy
Jr
2
Solution to 5.11.6: The integrand is absolutely integrable, because it is boundedin absolute value by I near the origin and by I/1x13 away from the origin. We
have
sin3 x = — = — e_3iX — 3e1X + 3e_1x)
= —
For 0 <s <R, consider the contour C€,R = FR U Ye U [—R, —s] U [s, RJ
iR
by Cauchy's Theorem [MH87, p. 152],
3eZ —dz=
zJ rR
— e3iZ 1 (2+ 3(iz)2z3
2
f — e3i2I dz f0 —
+ 0(1)) iee'0 dO
2i 0
= + 3i j dO + 0(s)
=O—3iri+ Ofr)-÷ —3ir as s -÷0.
o=fCS.R ÷1 ÷1+1[e,R]
The integral over is bounded in absolute value by (4R3)(27rR) sinceand 1e319 are bounded by 1 in the upper half-plane, so it tends to 0 as R —* oo.To estimate the integral over we note that
Hence,
— (3iz)2 + 0(z3))
(z-÷0).
Thus,/ —8 3eix — e3" R 3eiX — \urn I ( dx + ( dx i = 3iri.
x3 "8 /'
The integral is one-fourth the imaginary part of the preceding limit, so
1(z) = (z2±i)2Integrate f over a closed semicircular contour CR = [—R, R] U with radiusR in the upper half plane.
The portion [—R, R] along the axis gives
çR x3sinxI dx.
J_R(X2+1)2This integral converges as R -÷ 00 to
x3sinxI dx.
To establish the convergence, note that integration by parts gives
R(— cos x)x3
(x2+l)2 _R+
and(—cosx)x3
(x2+l)2 lxias x-+oo,
Solution to 5.11.7: Let
(z+i)2(z—i)2
0 R
R
LR (x2 + 1)2(cosx
whose imaginary part is
+ i sin x)dx
çR x3smxI dx=
J_R(x2+l)2dx
and the second term is integrable by the Comparison Test [Rud87, p. 60], O(1/x2).The real part also converges by a similar reasoning, and since the integrand is odd,it converges to zero.
The integral of f over FR is treated as follows. Let z = Re'9 for 0 0 ir.Then
p iz 3 pir iR(cos0+isin0) 3e z e •n sO
JrR (z2 + 1)2Z
— Jo (z2 + 1)2
This is bounded above in absolute value for large R by
A
for a constant A and this integral tends to zero as R -÷ oo (Jordan's Lemma[MH87, p. 301]). Finally,
I f(z)dz = 27ri Res(f, i)JCR
and since we have a second order pole, we have
Res(f,i) =L4e
Thus,
and so the required integral is2e
Solution to 5.11.8: Using an argument similar to the one in Problem 5.11.7 withthe function
z1(z)— (z2+1)2'
we get
f°° xsinx d etz
(x2 + 1)2dx = (2iri Res(f, i)) FJt2lrid_
!L\ 4e) 2e
Solution to 5.11.9: Consider the complex integral
fR z(z2 +
J f(z)dz =CR 2e
where the contour C8, R is the contour described below oriented counterclockwise.The integrand has simple poles at 0 and ia.
sinxNotice that =
x(x2 + a2) (X(X2 + a2)) for real x. By Jordan's Lemma
(MH87, p. 301], we have
f I
JrR z(z2 + a2)dZ JrR (z2 + a2)dZ o(1)
so, at the limit, the contribution from the bigger semicircleis zero. Using the Residue Theorem [MH87, p. 280], we have
(IRI -÷ 00)
to the integral above
If (+a2)z(z2
dz = 2jri Res
I=2iri. . +7ti—ia(2ia) a2
1
etz — 1Solution 2. Consider the function which has a removable singularity at
z(z2 + a2)sinx /
the origin and a simple pole at ia. Again, we have = Ix(x2 + a2) \x(x2 + a2)
By the same argument as above, but with one less pole, we can use the followingcontour
since f is even we have,
and get
therefore,
Solution to 5.11.10: We have
(sinx)(x + 3i)x2+9
çR sinxI
J_RX—31We evaluate these integrals over the contour CR = FR U [—R, R]:
As z2 +9 has simple zeros at ±3i, by the Residue Theorem [MH87, p. 280],
f dz = 2,ri Res (f, 3i) =CRZ +9
By Jordan's Lemma [MH87, p. 301], we have
LIFRdzl+9
I JrR
!zIle'92IzI —9
irRR2 —9
= o(1) (R-+oo)
0 R
Cl_dx=2iriRes+ a2)
I1,,z(z2 + a2)'
zain
a21 —
I —a1= (1—e2a2'
sinxx —3i
So
—
fR xeix
J-Rdx.
x2+9
—R 0 R
so, in the limit the integral along the upper half-circle contributes nothing.We can evaluate the second integral in the same way, getting
I' e3yrJcRz2+9dz= 3
Again, in the limit, the upper half-circle contributes zero, so
çR sjnxInn I dx=e ir.
R—+ooJ_R x — 3i
Solution to 5.11.11: Consider the function f defined by
eIZf(z)=
z(z—7r)
By Cauchy's Theorem [MH87, p. 152],
f(z)dz =0
where CC,R is the contour TR U U [e, ir — e} U U [jr + s, R].
The poles of f at 0 and are simple; therefore,
lim f f(z)dz = —in Res (1(z), 0) = is-÷O Jy0
urn I f(z)dz = —in Res (f(z), in) = i.
We also have, for Izi = R,
I=o(1r2
z(z—ir)(R—÷oo)
iR
so
urn I f(z)dz =0.R-+ooJrR
Taking imaginary parts, we obtain
too sjnx dx=—2.J_oox(x
Solution to 5.11.12: The integral equals / dx. For R> 1 let(1 +x2)3
be the contour consisting of the interval [—R, RJ on the real axis and the tophalf of the circle Izi R, with the counterclockwise orientation. The function1(z) = eiz/(l + z2)3 is analytic on and inside FR except for a pole of order 3 atz = i. By the Residue Theorem [MH87, p. 280],
I f(z)dz = 2iri Res (1(z), i)
Writing 1(z) = , we see that(z — i)3 (z + i)3
d2 /2Res(f(z),i) =
d / ietZ 3e1Z
= dz
(z + (z + (z + i)5 )f—i 6i 12 \ e1 /1 3 3"
7
in.Hence f(z)dz = — Now, since is an odd function on R,
JFR 8e
f(z)dzcosx
frR = J_R (1 + x2)3dx + ( f(z)dz,
SR
where SR is the top half of FR. As R —+ 00 the first term on the right convergesto the integral we want, and the second term converges to 0 because Jf(z)Ii/(1z12 — in the upper half-plane, and we conclude that the value of theintegral is 7ir/8e.
Solution to 5.11.13: Note that 1 + .x + x2 = (x + + so the denominator
does not vanish on the real axis. The given integral is I = 9t I 1+x+x2As
I1 for z in the upper half-plane, and since the denominator has degree
2, we may close the Contour in the upper half-plane to get I = 9t(2iriR), where
R is the sum of the residues of f(z) = + + in the upper half-plane. In
this domain f has just one singularity, a simple pole at + with residue•11 I •k
R = = . Therefore
k—.
Solution to 5.11.14: An argument similar to the one used in Problem 5.11.11 withthe contour around the simple poles —1/2 and 1/2, gives
/1
1dx = i (42_ 1' —
+ Res— 1'I
°° cos(irx)
Ij j\
Solution to 5.11.15: Consider the following contour CR and the function
1
1 dzkf,rR
R4 —Idzt
R4 — 1
iR
—R 0 R
which approaches 0 when R -+ oo (note that for z E I = 1).
Taking the real part and using the fact that for these roots 1/4 = —zo, we have
(00 cos nx= tim
/)x4 +1 dz
R-4oo +1 dz
/= (Res
+ 1 ) + Res+
I I
= (43 1.14 +4z3 Le3wu14)
n
= cos—+sin—).
Solution to 5.11.16: Consider the same contour CR as in Problem 5.11.15; whichI 3iri
encircles two of the poies and er), and the function
z2 + 1f(z)=z4 + l•
We have
1
+dz = (zir i (Res + Res
(z2+1I z2+1=lri I
4z3' +
4z3I
z=et I
Solution to 5.11.17: We have
p00 cos3x °° 2dxdx=
J_00a2+x2 L a2+x2(00 + + 3e1x + 3ex
dx.1—00 a2+x2
1
J°0e3tX 00 e1x
+x...
=
Ii (a), let rR (R > a) be the contour consisting of the interval[—R, R] on the real axis plus the semicircle in the upper half-plane with center 0
and radius R, oriented counterclockwise. By Cauchy's Integral Formula [MH87,p. 167],
f d* = IrR Z2 + a2 i JrR (z + ia)(z
• dz—ia)
z=ia
,re3a
a
The contribution to the integral from the semicircle is o(irR
) (as Je319 1R2—a2
for 0), so it tends to 0 as R -+ oo. The contribution from the interval
[—R, R} tends to Ii (a) as R —+ 00. It follows that ii (a) = • Exactly the
7re0same reasoning shows that 12(a) = • We obtain
a
1(a) + 3e_a).4a
Solution to 5.11.18: This integral converges, by Dirichiet's Test [MH93, p. 287],since
monotonically and
xlim =0
x-+±oox2 +4x +20
j'13
sinxdx=O(1) (a-+—oo,fl--+oo).Ja
= ICRf(z)eiz dz where 1(z) = +
z
4z+2O
The curve CR is the contour
—R 0 R
Consider the integral
where FR is the semicircle of radius R> 0. The integral I is equal to the sum ofthe residues of g(z) = f(z)etz inside of CR. f has poles at —2 ± 4i, of whichonly—2 + 4i lies inside CR. Hence,
urnz—+ —2+4i Z — (—2 — 4i)
= + 4i)e42'
We have, by Jordan's Lemma1 [MH87, p. 301], when R -÷ 00,
Rir
00
-00f(x)sinxdx = (L:
f dz)CR
= + 4i)e_4_21)
= '(2cos2+sin2)
Solution to 5.11.19: Consider the following contour avoiding the pole of the func-tion
we have
z +f(z)=
Izi + leizi
R3
which approaches 0 when R -÷ oo (note that for z E FR, leizi = 1); andaround 0,
(iz)2+iz+
I =2iriRes(g,—2+4i) =bri
So,
I I g(z) dz'R
frRIdzI
LFrR IR2—4R20
Z + dzkf ir(R -F- 1)
R2
z + ieZ 1
z+i(1
where g is analytic. Then
_____
Jrt z + ie1Z P dz+ I g(z)dz = + g(z)dz.
z3
Now since
I g(z)dz max Ig(z)Iire 0 as s -÷ 0
the integral along the real axis will approach Taking the real part, one gets
f°°x—sinx lit itI dx = —— = —.
Jo x' 22 4
Solution to 5.11.20: Denote the given integral by I. Consider the function1(z) = (1 + z + z2Y2. I has two poles of order 2 at z (—1 ±We evaluate the contour integral
= ICR
By the Residue Theorem [MH87, p. 280], for R large enough,
I' = 2jri Res2
As R tends to infinity, the integral along the upper half-circle tends to 0 since thedenominator of 1(z) has degree 4 higher than the numerator. Hence, I' tends to Ias R tends to infinity. Therefore,
—2
)
iR
—R 0 R
I=2iriRes 1(z),
=dz
4ir
Solution to 5.11.21: Letting t2 = x, we get
,'OO p 00X dx=21 dt.j0 i+x j0 i+t2
Consider the integral,'I dz,
JC€,R 1+zwhere C8, R is the contour IT?? U U [—R, —€1 U [e, R]:
where the small circle, has radius s, and the large circle, rR, radius R. Sinceis defined to be we choose as our branch of the logarithm that
with arguments between —jr/2 and 3ir/2 (that is, the one with the cut along thenegative imaginary axis).
The integrand has a simple pole inside C6 R at i, so, by the Residue Theorem[MH87, p. 280], the integral is equal to
2jri Res (f, i) = =
As e and R tend to 0 and infinity, respectively, the integral on [s, R] tends tothe desired integral. On the segment [—R, —s], we make the change of variablesy = —z, getting
6 R
LR 1 + z2dz = f + 2 dy,
so this integral tends to a constant multiple of the desired integral. On FR,a cal-culation shows that = so if we assume that 0,we have
Iand
fi
JF'R I+.Z R2—1
This tends to 0 whenever 9la <1.On Ye' essentially the same estimate holds:
dz1 —e
and this tends to 0 whenever > 0.So we have
p00 2a—I
(1 — I dt =J0
Dividing through, we get
—
2sinira
Therefore, twice this is our answer, subject to the restrictions 0 < < 1 and? 0. However, if 0, we may replace a by and obtain the above
equality with Then, by taking the complex conjugate of both sides, we see thatwe can eliminate the second restriction.
Solution 2. Considering the slightly more complex contour of integration below,we can do away with the change of variables. So consider the integral
pI dzJc z+l
where C is
The origin is a branch point and the two straight lines are close to the x—axis.The integrand has one simple pole at z = —1 inside C. Thus,
a—I / Z = 2yri urn Z (z + 1) =f Z dz=2riRes(Jcz+l \z+l )On the other hand, the integral can be split in four integrals along the segments of
the contour:
a—i a—i a—I a—i
[ Z dz+ IZ dz+f Z dz+f Z
dzJrRZ+l JYe.Z+1 t1Z+l e2z+l
2ir (Re'0 )dY_ 1 0
+1=
Re10iR e'0d0 + ise'0 dO
XR a—I f8j8 x+l R
Taking the limits as s —+ 0 and R -÷ oo, we get
00 Xa—i 0(a—1)iridx+ dx—2irief0x+1 100 x+1 —
orco a—i
(1 — e2 a_I)) f dx = 2irio x+l
Hence,°°
I dx= =o x + 1 1 — sin
Solution to 5.11.22: Making the substitution x = the integral becomes
I dyoo —1/4
l+Y
which is equal to by Problem 5.11.21 with a =4
Solution to 5.11.23: Making the substitutionx = the integral becomes
1(00
which is equal to by Problem 5.11.21 with1
a—5-
Solution 2. We integrate the function 1(z) = 1 /(1 + z5) around the counterclock-wise oriented contour FR consisting of the segment [0, R], the circular arc joiningthe points R and R and the segment {R 0], where R> 1.
The function f has one singularity inside FR, a simple pole at Theresidue is given by
Res(f —(1 + z5) — 5dz z=iri/5
By the Residue Theorem [MH87, p. 280], 1 f(z)dz2iri
The contribution to the integral from the segment [0, RI equals Iand the contribution from the segment [R 0] equals
1 1
— I d(t — 1 ' diJo 1 + (t — Jo 1 + t5
=1R
dxJo 1+x5
The contribution from the circular arc is so it tends toO as R 00.
Taking the limit as R -+ oo, we thus obtain
(1 — I dx = 2iri
Jo 1+x5 5e41nh/5'
or
2jri5(e4lni/5 — =
— 27ri —
— —
Solution to 5.11.24: Observing that the function is even, doubling the integralfrom 0 to 00 and making the substitution y = we get
f°° dx —(°° dx — 1 (°° d —
1 +_2j
+ — n Jo y + I — n sin ir/2n
by Problem 5.11 .21.1K
Solution 2. The 2nth roots of —1 in the upper half-plane are Zk = e T' for
0
1
1+x52iri
5(ehlu/5 — e"/5)
Thereforef00 dxI =2nzI 1.Lr2fl
',—00 I
iR
Ic=0
Since Zk is a simple pole of the integrand,
giving
1 1 Z—Zk 1 ZkRes
+'Zk) = lim = =
1 + 2n
a—dx
I-00 xk=O
jjt r 1 — 2jri , 1= =
l—e'n
Solution 3. Let 1(z) = 1
1and consider the contour in shape of a slice of
pizza containing the positive real axis from 0 to R, the arc of circle of angle jr/nand radius of the circle from the end of the arc back to 0, oriented in this way.
FR
f has one singularity inside the contour, which is a simple pole at There-fore, the integral of I around such a contour equals, by the Residue Theorem[MH87, p. 280],
/ 1•iri/2n
2iriRes( 1=I
—R 0 R
R
We have, then
1 1
+dx + I iRe'9d9 + I 1 +
i
i ++dx + iR / e'0d0.
The integral on the arc approaches zero as R —+ 00, therefore, we get
JOO1 =
dx=—2— I n(1 _elnu/h1)
1Solution 4. Consider the complex function 1(z) = • Its poles are all sim-
+ 1pie and located at the points
Ck = = k = 0, 1, . .. , 2n — 1
and since k and n are integers, the roots are never real and calling the primitiveangle a = the ones located above the real axis are
k=0,1,...,n—1.
Computing the residue of f at these points we have:
Res( k=0,1,...,n—1
so
1 1 (2k+I)(1—2n)
Res (Z2fl÷1 = 2n
=(2k+IX-2n)
2n
=2n
—— 2n
Using the upper-semi-circle as the contour of integration, we have to add then—i
residues on all poles in the upper-half-plane. Using the fact that— 1—z
and applying the Residue Theorem:
n—iR iRes ( 2n+1
11Rdx +
JCRdz =2
k=O
=k=O
in 1 —= ——e
1 — e12.a
— in — 1 — in 2i
— n ea — — n sin a
Solution to 5.11.25: Let
The answer is 21, where 1 = I f(x) dx.Jo
For R> 1 let FR be the straight line path from 0 to R followed by the arc ReItfor t E [0, 2ir/n], followed by the straight line from to 0.
Let = The poles of f are form E Z,so the only pole inside FRis and f has a simple pole at with residue By the Residue Theorem,we have,
!rR
2ini3
The integral over the first line tends to I as R -÷ 00, over the arc of circle partit approaches 0, since the integrand is while the arc length is 0(R), andthe integral over the last line segment tends to as the substitution z =shows. Thus,
=
Now,
sin—= =n 2i
so
21
Solution to 5.11.26: We have
Lex
2
1 f°° logudu2j0 (u—l/u)u
1 f°°du
where we used u = ex. Integrate
1(z)= logz
z —1over the contour r = rR u Y.i u u
iR
By Cauchy's Theorem [MH87, p. 152], we have
I f(z)dz=0
[ f(z)dz = Res(f, —1)—
Jy_1 — i-.
f(z)dzlog 1
fyi 2=0
f RlogR(z)dz-+0 as R-÷oo since urn =0
frR
[ - slogsf(z)dz -+ 0 since urn = 0.
"YO
I =8
Solution to 5.11.27: The roots of the denominator are I ± and for large lxi,the absolute value of the integrand is of the same order of magnitude as Itfollows that the integral converges. For R > 2, let
1Rf ezz2 —2z+4
dz,
also,
So we get
and£ log lUId =
where the contour CR consists of the segment [—R, R] together with the semicir-cle FR in the lower half-plane with the same endpoints as that segment, directedclockwise.
0 R
The integrand has only one singularity inside CR, a simple pole at 1 —By the Residue Theorem [MH87, p. 280],
IR=—2yriRese IZ
\z2—2z+4
The residue, since we are dealing with a simple pole, equals
/ etZlim
V
e1Z
z— 1
Hence, JR = . SinceI is bounded by 1 in the lower half-plane, we
have, for large R,
Hence,
fez
FRZdzk
IR2—2R—4
(R-+oo).
• •glvlng/=
Solution to 5.11.28: The integral converges absolutely since, for each 0 < s < 1,we have logx = o(f) (x —+ 0+).
Consider the function
logzf(z)
— (z2 + 1)(z2 +4)
—R
1Rdx+ [
—R x2 — 2x +4 J TR — 2z +4 dz > I (R-÷oo),
and, for 0 < 8 <R, consider the contour = FR U U [—R, —e] U [e, R].
On the small circle, we have
iR
IlogsI+ir
I
2
(e-÷0)
and on the larger one,
frR (R2 — 1)(R2R=o(1) (R—÷ oo).
Combining with the Residue Theorem [MH87, p. 2801 we get
0
-00
toof(z)dz+ I
Jof(z)dz = 2jri (Res(f, i) + Res(f, 2i)).
Res(f, 2i) = Jim f(z)(z — 2i) =—12i
We also have
and
Iog(—x) +d
(x2+1)(x2+4)x
foo logx foo=1 dx+IJo (x2 + 1) (x2 +4) Jo
,'OO
f(z)dz = IJO
We have
and
irRes(f, i) = Jim f(z)(z — i)
z—+i —i-i
f(z)dz =
It24
0
-00
ilog212
In
00
Jo
logx
(x2+1)(x2 +4)
dx+4)(x2+1)(x2
using the same contour and singularities to evaluate
(00 yrjI dx
J0 (x2+1)(x2+4)
weget ,so
coo logx ir2i •(n. ilog2
Jo (x2+1)(x2+4) 12 — k12 12 24
andlogx n.log2
IJo (x2+l)(x2+4) 12
Solution to 5.11.29: For 0 <s < 1 < R, let Ce,R denote the contour picturedbelow. Let log z denote the branch of the logarithm function in the plane slit alongthe negative imaginary axis.
By the Residue Theorem [MH87, p. 2801, since i is a simple pole of the integranci,we have
f (logz)2
JC€R Z +1
(logz)2=2jri limz-+i Z+j
2n.i2i
4
For x negative, we have logx = log lxi + jri, so (logx)2 = (log ix 1)2 ir2 +2,ri log lxi. Thus, the contribution to the integral from the interval [—R, —e]
and
— + 2iti logx R (logx)2 —dx= ( dx+fx2+i j8 x2+1
L1rR
+ [z2+1
= o(i)(log R)2(irR)
R2 — 1(R—÷oo)
I (logz)2
lYe
(log s)2(irs)= o(l)s—i (e —4 0).
In the limit we then obtain, considering the real parts,
Jo x2+1
Solution to 5.1130: For X = 0, we have
(sechx)2cos).x =
so we consider the function f given by
1(z) =(eZ
This function has a simple pole inside the contour
C—[—R,R]U{R, R+iri]U[R+iri,
equals
(logx)2
Also,
(log z)2dz
z2 + 1
Iogxdx
x2+1
fR+2iri I
4
2J0
(logx)2 it3+
"00
Jo
and
dxx2+1
it3 it3
it3
8
f(sechx)2dx = tanhxl8° = 1.
pictured below:
We have
—R 0 R
and
—
R dxf(z)dz =
— f-R (_ex — e_x)2
pR=
J_R (ex + e_x)2
JR. R+iri]
similarly, we get
I
=
(eR — e—R)2
f(z)dz=o(1)
Therefore,
Thus,
/ jri\(1 —
i_co (ex + e—')2= 2jri Res -i)
(00 Air/2J
(sech x)2 cos Ax dx =sinh(Xir/2)
— Air
2
Solution to 5.11.31: If b = 0, the integral is well known:
00
0dx
=2
and can be computed either by doubling it up
p00
IJo
2/100
I eYdy=j IJo Jo Jo
dx dy
•iri/2
iridx
=o(1) (R—*cx)
and converting to polar coordinates or by using the Residue Theorem applied tothe function 1(z) = e—z2 and the following contour, as shown by [Cad47, Mir49].For the details, see [MH93, p. 321-322]. We will use this result to compute thefull integral (for other values of b) below.
Now consider the function 1(z) = e2. Ifb > 0, we will consider the integralover the rectangle contour of height b and if b <0 the symmetric one below thex—axis.
Ti
—R 0 R
= — J dx—R
R= I (cos(2bx) + i sin(2bx)) dx
J—R
FR
R
Since is entire, the integral around the contour is zero, for any value of R. Ifb <0 draw the contour below the x—axis. Evaluating the integral over each oneof the two sides parallel to the x—axis, we have
pI f(z)dz= IJi J—R
I f(z)dzJill
ç_R
JRdx
R
= I cos(2bx) dx.J—R
Along the vertical segments (II and IV), If(z)I = J =
then 1111 11 and fI are bounded by b fixed, this goesto 0 as R oo. Letting R -÷ oo, the integral along the circuit becomes
n00 nOO
I dx — I e_X2 cos(2bx) dx = 0J—oo J—oo
Now using the pre-computed integral the result follows:
11001
/ cos(2bx) dx = — / cos(2bx) dxJo 2
"00— e_b2 / e_X2 dx
J-oo
2
Solution to 5.1132: The integral is the real part of"00 "00 "00
I= J
dx= J
dx= J
dx0 0 0
Consider the contour 1' in shape of a slice of pizza, like the one in Solution 3 ofProb. 5.11.24, with angle ir/8. By Cauchy's Theorem [MH87, p. 152],
IJrOn the arc the integral will approach 0 when the radius R —÷ oo, because theintegrand is bound in absolute value by I = Thus
1100 1100
0 = / dz — IJo Jo
or equivalentlyE'OO
0 = I du —Jo
so I = and taking the real part, the integral is equals to
(2+
Solution to 5.1133: Denote the line segment in C from zo to Zi by [zo, zi]. Leta, b > 0, and C1 = [b, —a], C2 = [—a, —a — iy], C3 = [—a — iv, b —iy], C4=[b—iV,b].
—a 0 b—
____
—a—yi yi b—yi
We integrate 1(z) = round the rectangularcontour= Cj+C2+C3+C4.(The figure shows the case y > 0.)
The reverse of Ci is parametrized by z = t, —a t b. Thus
f(z)dz =
The curve C2 is parametrized by z = —a — ity, 0 t 1. Thus
1C2 1(z) dz f dt = 101 dt
so
0, asa 00.
The curve C3 is parametrized by z = r — iy, —a t b. Thus
Ic3 1(z) dz = f t—1}')2 dt.
The opposite of C4 is parametrized by z = b — ity , 0 t 1. Thus
f(z)dz = iy 101 e_ dt =f1
dt
therefore
asb—÷ 00
By Cauchy's Theorem [MH87, p. 1521, the integral of f round the rectangular
contour is zero. Thus— 1b
dt + 1C2 1(z) dz + f dt + j 1(z) dz =0.
Let a, b —* oo. We deduce that
t°° 1 1 2I dt= I dt=1
proving the result.
6
Algebra
6.1 Examples of Groups and General Theory
Solution to 6.1.1: 1. The set G is the set of all invertible 2 x 2 real matrices. IfA, B E G, then IABI = IAIIBI 0 and thus AB E G. Matrix multiplicationis associative. The identity matrix I has determinant one, and thus belongs to G.Finally if A E G then A is invertible, with JA1
I= 1/IAI 0, and thus G is
closed to inverses. This proves that G is a multiplicative group.2. Let
A=(a I)b(1 0)
Then
a b
B A Ca b and A a the otherhand any scalar matrixcommutes with every element of G. We conclude that the center of G is the set ofall real 2 x 2 matrices al with a 0.3. if A, B are orthogonal then (AB)(AB)t ABBA1 = I, so 0 is closed undermultiplication. The identity matrix is orthogonal. Finally, if A is orthogonal thenA1 = A' and A_1(A_1)' = A'A = I, thus 0 isasubgroupofG.
(1
(110
show that c = 0. And the products
fi10
\01
o
1 11=100 1/a b\ /1I cI=I00 1)
a a+b1 1+c0
show that a 0. Finally the products
fl10
0
1
0
0
0
0
1
0
x = x 0 and
1 1 o\fo i\1i o\ fi 1X
—1
is not orthogonal. This proves that 0 is not a normal subgroup of G.4• = \ (0} is an abelian multiplicative group. The map det: G —÷ IR* is agroup homomorphism which is clearly surjective, and nontrivial.
Solution to 6.1.2: 1. The set G is a subset of the multiplicative group GL3 (IR) ofnonsingular matrices. It is easy to verify that the product of two elements of G hasones on the diagonal, and zeros below the diagonal. Moreover the identity matrixbelongs to G. To verify closure to inverses, let A E G. Then the characteristicpolynomial of A is (x — By Cayley—Hamilton's Theorem [HK61, p. 194],(A—I)3 0,therefore A' = A2 —3A+31. From the closure property, A2 E G,and it is easy to see that A2 — 3A +31 has ones on the diagonal, and zeros belowthe diagonal, so A1 E G. Thus G is a group.
(I a b\2. Let A = 0 1 c be in the center of G, so that A commutes with every\001)member of G. Then the products
a1
0
b\flcilOiJko
I
1
0
o\ fl01=101)
1+a1
0
bc1
1
1
0
0\ (10110
a1
0
b\ (1cI=I01) \o
I+a1
0
b+cC
1
a bI 1+c0 1
a b+x1
0
show that the center of G is the set of all matrices
fIOx\10 1 XER.\ooiJ
Solution to 6.1.3: 1. S3 is a nonabelian group.2. The Klein four group V = x Z2 is a finite abelian group that is not cyclic.3. Z is an infinite additive group; the subgroup 57L is a subgroup of index 5 in Z.4. and Z2 x Z2 are nonisomorphic finite groups of order 4.5. S3 has a subgroup H = (a), where a is any transposition in S3, which is notnormal in S3.6. The alternating group A5 is nonabelian and has no normal subgroups other thanthe full group and the identity.7. The additive group Z has a normal subgroup H = 3Z; the factor group Z/37Zhas order 3, and is not isomorphic to any subgroup of 7L.8. The left (also right) cosets of H in G form a partition of G, and for any g G,
gH = H iffg E H. Let H have index 2 in G. Then there are two left cosets Hand G \ H, and these are also the two right cosets of H in G. Fix x E G. If x HthenxH If = Hx.Ifx HthenxH = G\H = Hx.ThisprovesthatHisnormal in G.
Solution to 6.1.4: 1. If x, y G(R) have inverses x', y' E R respectively, thenxy(y'x') = (y'x')xy = 1 proving that xy has an inverse. The identity 1 hasitself as an inverse . If x has inverse x', then x' has inverse x. Associativity ofmultiplication is given. Therefore G(R) is a multiplicative group.2. Let x = a + ib have inverse y = u + iv, with a, b, u, v integers. Then xy = 1,
and taking moduli Ixilyl = I and 1x121y12 = 1. Since a, b, u and v are integers,1x12 = a2 + b2 = 1, giving a ±1, b = 0 or a = 0, b = ±1. The units are1, i, —1, —i. This group is cyclic, generated by i, and so isomorphic to
Solution to 6.1.5: Let G be the group of symmetries of the network and let D bethe triangle with vertices P0, P2, and P3.
Let a and be 180° degree rotations about the midpoints of the segments PoP3and P0P1, respectively, and let t be a reflection in the line extending the segmentPoP2. Let x be a point in the plane. With this notation, it is easy to see that:
afi(x)=x+(0, 1)arac(x)=x+(2,O).
Claim 1: a, t generate the group G.Proof. Note that the four triangles D, a(D), ar(D), and ata(D) tile a one-by-two rectangle. Applying the symmetries af3 and arat we may tile the plane withcopies of this rectangle. Finally, any element of G which fixes D setwise must bethe identity.
Note also that a, 6, and r satisfy the relations:
a2 462 = i:2 = 1
T] = I
the second relation holds because the line of reflection of r is parallel to the trans-
lation given by a46.We will show that the abstract group G' with presentation
is isomorphic to G. We first prove that G' surjects onto G via the natural map by
finding a single preimage for every element of G.
Claim 2: The set of elements N = in G', where p and q areintegers and where Z = 1, a, at, or ata, injects and surjects onto the elementsof G.Proof. Surjectivity follows from Claim 1. Injectivity is also clear as there is abijection between N and the triangles of the network. We will refer elements ofN as being in normalform.
Claim 3: Any word in a, 46, and r can be put in normal form using the relationsof G'.Prooft To do this we first apply the recursive algorithm below. Note that the rela-tion [cr46, ti = 1 can be rewritten to obtain t46 = 46ara and ra46 = c43t. We aregiven a word of the form (af3Y W(a, 46, r):
Step 1: Move all in W to the left:
=a16 —usestep2.
Step 2: Move all in the new word to the left using:
aa16 — delete the aa and go back to step 1.ta,6 =
— go back to step one and move the fi which is on the left.
This procedure can be shown to terminate by applying induction to the correctlexicographic order. Thus we have shown that any word can be put in the form
W(a, r), where W is a word in a and r only. But, since a2 = = 1, wemay assume that W(a, r) = (aT)r Q, where Q = 1, a, or r. It is straightforwardto place W in the desired normal form by considering the possible values of r andQ. It follows that G' is isomorphic to G. This group is isomorphic to the pmggroup of [CM57, p. 45].
Solution to 6.1.6: If a, b E G, then a > 0 and b 1, so E G. Therefore,the operation * is well defined.
Identity. The constant e is the identity since a * e = = a1 = a and
Associativity. We have
(a * b) * = = lIbI = = a * (b * c).
Invertibilily. Since a 1, exists and is an element of G. A calculationshows that a * = * a = e.
Solution 2. The map log: G -÷ R \ (0) is a bijection that transforms the operation* into multiplication; that is, log(a * b) = (log a)(logb). Since R \ (0) forms agroup with respect to multiplication, G is a group with respect to *.
Solution to 6.1.7: 1. Let a E G. The centralizer Ca = (x GIxax1 = a) of a
is a subgroup of G.We have,
xax' = yay1 y1xa = E Ca XCa YCa,
which shows that two conjugates xax1, of a are equal if and only if theleft cosets x YCa are equal. Therefore the number of distinct conjugates of ais equal to the number of distinct cosets of Ca in G. Thus IC(a)I divides IGI.2. Each element a E can be written as a product of disjoint cyclesa = o1 . . . this representation is unique apart from the ordering of the cycles.Include the cycles of length 1, and denote the length of by which we assume
are in increasing order. The cycle pattern of a is then the sequence tn, . .. , Tjj.Thus the elements of
53 = ((1)(2)(3), (1)(2 3), (2)(1 3), (3)(1 2), (1 2 3), (1 3 2))
have cycle patterns (1, 1, 1), (1,2), (1,2), (1,2), (3), (3). In two permuta-
tions are conjugate if and only if they have the same cycle pattern. The conjugacy
classes in S3 are therefore the classes (e), ((1 2), (1 3), (23)) and ((1 23), (1 32)).This shows that conjugacy classes do not all have the same number of elements.
3. Denote the order of G by n.The conjugacy class of the identity has one element,
namely the identity. The second conjugacy class then has n —1 elements, and thus
n — 1 divides n. We deduce that n = 2.
Solution to 6.1.8: Fix an element a E G different from the identity, and considerthe map ,o: G \ (e) —÷ G \ (e) defined by ço(c) = c1ac. The map is onto, so,since G is finite, it is one-to-one. As ço(a) = a and a2aa2 = a, it must be thata2 = e. Thus, all elements of G, other than the identity, have order 2. Then, if aand b are in G, we have
ab = a(ab)2b = a2(ba)b2 = ba
in other words, G is commutative, and it follows that G has order 2.
Solution 2. Since G is finite, it has an element of prime order p. Hence, everyelement of G, other than the identity, has order p. Since G is a p—group, it has anontrivial central element. Therefore, all elements are central; in other words, Gis abelian. Hence, G has order 2.
Solution 3. By our hypothesis, any two nonidentity elements are conjugate. Hence,there are two classes: The class containing the identity and the classcontaining all the other elements of G. Letting n be the order of G, we see that thesecond conjugacy class must contain n — 1 elements. But, by the class equation,we know that the order of any conjugacy class divides the order of the group, so(ii — 1)In. Solving for n > 0, we see that the only possible solution is n = 2.
Solution to 6.1.9: Since G is finite, every element of G has finite order. Since anytwo elements of G \ (e) are related by an automorphism of U, all such elementshave the same order, say q. Since all powers of an element of G \ (e) have order qor 1, q is prime. By Sylow's Theorems [Her75, p. 911, the order of G is a power ofq. Therefore the center Z of G contains an element other than e. Since the centerof G is invariant under all automorphisms, Z is the whole of G, i.e. U is abelian.
Solution to 6.1.11: Consider the group G' = (alzbrn 0 ( n r — 1, 0m s — 1). As the order of any element of G' divides the order of G', we haveIG'I = rs. This shows that is never the identity for 0 < k <rs. Clearly wehave = = e, so the order of ab is rs.
Solution to 6.1.12: 1. Suppose X C Y. If z C(Y) then zy= yz for all y E Y,and thus zx = xz for all x E X, so Z E C(X).
2. Let x E X. If z E C(X) then zx = xz; thus x commutes with each member ofC(X), that is, x E C(C(X)).3. Replace X by C(X) in (2): we get that C(X) c C(C(C(X))). Further, using(2) in (1), we conclude that C(C(C(X))) C C(X).
Solution to 6.1.13: Let g, h E H and letx E D \ H. Then
= xhx = = =and we can conclude immediately from this that H is abelian.
Let x E D \ H. We have [D:H]=2, so E H. By hypothesis, xx2x1 = x2or = 1. Therefore, x has order 1,2, or 4. But n is odd, so 4 does not divide theorder of D, so, by Lagrange's Theorem [Her75, p. 41], x cannot have order 4. Byour choice of x, x 1, so x cannot have order 1. Hence, x has order 2.
6.2 Homomorphisms and Subgroups
Solution to 6.2.1: Let f: Z2 x Z2 -÷ S3 be a homomorphism. Then(Z2 x Z2)/kerf imf. Denote I kerfl by k, and imf I by 1. Ihen k = 1,2 or4, and 1 = 1,2, 3 or 6. Therefore 4/k = 1. The only solutions are k = 4,! = 1
and k = 2,! = 2. The first case is realized by the map which sends each elementof Z2 x Z2 to the identity of S3. Now
Z2 x Z2 {(0,0), (0, 1), (1,0), (1, 1)),
S3 = {e,(l 2),(1 3),(23),(1 23),(l 32)}.
Each element of Z2 x 7Z2, not the identity, has order 2; denote these by Zi, Z2, Z3.
The cycles in S3 of length 2 have order 2; denote these by y2, We thereforecon sider the map f given by
f(0, 0) = f(zi) = e, f(z2) = f(z3) =It is routine to verify that f is a homomorphism. By 'varying' the zs and the ys wefind nine such homomorphisms. Thus there are ten homomorphisms from Z2 x Z2to 53.
Solution to 6.2.2: 1. Let Cg denote the conjugacy class of g and Zg the central-izer of g. Using the Orbit Stabilizer Theorem [Fra99, Thm. 16.16], [Lan94, Prop.1.5.1], I Zg I I Cg = G I for every element g. Hence >gEC I Zg I = I G I for everyconjugacy class C, and IXI = >gEG iZgI = CIGI.2. In this case G = S5. so IGI 5! = 120. The number of conjugacy classesis the number of partitions of 5, namely 7. So there are 7• 120 = 840 pairs ofcommuting permutations.
Solution to 6.2.3: A matrix A is a solution ofx3 = x2, or equivalently x2(x —I) =0, if and only if all its Jordan blocks are solution to the same equation. So each
Jordan block must have eigenvalues 0 and 1, and the possibilities are (0), (1),
conjugacy type of a matrix is determined by the multiplicity of the Jordan
blocks. Let a, b, c be the multiplicities of the blocks above, respectively. Then the
answer is the number of nonnegative integer solutions of a + b + 2c = 5. For
fixedc E (0,1,2),thereare6_2csolutionstoa+b5—2C.Thustheaflswer'S
(6—2.0)+(6—2.1)+(6—22) 12.
Solution to 6.2.4: Let n = [C* : H]. By Lagrange's Theorem [Her75, p. 41], theorder of any element of C*/H divides n. So x E C*. Therefore,
=CcH.Solution to 6.2.5: 1. The number of conjugates of H in G is IG : N(H; G)l whereN(H; G) = (g G I g1Hg = H) is the normalizer of H. As H C N(H; G),wehaveiG: HI (G : N(H; 0)1.2. By Problem 6.4.19, there is a normal subgroup of G, N, contained in H, suchthat G/N is finite. By Part I we can find a coset Ng E 0/N such that Ng is notcontained in any conjugate Hy/N of H/N in G/N. Then g 0 Hy foranyyEG.
Solution to 6.2.6: We will prove a more general result. Let G = (g), IGI = n, andcr E AutG. As a(g) also generates G, wehavea(g) = gk forsome 1 k <n,(k, n) = 1. Conversely, x is an automorphism of G for (k, 1) = 1. Letbe the multiplicative group of residue classes modulo n relatively prime ton. If kdenotes the residue class containing k, we can define 4 : Aut G —÷ by
if a(g)=gk.
It is clear that C1 is an isomorphism. As is an abelian group of order p(n)(çp is Euler's totient function [Sta89, p. 77], [Her75, p. 43]), so is Aut G. When nis prime, these groups are also cyclic.
Solution to 6.2.7: 1. If g1.p(g) = h1ço(h) for some g, h E G, thenço(g)ço(h)' = so, by hypothesis, gh1 = 1 and g = h. Thus, thereare IGI elements of that form, so they must constitute all of G.2. Using Part 1, we have, forz = E G,
q'(z) = çp(g')ço2(g) = =
so g i—k g' is an automorphism of G, which implies that G is abelian. For anyz E G, z 1,wehavez'is odd.
Solution to 6.2.8: Let G be the group. If G is not abelian and a is an element not inthe center, then the map x a1xa is the desired automorphism. If G is cyclic,say of order m, and n is an integer larger than 1 and relatively prime to m, then
the map x i-* is the desired automorphism. If G is any finite abelian group,then, by the Structure Theorem [Her75, p. 1091, it is a direct product of cyclicgroups. If one of the factors has order at least 3, we get the desired automorphismby using the preceding one in that factor and the identity in the other factors. Ifevery factor has order 2, we get the desired automorphism by permuting any twoof the factors.
Solution to 6.2.9: We show that u is an isomorphism from kerf onto kerg. Lety E kerf. We have
g(u(y)) = u(y) — u(v(u(y))) = u(y — v(u(y))) = u(f(y)) = u(0) = 0.
Hence, u maps kerf into kerg. Lety E kerf with u(y) = 0; then
0 = 1(y) = y — v(u(y)) = y.
Thus, u is injective. Letx E kerg. Then 0 = g(x) = x—u(v(x)), so u(v(x)) = x.Therefore, u is onto if v(x) E ker f. However, an argument identical to the firstone shows that this is the case, so we are done.
Solution to 6.2.10: Given an element 13 G, by the surjectivity of h, there is anelement y E G such that
y — u(v(y)) = u(/3)
now build a = fi + v(y), for this element
f(a) = + v(y) — v(u(fl + v(y)))
= 13 + v(y) — v(u(f3) + u(v(y)))
=/3
showing that f is surjective.
Solution to 6.2.12: 1. Suppose this is not the case, then there exists hi E H1 \ H2and h2 E H2 \ H1. Since they both belong to the group H1 U the ele-ment h1h2 E H1 U H2. In both cases, if E H1 we get a contradictionwith h2 = H1 or if hIh2 Ff2 we get the contradiction with
h1 = E 112.
2. Consider the product group G = Zr', and the subgroups
H1 = {(X1, ... , E G1x1 = 0).
ThenHiU...UHn_1G\{(1,1,...,1)).IetHn={(Xi,...,xn_1)EGk1+X2 = 0). Then (1, 1,..., 1) E so H1 U H2• • = G, no H1 is contained in
any other, since they are distinct groups of the same order.
Solution to 6.213: 1. The statement is false. Let G = S3 = (a, b a3 = b2 =1, ha = a2b) and H = {1, b), so that G : HI = 3. Let x = ab E G. Then
2. The statement is true. The n + 1 cosets H, Ha, Ha2,..., Ha's are not all
distinct. There are integers 0 i < j n such that Ha' = Hai. This meansthat a3' E H, proving the result.
Solution to 6.2.14: For each nonzero rational s, the map f: Q —* Q given byx i-+ sx is a bijection and satisfies f(x + y) = f(x) + f(y). Thus f is anautomorphism of Q.
Conversely, let f be an automorphism of Q. Then f(O) = 0, and s = f(l) is
a nonzero rational. The inductive step f(n + 1) = f(n) + f(1) = ns + s provesthat f(n) = ns for n 1. Further as 0= f(n + (—n)) = f(n) + f(—n) we seethat f(—n) = —f(n) = —ns for n 1. Thus f(n) = ns for n E Z. Now letx = rn/n where n > 0 and rn E Z. Then rns = f(rn) = f(nx) = nf(x). Thusf(x) = sx.
Solution to 6.2.15: If the origin were a limit point of F then F would be densein R, since I' contains all integer multiples of its elements. But then F wouldequal IR, since it is closed, which is impossible because F is countable as a setF = (mci + nfl I rn, n Z). Hence the origin is not a limit point of F.
Therefore F contains a smallest positive number y. If x is in F and n is thelargest integer such that ny x, then x — ny is in F, and 0 x — ny < y,implying that x — ny = 0. Therefore F = yZ, from which the desired conclusionis immediate.
Solution to 6.2.16: Suppose H is a proper subgroup of G of finite index. LetN = G/H, so that N is a finite non trivial abelian group. Let p be a primedividing the order of N, and n N an element of order p. Suppose n is the imageof the class of a/b in G. Let be the image of the class of a/p"b. Then hasorder because = n, so pk+lnk = 0 and dnk does not equal zero forany proper divisor d of This is not possible since every element of N hasorder dividing INI.
Solution to 6.2.17: The only homomorphism is the trivial one. Suppose p is anontrivial homomorphism. Then ço(a) = rn 1 for some a, m Q. We have
a a a a a
but rn is not the power of a rational number for every positive n. For example
3/5 = 1/5 + 1/5 + 1/5 but Q+
Solution to 6.2.18: For i = 1, 2 let S, denote the set of right cosets of H,, G/H,.Then G acts on S, on the left. Hence G acts on S1 x 52, and the stabilizer of thetrivial pair (H1, H2) of cosets is H1 fl H2 Hence [G : H1 fl H2] is the size of theG-orbit of (Hj, H2) E S1 x 52, so [G: H1 fl H2] 9. On the other hand, H1 andH2 are distinct subgroups of the same index, so Hj is not contained in H2, and[G : H1 fl H2] is a proper multiple of [G: H1] = 3. Thus it equals 6 or 9.
Both are possible. The index is 6 if H1 = (1, (12)) and H2 {1, (13)) in thesymmetric group G = S3. The index is 9 if H1 and 1-12 are distinct subgroups oforder3inG=Z/3x7Z/3.
Solution to 6.2.19: By assumption, H has only finitely many right cosets, sayH, x1H, . . . , whose union is G. Hence, K is the union of the setsK fl H, K fl H, ..., K 1) some of which may be empty, sayK = (K fl H) U (K fl H) U U (K fl H) (the notation being so cho-sen that K fl H = 0 if and only if j > m). If K fl Xj H 0, then we mayassume is in K (since y is in xH if and only if yH = xH). After making thisassumption, we have K = so that K 11 = flx3H = fl H),whence
K (K fl H) Ux1(K fl H) U •.• UXm(K fl H)
This shows KflH has only finitely many rightcosets in K, the desired conclusion.
Solution to 6.2.20: For each g E G, let Sg = {a E G I ag = ga). Each is anontrivial subgroup of G, because g E S8 and Se = G. The intersection of all Sg,g E G, is the center of G. So H is a subset of the center of G.
Solution to 6.2.21: Let N = 6k• Let = x for 0 i k. These groupsare not isomorphic because the center of G1 is
Solution 2. Let N = 22k• Let G, = C2i x for 0 i k. These groups arenot isomorphic because the exponent of C, is
Solution to 6.2.22: Since G is finitely generated, Hom(G, Sk) is finite (boundedby (k!)"), where Sk denotes the symmetric group on k numbers 1,2, ... , k. Forany subgroup H of index k in G, we can identify G/H with this set of symbols,sending the coset H to 1. Then the left action of G on G/H determines an elementof Hom(G, Sk) such that H is the stabilizer of 1. Thus, the number of such H'sis, at most, (k!y2.
6.3 Cyclic Groups
Solution to 63.1: 1. Let G be the subgroup of Q generated by the nonzeronumbers ai, ... , a,., and let q be a common multiple of the denominators of
ar. Then each a, has the form pj/q with p3 E 74 and, accordingly,G = where is the subgroup of Z generated by pi,..., Pr. Since allsubgroups of Z are cyclic, it follows that G is cyclic, that is, G = where p isa generator of Go.2. Let ir : Q —÷ Q/7L be the quotient map. Suppose G is a finitely generatedsubgroup of Q/Z, say with generators b1, ... , Let G1 be the subgroup of Qgenerated by 1, ir1(bi), . .. , ir 1(br). Then ir (Gj) = G, and G1 is cyclic byPart 1. Hence, G is cyclic.
Solution to 6.3.2: The subgroup of Q /Z generated by the coset of 1/: is a cyclicsubgroup of order:, say. Let H be any cyclic subgroup of order t, and r +Z beone of its generators, where 0 <r < 1. Then tr = p is an integer, and r = p/t.This implies that the coset of r is in Ho, hence that H is a subset of Ho. Since Hand Ho both have order t, we have H = Ho, as desired.
Solution to 6.33: Suppose G = {O, . . .} is countable, and, for eachn E N, .. ., gj is contained in an infinite cyclic subgroup of G. Then thesubgroup generated by (gi, ... , gñ}, is infinite cyclic, and C
Let be generated by We have = for some nonzero E Z.Let = qiq2 and let con : —÷ Q be the morphism verifying
= 1/rn. Then we have
qn 1
= = = — =rn+I
Since generates co,2 = c°n÷i Thus there is a unique homomorphism
ço: G —÷ Q such that = for all n N. As each is injective and thesubgroups cover G, ço is an isomorphism from G onto a subgroup of Q. The
converse is clear.
Solution to 6.3.4: 1. Let G = (c). We have a = c' and b = cS for some positiveodd integers r and s, so ab = with r + s even, and ab is a square.2. Let G = Q 'i', the multiplicative group of the rational numbers, a = 2, and
b = 3. Then ab =6, and none of these is a square in G.
Solution to 6.3.6: Let e be the identity in G, and let N (e, a) be the normalsubgroup of order 2. If x E G, then E N and certainly does not equale, so it equals a. Thus, xa = ax for all x E G. The quotient group G/N hasorder p and so is cyclic. Let x be any element not in N. Then the coset of x inG/N has order p, so, in particular, the order of x itself is not 2. But the orderof x divides 2p, so it must be p or 2p. In the latter case, G is the cyclic groupgenerated by x. In the former case, since xa ax, we have (xa)P = = a,so (xa)2P = a2 = e, and xa has order 2p, which means G is the cyclic groupgenerated by xa.
Solution to 63.8: It follows that every Sylow subgroup is normal. From this itfollows that if a and b are in distinct Sylow subgroups a and b commute. Indeedabatb' lies in the intersection of the two groups which is trivial. Hence, itsuffices to establish the result when G has order for a prime p. Let a E Gbe an element of largest order, say p1'. Let H = (a). If b E G and b has orderp1
pk, then b E H because l(b)I = p' and H contains the unique subgroupof G of order p'. Thus if c g H then C has order strictly bigger than whichcontradicts the maximality of the order of a. Thus H = G and G is cyclic.
6.4 Normality, Quotients, and Homomorphisms
Solution to 6.4.1: 1. For each x G the map h i-* xh is a bijection betweenH and the left coset xH. Thus, IHI IxHI. Further, the left cosets of H forma partition of G, arising from the equivalence relation on G given by x y iff
H. Thus, the number of left cosets of H in G is given by IGI/IHI. Thesame holds for right cosets: here the partition of G is given by the relation x "-' y1ff H. This proves the result.2. The group of symmetries of the square is the dihedral group D4 of order 8. Takethe square with vertices P4 = {1, i, —i, —i) in the complex plane. Let r denotethe rotation of the about the origin through the angle ir/2. Then (1, r, r2, r3) is anormal subgroup of D4. Let s be the reflection in the real axis. The elements ofD4 are
1, r, r2, r3, s, rs, r2s, r3s.
Further, s2 = 1, and rs is also a reflection, so (rs)2 = 1 or sr = r1s = r3s. Weobtain the presentation
= (r,s = 1 =s2,srLetH=f1,s)ThenrH =fr,rs)
Solution to 6.43: The subgroup H is normal only if aHa1 = H or aH = Hafor all a G. Since H has only one left coset different from itself, it will sufficeto show that this is true for a fixed a which is a representative element of thiscoset. Since H has the same number of right and left cosets, there exists a b suchthat H and bH form a partition of G. Since cosets are either disjoint or equal andHflaH =Ø,wemusthavethataH=Hb.Butthena €Hb,soHb=Ha.
Solution to 6.4.5: The number of Sylow 2-groups is odd, and divides 112, there-fore it divides 7, thus, it is 1 or 7. If there is only one Sylow 2-group, it's normal,so we may assume otherwise.
Let G act on the set of seven 2-groups by conjugation, so it maps into S7. Thisis a nontrivial map, since the action is transitive. If it is not an injection, then thekernel is a nontrivial normal subgroup, so we may assume it is an injection.
Since 16 divides IGI and does not divide IA7 the image of G is not entirelyinside A7. Therefore the composite of G —÷ S7 -+ S7/A7 = Z2 is onto, and itskernel is a normal subgroup (of index 2).
Solution to 6.4.6: The groups of order 6 are (1), Z2, Z3, Z4, Z2 x Z2, Z5,Z6, and the symmetric group S3. Suppose that G is a group with the mentionedproperty. Let S be a Sylow-2 subgroup of G. Some Sylow-2 subgroup of G mustcontain (a subgroup isomorphic to) Z4, and all Sylow 2-subgroups are conjugate,so S contains Z4. Similarly S contains Z2 x Z2. Thus S cannot equal either, so8 divides the order of S, which divides the order of G. On the other hand, G hassubgroups of orders 3 and 5, so 3 and 5 also divide IGI. Since 8,3 ,5 are pairwiserelatively prime, their product 120 divides IGI. Thus IGI ? 120.
On the other hand, the group G = x Z5 x S3 of order 120 has the property:
Z2 is isomorphic to a subgroup of 74, and Z2 and Z3 are isomorphic to subgroups
of S3. so Z2 x Z2 and Z6 Z2 x Z3 are isomorphic to subgroups of G. The other
desired inclusions are obvious.
Solution to 6.4.7: Denote by h, H (and k, K) a coset of H (and K) and suppose
G U U k,KU U k8K.
Since all of the cosets of K are equal or disjointand since the index of K in G is
infinite, there is a k E K such that
kK Ch1HU"UhrH.
Therefore, for 1 i
Ckjk1hiHU"Ukjk1hrH.
This implies that G can be written as the union of a finite number of cosets ofH, contradicting the fact that the index of H in G is infinite. Hence, G cannot bewritten as the finite union of cosets of H and K.
Solution to 6.4.8: 1. First, note that the multiplication rule in G reads
(a b \ (ai b1 \ (aai +baj'0
which gives = j'). This makes it clear that N is a subgroup, and
is in N, then
(a b \'(i b \_(a' —b\('a b+/3a'a—') a a1
proving that N is normal.By the first equation, the from G onto]R÷ (the group of positive reals under
multiplication) given by a a homomorphism whose kernel is N(which by itself proves that N is a normal subgroup). Hence, G/N is isomorphicto R÷, which is isomorphic to the additive group R.2. To obtain the desired normal subgroup majorizing N, we can take the inverseimage under the homomorphism above of any nontrivial proper subgroup of ILF.If we take the inverse image of Q the group of positive rationals, we get theproper normal subgroup
of G, which contains N properly.
Solution to 6.4.9: H1 {m(1, 2) + n(4, 1) rn, n E Z). By inspection, we seethat a transversal for the quotient group G1, a set containing one representativeof each coset, consists of the lattice points inside the parallelogram with edgesconnecting (0,0), (1, 2), (5, 3) and the origin. A glance at the figure shows that
G1 = {(O, 0), (1, 1), (2, 1), (3, 1), (2, 2), (3, 2), (4, 2))
where we have abbreviated (a, b) = H1 + (a, b). Thus G1 is an abelian group of
order 7, isomorphic to the cyclic group Z-j. In the same way we see that
G2 = ((0,0), (1, 1), (1,2), (2,2), (2,3), (3, 3), (3,4)),
so that G2 is isomorphic to the cyclic group Z7. Thus G1 G2.
Solution to 6.4.10: Let H = (1, b) be a normal subgroup of order 2. The group
G contains an element of order 10, or G contains an element of order 5, or allelements (not 1) of G have order 2. If G has an element of order 10, or if x2 = 1
for all x E G, then G is abelian. When G has an element of order 5, a, let K =(1, a, a2, a3, a4). Since IHI, IKI are coprime, we have H fl K = 1. Thus b Kand G = K U Kb = (1, a, a2, a3, a4, b, ab, a2b, a3b, a4b). Now (a, ab) =aH = Ha = (a, ba) whence ab = ba. It is now easy to see that G is abelian, foreach element of G has the form a'b3, and aibjakII = =
Solution to 6.4.11: Let x E N1 andy E N2, since N1 is normal, E
N1 and by the same token (xyx')y' E N2. Soxy = yx. A similar argument works for the other pairs of subgroups, showing
that elements of N1 commutes with elements of Nj for all i j. Now supposethat x, y E Ni. Since Ni is normal in N2N3 = G, there exist X2 N2 andX3 E N3 such that x x2X3. By the previous remarks, y commutes with X2 andX3 SO x2X3y = yx2x3 = yx, and we can now see that the elements ofcommutes will all elements of for all i and j, since G = N1 N2, G is abelian.
G2
Applying the Second Isomorphism Theorem [Fra99, §3.1], we obtain,
N2 N2 N1N2 G N3N1N "S' — = — = N3.2— {e) — N1 fl N2 Nj Nj Ni
The other two isomorphisms N1 and N1 N2 are obtained in the same
way.
Solution to 6.4.13: Let {ai H, a2H, . . ., au H) be the set of distinct cosets of H.G acts on on this set by left multiplication and any g E G permutes these n/rncosets. This group action defines a map
ço:G-+Su
from G to the permutation group on n/rn objects. There are two cases to considerdepending on being injective or not.
If is not injective, then is a normal subgroup K fe); and K G, aswell, because if g H, gH H; so g is not a trivial permutation.
is injective, then = n, and isa subgroup of But
:co(G)] = = <2
So : = 1, that is, G is isomorphic to S,i. and in that case, Au is anontrivial normal subgroup.
Solution to 6.4.14: Let H be a subgroup of G of index 3. G acts by left multiplica-tion on the left cosets {gH) of the subgroup H. This gives a homomorphism of Ginto the symmetric group of degree 3, the group of permutations of these cosets.The subgroup H is the stabilizer of one element of this set of cosets, namelythe coset 1 H. This homomorphism cannot map onto the entire symmetric group,since this symmetric group has a subgroup of index 2, which would pull back toa subgroup of G of index 2. Thus, it must map onto the cyclic subgroup of order3, and the group H is then the kernel of this homomorphism.
Solution to 6.4.15: For each element g E G consider the inner automorphism*g(h) = ghg* Defined in this way the map g i—+ defines a homomorphismG —* A. It is nontrivial because G is not abelian, and injective since G is simple.Let B be its image, so G B. If E A and g, h E G, we have
= o!(ghg1) = =
*a(g)OCdnA.ThUs,aOlfrgoa'
Solution to 6.4.16: 1. The cycles of are the right cosets (g)x of the subgroup(g) in G, so the lengths of each one is the order of g, and the number of cycles is[G: (g)].
If the order of g is odd, then each cycle has odd length, so each cycle is even,and so is Ag. If the order of g is even, then each cycle is of even length and,therefore, odd. Also,
IGI
is odd. As [G; (g)] is the number of cycles, Ag is odd.2. Let ço : G —* (—1, 1) be defined by g p(g) = sign of Ag. ço isamorphismand, by Part 1, its kernel is N. So N is a normal subgroup of G with index 1 or 2.By Cauchy's Theorem [Her75, p. 6]], G has an element of order 2, which is notin N, so N has order 2.
Solution to 6.4.17: Let g = xyx1y1 be a commutator. It suffices to show thatconjugation by g fixes every element of N. As N is cyclic, Aut(N) is abelian, and,because N is normal, conjugation by any element of G is an automorphism of N.Let çox be the automorphism of conjugation by x. We have cox coy = coy cox. Hence,forn E N, gng' = cox oçoy oçç1 ociç'(n) = n.
Solution to 6.4.18: We will show that if the index of N in G is not finite and equalto a prime number, then there is a subgroup H properly between N and G. Sinceany nontrivial proper subgroup of GIN is the image of such a subgroup, we needonly look at subgroups of G/N.
Suppose first that the index of N in G is infinite, and let g be an element ofG/N. If g is a generator of G/N, then GIN is isomorphic to Z, and the elementg2 generates a proper nontrivial subgroup of G/N. Otherwise, g generates such asubgroup.
Suppose that the index of N in G is finite but not a prime number. Let p beany prime divisor of the index. By Cauchy's Theorem [MH87, p. 152], there is anelement of order p in G/N. This element cannot generate the whole group, so itgenerates a nontrivial proper subgroup of G/N.
Solution to 6.4.19: Let the index of A in G be n. G acts by left multiplication onthe cosets gA, and this gives a homomorphism into the group of permutations ofthe cosets, which has order n!. The kernel, N, of this homomorphism is containedin A, so the index of N in G is, at most, n!.
Solution to 6.4.20: 1. False. Consider the integers Z and the subgroup of themultiples of n > 1, then the quotient H is finite cyclic but G H x Kbecause the product has torsion and Z is free abelian.2. False. If H is a countable direct product of infinite cyclic groups and G is a freeabelian group mapping onto H with kernel K, G is not isomorphic to the directproduct of H and K. For if G were isomorphic to H x K, H would be isomorphicto a subgroup of G. All subgroups of G are free abelian groups, but H is not.
6.5 Sn, An,
Solution to 6.5.1: The matrix X = is in G if and only if
IAI = ad — bc = 1. If a = 1 then d = 1 + bc and X is one of
11 o\('l o\('I i\(i 1
o
If a = 0 then bc = 1 and X is one of
10 (0 1
1
HenceIGJ=6.
A=(?
It is easily verified that A3 = I = B2, BA = A2 B. The subgroup (A) has index2 in G. Moreover, B (A) since B has order 2 and elements of (A) have order1 or 3. Therefore we have the coset decomposition
G = (A)U(A)B —_ (I, A, A2, B, AB,A2B).
The relations A3 = I = B2, BA = A2B completely determine the multiplica-tion in G.
Now 53 has an element a = (1 2 3) of order 3, and an element = (1 2) oforder 2. Moreover 16a = a2/3. As above
== (a, /3 I
a3 = 1= /32 =
Thus the correspondence A' &/33 is an isomorphism between G and S3.
Solution to 6.5.2: Think of 54 as permuting the set (1, 2, 3,4). For I ( i 4,let G1 c 54 be the set of all permutations which fix i. Clearly, is a subgroupof S4; since the elements of G, may freely permute the three elements of the setf 1,2, 3,4) \ {i), it follows that G, is isomorphic to S3. Thus, the G1's are thedesired four subgroups isomorphic to 53.
Similarly, for i, j E (1, 2, 3,4), i j,let be the set of permutations whichfix i and j. Again H11 is a subgroup of 54, and since its elements can freelypermute the other two elements, each must be isomorphic to S2. Since for eachpair i and j we must get a distinct subgroup, this gives us six such subgroups.
Finally, note that 52 is of order 2 and so is isomorphic to Z2. Therefore, anysubgroup of S4 which contains the identity and an element of order 2is isomorphicto S2. Consider the following three subgroups: (1, (1 2)(3 4)), (1, (1 3)(2 4)), and(1, (1 4)(23)). None of these three groups fix any of the elements of {1, 2, 3,4),
so they are not isomorphic to any of the Thus, we have found the final threedesired subgroups.
Solution to 653: Let a be a 5-cycle and r a 2-cycle. By renaming the elementsof the set, we may assume a = (1 2 3 4 5) and r = (a b). Recall that if(i1 j2 •• i,,) is a cycle and a is any permutation, then
a(i1 i2 = ) cr(i2) •..
a act repeatedly on r as above, we get the five transpositions (a + i b + i),I (i ( 5,whereweinterpreta+itobea+i—5ifa+i > 5.Fixingi suchthata + i —5 = 1 and letting c = b + i, we see that G contains the five transpositions(1 c), (2c+ 1),..., (5 c+4).Sincea 1.
Let d = c — 1. Since d does not equal 0 or 5, these five transpositions can bewritten as (c + nd c + (n + I )d), 0 ( n (4. By the Induction Principle [MH93,p. 7] the equation above shows that G contains the four transpositions
(1 c+nd)= (1 c+(n— l)d)(c+(n— 1)dc+nd)(l c+(n— l)d),
1 ( n ( 4. Since they are distinct, it follows that G contains the four transpo-sitions (1 2), (1 3), (1 4), and (1 5). Applying the equation a third time, we seethat(i j) = (1 i)(1 j)(l i)isanelementofGforalliandj,1 ( i,j (5.Hence, G contains all of the 2-cycles. Since every element in can be written asthe product of 2-cycles, we see that G is all of
Solution to 6.5.5: We use the notation to denote the cycle pattern ofa permutation in disjoint cycles form; this means that the permutation is a prod-uct of cycles of length 1, a2 cycles of length 2,.... (Thus the pattern of, say,(8)(3 4)(5 6)(2 7 1) E S8 is denoted by 112231.) The order of a permutation indisjoint cycles form is the least common multiple of the orders of its factors. Theorder of a cycle of length r is r. The possible cycle patterns of elements of S7 are
1521; 1322 1123;j431; 122131; 2231; 1132;
112141; 3141
1251; 2151;
1'6'71
We see that possible orders of elements of S7 are 1,2,3,4,5, 6,7, 10, 12.
Solution to 6.5.6: 1. a = (1234)(56789) has order Icm(4, 5} = 20.
2. If a = (k is written as a product of disjoint cycles 0j of length then
the order of a is lcm{rj, .. . , rk). In order that this 1cm is 18, some must haveafactor2.Asl (9andri +.••+rk (9thisis
impossible.
Solution to 6.5.7: The order of a k—cycle is k, so the smallest m which simulta-neously annihilates all 9-cycles, 8-cycles, 7-cycles, and 5-cycles is .32.5.7 =2520. Any n—cycle, n 9, raised to this power is annihilated, so n = 2520.
To compute n for A9, note that an 8-cycle is an odd permutation, so no 8-cyclesare in A9. Therefore, n need only annihilate 4-cycles (since a 4-cycle times atransposition is in A9), 9-cycles, 7-cycles, and 5-cycles. Thus, n = 2520/2 =1260.
Solution to 6.5.8: We have 1111 = 11 x 101, the product of two primes. So Gis cyclic, say G = (a). From 1111 > 999, it follows that a, when written as aproduct of disjoint cycles, has no cycles of length 1111. Therefore, all cycles of ahave lengths 1, 11, and 101. Let there be x, y, and z cycles of lengths, respectively,1, 11, and 101. If x > 0, then a has a fixed point, and this is then the desired fixedpoint for all of G. So assume that x = 0. Then lix + 101z = 999. It followsthat 2z 9 (mod 11), so z 10 (mod 11), and, therefore, z 10. But then999 = ily + lOlz 1010, a contradiction.
Solution to 6.5.9: For o in let 8a 0 if ci is even and = I if ci is odd. Letr denote the cycle (n +1 n +2) in Sn+2. and regard the elements of as elementsof Sfl+2 in the obvious way. The map of into Sn+2 defined by a i-÷ isthen an isomorphism of onto its range, and the range lies in An÷2.
Solution to 6.5.10: Call I and j (1,2, . . . , n} equivalent if there exists oEGwith cr(i) = j. (This is clearly an equivalence relation.) For each i, the set
= (ci E GIcr(i) = I) is a subgroup of G, and the map G -+ (1,2, . . . , n),
ci i-* a(i), induces a bijection from the coset space GIGI to the equivalence classof 1. Hence, for each i, the size of its equivalence class equals [G: G], which isa power of p. Choosing one i from each equivalence class and summing over i,one finds that all these powers of p add up to n, since p does not divide n, oneof these powers has to be p0 = 1. This corresponds to an equivalence class thatcontains a single element i, and this i satisfies ci(i) = i for all a E G.
Solution to 65.13: To determine the center of
=(a,bla'3 —b2= l,ba=a'b)(a is a rotation by 2ir/n and b is a flip), it suffices to find those elements whichcommute with the generators a and b. Since n 3, a' a. Therefore,
= a(arb) = (a'b)a = ar_lb
so a2 = 1, a contradiction; thus, no element of the form a'b is in the center.Similarly, if for 1 s <n, a9b = a_sb, then = 1, which is possibleonly if 2s = n. Hence, a3 commutes with b if and only if n = 2s. So, if n = 2s,thecenterofDn is is (1).
Solution to 65.14: The number of Sylow 2-subgroups of is odd and dividesn. Each Sylow 2-subgroup is cyclic of order 2, since 21 is the largest power of 2
dividing the order of the group. By considering the elements of as symmetriesof a regular n—gon, we see that there are n reflections through axes dividing then—gon in half, and each of these generates a different subgroup of order 2. Thus,the answer is that there are exactly n Sylow 2-subgroups in when n is odd.
6.6 Direct Products
Solution to 6.63: Suppose Q was the direct sum of two nontrivial subgroups AandB.Fixa =ao/aiandb=bo/bi,where the a- 's and b 's are nonzero integers. Since A and B are subgroups, na E Aand nb E B for all integers n. In particular, (aibo)a E A and (biao)b B. But
(aibo)a = (aibo)ao/ai = aobo = (b1ao)bo/bi = (biao)b.
Hence, A and B have a nontrivial intersection, a contradiction.
Solution to 6.6.4: Let Cm denote the cyclic group of order m for m E N. Bythe Structure Theorem for abelian groups [Her75, p. 109], if G is a finite abeliangroup, there exist unique nonnegative integers npr(G) for each prime number pand each nonnegative integer r such that G is isomorphic to
If H is another abelian group, G x H is isomorphic to
flfl flHence, x H) + Now this and the fact that A x B isisomorphic to A x C yield the identities (B) = flpr (C) for all primes p andall nonnegative integers r. We conclude B and C are isomorphic.
Solution to 6.6.5: Let 213 : A -+ G1 x G2 be the natural projection map. Wehave kerlr3 = {(1, 1, g3) E A). Let N3 = {g3 1(1, 1, g3) E kerlr3). Since kerlr3is a normal subgroup of A, N3 is normal in G3. Let A' = G1 x G2 x G3/N3.Since 213 is onto, for any G1 x G2, there exists 53 E G3 such that
(gi, g3) E A, and, thus, (gi, E A'.Define the map G1 x G2 —÷ G3/N3 by gz) = where is such
that (Si. g3) E A. This is well defined, for if ga, g3) and h3)
are both in A, then (1, 1,g3h31) E A, so N3, which, in turn, im-
plies that = h3. The map is clearly a homomorphism. Furthermore, since
211 : A —÷ (32 x G3 is onto, ifg3 E E G1 G2such
that (gi, g3) E A. Thus, = so p is onto.
Therefore, x {1)) and ço([1} x G2) are subgroups of G3/N3 which com-mute with each other. If these two subgroups were equal to one another and toG3/N3, then G3/N3 would be abelian. As G3 is generated by its commutatorsubgroup, this would imply G3/N3 to be trivial or, equivalently, G3 = N3, whichwe assumed not to be the case. So we may assume that (piff) x G2) G3/N3.Pick E G3/N3 \ p((l} x Ga). Since ,t2 : A —÷ G1 x G3 is onto, there existsa E G2 such that (1, g3) E A. Hence, g,(1, = contradicting ourchoice of
Therefore, we must have that N3 = G3, so (1) x (1} x G3 C A. Similarshow that (1) x G2 x (1) and G1 x (1) x (1) are contained in A and,
thus, A = G.
Solution to 6.6.6: Yes, we will show that there exist a Y C C such that C = A+Y.Choose I = X fl C, then A fl I C A fl X =0. Now to see the decomposition,observe that for all c e C C B, there exist an a E A and x E X such thatc = a + x, but x = c — a E C so x e C fl X = I, thus showing that for all c E Cthere exist an a E A and x E I such that c = a + x.
Solution to 6.6.7: The obvious isomorphism F: Aut(G)xAut(H) —÷ Aut(GxH)is defined by
= (aG(g),ajj(h)).Since aG and aH are automorphisms of G and H, it is clear that F(CrG, aR) is anautomorphism of G x H. Let us prove now that F is an isomorphism.
• F is injective: Let F(aG, aH) = idGxH. Then ac = idG and aH = idjjby definition of F. Hence, ker F is trivial so F is injective.
• F is surjective: Choose a E Aut(G x H). Define aG, aH by
aG(S) = YrG(a (S. id11)) and ajq(h) = lrH(a(idc, h))
where irc and 7Th are the quotient maps. Thus,
a(g, h) = (aG(S), aH(h)) — F(aG, aH)(g,
Since the situation is symmetric between G and H; and G is finite, weneed only show that is injective. Let aG(S) = idG. Then a(g, idH) =(idG, h) for some h E H. Suppose n = IGI. Then
(idtj, h't) = a(g, = idjj) = (IdG, idH).Hence, hr? = idH, so the order of h divides fGf. But, by Lagrange's The-orem [Her75, p. 41], the order of h also divides IHI, which is relativelypnme to GI, so the order of h is one and h = id11.
Solution to 6.6.8: We only have to determine where the automorphism send itsgenerators (1,0) and (0, 1), and these can be mapped into (a, b) and (c, d), re-spectively, if and only if,
a pZ/p274 c pZ/p2Z and d 0 €
If is an automorphism that takes (1,0) to (a, b) and (0, 1) to (c, d), then the(a, b) must not be killed by p, so a p7L/p27L and (c, d) must be killed by p,so c E pZ/p2Z. Moreover (c, d) should not be multiple of p(a, b) = (pa, 0), so
On the other hand, if (a, b) and (c, d) satisfy the conditions, then it is easy tofind an automorphism, mapping the generators there, since (a, b) is killed by p2and (c, d) is killed by p. The condition on a implies that (a, b) has order p2.If (c, d) were a a multiple of (a, b), then since c E pZ/p27Z, the element (c, d)would be a multiple of p(a, b) = (pa, 0), which is impossible, since d 0 E
This makes the > p2 and by the Lagrange's Theorem the number has tobe p3. Thus is surjective, but G is finite, so then is injective and then anautomorphism.
We need now to count all possibilities for a, b, c and d that satisfy the condi-tions. First there are p2 — p choices for a, p total for b, p for c and p — 1 choicesfor d, and these are all independent, so in total there are p3 (p _1)2 automorphismsof
6.7 Free Groups, Generators, and Relations
Solution to 6.72: Suppose Q is finitely generated, with generators
E7Z.
Then any element of Q can be written as where the n,'s are integers.This sum can be written as a single fraction with denominator s = • /3,.
Consider a prime p which does not divide s. Then we have i/p = r/s for someinteger r, or pr = s, contradicting the fact that p does not divide s. Hence, Qcannot be finitely generated.
Solution 2. Suppose Q is finitely generated, then using the solution to Part 1 ofProblem 6.3.1, Q is cyclic, which is a contradiction.
Solution to 6.73: Let A be the matrix
/15 3 0A=I3 74
\18 14 8
that represents the relations of the group G, following [Hun96, p. 343] we perform
elementary row and column operations on A, to bring it into diagonal form.
/ 15 3 o\ (15 3 o\ f3 3 o\ f3 3 0
13 7 41-÷13 741-÷137 44k 18 14 8) \12 0 0) \12 0 0/ \12 0 0
/3 3 o\ /0 3 0\ /12 0 0
—+1 0 0 0 0 41-÷I 0 3 0
0 0) \12 0 0) \O 0 4
1. G is then the direct product of two cyclic groups Z12 x Z12.
2. G is also the direct product of the cyclic groups of prime power orders, namely
Z3x7L3xZ4x7L4.3. Using the decomposition G Z12 x Z12, we can see that for any element(p, q) of order 2 of G, p and q have to be either 0 or 6, so it has three elementsof order 2, which are (6,0), (0,6) and (6,6).
Solution to 6.7.4: x5y3 = x8y'5 implies x3y2 = 1. Then x5y3 = x3y2 and
x2y = 1. Hence, x3y2 = x2y, so xy = 1. But then xy = x2y, so we have that
x = 1. This implies that y = 1 also, and G is trivial.
Solution to 6.7.5: We start with the given relations
and b1a2b=a3
that we will call first and second, respectively. From the first one, we get= b6 and
a2b4a2 = = (a1b2)b4a = b3(cr'b4a) = b3b6 = b9
using the relation just obtained and the second one, we get
b9 = (a2)b4a2 = = ba3b1b4ba3b1 = ba3b4a3b1
and we conclude thata3b4a3 = b9 = a2b4a2
from which we obtain ab4 = b4a. This, combined with the square of the firstrelation, (a'b4a = b6), gives b2 = 1 and substitution back into the first showsthat b = 1; then, substituting that into the second relation, we see that a = 1, sothe group is trivial.
Solution 2. Using the same labels for the first and second relations, as above, wecan conjugate the first one by a2 to obtain the new relation ab2a1 = a2b3a2.At the same time multiplying the second relation by b on the left and a2 on theright, we get a2ba2 = ba and multiplying three copies of it back to back we geta2b3a2 = (ba)3, Getting these two last relations together we see that
ab2a1 = (ba)3 *
and similarly we can show that
ba2b1 = (ab)3 *
Now consider this last relation (*):
= (ab)3
= a(ba)2b
= a(ab2a2b1)b using (*)
= a2b2a2= (a2b)ba2= (ba3)ba2 using second relation
= ba(a2ba2)= baba using second relation
= (ba)2
so ab1 = ba and similarly we can show that = ab. These two togethercombined show that a2 = b2, and substituting back into the second relation, wesee that a = 1 and the group is trivial.
Solution 3. We have
= a1b3a = a'b'(lr2a) = a'lr'ab3= a11'3a = (a'b2)b1a = b3a1b1a
b2a2b2 = b1a3b = b1a1(a2b) == = =
a2b2a2 = a1b3a = == a'b3a = (cr'b2)ba = b3a1ba
b2a2b2 = b1a3b = = b1aba3
= b1a3b = (b1 a2)ab = a3b1ab
so we get
a2b2a2b2 = (a2b2a2)b2 = a1b1ab3b2 == a2(b2a2b2) = a2a3b1ab = ab1ab
b2a'2b2a2 = (b2a2b2)a2 = b1a1ba3a2 = b1a1ba'= b2(a2b2a2) = b2b3a1ba =
therefore, ba1ba = baba1 or a'ba = and a2ba2b1 = 1. Thus,ba2 = a2b so the original relations give b2 = b3, a2 = a3 so a = b = I and thegroup is trivial.
Solution 4. By induction we show that
=
in particular a3b8a3 = b27, and then substituting the second relation in, we get= b27, but we know from the general relation obtained by induction
that a2b42a2 = b'8, sob18 = b27 orb9 = I. Similarly we obtain a9 = 1, now
using the general relation again for n = 9 and m = 1 we get
b29 = a_9b29a_9 =
but since 39_ 1 (mod 9), we have b = 1 and the group is trivial.
Solution 5. In this one we will go further and show that the group
(a,bla =[am,b'2], b =[b',a"])
is trivial forall m, n, p. q E Z. Here[x, yJ = and form p = 2 and
n = q = 1, we have our original group.The two relations can be written as
=
=
By induction and raising the first relation to the shows that
= k E Z
and we can write
b_k(P+albk(P+D = k, 1 E Z
and in the same fashion we can obtain from the second relation
i tNow taking k = n, 1 = and r = p + 1 in the two equations above andequating the right-hand sides we get
=
and the relation t by itself is
=
so raising this last relation to m + 1 and equating with the left-hand side of theprevious one we end up with
= =
which shows thatamP=l
Now aml' = (am)mP' so substituting back into the first relation we see that
a(m+Dm = 1
that is,
= 1
and using again
= 1
continuing in this way we eventually get to am° = 1, that is a = I and the groupis trivial.
Solution to 6.7.6: Since a2 = b2 = 1, every element of G can be expressedin (at least) one of the forms ab...ab, ba...ba, ab...aba, ba...bab. Butab..ab = forsomen 0,ba...ba = (ha)" = (ab)",ab...aba =(abY'a, and ba . . . bab = = Let H be the cyclic subgroupof G generated by ab. Then G H U Ha, so either H is a cyclic subgroup ofindex 2 in G or G = H is itself cyclic. In the latter case, G is either infinite cyclicor finite cyclic of even order, so G has a cyclic subgroup of index 2.
Solution to 6.7.8: We use the Induction Principle [M1193, p. 7]. For n = 1, theresult is obvious.
Suppose gi, . .., generate the group G and let H be a subgroup of G. IfH C (ga, .. . , ga), by the induction hypothesis, H is generated by n —1 elementsor fewer. Otherwise let
EH
be such that rn i I is minimal but nonzero. We can assume, without loss of gener-ality, that m i > 0. For any z H,
there are integers q and r such that k1 = qmi + r and 0 r <mt. Then theexponent of in is r, and, by the choice of mi, we obtain r =0. Hence,
H=(y,K) where
K is generated by, at most, n — I elements, and the
result follows.
Solution to 6.7.10: By the Structure Theorem for finite abelian groups [Her75,
p. forl j 1,suchthatAis isomorphic to Zm1 We identify A with this direct sum. Clearly
m = mk. Therefore, S must contain the elements of the form (0,0,..., 1) and(0, ... , 0, 1,0,..., 1), where the middle 1 is in the j' position, 1 j k — 1.
Hence, using elements in S. we can generate all the elements of A which are zero
everywhere except in the ith position. These, in turn, clearly generate A, so S
generates A as well.
6.8 Finite Groups
Solution to 6.8.1: Any group with one element must be the trivial group.By Lagrange's Theorem [Her75, p. 41], any group with prime order is cyclic
and so abelian. Therefore, by the Structure Theorem for abelian groups [Her75, p.109J, every group of orders 2,3, or 5 is isomorphic to Z2, 7L3, or Z5, respectively.
If a group G has order 4, it is either cyclic, and so abelian, or each of its ele-ments has order 2. In this case, we must have 1 = = abab or ba = ab, sothe group is abelian. Then, again by the Structure Theorem for abelian groups, agroup of order 4 must be isomorphic to Z4 or 7L2 x 7L2. These two groups of order4 are not isomorphic since only one of them has an element of order 4.
Since groups of different orders can not be isomorphic, it follows that all of thegroups on this list are distinct.
Solution to 6.8.2: By the Solution to the Problem 6.8.1 all groups of order up to5 are abelian and the group of symmetries of the triangle, 1)3, has 6 elements andis nonabelian.
Solution to 6.83: Let G be a group of order 6. By Cauchy's Theorem there existan element a of order 3, and an element b of order 2. H = (a) is a subgroup of Gand, since b has order 2 and 2 does not divide b H. Therefore
G = HUHb= {1,a,a2,b,ab,a2b).
The multiplication of G is determined if we know which of the six elements of Gis equal to ba. We find that
ba 1 else a = = b,else b=1,else b=a,else a=1.
Two cases remain.Suppose lxi = ab. Then = for all n 1. The order of ab is 2,3
or 6. Now (ab)2 = a2b2 = a2 1, and (ab)3 = a3b3 = b 1. Hence ab hasorder 6. Thus G is cyclic, and G Z6.
Suppose ba = a2b. In this case G S3. For 53 has an element a = (1 23) oforder 3, and an element = (1 2) of order 2 which satisfy 46a = Since (a)has index 2 in S3 and (a), it follows that
S3
Then the correspondence is an isomorphism between G and S3.
Solution to 6.8.4: 1. The Klein four-group V = x 7L2 is a noncyclic groupof order 4, by the arguments of the Solution to Problem 6.8.1 this is the only
one of order 4. Alternatively to show uniqueness up to isomorphism, let G be anoncyclic group of order 4. Then the order of any non-identity element is equalto 2, whence g = g E G. Thus G is abelian. For if x, y E G thenxy = = (yx)' = yx.
LetG =(1,a,b,c),andconsiderab.We see that
else b=a1=a,else b=1,else a=1.
Hence ab c. The multiplication of G is then completely determined using the
fact that each element of G appears exactly once in any row or column in themultiplication table. This completes the proof of uniqueness.2. Let G = (1, a, b, c) be the noncyclic group of order 4. If E Aut(G), then
= 1 and the restriction of to X = (a, b, c}, is a permutation of X, i.e.,(pE E Aut(G), f : i-* visagroup homomorphism Aut(G) S3.
If E Aut(G), then ço(l) = iI'(l) and ço, differ on X, so f is injective.To show that f is surjective, let o be a permutation of X, and let 3 be the
extension of o- to G given by 3(1) = 1. Then 3 is a bijection G —* G. It is left toshow that 3 is a homomorphism.
Let x, y E G. If x = 1 or y = 1 then 3(xy) = 3(x)â(y), since 3(1) =1. ifx = y, then ô(xy) 3(x2) = 3(1) = (3(x))2 = &(x)&(y), since g2 = 1 for allg E G. Finally for the case when x, y are distinct elements of X we note that theproduct of two of a, b, c is the third, i.e.,
ab=ba=c, bc=cb=a, ca=ac=b.
Hence &(a)&(b) = 3(c) = â(ab). The other cases are similar. Thus the mapf : Aut(G) S3 is an isomorphism.
Solution to 6.8.5: By the Structure Theorem for finitely generated abelian groups[Her75, p. 109], there are three: Z8, Z2 x Z4, and Z2 x Z2 x Z2.
1. (Z15)* = (1,2,4, 7, 8, 11, 13, 14}. By inspection, we see that every ele-ment is of order 2 or 4. Hence, (Z15)* Z2 x
2. (Zii)* = (1, 2,..., 16} = (±1, ±2, ... , ±8), passing to the quotient= (1, 2, ..., 8) which is generated by 3, so (Z17)* 7L8.
3. The roots form a cyclic group of order 8 isomorphic to Z8.
4. F3 is a field of characteristic 2,so every element added to itself is 0. Hence,
5.
Solution to 6.8.6: Order 24: The groups S4 and 53 x Z4 are nonabelian of order24. They are not isomorphic since S4 does not contain any element of order 24but S3 x Z4 does.
Order 30: The groups D3 x Z5 and 1)5 x Z3 have different numbers of Sylow2-subgroups, namely 3 and 5, respectively.
Order 40: There are two examples where the Sylow 2-subgroup is normal: Thedirect product of Z5 with a nonabelian group of order 8. Such order 8 groupsare the dihedral group of symmetries of the square (which has only two elementsof order 4), and the group of the quatemions f±1, ±i, ±f. ±k) (which has sixelements of order 4). There are also several other examples where the Sylow 2-subgroup is not normal.
Solution to 6.8.7: Consider the following eight groups of order 36:
xA1.
Thefirst four are abelian andpairwise nonisomorphic because each pair has eitherdistinct 2-Sylow subgroups or distinct 3-Sylow subgroups. They are not isomor-phic to the last four, since the latter are not abelian.
Of the last four, only Z2 x 1)2.9 has a cyclic 3-Sylow subgroup, only Z3 x A1has a normal 2-Sylow subgroup, and only S3 x 53 has trivial center. Thus the lastfour are pairwise nonisomorphic.
Solution to 6.8.8: Let G be a group of order p2. The conjugacy classes of Gform a partition of G, as the ones associated with the equivalence relation x y1ff there is z such that y = zxz* Let Z denote the center of G. Then x E Ziff C(x) = (x}, where C(x) denotes a conjugacy class. Thus IZI is the numberof singleton conjugacy classes. Now IC(g)I is a divisor of IGI, for each g G.Denote the non-singleton conjugacy classes by C1,..., Ck, where k 0. ThenIGI = IZI + Ci I + + ICkI. If IZI = 1 we obtain the contradiction that pdivides I. Thus IZI> 1.
If G is cyclic then G is abelian. So we suppose that G is not cyclic. Therefore,
each element of G, except 1, has order p. Select x 1 in Z, and y (x). Then
1(x) I = p = J and thus, by Lagrange's theorem, (x) fl (y) = (1). We deducethat the elements x' yf, (1 i, j p) are distinct, and therefore (x) (y) = G. As(x) C Z we see that each element of (x) commutes with each element of (y).
Write H = (x), K = (y); H x K is the direct product group. We define abijection f: H x K G given by f(h, k) = hk. This map is an isomorphism as
f((h, k)(h', k')) = f(hh', kk')
= hh'kk'
= hkh'k' (u E H commutes with v E K)
= f(h, k)f(h'k').
Now H and K are both isomorphic to the cyclic group Z,, of order p. Hence Gis isomorphic to the abelian group x Zr,, which proves that G is abelian. (Infact, this proves that a group of order p2 is either Z,,2 or x
Solution to 6.8.9: The number of Sylow 3-subgroups is congruent to 1 mod 3and divides 5; hence, there is exactly one such subgroup, which is normal in thegroup. It is an abelian group of order 9. The abelian groups of this order are thecyclic group of order 9 and the direct product of two cyclic groups of order 3.
The number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 9;hence, there is exactly one such subgroup, which is the (normal) cyclic group oforder 5. The Sylow 3-subgroup and the Sylow 5-subgroup intersect trivially sotheir direct product is contained in the whole group, and a computation of theorder shows that the whole group is exactly this direct product. Therefore, thereare, up to isomorphism, two possibilities
Z9xZ5, Z3xZ3xZ5.
Solution to 6.8.10: Every element of G of order 7 generates a Sylow 7-subgroup.Therefore G contains more than one Sylow 7-subgroup. The number of Sylow7-subgroups of G is congruent to I modulo 7, and any two of them have a trivialintersection. Hence the number must be 8 (since G, with 56 elements, is too smallto accommodate 15 or more Sylow 7-subgroups). Thus G has 8 x 6=48 elementsof order 7.
If P is a Sylow 2-subgroup of G, then P is contained in the complement ofthe set of elements of order 7, and that complement has cardinality 56—48= 8.Therefore G has only one Sylow 2-subgroup.
Since G has a unique Sylow 2-subgroup P, the subgroup P is normal in G. Letg be an element of G of order 7, and let g act on P by conjugation: h i-÷(h E P). If this automorphism were the identity, then P would commute with thesubgroup generated by g, and it would follow that G is the direct product of Pand the subgroup generated by g. In that case G would have only 6 elements oforder 7, a contradiction. Hence the automorphism of P induced by g is not theidentity. That automorphism therefore has order 7 . Being a permutation of the7-element set P\(1), it must be a cyclic permutation of that set. It follows that allelements of P\ (I) have the same order. Since P contains an element of order 2,all of its nonidentity elements must have order 2.
Solution to 6.8.11: The prime factorization of the order is 80 = . 51• Sincethere are five partitions of 4 and one of 1, by the Structure Theorem for finiteabelian groups [Her75, p. 109], there are five non-isomorphic groups of order 80,
which are:
Z2xZ2XZ2xZ2xZ5Z2xZ2xZ4xZ5Z2xZ8xZ5Z4xZ4xZ5Z16xZ5
Solution to 6.8.12: If p =2, then the group is either cyclic and so isomorphic toZ4, or every element has order 2 and so is abelian and isomorphic to Z2 Z2; fordetails see the Solution to Problem 6.8.1.
Now suppose p > 2 and let G have order 2p. By Sylow's Theorems [Her75, p.91], the p-Sylow subgroup of G must be normal, since the number of such sub-groups must divide 2p and be congruent to 1 mod p. Since the p-Sylow subgrouphas order p, it is cyclic; let it be generated by g. A similar argument shows thatthe number of 2-Sylow subgroups is odd and divides 2p; hence, there is a unique,normal 2-Sylow subgroup, or there are p conjugate 2-Sylow subgroups. Let oneof the 2-Sylow subgroups be generated by h.
In the first case, the element ghg1h1 is in the intersection of the 2-Sylowand the p-Sylow subgroups since they are both normal; these are cyclic groups ofdifferent orders, so it follows that ghg'h' = 1, orhg = gh. Since g and h mustgenerate G, we see that G is abelian and isomorphic to Z2
In the second case, a counting argument shows that all the elements of G canbe written in the form g'hJ, 0 i < p, 0 j <2. Since all the elements ofthe form g' have order p. it follows that all the 2-Sylow subgroups are generatedby the elements g'h. Hence, all of these elements are of order 2; in particular,ghgh= 1,orhg = (g,h Ig" =h2 = 1,hg=g1h)andsoG is the dihedral group
Solution to 6.8.13: By Cayley's Theorem [Her75, p. 71], every group of order nis isomorphic to a subgroup of so it is enough to show that is isomorphicto a subgroup of 0(n). For each a E Sn, consider the matrix Aa = (ajj), whereaa(i) = I and all other entries are zero. Let q be defined by a i—± = Aa.The matrix Aa has exactly one I in each row and column. Hence, both the rowsand columns form an orthonormal basis of R'7, so is orthogonal. mapsinto 0(n). Let = and = Then
An element of this matrix is I if and only if i cr(k) and k r(j) for some k;equivalently, if and only if i = a(r(j)). Hence, Ca(v(j))j = 1 and all the otherentries are 0. Therefore, (Cjj) = so q, is a homomorphism.
If Aa equals the identity matrix, then a(i) = i for 1 i n, so a is theidentity permutation. Thus, has trivial kernel and is one-to-one hence anmorphism.
Solution to 6.8.14: Suppose P and Q are Sylow subgroups corresponding todifferent primes. If x E P and y E Q then Xyx" E Q (since Q is normal),therefore [x, y] = E Q. Similarly, using the normality of P. [x, y] EP. But P fl Q = (1) since P and Q have coprime orders, so xy = yx. The unionof the Sylow subgroups generate a subgroup of G, H say, whose order is divisibleby the order of each Sylow subgroup, therefore by the order of G, and we getG = H. Since G is generated by a set of commuting elements, it is abelian.
Solution to 6.8.15: Any counterexample is clearly nonabelian. In the symmetricgroup S3 we have S = A3, a subgroup. But if G = S4, then S contains the squaresof all 4-cycles, hence all transpositions. But the transpositions generate S4, so ifS were a subgroup it would equal 54. However 4-cycles are not in S since S hasno elements of order 8.
Solution 2. Let G be any simple nonabelian group of even order. Since S is obvi-ously invariant under conjugation = S for all a), if it were a subgroup itwould be either {11I or G. If S = (1) then abab = aabb for all a, b, implying thatG is abelian, a contradiction. If S = G then the squaring map is surjective, henceinjective (since G is finite), contradicting the fact that a group of even order haselements of order 2. Hence S is not a subgroup.
Solution to 6.8.16: For each element x E X, G I gx = x} is asubgroup of G of index IXI, by transitivity. Since the identity is in eachUXEX contains at most
IXI(IGI/IXI — 1) + 1 = IGI — IXI + I
elements. Therefore at least IXI — I elements of G fix no elements of X.
Solution to 6.8.17: Suppose b E G is not in the center of G. Let P =Then G clearly acts transitively on P. The cardinality of P equals (GI/IC(b)Iwhere C(b) is the centralizer of b. Since fe, b) C C(b), b e and C(b) G,the result follows.
Solution 2. Since G is non-Abelian its order is not prime, so it has a nontriv-ial proper subgroup H (for example, a subgroup of prime order, obtained eitherfrom a Sylow Theorem [Her75, p. 91], or as a subgroup of the cyclic subgroupgenerated by a nonidentity element of G). Then G acts transitively by left multi-plication on the left cosets of H.
Solution to 6.8.18: 1. For any set X let S1y be the group of bijections a : X -÷ X.If X is a finite set with n elements then is isomorphic to the group of allpermutations off!, 2, ... , n).
For any group G and any element g E G, define a mapping : G —÷ G by(x) = gx for all x E G. For any g, h, x G, ((pg . ) (x) cog ('ph (x)) =
cog(hx) = g(hx) = (gh)x cogh(x). This shows that c°gh = c°gIt follows that, for each g E G, : G -÷ G is a bijection, with inverse
In other words ço8 E SG for all g E G, defining a map 'p : G SG• Since
ço8h = . ço : G -÷ SG is a group homomorphism. If cog = (Ph for twoelements g,h E G then g = pg(l) = Ph(i) = h, showing that p: G -÷ SG
injective. Therefore, G is isomorphic to the subgroup q(G) of SG.2. Since any group of order n embeds in it suffices to embed into the groupof even permutations of n + 2 objects. Let s : Sn -÷ be the homomorphismthat maps even permutations to 0 and odd permutations to 1.
Define 8 : -÷ Sn+2 by 0(a) = a . (n +1, n + Since the transposition(n +1, n +2) commutes with each element a E 5n. 0 is a homomorphism, clearly
injective. Since a and (n + 1, n + have the same parity, their product 8(a)is even.
Solution to 6.8.19: Case 1. Suppose a and b commute. Then the elementso i <p, 0 j <p are distinct, have order p (if i, j are not both zero).Case 2. Suppose a, b do not commute. Then the p2 —1 elements and aC,
o i < p, 0 < j < p, 0 <k <p are distinct and have order p. Indeed, ifa homomorphism from (a) into Aut(b). Since
the first group has order p and the second has order p — I this morphism must betrivial, which is impossible since a and b do not commute. We havebecause (a) and (b) are distinct groups.
Solution to 6.8.20: Suppose G contains an element whose order is not 2. Then themap g F-± —g is an automorphism of G of order 2, and it follows by Lagrange'sTheorem [Her75, p. 41] that Aut(G) has even order. By the Structure Theoremfor finite abelian groups [Her75, p. 109], it only remains to consider the groupsG = (Z2)' (r = 1,2, . . .), plus the trivial group. If G is trivial or r = 1, thenAut(G) is trivial. If r 2, then G has the automorphism
(X1,X2, X3,X4, ... ,Xjr) (x2,xl, X3,X4, .. . ,
of order 2, so Lagrange's Theorem again shows that Aut(G) has even order.So Aut(G) has odd order if and only if G is trivial or G Z2. In both
Aut(G) is trivial. Observe that it is not necessarily true that Aut(G x H) =Aut(G) x Aut(H) even if G and H are cyclic of orders equal to distinct pow-ers of a prime.
Solution to 6.8.21: Let p be a prime dividing n and ç9(n).The expression for ço(n)shows that either or there is a different prime number q such that qln and
— 1). If p2In, then x x is a noncyclic group of order n. Inthe second case, let G be the subgroup of GL of matrices of the formwhere = 1. Since 1F is cyclic of order q — 1, there are p solutions to = 1
inFq. Therefore IGI = pq.IfaP = 1 andaso G is not abelian. Therefore, G x has order n and is not abelian, so it isnot cyclic either.
6.9 Rings and Their Homomorphisms
Solution to 6.9.2: 1. Let N = If A, B E R, then (A + B)N = AN + BN =NA + NB = N(A + B), so A + B E R. If A, B E R then (AB)N A(BN) =A(NB) = (AN)B = (NA)B = N(AB), so AB E R. And it is trivial to verifythat the matrix, —J also commutes with N, so it belongs to the R, implying thatR is a subring.
2. A simple calculation shows that the matrix A = belongs to R if a =a + c, a + b = b + d, c + d = d, that is, 1ff A has the form Define
a Q-algebra homomorphism : Q[x] -÷ R by mapping x to Clearly= (0 = 0, so induces a homomorphism Q[x]/(x2) —÷ R. Since
i/i(a+bx) = a)' this homomorphism is an isomorphism, and the result follows.
Solution to 6.9.4: Let ço C C be a ring homomorphism and= (1,0,0,..., 0), = (0, 1,0, . .., 0),..., = (0,0,0,..., 1), then
60
that is a nontrivial homomorphism, then 0 for some iand in this case ço(ej) = ço(ejej) = ço(ej)ço(ej) and ço(e,) = 1. At the same time0 = ço(ej ej) = ço(ej )go(e3) we conclude that ço(ej) = 0 for all j i, and ço isdetermined by its value on the coordinate.
ço(xi, .. . ,Xj, = 60(0,... ... ,0)ço(e1)
= ço(O, ...,0)
So for every homomorphism : C C we can create n such homomorphismsfrom C to C by composing ço(x i, . . . , = o (In (xi, ... , where isthe projection on the coordinate, and the argument above shows that all arisein this way. We observe here that is probably the best that can be done, sincehomomorphims from C to C cannot be easily classified.
Solution to 6.9.5: Let R contain k elements (k <oo) and consider the ring
S=RxRx•••xRkcopies
Let R = {ri,...,r*J and a = (rl,...,rk) E S. Now consider the collec-tion of elements a, a2, a3,.... Since S is also a finite ring, by the PigeonholePrinciple [Her75, p. 127], there exist n and m sufficiently large with a'3 = atm.
Coordinatewise, this means that = rr for I i k, and we are done.
Solution to 6.9.6: Ifax = 0 (or xa = 0) with a 0, then axa = Oa or aO =0.If b is as in the text, then a(b +x)a = a, so, by uniqueness of b, b = b +x and
x =0. Thus, there are no zero divisors.Fix a and b such that aba = a. If x E R, then xaba = xa and, as there are
no zero divisors, xab = x, so at' is a right identity. Similarly, abax = ax implies
bax = x and ba is a left identity. Since any right identity is equal to any rightidentity, we get ab ba = 1. Since b = a1, R is a division ring.
Solution to 6.9.7: Since (R, +) is a finite abelian group with p2 elements, by theStructure Theorem for finite abelian groups [Her75, p. 109], it is isomorphic toeither or x In the first case, there is an element x E R such that every
element of R can be written as nx, for 1 n p2. Since all elements of thisform commute, it follows that R is abelian.
In the second case, every nonzero element must have additive order p. Letx R be any element not in the additive subgroup generated by 1. Then it toomust have additive order p. Thus, a counting argument shows that every elementofRcanbewrittenintheformn+kx, 1 p, 1 k p. Sinceallelementsof this form commute, it follows that R is commutative.
Solution to 6.9.8: Assume R is not the zero ring, and let A be the center of R.Then 0, 1 are distinct elements of A. Hence #A 2. The quotient of additivegroups R/A has less than 8/2 = 4 elements, so it must be cyclic. Thus thereexistsx R such that R = {a +mx I a E A, m E Z}. Such expressions commute(so in fact A = R).
The result is the best possible, the ring of upper triangular 2 x 2 matrices overthe field of two elements is an example of a noncommutative ring of order 8, since
11 o\Io i\ 10 i\fi o\lo o) (mod2).
Solution to 6.9.9: R is trivially an additive subgroup and closed under multipli-cation and since C is a field, any subring is an integral domain, finishing the firstpart. Now consider two factorizations of the integer 10 in R, 10 = 2. 5 and
10=(1+3i).(1—3i).Thenormizf2=a2=9b2ofanyzERisaninteger,and if 1z12 <9 then b = 0, so z is a real integer. This implies, in particular, that2 has no non-trivial factorizations in R. If R were a unique factorization domain,then 2 would divide 1 + 3i or 1 — 3i, but can't since (1 ± 3i)/2 are not in R.
Solution to 6.9.10: One can embed R in the field of quotients of R; then the finitesubgroup of R* is a finite subgroup of the multiplicative group of the field; it is afinite abelian group, and so can be written as a direct product of for variousprimes p. If there are two such factors for the same p. then there are at least pelements of the field satisfying the equation — I = 0. However, in a field,due to the uniqueness of factorization in the polynomial ring, there are, at most,n solutions to any n's' degree polynomial equation in one variable. Thus, in thefactorization of our group, each prime p occurs at most once, therefore, any suchgroup is cyclic.
Solution to 6.9.11: Consider the map a : R —÷ R defined by a(x) = ax. Ifa(x) = a(y) then ax = ay, so a(x — y) = 0. Since a is not a left zero divisorthis implies x = y. Therefore a : R —÷ R is one-to-one. Since R is finite, a isalso onto, so there is an element b E R such that ab = 1.
Similarly, using the fact that a is not a right zero divisor, there is an elementC E R such that ca = I. Therefore b = (ca)b = c(ab) = c satisfies ab=ba=1.
Solution to 6.9.12: 1 2 : There exist such that uvl = uv2 = 1; thus,u(vi — 1)2) = 0 and u is a zero divisor.
2 3 : Suppose that u is a unit with inverse v. If the uv = 0 then
w = (vu)w = v(uw) = vO = 0 and, therefore, u is not a left zero divisor.3 1: Let v be a right inverse for u, that is, uv = 1. Since u is not a unit
vu 1 implying vu 1 0. Now consider the element v' = + (vu — 1)and we have
uv'=uv+u(vu— 1)=1—(uv)u—u=1+u-u=1
showing that u has more than one right inverse.
Solution to 6.9.13: The identity element of such a ring would belong to the ad-ditive group of the ring, which is a torsion group; thus there is some finite n suchthat if you add the identity element, 1, to itself n times, you get 0. In other words,the ring would have some finite characteristic n. But this implies that the additiveorder of every element of the ring divides n, and this is false, for example, for theelement 1/(n + 1).
Solution to 6.9.14: Consider separately the cases: a> 1, a = 1, a < 1.
• If a > 1 then a — 1 > 0; multiplying this by a (which is> 0) we get
a2 — a > 0. Adding 1, we get a2 — a + 1> 1 > 0.
• Ifa=lthena2—a-t-1=1>0.
• If a < 1, then 1 — a > 0. If a 0, then a2 > 0; the sum of posItives is
positive, so a2 — a + 1 > 0. If a = 0 then a2 — a + 1 = 1 > 0.
Solution to 6.9.15: The degree 2 polynomial x2 + y2 — 1 does not factor intothe product of two linear ones (since the circle x2 + y2 = 1 is not a union oftwo lines). This implies that the ideal (x2 + y2 — 1) is prime and, thus, the ringR = Q [x, y]/(x2 + y2 — 1) is an integral domain.
Consider now the stereographic projection (x, y) i-÷ (1, y/(x + 1)) (at halfof the speed of the standard one, in order to make the expressions simpler) ofthe circle from the point (—1,0) to the line (1, t). It provides a homomorphismt = y/(x + 1) of Q (t) to the field of fractions of R. The inverse homomorphismis given by the formulas x = (1 — t2)/(1 + t2) and y = 2t/(1 + t2).
Solution to 6.9.16: For any set S of primes, let be the set of rational numbers
of the form a/b where a, b are integers and b is a product of powers of primes inS. It is clear that contains 0 and 1, and is closed under addition, multiplication,
and additive inverses. Hence is a subring. If S and T are distinct sets of primes,
say with p E S \ T, then i/p E but i/p so Since there areinfinitely many primes, we obtain at least subrings in this way.
On the other hand, as Q is countable, its number of subsets is thereforethe set of subrings of Q has cardinality 2K0•
6.10 Ideals
Solution to 6.10.1: For two matrices in A we have
(a b\f bi\_(aai abi+bci\ko
therefore the sets of matrices of any of the forms and are idealsin A. Call these ideals 21k, 212, :13, respectively. We'll show that, together with fO)and A, these are the only ideals in A.
Consider any ideal 21 (0). As A contains all scalar multiples of the identitymatrix, 21 is closed under scalar multiplication. From the equalities
(1 0\(a b\fO o\ (0 b\0)'
(a b\(0 1\ (0 a\ (0 1\(a b\(0 c\0)'
it follows that J contains a nonzero matrix in hence that 21 contains 21i. Suppose21k. Every matrix inJ is the sum of a diagonal matrix and a matrix in i.
There are three cases:
• All diagonal matrices in 21 have the form Then 21 contains all such ma-trices, since it contains a nonzero one, and we conclude that21=212.
• All diagonal matrices in 21 have the form Then, by a similar argument,21=213.
• 21 contains a matrix of the form with a 51 0 c. Since
(1 O\(a 0\_(a 0'\ (0 0\(a 0'\_(0 0\l)\0 c)k0 c,J'
we conclude in this case that 21 contains both 12 and 213, hence that:1 = A.
Solution to 6.10.2: Assume that:1 is a nontrivial ideal. Let be the n x nmatrix with 1 in the (i, f)1Z position and zeros elsewhere. Choose A E 21 suchthat a = 0. Then, for 1 k n, MkjAMJk is a matrix which has a in the(k, k)" entry and 0 elsewhere. Since 21 is an ideal, MkIAMJk E 21. The sum ofthese matrices is al and so this matrix is also in 1 However, since F is a field, ais invertible, so =
Solution to 6.103: We have see in the Solution to Problem 6.10.2 thathas no nontrivial proper ideals. (F) is an F—vector field, and if we identify Fwith (a21 I a F), we see that any ring homomorphism induces a vector space ho-momorphism. Hence, if and M(n+1)x(fl+1)(F) are isomorphic as rings,they are isomorphic as vector spaces. However, they have different dimensions n2and (n + 1)2, respectively, so this is impossible.
Solution to 6.10.4: Since the kernel of h is a two sided ideal in R, it suffices toshow that every ideal 3 in R is either trivial (h is injective) or all of R (h is zero).
Assume 3 is a non-trivial two sided ideal in R. Suppose A 3 with 0for some 0 i, j n. If we multiply A on the left by the elementary matrixwe get the matrix B = which has only one nonzero entry = ajj). If wemultiply B on the left by (l/a,j)I, we get therefore E 3. Multiplyingon the left and on the right by elementary matrices, we produce all elementarymatrices, so these are in 3. As they generate R we have 3 = R.
Solution to 6.10.5: Each element of F induces a constant function on X, and weidentify the function with the element ofF. In particular, the function 1 is the unitelement in R(X, F).
Let 21 be a proper ideal of R(X, F). We will prove that there is a nonemptysubset Y of X such that 21 = (f R(X, F) I f(x) = 0 for all x Y) = 2Fy.Suppose not. Then either 21 C Yy for some set Y or, for every point x E X, thereis a function in 21 such that (x) = a 0. In the latter case, since 21 is an idealand F is a field, we can replace f, by the function so we may assume that
(x) = 1. Multiplying by the function g,, which maps x to I and all otherpoints of X to 0, we see that 21 contains g, for all points x E X. But then, since Xis finite, 21 contains g, 1, which implies that 21 is not a proper ideal.
Hence, there is a nonempty set Y such that 21 C 21y. Let Y be the largest suchset. As for every x 0 Y, there is an €21 such that (x) 0 (otherwise wewould have 21 C IYU(X)) by an argument similar to the above, 21 contains all thefunctions g,, x Y. But, from these, we can construct any function in 21y, so
2Jy c21.Let 21 and 3 be two ideals, and the associated sets be Y and Z. Then 21 C 3 if
and only if Z c Y. Therefore, an ideal is maximal if and only if its associatedset is as small as possible without being empty. Hence, the maximal ideals areprecisely those ideals consisting of functions which vanish at one point of X.
Solution to 6.10.6: If f, g E E(U, W) then so is the homomorphism f + g sinceits image on U is contained in fU+gU. If Y is a subspace and h E(W, 1'), then
h o f E E(U, Y) since fU W + X for some finite dimensional subspace Xand h(f(U)) C h(W) + h(X). Thus E(W, Y)E(U, W) E(U, Y). From thiswe see that E(U, U) is a ring with left ideal E(V, U) and right ideal E(U, 0).Also, E(U, U) ç E(V, V) = E(0, 0) so these latter two sets are also right andleft ideals. The conclusion follows.
Solution to 6.10.7: Let = (aPz — 1, atm — 1) and 3 = (ad — 1). For n rdthe polynomial x" — 1 factors into — + + •• + + i.Therefore, in R, a'1 — 1 = (a'1 — + + • + ar + 1). A similaridentity holds for atm — 1. Hence, the two generators of are in 3,50 C 3.
Since d = gcdfn, m}, there exist positive integers x, y such that xn — ym = d.
A calculation gives
ad — 1 = ad — 1 — += — 1) + a" — 1
=_ct!(am —
+(a" — + • +'? + 1).
Hence, ad — I is in so 3 C and the two ideals are equal.
Solution to 6.10.8: 1. If there is an ideal R of index, at most, 4, thenthere is also a maximal ideal of index, at most, 4, say. Then is a fieldof cardinality less than 5 containing an element a with a3 = a + 1, namelya = a + By direct inspection, we see that none of the fields F2, F3, F4contains such an element. Therefore, t = R.
2. Let R = a = 2 (mod 5), and Y = {O}.
Solution 2. Since has order less than 5, two of the elements 0, 1, a, a2, a3have the same image in R/J. Then 1 contains one of their differences, that is, oneof
1, a, a2, a3, a — 1, a2 — 1, a3 — 1, a(a — a), a(a2 — 1), a2(a — 1).
But all these elements are units, since
a(a—1)(a+I)=a(a2—1)=1, a3—1 =a.Therefore, contains a unit, so 3 = R.
Solution to 6.10.9: Let J be such an ideal. Consider R -÷ the quotientmap. Since is a three element ring with 1, it must be isomorphic to Z3. Ifu E R* is a unit then so is ç9(u). Hence, ço(u) ±1, and ço(u2) = 1. As thesquares of the units generate the additive group of R, this uniquely determines çôso there is, at most, one such J.
Solution to 6.10.10: Let 9Jt be the maximal ideal, so in particular 2 ga a R is not contained in any maximalideal, so (a) = (1), and a is a unit. By assumption, R* is trivial, so 1 + 9Jt has atmost one element, and has at most one element. Thus 9R = {0), and R is afield. Hence I = #(R*) = #R — 1, and #R = 2. Therefore R is the field of twoelements, Z2, which can easily be verified to satisfy the condition.
Solution to 6.10.11: It suffices to show that if ab — ba is any generatorof is a (bc) — (bc)a
is an element of 1 Further, since is a left ideal, b(ca — ac) = bca — bac is anelement of J. Therefore, abc — bac = abc — bca + bca — bac is in 1, and we aredone.
Solution to 6.10.12: Using the direct sum decomposition, there are elementssuch that
1=UI+U2+"+Un.If E 1i then
Therefore (aiui —al)+a2u2 + = 0. SincetheithandR =ai =0for j 1. Similarly we get a1u1 = a, and = 0 for j i.
Solution to 6.10.13: The fact that this map is a ring homomorphism follows fromthe fact that the inclusion map R -+ S is a ring homomorphism mapping mR intomS. Let I = an +bm, with integers a, b. Let r E R be in the kernel. Then r = msfor some s 5, so r = (am + bn)r = m(ar + bns). Since R has index n in S.we have ns R and so r E mR. This shows that the map is an injection. Nowsupposes E S.Thens = (am+bn)s modmS.Asns E R,themapisasurjection.
Solution to 6.10.15: We have 3 = (i) and 3 = (j), for some i, j R. Supposelirstthatl-)-3 = R. Then 1 E 1 = ri+sj forsomer,s E R.Therefore,
the greatest common divisor of i and j is 1. Now Xl and 3 fl 3 are both ideals
and, clearly, have generators ij and k, respectively, where k is the least commonmultiple of i and j. But the greatest common divisor of i and j is 1, so ij = k, andJ3 = 1 fl 3. Since every implication in the previous argument can be reversed, if
= J fl 3, then 3 + 3 = R and we are done.
6.11 Polynomials
Solution to 6.11.1: If P(z) is a polynomial of degree n with a as a root, then
z" P(1/z) is a polynomial of degree at most n with i/a as a root, multiplying by
an appropriate term zk, we have a polynomial of degree n.
Solution to 6.11.2: We have
x3+2x2+7x+1 =(x—aI)(x—a2)(x
Equating coefficients we get
ai +a2 +a3 = —2
ala2+a2a3+a3al =7.
Therefore
a? + + = (aI + + a3)2 —2 (aIa2 + + a3aI)
= -10.
Fori = 1,2,3
Adding these equations we get
(a? +a2+aS)43 =0,
hence,
a primitive seventh root of unity, we have
=0.
Dividing this by we get
1=0.
= +c2)+2and +c')3 = +ç.'),the above equation becomes, letting a = +
- a3+a2—2a—I=O.
Solution to 6.11.4: 1. Let x = + Squaring and rearranging successively,we get
x2
x4—24x2+4=0
This calculation shows that + is a root of f(x) = — 24x2 +4.2. 11 f had a linear factor, then it would have a rational root, but a calculationshows that none of ± 1, ±2 is such a root (if p/q in lowest terms is a root of
+ . • + ao then, pIao and qlan). Suppose now that for some a, b, c, d E Z,
f(x)=(x2 +ax +b)(x2 +cx +d).
Since the coefficient of x3 in f is zero, c = —a, so we have
f(x)=(x2+ax-f-b)(x2 —ax+d).
As the coefficient of x in f is zero, we get ad — ab = 0. If a = 0, then f(x)(x2 +b)(x2 +d) = x4 +(b +d)x2 +bd, but the equations bd = 4, b+d = —24
have no integer solutions. If b = d, then f(x) = (x2 + ax + b)(x2 — ax + b) =x4 + (2b — a2)x2 + b2, so b2 = 4 and 2b — a2 = —24, which also have nosolutions in Z.
Solution to 6.113: It is easy to see that is a zero of a monic polynomialp E Z[xJ (use the process described on Problem 6.11.4.) If it were a rationalnumber, it would have to be an integer, since its denominator would divide theleading coefficient of p. As + is strictly between 2 and 3, it must beirrational.
Solution 2. Suppose + E Q. Then Q = Q However, this
contradicts the fact that the fields Q and Q have degrees 2 and 3 over
Q, respectively, since by Eisenstein Criterion [Her75, p. 160], the polynomialsx2 —2 and x3 —3 are irreducible over Q.
Solution to 6.11.6: Suppose that w is a primitive kth root of unity and that( for I i k. Let P(z) = ao + aiz + ••• + a3z3 (j < k); we
have
p((S) = a,.WW = a,. (,ri=1 i=1 r=O r=O i=1
Since wk = 1, we have — 1 = Ofor 1 r j. Factoring and replacing 1 bywe get
0 = — +
r <k and w is a primitive root of unity, il 1. Therefore,
+ + • + iSubstituting this into the above equality gives
k k
! P(w') = ! = ao = P(0).
Solution to 6.11.7: By the Euclidean Algorithm, the vector space V = Q [x]/(f)has dimension d = deg(f) Therefore, the infinitely many equivalence classes
are linearly dependent in V, so we can let qj be a finite collection of rationalnumbers not all zero and satisfying
This means that
q2x2 + q3x3 + +••• = f(x)g(x)
for some nonzero g E Q [x].
Solution to 6.11.8: First, note that each polynomial p(x) in Z[x] is congruentmodulo x —7 to a unique integer, namely the remainder one obtains by using thedivision algorithm to divide x —7 into p(x). (Only integer coefficients arise in theprocess, because the coefficient of x in x —7 is 1.) if p (x) lies in then so doesthe preceding remainder. However, the only members of 1 fl Z are the integers thatare divisible by 15. In fact, if k is in 1 fl Z, say k = (x — 7)q(x) + 15r(x), thenk = 15r(7). Hence, we get a well defined map from Z[x]/3 into Z15 by sendingp(x) + to k + 15Z, where k is the remainder one gets when dividing p(x) byx — 7. The map is clearly a homomorphism. If p(x) is not a unit in J, then theremainder is clearly not divisible by 15, from which we conclude that the map isone-to-one. The map is obviously surjective. It is, thus, the required isomorphism.
Solution 2. The map ço : Z[x} -+ Z[xJ defined by ço (p(x)) = p(x +7) is a ringautomorphism and it maps J onto the ideal generated by x and 15. The quotientring Z[x]/ço(1) is isomorphic to Z15 under the map p(x) i-* p(O)+15Z, implyingthe desired conclusion.
Solution to 6.11.9: is prime. To show this we will prove that the quotient ringZ[x]/J is a field. Since 5 E this quotient ring is isomorphic toZ5[x]/(x3 + x +1). So it suffices to show that x3 + x + 1 is irreducible (mod 5).If it were reducible, it would have a linear factor, and, hence, a zero. But we canevaluate this polynomial for each x E Z5 as follows:
x x3-I-x+l
0 1
1 32 1
3 1
4 4
Since there is no zero, the polynomial is irreducible, and the quotient ring
Z5[x]
(x3+x+1)is a field.
Solution to 6.11.12: Case 1: p = 2. In this case, define a map of F2[x] into itselfby ço(l) = 1 and ço(x) = x + 1, and extend it in the obvious way. Since constantsare fixed and p(x + 1) = x, it is clear that this is a ring isomorphism. Further,
— 2) (x — 1)2 — 2 x2 + I = — 3; we see that ço maps the ideal
(x2 — 2) onto the ideal (x2 — 3). It follows immediately from this that the tworings F2[x]/(x2 —2) and F2[x]/(x2 —3) are isomorphic.Case 2: p = 5. By checking all the elements ofF5, we see thatx2 —2 and x2 —3are both irreducible polynomials in F5[x]. Therefore, the ideals they generate aremaximal and the quotient rings F5[x]/(x2 —2) and F5[x]/(x2 — 3) are fields.The Euclidean Algorithm [Her75, p. 155] shows that each is a finite field with 25elements. Since finite fields of the same order are isomorphic, the quotient ringsin this case are isomorphic.Case 3: p = 11. In this case, checking all the elements of F11 shows that x2 — 2is irreducible, but x2 — 3 = (x — 5)(x + 5) is not. Hence, the quotient ringFii[x]/(x2 — 2) is a field, whereas Fij[x]/(x2 — 3) is not, so the two quotientrings are not isomorphic in this case.
Solution to 6.11.13: Since x —3 is a monic, given any polynomial r(x) in Z[x],there exist polynomials t(x) and s(x) such that r(x) = t(x)(x — 3) + s(x) anddeg s(x) < deg(x — 3) = 1. Hence, s(x) is a constant, and so it is congruentmodulo 7 to some a, 0 a 6. Hence, r(x) — a = t (x)(x — 3) + (s(x) — a),and the right-hand term is clearly an element of 1
In the special case where r(x) = x2 + 5, we have, by theEuclidean Algorithm [Her75, p. 155], r(x) = t(x)(x —3)+a. Substitutingx = 3,
we get r(3) = a. Since we only need to know a modulo 7, we reduce r(3) mod7 using the fact that n7 n (mod 7), getting r(3) 34 + 32 + 32 + 5 6(mod 7). Hence, a = 6 is the desired value.
Solution to 6.11.14: Let q' : Z[x} Z13 be the unique ring homomorphismsuch that = 4. A polynomial a(x) E Z{x] is in the kernel of ço if and onlyif a(4) 0 (mod 13). This occurs if and only if a(x) (x — 4)46(x) (mod 13)
for some fl(x) E Z[x], i.e., exactly when a(x) = (x — 4)46(x)+ 13y(x) for somey(x) E Z[x], in other words if and only if a(x) E 1.
Set f(x) = (x26 + x + E Z[x]; then f(x) — m E if and only if— m) = 0, which holds if and only if f(4) m (mod 13).
By Fermat's Little Theorem [StaS9, p. 80], {Her75, p. 44], if a E Z is notdivisible by the prime p then I (mod p). This gives (426 + 4 +(42 +5) 8 (mod 13), and f(4) 8 (mod 13). So m = 8 is the uniqueinteger in the range 0 m 12 such that (x26 + x + — m E J.
Solution to 6.11.15: Let f(x) = —2, and denote by the real cube root of
2. Then Q is an algebraic extension of Q generated by a root of f, hence
isomorphic to K = Q [x]/(x3 — 2). Since Q C IR, and x3 —2 has only
one real root, x3 —2 does not factor completely over K.
Solution to 6.11.16: 1. Let K be the set off E R[x] for which f(2) = f'(2) =f"(2) = 0. K is an ideal in the ring R{x] if
(1) K is closed under addition and negation, and 0 K,
(ii) gf and fg belong to K whenever g E and f E K.
Condition (i) is easily verified. For condition (ii), let g E IR[xJ, f E K. Then
(gf)' = g'f + gf', (gf)" = g"f + 2g'f' + gf". We see that (gf)(2) =(gf)'(2) = (gf)"(2) = 0, so that gf, and also fg, belongs to K. Thus K isan ideal of REx].
If K is an (nonzero) ideal of R = IR[x], let a be a nonzero element of K oflowest degree. Ifb E K then thereexistq, r E R such thatb = aq+r withr = 0or deg(r) <deg(a). Now r = b — aq E K, so deg(r) deg(a). That is, r = 0and b = aq. We conclude that K = aR is generated by a. Two generators of Kare associates and there is a unique monic generator. Expand in powers of (x —2):
a = ao + ai(x —2) + a2(x — 2)2 + a3(x — +•• . + — 2)1z.
Then ao = a(2) = 0, ai = a'(2) = 0, = a"(2)/2! = 0. The monic generatoris(x—2)3.2. The function f(x) = x2 — 6x + 8 satisfies f(2) =0 = f'(3). If g(x) = x then(gf)'(3) 0. Thus the stated condition does not define an ideal.
Solution to 6.11.17: Let : Z[xJ -÷ Z[x] be any automorphism. Since ço isdetermined by the value of x, every element of Z[x] must be a polynomial inço(x). In order to get x in the image of çü, we see that must be of the form±x + a for some constant a.
Solution to 6.11.19: The multiplicative group of units has p — 1 elements.
Therefore, = I foreacha E —a =Oforalla Epolynomial — x has p distinct roots, thus factorizes as
xt' —x =x(x — l)...(x —(p—i)).
If f(x) = u is a unit and are distinct monic irreduciblepolynomials, then we see that the gcd of f and g is equal to the product of thedistinct linear factors of f.
Solution to 6.11.20: Let p . . . a, aremonic irreducible in F[x], and 0, Ui units of F. If y, = min(a,, /3k) thenh = aT' .. . is the god of p and q over F.The units of F[x] are the nonzero constants, and remain units in K[x]. If a, =
is the irreducible factorization of a' over K then p = fl, fi. isthe irreducible factorization of p over K, by uniqueness of factorization.obtain the irreducible factorizations of p and q over K by further factorizing the
over K.If a and b are distinct irreducible polynomials over F then their god over F is
1. There exist polynomials S. t over F with sa + tb = 1. Thus, if d e K[x} is thegcdofa andb over K, then dli, and sod is a unit ofK[x], andd = las d ischosen monic.
Let h, h' be the gcds of p and q over F, K respectively. Then h divides h'. Bythe above we cannot get any extra factors over K in h'. Thus the gcd of p and qover K is the same as their gcd over F.
Solution to 6.11.21: If the fraction p/q in lowest terms is a zero of +x9 +x8 ++x + 1, then p11 and qJl, so the possible rational zeros are ±1. A calculation
shows that neither of these is a zero, so the given polynomial is irreducible overQ . Now —1 is a zero of the second polynomial, so it is reducible over Q.
Solution to 6.11.22: Note that 539 = 72. 11, 511 = 7 23, and 847 = 7. 112.
Thus, all the coefficients except the leading one are divisible by 7, but the constantterm is not divisible by 72• Since 7 is a prime, by Eisenstein Criterion [Her75, p.160], the polynomial is irreducible in Z[x].
Solution to 6.11.23: By the Gauss Lemma [BML97, p. 85], an integral polyno-mial that can be factored over rationals can be factored into polynomials of thesame degree over the integers. Since ±1 are not roots (and they are the only onespossible because ao = = 1), there are no linear terms and the only possiblefactorizations are in polynomials of degree 2.
• (x2+ax-t-1)(x2+bx+1)
• (x2+ax—1)(x2+bx—1)
In the first, case we get x4 + (a + b)x3 + (2 + ab)x2 + (a + b)x +1, which impliesthat the coefficients of the terms of degree I and 3, are the same, a contradiction.The other case is analogous, showing that the polynomial is irreducible over Q.
Solution to 6.11.24: We will use Eisenstein Criterion [Her75, p. 160]. Letx =y+1.Wehave
+(y+ 1
—
— (y+1)-1
y
z—yP1 +py"2+"+p.
Since the prime p divides all the coefficients except the first and p2 does notdivide the last, it follows that the polynomial is irreducible in Q[x]. Therefore, thegiven polynomial must also be irreducible, since if it were not, the same changeof variables would give a factorization of the new polynomial.
Solution to 6.11.25: Put x = y + 1 to get
+5x=(y+1)5—1
+5y+5.
The coefficients of y3, y2, y, and 1 are integers divisible by p 5 and the constant
term is 10, which is not divisible by p2. Thus, by Eisenstein Criterion [Her75, p.
160], f is irreducible over Q.
Solution to 6.11.26: By the Gauss Lemma [BML97, p. 85], it is enough to show
that f(x) is irreducible over the integers. Suppose f(x) = g(x)h(x), where g(x)
and h(x) have positive degrees and integer coefficients, say
h(x) = CkXFC + Ck_1X' + ... +Co
The we can see that boco = 25, which can only happens in two ways, eitherbo=25,co=lorbo=5,co=5.Case 1: bo = 25, = 1.
Then 75 = biCO + cibo = hi + 25ci, so 25 divides b1. Hence —100 = b2co +biCi + = 1,2 + bici + 25ci, so25 divides b2. Continuing in this way, wefind that 25 also divides 1'3 and b4. However 16, the leading coefficient of f(x),is given by
16 = b4cl +b3c2 +b2C3 +bic4 (*)
(since j <5, k <5), and 25 does not divide 16. Therefore Case 1 is impossible.Case 2: b0=5,c0 =5.Then 75 = 5b1 + —100 = 51'2 + b1c1 + 5c2. The first of these equalities
tells us that if either bi or Cl is divisible by 5 then both are divisible by 5. By thesecond equality, 5 divides bi Cl. Hence 5 divides both b1 and Cl.
Similarly, we have 50 = Sb3 + b2c1 + blc2 + 5C3 and —125 = Sb4 + +b2C2 + bic3 + 5b4. Exactly the same reasoning as before shows that 5 divides b2and C2, again a contradiction with (*), since 5 does not divide 16.
Solution to 6.11.27: Let 1(x) = x3 + x + 2. A calculation shows that 2 isa zero of f(x) over Z3, but 0 and 1 are not. Hence, we get the factonzationf(x) (x — 2)(x2 + 2x +2) = (x — 2)g(x). Clearly, 0 and 1 are not roots ofg(x) since they are not roots of f(x); another calculation shows that 2 is not aroot of g(x). Hence, g(x) is irreducible, and the above factorization is the desired
one.
Solution to 6.11.28: Let 1 x4 + x3 + x +3. A calculation shows that f hasno zeros in Z5,so it contains no linear factors. Consider a product of two monicquadratics.Now3 = 3.1 =2-4..Iff =(x2+ax+1)(x2+bx+3),then,equatingcoefficientsofpowersofx,wegeta+b = 1,4+ab = Oand3a+b = 1.This equation has no solutions in Z5. If f = (x2 + ax + 2)(x2 + bx + 4), then
a+b= 1,1 +ab=Oand4a+2b= I.Thisequationhasnosolutionsinz5.Thus x4 + x3 + x +3 is irreducible in Z5[x].
Solution to 6.11.29: 1. There are five distinct monic irreducible polynomials ofdegree 1: x, x + 1, x + 2, x + 3, x + 4. Multiplying these in pairs we obtainreducible monic polynomials of type p2 or type pi P2 with P1 P2. There are5 + = 15 distinct reducible monic polynomials of degree 2. The number ofmonic polynomials x2 + ax + b is 25. Therefore there are 10 monic irreduciblepolynomials of degree 2.2. There are 5 monic irreducible polynomials of degree 1 denoted p, and 10 ofdegree 2 denoted q. A reducible polynomial of degree 3 has the form p3 (5 ofthese), or (5 x 4 = 20 of these), or = 10 of these), or pq(5 x 10= 50 of these). There are 53 = 125 monic polynomialsx3 +ax2 +bx +cof degree 3. Thus there are 125 — 85 = 40 monic irreducible polynomials ofdegree 3.
Solution to 6.1130: x4+ 1 = (x2+ 1)2 —2x2 = (x2 —ñx+1 is reducible over the real numbers.
The above expression is the irreducible factorization of x4 + 1 over IL Theirreducible factorization over Q can be obtained from this by suitably groupingfactors and can only be x4 + 1 itself. Thus x4 + 1 is irreducible over the rationals.
The field F16 has characteristic 2; that is, 1 + 1 = 0. Thus x4 + 1 = (x +is reducible over F16.
Solution to 6.1131: In this solution we use the fact that a polynomial of degree2 or 3 is reducible over a field if it has a root there.
Decomposing x4 —4:
• over R, we have
where x2 + 2 is irreducible.
• over Z, we havex2—4=(x2—2)(x2+2)
where both x2 —2 and x2 + 2 are irreducible.
• over Z3, we have
x4_4=x4_1=x_1xx3+x2+x÷1)=(x_1)(x+1)(x2+1)
and x2 + 1 is irreducible.
Decomposing x3 —2:
• over R, we have
and x2 + + is irreducible.
• over 7L, x3 —2 has no roots, and is irreducible.
• over Z3, we have
x3—2=x3+l = (x+1)(x2—x+l) = (x+1)(x+1)(x+1) = (x+1)3.
Solution to 6.1132: Suppose, to the contrary, that xl' — a has nontrivial fac-tors f(x) and g(x) in Fix]. Let K be a splitting field of — a. Then, in K,there are elements al,...,ap such — a = (x — aI)••(x — ap). We
may assume without loss of generality that f(x) = (x — aI)• • (x — ak) and
g (x) = (x — ak+1) .. (x — ap). Therefore, A = ai • • ±f (0) and
B = ak+1 . . = ±g(0) are both elements of F. Further, since the a/s arezeros of — a, a for all j. Hence, = and B!' = Since k andp are relatively pnme, there exist integers x and y such that kx + py = 1. Letr = —y and s = x + y. Then /B' is an element of F and
a a power, contradicting our assumptions. Therefore, xP — a mustbe irreducible over F.
Solution to 6.1133: Suppose g(x) (or h(x)) is not irreducible. Then x4 + 1 hasa linear factor and, hence, a zero. In other words, there is an element a E witha4 = —1. It follows that a8 = 1, and since a4 I (p is odd), a has order 8 in themultiplicative group Z, of the field with p elements. But is a group of orderp — 1 2 (mod 4), 808 cannot divide p — 1, and we have a contradiction.
Solution to 6.11.34: Let deg f = n. Then the collection of all real polynomialsof deg n — I is an rn—dimensional vector space. Hence, any collection of n + Ipolynomials of degree n —I is linearly dependent. By the Euclidean Algorithm[Her75, p. 155], we have
= qo(x)f (x) + ro (x) with deg TO <nx2 = qi(x)f(x)+ri(x) withdegr1 <n
= <n.
The polynomials ro, ..., are linearly dependent, so, for some E IR, we have
0.
Therefore,
p(x) = =
and f(x)Ip(x).
Solution to 6.11.35: Fix f R \ F of least degree n, say. Choose polynomialsfl,f2,..,fn—J j (modn)andsuchthateach isof least degree with this property, 1 j n — 1, if such a polynomial exists,otherwise take = 0. Let = f. We will prove that R = F[f1, ...,Suppose not, and fix g E R \ F[fi, 12, ..., In] of least degree, and supposedeg g j (mod n). For some k 0, deg g = Hence, g — isof lower degree than g for some a E F, and, by the minimality of g, must lie inF[f1, 12, ..., However, this implies that g E F[f1, ..., as well.
6.12 Fields and Their Extensions
Solution to 6.12.1: Since the 2 x 2 matrices over a field form a ring, to show thatR is a commutative ring with 1 it suffices to show that it is closed under additionand multiplication, commutative with respect to multiplication, and contains I.The first and the last are obvious, and the second follows almost as quickly from
(a —b\(c —d\ — (ac — bd —ad — bc\ — (c —d\(a —bkb a)k.d a
a nonzero element in R is given by
1 (aba
whichliesin Rprovidedthata2+b2 0.1fF —Q,thena2+b2 > Ofora andb not both equal to 0; hence, every nonzero element of R has an inverse, so R is afield. 1fF = C ,then the matrix
(i —1
i
has no inverse since i2 + 12 = 0. Therefore, in this case, R is not a field. Similarly,if F = Z5, we have that 22 + 12 = 0, so there exists a noninvertible matrix inR and so R is not a field. Finally, if F = Z7, the equation a2 + b2 = 0 has nononzero solutions, so every nonzero element of R has an inverse and R is a field.
Solution to 6.12.2: Let R be a finite integral domain. Let 0 b E R and enumer-ate the elements of R by C2 ... , Since R, has no zero divisors, cancellationshows that the elements are distinct. Since there are n of them, it follows thatthere is an element c10 such that = 1. Hence, c,0 is the inverse of b and weare done.
Solution to 6.12.3: Since a and b are algebraic over F, there exist integers nand m such that [F(a) : F] = n and [F(b) : F] = m. Because b is algebraic
over F, it must also be algebraic over T = F(a) of degree, at most, m. Hence,
[T(b) : F(a)] m, which implies
[T(b) : F] = [T(b) : F(a)][F(a) : F] nm.
Therefore, T(b) is a finite extension of F. T(b) contains a + b and so containsF (a + b); the latter must, therefore, be a finite extension of F, so a + b is algebraic
over F.
Solution to 6.12.4: From the fact that the group is finite, we can see that all ele-
ments are in the unit circle, otherwise consecutive powers would make an infinite
sequence.All elements of the group are roots of unity (maybe of different degrees), since
a high enough power will end up in 1 (the order of an element always dividesthe order of the group). We will prove that they are all roots of unity of the samedegree. Let a be the element with smallest positive argument (arg a E (0, 2ir)).We will show that this element generates the whole group. Suppose that there is
an element that is not in the group generated by a. There is a p E N such that
arga1' <argaP+i
therefore,<arga
which contradicts the minimality of arg a. We conclude then that the group isgenerated by a. See also the Solution to Problem 6.12.5.
Solution to 6.12.5: Let G be a finite subgroup of F* of order n. By the StructureTheorem for finite abelian groups [Her75, p. 109], there are integersmi 1m21 . . . lmk such that G is isomorphic to x ... x Zmk. To show that G iscyclic, it suffices to show that mk = n. Suppose that <n. From the structureof G we know that gmk = 1 for every g E G. Hence, the polynomial Xmk — 1
has n roots, contradicting the fact that a polynomial over a field has no more rootsthan its degree. Hence, mj = n and G is cyclic.
Solution to 6.12.6: Let K+ denote the additive group of K. By the StructureTheorem on finitely generated abelian groups [Her75, p. 109], K÷where n 0 and T is a finite abelian group. Then 2K÷ (2T). Ifn > 0, then 2K is an ideal of K not equal to (0) or K, but this contradicts theassumption that K is a field. Thus n =0, and K T is finite.
Solution to 6.12.7: Since x3 —2 is irreducible over Q [x], (x3 — 2) is a maximalideal in Q [x], so F = Q [x]/(x3 — 2) is a field. Using the relationship = 2,we get that every element of F can be written in the form a + + cx12, wherea, b, c E Q. Further, such a representation is unique, since otherwise we could
0.OnpullingbacktoQ[x],wefind that x3 —2 divides a + bx + cx2, a contradiction.
Consider the map q.' : F -+ F given by q(a) = a if a E Q, and =Since = 2 and = 2 this extends to a ring epimorphism in the obviousway. ills also one-to-one, since = Othen
a = b = c =0. Hence, F is the isomorphic image of a field, and it is field.Further, by the isomorphism we see that every element can be expressed
uniquely in the desired form. In particular, (1 — — — = I.
Solution to 6.12.8: Consider the ring homomorphism q.' : Z —÷ F that satisfies= 1. Since F is finite, kerq, 0. Since F is a field, it has characteristic p.
where p is a prime number. Hence, ker q.' contains the ideal (p), which is maximal.Therefore, ker q.' (p), and the image of Z is a subfield of F isomorphic to Zr,.Identify with this subfield. Then F is a vector space over Z,,, and it must beof finite dimension since F is finite. Let dimF = r. A counting argument showsthat F has pr elements.
Solution to 6.12.9: Let a be any element of K. Then L contains a2 and (a + 1)2,hence it contains
(a+1)2—a2—1=2a.
Since 2 0, L contains a.If K is the finite field of order 2', then the multiplicative group of K is cyclic
of order T' — 1, an odd number. The homomorphism a a2 on this group hasa trivial kernel, hence it is surjective, therefore, every element of K is a square. Inthis case, then, L = K.
To obtain a field of characteristic 2 in which the equality L = K can fail,consider K = F2(x), the field of rational functions with coefficients in F2, thefield of order 2. For r(x) in F2(x) we have r(x)2 = r(x2), so that the subfieldL = (r(x2) r(x) E K) is the desired counterexample.
Solution to 6.12.10: Since F has characteristic p. it contains a subfield isomorphicto which we identify with Z,,. For j E a +j is an element of F(a). Usingthe identity (a + j)P = + jP, we get
f has p roots in F(a), which are clearly distinct.
Solution to 6.12.11: The polynomialx'0—2 is irreducible over Q ,by Eisenstein'sCriterion. Hence
Q (Wi) =Q[x]/(x10—2)
has degree 10 over Q.Let be an automorphism of K. It is well known that q acts like the identity
onQ.Leta=in K. Therefore, either = a or = —a. In the first case q is the identity.It is easy to see that p(a) p(—a) (p E Q [x]) defines an automorphism of K.Thus, K has exactly two automorphisms.
Solution to 6.12.12: Notice that Q [xl is an Euclidean domain, so the ideal (f, g)is generated by a single polynomial, h say. f and g have a common root if andonly if that root is also a root of h. We can use the Euclidean Algorithm to findh = x2 + 3x +1, which has roots —3/2± Therefore, f and g have exactlytwo common roots, both in Q
Solution to 6.12.13: Let x be an element of F that is not in Q . Then x satisfies anequation ax2 + bx + c = 0 with a, b, and c E Q . Completing the square, we see
that (x + E Q, whereas (x + Q . Let = (x + As E Q, we can
write it as where c, d, m E Z and m is square free (i.e., m has no multipleprime factors). As F = Q = Q (s), we get an isomorphismF -÷ Qby sending r + s to r + The uniqueness of m follows from the factthat the elements of F that are not in Q but whose squares are in Q are those ofthe form k E Q.
Solution to 6.12.14: Let P1 = 2, = 3, ..., be the prime numbers and= Q Claim: The fields F
were isomorphic to then there would exist r E such that r2 = p,. Writesuch an r in the form
a,bEQ.
Thenr2 = a2 + + = p,
if and only if ab = 0. Therefore, either p, = c2 or pj = c2 for some c E Q,which contradicts the primality of p, and Pj•
Solution to 6.12.15:
1 0cos sin 0
sin 0
All the possibilities can occur. For example
dimF9Eo=1 if
dimF0Ee=2 if 0=irdimF9Eo=3 if
In the last example, 4x3 — 3x + 1/2 or 8x3 — 6x + 1 is irreducible because ±1±1/2, ±1/4, and ±1/8 are not roots of the above polynomial.
Solution to 6.12.16: We first prove, using the Induction Principle [MH93, p. 7],that there are n algebraically independent ([BML97, p. 73]) real numbers. Thecase n = 1 is obvious. Suppose ai, ... , i are algebraically independent. ThenQ(ai, .. . ,an_I) is countable, and so is the the set of numbers algebraic over it.Let an be transcendent over Q(a1, ... , an_I). Then clearly ai, . .. ,an_I, an areindependent too.
Let ci': Rbedefinedby
... , p(ai, ...
is clearly a homomorphism. If
pI(al,...,an) — p2(aI,...,an)
then — q2pI)(aj, ... , = 0 and the independence of the a's givesP1 /qi = p2/q2, so is injective, therefore it is an isomorphism over its range.
Solution to 6.12.18: We'll show that K[t'2, ti'] is a unique factorization domain,UFD for short, if and only if one of a and b divides the other. Let d = gcd(a, b),a' = a/d and b' = b/d. Then tb] is the image of K[ucA', ub'] under theinjective K—algebra homomorphism K[u} —÷ K[t] mapping u to so we mayreduce to thecase thatgcd(a, b) = 1.Ifa=l orb=1, t1'] = K[t]isaUFD. It remains to show that if a, b> 1 and gcd(a, b) = 1, then K[ta, ti'] is nota UFD.
First we show that ta is an irreducible element of tb]. Since t is an irre-ducible element of the UFD K[t] with unit group K*, the only ways to factor taintononunitsofK[t] are as for some c E K* and 0 <rn <a. Butm and a — m cannot both be nonnegative integer combinations of a and b (theywould have to be positive multiples of b, and their sum then could not be a), soCtm and c_lta_m cannot both lie in K[ta, ti']. Thus t" is irreducible in K[ta, tb].Similarly tb is irreducible in K[ta, t'i. Moreover t" and tb are not associate in
t1'] because even in K[t] it is not true that each of them divides the other.Now tab can be factored in tb] either as or as violating uniquefactorization.
As an alternative to the argument in the previous paragraph, one could note thatsince gcd(a, b) = 1, there exist r, s E Z with ra + sb = 1, so the fraction fieldof t1'] (viewed as a subfield of K(t)) contains t = even thoughK[ta, ti'] obviously does not contain t if a, b> 1. Butt is integral over tb]
because it satisfies the polynomial equation — ta = 0. Hence K[ta, ti'] is notintegrally closed (in its fraction field). UFD's are integrally closed, so tb] isnot a UFD.
Solution to 6.12.19: 1. A 2 x 2 matrix over F is invertible if and only if thefirst column is nonzero and the second is not a F—multiple of the first. This givesFl2 — 1
= p2Ji — 1 possibilities for the first column of an invertible matrix, and,
given the first column, — Fl = p2n — for the second, hence the result. Seealso Problem 7.1.3 for a more general solution.2. The map F -÷ G sending a to ) is easily checked to be an injective grouphomomorphism. Its image is a subgroup S of G that is isomorphic to the additivegroup ofF and that, consequently, has order p's. By 1., this is the largest power ofp dividing the order of G. Hence, S is, in fact, a p-Sylow subgroup of G. Since allp-Sylow subgroups of a finite group are conjugate (and hence isomorphic), thisimplies the result.
Solution to 6.12.20: Assume g E has order p1. and let m(x) be theminimal polynomial of g —1. By Fermat's Little Theorem [Sta89, p. 80], [Her75,p. 44], (x — 1)P = xP —1 (mod p), and iteration gives (x —
1)Pt = —1 (modp) for all positive integers £. Since
gpk— I = 0, also (g — = 0, so m(x)
divides (x — By the same token, m(x) does not divide (x —1 )pk_1
Thereforem(x) = (x — 1)) with p1'1 <j pk• But j = deg m(x) n, so the desiredinequality follows.
For the other direction, assume first pk• Let g be the n x n matrixwith 1 in each position on the main diagonal and immediately above the maindiagonal, 0 elsewhere. Then g — I is nilpotent of index n, so
0=(g__1)Pk =gpk_1,
It follows that the order of g divides but does not divide so it must equal
Finally, if n > we can get a g of order p1' by taking the direct sum of thepreceding g (corresponding to n = p1') and an identity matrix of size (n pk) x(n _pk)
Solution to 6.12.21: 1. We can write down the general commutator:
'a b \ Ic d \ —b\
( ac ad+bc')
1 —cd—abc2 +a2cd+ab1 )
The upper-right entry can be rewritten as ab(1 — c2) — cd(1 — a2). This is 0 ifF is the field of two or the field of three elements. Otherwise we can take d = 0and c such that 1 — c2 0. Then, by suitably choosing a and b, we can makethe upper-right entry equal any element of F. Conclusion: the commutators arethe matrices with x in F, except in case F is the field of two or the field ofthree elements, in which case the identity is the only commutator. In either case,the commutators form a group, which is the commutator subgroup.
2. Except in the trivial cases p = 2 or 3, k = I, the commutator subgroup isisomorphic to the additive group of F, which is a vector space of dimension kover the field of p elements. The fewest number of generators is thus k.
Solution to 6.12.22: The zero element of F,, obviously has a unique square rootand a unique cube root. Let F denote the multiplicative group of nonzero ele-ments of F,,. It is a cyclic group of order p — 1. Since p — I is even, the homo-morphism x —+ x2 of F into itself has a kernel of order 2, which means that itsrange has order (p — 1)/2. There are, thus, 1 + (p — 1)/2 elements of withsquare roots.
If p — 1 is not divisible by 3, the homomorphism x -÷ x3 of F into itself hasa trivial kernel, and so every element of has a cube root. If 3 divides p — 1,
then the preceding homomorphism has a kernel of order 3, 50 its range has order(p — 1)/3. In this case, there are 1 + (p — l)/3 elements of F,, with cube roots.
Solution to 6.12.23: All functions are polynomials. A polynomial with the value1 at 0 and 0 elsewhere is p(x) = 1 — xe'; from this one, we can construct anyfunction by considering sums f1 . p(x —xe). Thus, there are such functions,and that many polynomials. Another way is to observe that all polynomials ofdegree, at most, q — 1 define nonzero functions unless the polynomial is the zeropolynomial.
Solution to 6.12.24: Let K be that subfield. The homomorphism of multiplicativegroups F* K* sending x to x3 has a kernel of order, at most, 3, so IK* IIF*I/3, that is, (IFI — 1) 3. Also, if the extension degree [F : K]equals n, then n 2, so IFI = IKI?z 1K12, and (IFI — — 1) IKI + I,with equality if and only if n =2. Thus, 3 IKI +1, which gives IKI = 2, n =2,and IFI=22=4.
Solution to 6.12.25: As A has dimension at least 2 as a vector space over C,it contains an element a which is not in the subspace spanned by the identityelement 1 A. Since A has finite dimension over C, there exists a complexnumber A such that (a — Al)x = 0 for some nonzero x E A. Let f be the idealgenerated by b = a — Al, and 3= fx E A
Ibx = 0} the annihilator of b. We have
J fl 3 = (0) since all the elements of 1 fl 3 have zero square. As the dimensions ofand 3 add up to the dimension of A we must have A = 3 as vector spaces
over C. Since 1 and 3 are ideals in A, A = 3 as rings. Let 1 = e + f withe E land f E 3. Then e2 = e, = f and neither of them is zero or 1.
Solution to 6.12.26: Suppose that A* has order 5. We have —1 = 1 in A,since otherwise A* would have even order. Hence we have a homomorphism
: F2[x] —÷ A sending x to a generator of A*. The kernel of is generatedby a polynomial dividing x5 —1 in F2[x], but not dividing x —1, since the imageof x is not 1. Now x5 —1 = (x—l)(x4+x3+x2+x+ 1), and the latter is irreducibleover F2, since otherwise it would have a factor of degree 2, and F2 or F4 wouldcontain a nontrivial fifth root of unity. Hence ker q.' equals (x4 + x3 + x2 + x +1)
or ((x — 1)(x4 + x3 + x2 + x + 1)) and A contains a subring B = F2[x}/kercp
isomorphic to or F2 x F24. In either case, IB*I = 15, a contradiction.
Solution to 6.12.27: The relations imply ab + ba 1. In particular a 0 b.
Multiplying the equality ab + ba = 1 (from either side) by a gives aba a, so
ab 0 ba. Multiplying the same equality by b gives bab = b. The three-
equalities ab + ba = 1, aba = a, bab = b, imply that the four elements a, b,ab, ba generate R as a vector space over K. It is asserted that these four elements
are linearly independent. In fact, suppose u, v, w, x are in K and
ua+vb+wab+xba0.Multiplying on the iight by b, we get uab + xb = 0. Multiply this on the leftby a to get xab = 0, whence x 0. Thus uab = 0, whence u = 0. Similarly
v = w 0. Thus R has dimension 4 over K.Let T be the quotient of the noncommutative ring K[X, Y] by the two-sided
ideal generated by X2, Y2 and (X + Y)2 —1. It follows from the above that T hasdimension 4 and maps onto R, hence is isomorphic to R.
Now letA=
and B = (? Then A2 = B2 = Oand(A+B)2 =
1 and the matrices A, B, AB, BA span M2(K). It follows from the above that
6.13 Elementary Number Theory
Solution to 6.13.1: Let the six people be Aline, Ma, Laura, Lucia, Manuel, andStephanie. Fix one of them, Manuel, say. The five girls form two sets: X (Manuel'sfriends) and V (the others). By the Pigeonhole Principle (Her75, p. 1271, one ofthese sets has cardinality at least 3. Suppose it is X that contains at least threeelements, say (Aline, Laura, Stephanie) C X. If Aline, Laura, and Stephanie arepairwise strangers, we are done. Otherwise, two of them are friends of each other,Stephanie and Aline, say. Then Manuel, Aline, and Stephanie are mutual friends.If the set with three or more elements is Y, a similar argument leads to the sameconclusion.
Solution to 6.13.2: Suppose not. Changing signs, if necessary, we may assumethat M > 0 where M is the maximum of the Um,n 's. Define the set
A = ((m,n)Iumn = M) C (2,3, ...,N 1) x (2,3, ...,M— 1).
and choose (m, n) E A with m minimal. Since (m — 1, n) g A, we have
1 1+Um+1,n + Um,n_1 + Um,n+J) < + M + M + M) = M Um,n,
which contradicts the relation.
Solution to 6.13.3: There are possibilities for the range, so the answer isiON, where N is the number of surjective functions from (1, 2, 3,4,5) to a given3-element set. The total number of functions (1,2, . . . , 5) —+ (1,2,3) is 35, fromwhich we subtract (the number of 2-element subsets of (1, 2, 3)) times(the number of functions mapping into that subset), but then (according to thePrinciple of Inclusion-Exclusion) we must add back (the number of functionsmapping into a 1-element subset of (1,2, 3)). Thus
150
and the answer is iON = 1500.
Solution to 6.13.5: Let x = O.ala2... in base 3. If aj = I for some j, choosethe smallest such j, and define
x_ =0.aIa2...aJ_1022222..,
These are the numbers from the Cantor set closest to x. Then f(x_) = f(x+) sof is constant on [x_,
It suffices then to prove the inequality forx = y = withaj. E (0,2). Let j with a3 b3. Then x — Onthe other hand, we have,
If(x) — 1(Y) = =2.i>A.
Combining, we obtain,
11(x) — f(y)( 2• 2 —
Solution to 6.13.6: If a = 0, the congruence has the trivial solution x = 0. For1 a p — 1, if x2 a (mod p), we have
(p — x)2 = p2 — 2xp + x2 a (mod p)
so, for a 0, there are two solutions of the quadratic congruence in each com-plete set of residues mod p. We conclude, then, that the total number is
1 + (p — 1)/2 = (p + 1)/2.
Solution to 6.13.7: By Fermat's Little Theorem [Sta89, p. 80], [Her75, p. 44],
we have, raising both sides of the congruence —I x2 (mod p) to the power(p—l)/2,
(—1) 1 (mod p)
which implies that (p — 1 )/2 is even, and the result follows.
2
Solution to 6.13.8: Let f(n) = 21z +n2. It suffices to show that f(n) is compositeif n i (mod 6) for i 3. If n is even, then f(n) is an even number largerthan 2, a composite. Let n = 6k + 1 for some integer k. We have
f(n) = 26k+1 + 36k2 + 12k + 1
I
(mod3),
so f(n), being a multiple of 3 larger than 3, is a composite. For n = 6k + 5 weget
f(n) = + 36k2 ÷ 60k +25+ 25
(mod3),
and again f(n) is composite.
Solution to 6.13.9: 1. See the Solution to Part 1 of Problem 6.1.4.2. The congruence ka 1 (mod n) is equivalent to the equation ka = mn + 1for some integer m. As all integer linear combinations of k and n are multiples ofgcd{k, n), the first congruence has a solution iff gcd(k, n} = 1.
3. Let ço be Euler's totient function [Sta89, p. 77], [Her75, p. 43]. As is multi-plicative, we have
ço(n) = go(p)ço(q) = (p — 1)(q — 1).
Solution to 6.13.10: Let p(t) = 3t3 + lOt2 — 3t and n/rn E Q; gcdfn, m} = 1.
We can assume m ±1. If p(n/m) = k Q, then m13 and njk. Therefore, wehave rn = ±3.
Suppose m =3. We have
= (ç + -This expression represents an integer exactly when n2 + iOn = n(n + 10) 0(mod 9). As gcd{n, 3} = 1, this means n + 10 0 (mod 9), that is, n 8(mod 9).
A similar argument for the case m = —3 shows that the numbers n/(—3) withn 1 (mod 9) produce integer values in p.
Solution to 6.13.11:
(1/2\ —(n—i)) — •3.5...(2n.... 1)\n) — 2"n!
12n— n !)2 = —. fl
Solution to 6.13.12: A counting argument shows that the power of 2 which di-vides n! is given by
where Lxi denotes the largest integer less than or equal to x. Since=
(n!)to show that is even it suffices to show that
EL]Suppose2' Fork r, there is an rk,O <1, such that
or
so
and equality holds if and only if n is a power of 2. For k = r + 1, we have that= 0 while = 1. Finally, for k > r, the terms in both sums
are 0. Hence, we see that the above inequality holds. Further, we see that the leftside is 2 or more greater than the right side (i.e., is divisible by 4) if and only nis not a power of 2.
Solution to 6.13.13: We may assume that n > 3. Converting the sum into a singlefraction, we get
n!/1+n!/2+..+n!/n
Let r be such that 2'In! but 2r+I does not divide n!, and s be such that2 less than or equal to n. Since n > 3, r > s > 0. The only
integer in 1, . . . , n, divisible by is 2. Hence, for 1 k n, n!/k is divisibleby and every term except 1 is divisible by So
— — 2Sk
for some integers j and k. The numerator is odd and the denominator is even, sothis fraction is never an integer.
Solution 2. Let n 4. The Bertrand's Postulate asserts that there is at least oneprime number in the interval (n/2, n], [HW79, Chap. 22]. Let p be the largestsuch prime. Converting the sum into a single fraction, as above, we get
n!/1+n!/2+••+n!/nn!
Clearly p divides the denominator and all summands in the numerator except
n!/p, since p > n/2.
Solution to 6.13.14: Recall that if pi, pa,... is the sequence of prime numbers
andy=flp7',wehave
gcdfx, = 1cm (x, y) = flLet
c=J]prwe have
gcd{a, lcm(b, c)) = f1j
— VTI I Pl
1cm (gcd{a, b), gcd{a, c}).
Solution to 6.13.15: There are nine prime numbers 25:
p2=::3, p3=5, p4=7, p5=!!,p7=17, P8='9' P9=23.
By unique factorization, for each 1 a 25 there is an integer sequence
v(a) = with a = The 10 sequences v(a1) E Q9 must be
linearly dependent, so
10
>njvj(aj) =0
for all j, for some rational numbers tz which are not all 0. Multiplying by acommon multiple of the denominators, we can assume that the n,'s are integers.So
=
nivj(ai) = 1,
as required.
Solution to 6.13.16: Denote the given number by n and let n = a 13• By countingdigits, we see that n < 1026, so a < 100. As = 7934527488, we have8013 <n and a> 80. Note that n 9 (mod 10). The integers c < 10 such thatck (mod 1 (mod 10)
(modl0)soc=9.Hence,a=89or a = 99. As 3 does not divide n, 3 does not divide a. Hence, a = 89.
Solution to 6.13.17: Since 17 7 (mod 10),
A71717
(mod 10).
Since (7, 10) = 1, we can apply Euler's Theorem [Sta89, p. 80], [Her75, p. 43]:
7W(lO) I (mod 10).
The numbers k such that 1 k 10 and (k, 10) = 1 are precisely 1,3, 7, and 9,so go(10) = 4. Now 17 1 (mod 4), so 1 (mod 4). Thus,
717'7 (mod 10)
and the rightmost decimal digit of A is 7.
2323Solution to 6.13.18: As 23 3 (mod 10), it suffices to find 323
We have ço(lO) =4, where ço is the Euler's totient function, and,orem [Sta89, p. 80], [Her75, p. 43], Y 3S (mod 10) when rSo we will find 232323 (mod 4). We have 23 3 (mod 4),(mod 4). As —1 3 (mod 4),
3 (mod 4), and 33 7
Solution to 6.13.19: Let
(mod 10).by Euler's The-_=s (mod4).
so 232323
(mod 4), because
(mod 10).
We have
NoN1
= (0,1)U(4,5,6}U(11,12,13,14,15}U...= (2,3}U{7,8,9, 10}U(16, 17,18, 19,20,21}U...
N0flN1 =0, NoUN1
and, clearly, neither can contain an arithmetic progression.
Solution to 6.13.20: Consider the ring Zak_i. Since a> 1, (a,a E Zk_1. Further, it is clear that k is the least integer such that a1' 1 (mod
— 1), so k is the order of a in Hence, by Lagrange's Theorem [Her75,
p. 41], k divides the order of the group which is ço(a" — 1).
Solution to 6.13.21: Let N be the desired greatest common divisor. By Fermat'sLittle Theorem [5ta89, p. 80], [Her75, p. 44], we have
n13 (n6)2n (n3)2n n4 n2 n (mod 2).
Hence, 21(n'3 — n) for all n, so 2jN. An identical calculation shows that pINfor p E (3, 5,7, 13}. Since these are all prime, their product, 2730, divides N.However, 2 = 8190 = 3 . 2730, 50 N is either 2730 or 3 2730. As313
— 3 = 3(312— 1) is not divisible by 9, N = 2730.
Solution to 6.1322: Letk1 k2n=p1 P2
be the factorization into a product of prime powers (P1 <pz < ... < for n.The positive integer divisors of n are then the numbers
ii inp1 p2
It follows that d(n) is the number of r—tuples (ii, J2, ..., jr) satisfying the pre-ceding conditions. in other words,
d(n) = (k1 +1)(k2 + 1),
which is odd if each is even; in other words, if n is a perfect square.
Solution to 6.13.23: Only 2 and 8 do not have the required property. This can beproved in four steps as follows.
1. If the last digit of n is 0 or 5, every power of n has the same last digit, so inparticular does. Hence 0 and 5 have the required property.
2. Suppose n is even and not divisible by 5. Then is even and, by Fermat'sLittle Theorem [Sta89, p. 80], [Her75, p. 44], mod 5 is determined bymod 5 and n mod 4. By the Chinese Remainder Theorem (Sta89, p. 72] (nmod5,nmod4) canbeanypairin{1,2,3,4)x{0,2),son"canbelor4 mod 5, and n" has last digit 6 or4.
3. Suppose n is odd and not divisible by 5. Then ntZ is odd, and n?z mod 5determined by(n mod 5, n mod 4)in (1,2,3,4) x (1,3). Therefore mod5 can be anything in (1, 2,3,4), and has last digit 1,7, 3, or 9.
4. Finally, each of the values 0,1,3,4,5,6,7,9 occurs infinitely often, since thearguments above show that nlZ mod 10 has period 20.
Solution 2. Let = n mod 10 denote the last digit of n. If n = lOm + k with
____
For fixed_k, we get a sequence as m increases. Each term is the image under
x k'0x of the previous term, so once a value in the sequence is repeated,we can predict the rest of the sequence, which will be periodic. After computing
= kk and bk = k'0 for k up to 9, we can immediately write down, for each k,the sequence that starts with a* (or 1010 jfk = 0) and then iteratively multiply bybk and reduce modulo 10:
1010,2020,... = 0,0,0,0,0,0,... (bo = 0)i', 1111,... = 1,1,1,1,1,1,... (b1 = 1)
22,1212,...=4,6,4,6,4,6,.,. (b2=4)33,1313,...=7,3,7,3,7,3,.,. (L'3 =9)
(b4=6)55, =5,5,5,5,5,5,... (15 =5)66,1616,... =6,6, 6,6,6, 6,... (b6 =6)
= I. =
S! j — d ç = du — d jo o11oico
-!W" iopio Jo Jo Si ptmq .ioqio j = jo Si du
d
J = LX JO SUOUfljOS JO lUflOO 0;
.IflDDO IOU op U31J0 "°°° 6 'L '9 'ç 'i '0 SflU.J
(i =6q)
(j7=Sq)
(6= L4) = "'tiLt 'ti.
7
Linear Algebra
7.1 Vector Spaces
Solution to 7.1.1: 1. Is sufficient. The evaluation map that assigns to each polyno-mial its value at 1 is a nonzero linear functional on the linear space of polynomialsof degree at most 4. Like all such functionals, its null space, having codimension1, has dimension 3. The four polynomials being in that 3 dimensional null spaceare therefore a dependent set.2. Is not sufficient. The four polynomials 1, 1 + x, 1 + x2, 1 + x3 form a coun-terexample.
Solution to 7.1.2: Following by induction on k, we can easily see that the functionF(k) is a linear combination of the mn functions f(t)g(J) for 0 i <n, 0 j <m, with constant coefficients. Hence, the mn + 1 functions arelinearly dependent over C.
Solution to 7.1.3: 1. Every element of V can be uniquely written in the formaivi + . .• + where the v1's form a basis of V and the a's are elements ofF. Since F has q elements it follows that V has elements.2. A matrix A in (F) is nonsingular if and only if its columns are linearlyindependent vectors in F'1. Therefore, the first column A1 can be any nonzerovector in F'1, so there are q'1 — 1 possibilities. Once the first column is cho-sen, the second column, A2, can be any vector which is not a multiple of thefirst, that is, A2 cAj, where c F, leaving qfl
— q choices for A2. In gen-era!, the column A, can be any vector which cannot be written in the form
Cl A1 + c2A2 + • + Cl—I A,_.1 where E F. Hence, there are — possi-bilities for A1. By multiplying these together we see that the order of (F) is
then — 1)(q'1 — • —
3. The determinant clearly induces a homomorphism from (F) onto the mul-tiplicative group F*, which has q — I elements. The kernel of the homomorphism
is and the cosets with respect to this kernel are the elements ofwhich have the same determinant. Since all cosets of a group must have the sameorder, it follows that the order of is (F)I/(q — 1).
Solution to 7.1.4: Suppose v E V is nonzero. Let = [A E EndV I Av = v}.Choose w E V so that fv, w) is a basis. With respect to this basis, correspondsto the matrices of the form with a, b E F, so = q2. We have =if v and v' are F-multiples of each other, and otherwise fl = f I), since ifan endomorphism fixes a basis, it is the identity. There are (q2 — 1)/(q — 1) =q + I nonzero vectors in V modulo the action of F*, so the total number ofendomorphisms fixing some nonzero vector is (q + 1)q2 — q, where the —q isthere to avoid counting the identity q + I times. Thus the answer is q3 + q2 — q.
Solution 2. We are asked for the set of endomorphisms having 1 as an eigenvalue.There is a bijection between this set and the set of endomorphisms of determinantzero, taking A to A — 1. Therefore we need to count the number of solutions toad—bc=Owitha,b,c,dE F.
The number of solutions with a 0 is (q — 1)q2, since for each choice ofa E F* and b, c E F, solving the equation for d yields a unique solution. Thenumber of solutions with a = 0 equals q(2q — 1), since d is arbitrary, and thereare 2q — 1 pairs (b, C) with b = 0 or c 0 (there are q with b = 0, and qwith c = 0, but (b, c) = (0,0) has been double counted). Thus the answer is(q—1)q2+q(2q— 1)=q3+q2—q.
Solution to 7.1.5: If p is prime then the order of is the number ofordered bases of a two-dimensional vector space over the field Z , namely(p — I)(p2 — p), as in the solution to Part 2 of Problem 7.1.3 above.
A square matrix A over Zj,n is invertible when det(A) is invertible modulowhich happens exactly when det(A) is not a multiple of p. Let p(A) denote thematrix over Z,, obtained from A by reducing all its entries modulo p. We havedet(p(A)) det(A) (mod p), thus
A E GL2 1ff p(A) E GL2 (Z,,),
giving a surjective homomorphism
p : GL2 (Zj,n) GL2 (zr).The kernel of p is composed of the 2 x 2 matrices that reduce to the Identity
p +1) and the off-diagonal are drawn from the set that reduce to 0 modulo p.that is, (0, p, 2p, ..., p" — p). Both sets have cardinality so the order of
the kernel is (p171)4, and order of GL2 (Zr) IS
p4fl_4(p2— 1)(p2 — p) = p4fl•_3(p
— l)(p2 — I).
Solution to 7.1.6: The multiplicative group F,, of the field of elements iscyclic; let g be a generator. Then the map T : Fj,n -+ sending x to gx isan F p-linear map acting as a single cycle on the nonzero elements of Thematrix of T with respect to a basis for over has the desired property.
Solution to 7.1.7: Let F = (0, 1, a, b}. The lines through the origin can haveslopes 0, 1, a, b, or oo, so S has cardinality 5. Let be the line through theorigin with slope y. Suppose y E G fixes all these lines, to be specific say
_(x yy
W
Then
yLo=Lo
implies that
y(l, 0)' = (x, z)t = (c, 0)'
for some c 0. Thus, z = 0. Similarly, the invariance of implies y = 0 andof L1 implies x = w. Then det(y) = x2 = 1 and since F has characteristic 2, wemust have x = 1 and y is the identity.
Solution to 7.1.8: 1. F[x} is a ring under polynomial addition and multiplicationbecause F is a ring. The other three axioms of vector addition — associativity,uniqueness of the zero, and inverse — are trivial to verify; as for scalar multiplica-tion, there is a unit (same as in F) and all four axioms are trivial to verify, makingit a vector field.2. To see this, observe that the set (1, x, x2, .. . , form a basis for this space,because any linear combination will be zero, if and only if, all coefficients arezero, by looking at the degree on both sides.3. An argument as above shows that
aol +al(x =0
only if the coefficients are all zero.
Solution to 7.1.9: For any two finite-dimensional subspaces X and Y,
dim(X + Y) + dim(X fl Y) = dim(X) + dim(Y),
which you get by applying the Rank—Nullity Theorem [HK6 1, to the quo-tient map X -÷ (X + Y)/ Y (in this case the dimension of the kernel of the map
is dim(X fl Y), and the dimension of the range is dim(X + Y) — dim(Y)). Thus,
dim(U) + dim(V) + dim(W) — dim(U + V + W)
= dim(U) + dim(V) - dim(U + V) + dim((U + V) fl W))
dim(U) + dim(V) — dim(U + V)= dim(U fl V)
Observe that it is not true for subspaces of a vector space (U + V) fl W =(UflW)+(VflW).
Solution to 7.1.10: Let Y denote the given intersection. Then Y is a subspace of Vand, clearly, W c Y. Suppose that there exists a nonzero vector v Y\ W. Since vis not in W, a set consisting of v and a basis for W is linearly independent. Extendthis to a basis of V. and let Z be the (n — 1)-dimensional subspace obtained byomitting v from this basis. Then W c Z, so Z is a term in the intersection used todefine Y. However, v is not in Z, so v cannot be an element of Y, a contradiction.Hence, Y C W and we are done.
Solution to 7.1.11: We use the Induction Principle [MH93, p.7] on the dimensionof V. If dim V = 1, the only proper subspace is (O),so V is clearly not the unionof a finite number of proper subspaces.
Now suppose the result is true for dimension n — 1 and that there is a V,dim V = n, with
where we may assume dim Wj = n — 1, 1 i k. Suppose that there existed asubspace W of V of dimension n — 1 which was not equal to any of the W1 's. Wehave
W=U(WnW,).
But dim(W fl W,) n —2, and this contradicts our induction hypothesis.Therefore, to complete the proof, it remains to show that such a subspace
W exists. Fix a basis ... , of V. For each a E F, a 0. consider the(n — 1)—dimensional subspace given by
Wa = (aixi + + • +aan = 0).
Any two of these subspaces intersect in a subspace of dimension, at most, n — 2,so they are distinct. Since there are infinitely many of these, because F is infinite,we can find W as desired.
Solution 2. Suppose that V = After discarding superfluous V,'s, wemay assume that
forall
Then k 2, and there must be vectors vI, in V such that
(1) vIEVI\UVI and (2) V2EV2\UVI.i#2
Let (x.ç) be sequence of distinct, nonzero elements of the field. Then, for each s,the vector = vj + does not lie in V1 U V2 (if E V1, thenV2 = (U5 — vi)/x5 E V1 contradicting (1); similarly, U5 g V2.) It follows that, forall s, u5 E Since the vectors U8 are all distinct, it follows that, for somes 1,2,ucandu5'lieinVj.Butthenv2= (u1—u5')/(x3—x5') E
V
Solution to 7.1.12: Note first that if A and B are matrices and C is an invertiblematrix, then
AB = BA if = C1BCC1AC.
Also, if Di,..., are linearly independent matrices, so are the matricesWe may then assume that A is in Jordan Canonical
Form [HK61, p. 247].a 1 ... 0
A direct calculation shows that if A = is a k x k Jordan1
0 afbi b2 bk
block, then A commutes with = .. ...ko b1
Therefore, by block multiplication, A commutes with any matrix of the form
fBiB=I
k Br
where the 's have the form of B and the same dimension as the Jordan blocksof A. Since there are n variables in B, dim C(A) n.
Solution to 7.1.13: tr(AB — BA) 0, 50 S is contained in the kernel of thetrace. Since the trace is a linear transformation from (R) = R"2 onto R, itskernel must have dimension, at most, n2 — 1. Therefore, it suffices to show that Scontains n2 — 1 linearly independent matrices.
Let denote the matrix with a 1 in the (i, f)th coordinate and 0's elsewhere.A calculation shows that for i j, = MjkMk.j — MkJM,k, so is in S.Similarly, for 2 j n, M11 — = — Together, these
— I matrices are clearly a linearly independent set.
Solution to 7.1.14: Let f, g E S and let r and s be scalars. Then, for any v E A,(rf + sg)(v) = f(rv) + g(sv) A, since A is a vector subspace and f and g fixA. Similarly rf + sg fixes B, so if + sg E S and S is a vector space.
To determine the dimension of 5, it suffices to determine the dimension of thespace of matrices which fix A and B. To choose a basis for V. let A' denotea complementary subspace of A fl B in A and let B' denote a complementarysubspaceofAflBinB.Then,sinceA+B = V,r =a+b—nisthedimensionofAflB.Further,climA' =a—randdimB' =b—r.Takeonebasisineachofthe spaces A', B', and A fl B. The union of these bases form a basis for V. Sinceany endomorphism which leaves A and B invariant must also fix A fl B, its matrixin this basis must have the form
1*10 * 0\o 0 *
which has, at most, a2 + b2 + n2 — an — bn nonzero entries, so the dimension ofSisa2+b2+n2—an—bn.
Solution to 7.1.15: Suppose there are scalars such that
aox + a1Tx + .. + + . + am_ITm_lx = 0
applying to both sides, we get, since TO 0,
+aiTmx + .. .+akTm 1+kx÷ ... +am_lTm_l+m_lx = 0
so
and ao = 0. By the Induction Principle [MH93, p. 7] (multiplying by Tm—k—I)we see that all ak = 0 and the set is linearly independent.
Solution to 7.1.16: We may assume ai < •.. < Suppose that there areconstants Cl, ..., such that + . . + 0.If not all the coefficientsvanish, we have, without loss of generality, 0. Thus,
+ ...+cn_ie(an_i_an)t 0.
Letting t -÷ oo, and noting that —* 0 for 1 i n — 1, we get = 0,a contradiction.
Solution 2. This proof is by induction. Let P(n) denote the statement that for alldistinct real numbers al, . . . , the functions eU1t, ... , are linearly inde-pendent.
Then P(1) is true since 0.Assume P(n) is true for some fixed n 1, and let ai, .. . , be n + 1
distinct real numbers. Suppose for scalars A,, + ••• +
are distinct nonzero real numbers. Differentiating with respect to t, we obtainX2162efl2' + - + )'n+jfln+i efln1-It 0. By the induction hypothesis, X,j9, = 0for j = 2, . , n + 1. Thus A, = 0, j = 2, ... ,n + 1. Substituting in the firstequation above gives Xi =0, proving P(n + 1).
Solution to 7.1.17: Let P be the change of basis matrix from (as) to (b1).A straightforward calculation shows that I + 2P is the matrix taking (a1) to(a, + 2b1). Now (I + 2P)v = Ày implies that Pv = — l)v. So if A isan eigenvalue of 1 + 2P, then — 1) is an eigenvalue of P. and they corre-spond to the same eigenvectors. The reverse also holds, so there is a one-to-onecorrespondence between the eigenvalues of P and those of I + 2P. As (a,) and(b,) are orthonormal bases, P is orthogonal and therefore, all the eigenvalues ofP are ±1. But this implies that the only possible eigenvalues of I + 2P are 3 and—1. Hence, 0 is not an eigenvalue of! +2P, so it is an invertible matrix and, thus,
+ 2b,) is a basis. Further, det P = (_1)a where a and are the algebraicmultiplicities of —1 and 1 as eigenvalues of P. Thus, det(I + 2P)Since we are given that det P > 0, a is even and, thus, det(I + 2P) is positive aswell. Therefore, (aj + has the same orientation as (a,).
7.2 Rank and Determinants
Solution to 7.2.1: 1. Let A = and R, denote the row of A. Let r,1 r m, be the row rank ofA, and = (b,1, .. . , 1 i r, beabasisfor the row space. The rows are linear combinations of the S 's:
R,
= rS3. 1 i m.
For 1 I n, isolating the coordinate of each of these equations gives
au = k!1b1, + + kirbri
ami=kmibij+..+kmrbrj.
Hence, for I I n the column of A, C1, is given by the equation
C1
where K3 is the column vector (ki], ..., kmj)t. Hence, the space spanned by thecolumns of A is also spanned by the r vectors so its dimension is less than or
equal to r. Therefore, the column rank of A is less than or equal to its row rank.In exactly the same way, we can show the reverse inequality, so the two are equal.2. Using Gaussian elimination we get the matrix
103—20 1 —4 40020000 2000 0
so the four columns of M are linearly independent.3. If a set of rows of M is linearly independent over K, then clearly it is alsoindependent over F, so the rank of M over K is, at most, the rank of M overF; but looking at the Gaussian elimination process one can see that it involvesoperations in the sub-field only, so the rank over both fields is the same.
Solution to 7.2.2: Let v be a nonzero right null vector of A and u a left nullvector of A such that u v =0. Since A is diagonalizable, we can choose a basis{vi, v2, ..., v,j of RtZ consisting of eigenvectors of A. Since the null space of Ahas dimension 1, one of these basis vectors is a multiple of v, so we can assumeVI = v.LetA3 =2,and forj> 1,
uv3 = = 0.xi xi
Hence u is orthogonal to each of the basis vectors v1, ..., so u =0.
Solution to 7.2.3: As A' AV C At V. it is sufficient to prove that dim AtAV =dim A' V. We know that rankA = dim(Im A) and that
dim(Im A) + dim(ker A) = n
Similar formulas hold for A' A. Therefore, it is enough to show that ker A andker At A have the same dimension. In fact, they are equal. Clearly, ker A c ker A' A.Conversely, take any v E ker At A. Then
0 = (A'Av, v) = (Au, Au),
so IlAvil =0. Hence, v kerA and we are done.
Solution to 7.24: Since 1 — P — Q is invertible, P has the same rank as
P(1—P—Q)=P—p2—pQ—_pQ.
Similarly, Q has the same rank as
(1—P-Q)QQ_PQ_Q2_pQ,so P and Q have the same rank.
Solution to 7.2.5: The upper k x k square minor A(k, k) of A(m, n) is the Van-dermonde matrix, which determinant is —i). If k p, this determi-nant is not zero (mod p), which shows that rankA(m, n) min[m, n, p). Nowif A(m, n) has at most p distinct columns (mod p), so rankA(m, n) p. SincerankA(n, n) min{n, m), we have rankA (m, n) = min{m, n, p}.
Solution to 7.26: Let V = and consider the two quotient spaces AV/A2Vand A2 V/A3 V. The first has dimension rankA — rankA2, and the second hasdimension rankA2 — rankA3. The matrix A induces a linear transformation ofAV/A2V onto A2 V/A3 V. so dim(A2V/A3V) dim(A V/A2 V), as desired.
Solution to 7.2.7: 1. and 2. Since T is symmetric it is diagonalizable, so IRJ? canbe written as the direct sum of the eigenspaces of T. It suffices to show that anyeigenspace has dimension, at most, 1. For if this is the case, then the kernel hasdimension, at most, 1, and, by the Rank—Nullity Theorem [HK61, p. 71j1, T hasrank at least n —1, and there must be n distinct eigenspaces, so there are n distincteigenvalues associated with them.
Let A R, and consider the system of equations Tx = Ax. The first equationis + bix2 = Ax1. Since b1 0, we can solve forx2 in terms of xi. Supposethat we can solve the first i —1 equations for ... , in terms of xi. Then, sinceb, 0, we can solve the equation b1_1x1_1 + aix, + Ax1 for Xj+I interms ofxi. Therefore, by the Induction Principle [MH93, p. 71, we can solve thefirst n — 1 equations forx2, .. . , in terms of xi.
The last equation, + is either consistent with this oris not. If not, A is not an eigenvalue; if it is, then A is an eigenvalue and wehave one degree of freedom in determining eigenvectors. Hence, in either casethe associated eigen space has dimension, at most, 1 and we are done.
Solution 2. 1. The submatrix one obtains by deleting the first row and the firstcolumn is upper-triangular with nonzero diagonal entries, so its determinant isnonzero. Thus, the first n — 1 columns of T are linearly independent.2. By the Spectral Theorem [HK61, p. 335], {Str93, p. 235], RPZ has a basis con-sisting of eigenvectors ofT. If A is an eigenvalue ofT, then T —Al has rank n —1by Part 1, so ker(T — A!) has dimension 1. Since the eigenspaces span R" andeach has dimension 1, there must be n of them.
Solution to 7.2.9: 1. Write the characteristic polynomial of A, det(A — Al), as
+ +•• + Cr. Since the entries of A are integers, each Ck is aninteger, and Cr = det A. If A is an integer eigenvalue, then det(A — nI) = 0, so
det A = + Clfl'' +' + Cr.-Ifl
showing that n divides det A.2. Under the given hypotheses, n is an eigenvalue with eigenvector (1, 1,..., 1)1,
so Part 1 applies.
Solution to 7.2.10: Since A has rank n —lit has n —1 linearly independent rows,and the remaining row is a linear combination of those n — 1. Interchanging rowsand corresponding columns of A, we can assume without loss of generality thatrows 1 through n —1 of A are linearly independent. Thus, there are real numbers
Cl, . .. , Cpa_I such that, letting ajk denote the entries of A, we have
n—I
(*)
j=i
It is asserted that the matrix obtained from A by deletion of its last row and column
has rank n — 1. Assume not. Then there are real numbers b1, .. . , not all 0,
such thatn—i
>bjajk = 0, k=1,...,n—1. (**)
By (*) and the symmetry of A,
n—I
ajn=>ckajk, j=1,...,n.k=i
Hencen—I n—in—i n—i n—i
ajn = = Ck =0,j—l j=1 k=i *=I j=i
i.e., (**) also holds for k = n, contradicting the linear independence of the firstn — 1 rows of A.
Solution to 7.2.11: We use the Induction Principle [M1193, p. 32] in the order ofthe matrix. [In = 2,
X2)
which has determinant (x2 — xi).Suppose the result holds for all k <n, and let A be the n x n Vandermonde
matrix, see the Solution to Problem 7.2.11 or [HK61, p. 125]. Treating the inde-terminates xj, . .. , i as constants and expanding the determinant of A alongthe last row, we see that detA is an (n — 1)1h degree polynomial in which canhave, at most,n— I roots. If we letxn = x1 for! i n—i, A wouldhave twoidentical rows, so det A would equal 0. Hence, the x1 's are the roots of det A as apolynomial in In other words, there exists a constant c > 0 such that
n—i
detA = —xe).i=l
c is the coefficient of the term, which, when we expand the determinant, isequal to the determinant of the (ii — 1) x (n — 1) Vandermonde matrix. So, by theinduction hypothesis,
det A = fl — fl — = fl — xi).i=1 i>j
Solution to 7.2.12: 1. As shown in the solution of Problem 7.2.11, the determinantof the matrix is
fl(a, —af)i>j
which is nonzero if the are all different.2. The function f given by
n
— — ao)• . • (a1 — a,._i)b,(a1 — a÷i)•• • — an)
has degree n and takes f(a,) into b1. Now, if is another such polynomial of
degree n, the polynomialf(x) — ifr(x)
has degree n with n +1 different roots (the a1 's), soit has to be the zero polynomialand f is unique.
Solution to 72.13: Define the matrix C = A1 B, one can immediatly see that itis also unitary. And we can also verify that A + B = A(I + C). Since A is uni-tary, its eigenvalues have absolute value 1. Multiplying them together shows thatI det Al = 1. If ..., are the eigenvalues of C, listed with their multiplicities,then 1, so theeigenvalues of I+C are 1+ ..., so
ldet(I+C)l = II
Hencedet(A + B)I = I det All det(I + C)l
Solution to 7.2.14: Consider the function v(t) = (1, t, t2). To show that v(t1),v(t2), and V(t3) form a basis for R3 whenever the t1's are distinct, it will sufficeto show that the matrix which has these vectors as rows has nonzero determinant.But this matrix is
which is the 3 x 3 Vandermonde matrix, see the Solution to Problem 7.2.11 or[HK61, p. 1251. Its determinant is given by
(t3 — t2)(t3 — tl)(t2 — ti)
which is nonzero whenever the ti's are distinct.
Solution to 7.2.15: Let G be the matrix with entries
b
= f fi (x) (x)dx.
If the determinant of 0 vanishes, then G is singular; let a be a nonzero n—vectorwith Ga =0.Then
2
a' fj fj (x)dx= f dx,
a
so, since the f, 's are continuous functions, the linear combination f mustvanish identically. Hence, the set {f, J is linearly dependent on [a, b]. Conversely,if (f,) is linearly dependent, some fj can be expressed as a linear combination ofthe rest, so some row of G is a linear combination of the rest and G is singular.
Solution to 7.2.16: Identify with 1R4 via
ab
and decompose L into the multiplication of two linear transformations,
M2x2LA LB
1R4 M2x2
where LA(X) = AX and LB(X) = XB.The matrices of these two linear transformations on the canonical basis of 1R4
Is
f 1 0 2 o\ /2 0 0 010 1021 11400LA=11 030J andLB=10020\0—103) \0014
then det L = det LA det LB = (9 + 6 + 2(2 + 3)) . (2 32) = 26 . 52, and tocompute the trace of L, we only need the diagonal elements of LA . LB, that is,
trL =2+4+6+ 12=24.
Solution to 7.2.17: Let X = be any element of M3(R). A calculation gives
f X13T(X) = 2X22 3x23/2
\ 3X32/2 X33
,z n
It follows that the basis matrices are eigenvectors of T. Taking the product oftheir associated eigenvalues, we get det T = = 81/8.
Solution to 7.2.18: Since the minimal polynomial of A splits into distinct linearfactors, R3 has a basis (vi, 1)2, 1)3) of eigenvectors of A. Since det A = 32, twoof those, say v1 and v2, are associated with the eigenvalue 4, and one, 1)3, is asso-ciated with the eigenvalue 2. Now consider the nine matrices I i, j 3,whose column is the vector and whose other columns are zero. Since thevi's are linearly independent, the matrices are linearly independent in M3x3and form a basis of Further, a calculation shows that = whereXi = A2 = 4 and A3 = 2. Hence, M3x3 has a basis of eigenvectors of LA, so itfollows that tr LA =64+3 .2 = 30.
Solution to 7.2.20: We have
dimrangeT dim — dimkerT = 49— dimkerT
so it suffices to find the dimension of ker T; in other words, the dimension ofthe subspace of matrices that commute with A. Let E+ be the eigenspace of Afor the eigenvalue 1 and E_ be the eigenspace of A for the eigenvalue —I. Then
= E+ e E_. A matrix that commutes with A leaves E÷ and E_ invari-ant, so, as linear transformations on R7, can be expressed as the direct sum of alinear transformation on E÷ with a linear transformation on E_. Moreover, anymatrix that can be so expressed commutes with A. Hence, the space of matri-ces that commute with A is isomorphic to e and so has dimension16+9= 25. It follows that dim range T =49 —25 = 24.
Solution to 7.2.21: m > n: We write T = T1 T2, where T2 : Mnxm -+ Mnxnis defined by T2(X) = XB and —÷ is defined by T1(Y) = AY.Since dim Mnxm = nm > = dim the transformation T2 has a non-trivial kernel, by the Rank—Nullity Theorem [HK61, p. 71]. Hence, T also has anontrivial kernel and is not invertible.m < n: We write T = T2T1, where Tj : Mnxm -4 Mmxm is defined byT1(X) = AX and T2 : Mmxm -4 Mmxn is defined by T2(Y) = YB. Now
we have dim Mnxm = nm > m2 = dim Mmxm, so T1 has a nontrivial kernel,and we conclude as before that T is not invertible.
7.3 Systems of Equations
Solution to 73.1: Subtracting the first equation from the third we get xI = X7,
continuing in this fashion subtracting the kth equation from the (k + 2)th equation
we obtain
Xk = xk+6 for every k
and we are left with a maximum of six independent parameters in the system.Solving the first and second equations, separatedly, we get
X5 = —(xl +X3)
= —(x2 +X4)
so four free parameters is all that is required.
Solution to 73.2: 1. Through linear combinations of rows, reduce the system ofequations to a row-reduced echelon form, that is, a system where:
• the first nonzero entry in each nonzero row is equal to 1;
• each column which contains the leading nonzero entry of some row has allits other entries 0;
• every row which has all entries 0 occurs below every row that has a nonzeroentry;
• if rows 1, . . . , r are the nonzero rows and if the leading nonzero entry ofrow i occurs in column i = 1, ... , r, then <k2 <<kr.
This new system has a number of nonzero rows r m <n and it is easy to seethat it has nonzero solution.
Since the original system is equivalent to the row-reduced one, they have ex-actly the same solutions.2. Let V be a vector space spanned by m vectors ... , We will show thatevery subset S = .. . of V with n > m vectors is linear dependent.
Since ,6i, .••' span V, there are scalars in the field F such that
For any set ofscalarx1, ... inF, we have
.xIaj
)Jx, >JAjjflj
=j=l i=l
=i:=I j=l
Since n > m, the linear systems of equations
>Ajjxj=O,
has a nontrivial solution so the set is linear dependent, proving the assertion.
Solution to 7.3.3: The answer is yes. Writing the system of linear equations inmatrix form, we have Ax = 0, where A is an m x n matrix with rational entries.Let the column vector x = (Xl, ... , be a complex solution to this system,and let V be the Q -vector space spanned by the x,'s. Then dim V = p n.If ... , E C is a basis of V. then there is a rational n x p matrix B withBy = x (where y is the column vector (yi. •••, Substituting this into theoriginal equation, we get ABy = 0. Since y is composed of basis vectors, this ispossible only if AB =0. In particular, every column of B is a rational solution ofthe equation Ax = 0.
7.4 Linear Transformations
Solution to 7.4.1: 1. We need to show that vector addition and scalar multiplica-tion are closed in 5(E), but this is a trivial verification because if v = S(x) andw = 5(y) are vectors in S(E), then
v+w=S(x+y) and cv=S(cx)
are also in S(E).2. If S is not injective, then two different vectors x and y have the same imageS(x)=S(y)=v,so
S(x—y)=S(x)—S(y)=v-v =0
that is, x — y 0 is a vector in the kernel of S. On the other hand, if S is injective,it only takes 0 E E into 0 E F, showing the result.3. Assuming that S is injective, the application : 5(E) -+ E is well defined.
Given av + bw E S(E) with v = S(x) and w = S(y), we have
+ w) = +bS(y))= + by))
=ax+by= aS'(v)
therefore, is linear.
Solution to 7.4.2: Let fai, ... , be a basis for ker T and extend it to
basis for the range of T. It is obvious they span the range since = 0 for
j ( k. Assume
cTa=O,i=k+1
which is equivalent to
i—k+1
that is, a = cia, is in the kernel of T. We can then write a as
a = b1a1 and have
>bjaj — cTa1 = 0,i=1 i=k+1
which implies all c, = 0, and the vectors Tak+1, ..., form a basis for therange of T.
Solution to 7.4.3: Applying the Rank—Nullity Theorem [HK61, p. 71] (see Solu-tion to Problem 7.4.2) to the map TIT-1(x) : T'(X) —÷ X, we get
dim T1(X) = dimker TIT-1(x) + dimX,
but ker TIr_i (X) = ker T, so, by the same theorem, we have
Therefore
Solution 2. Let Q be the projection from W onto the quotient space Z = W/X.The linear map QT: V —÷ Z has null space T1 (X), therefore
dim V — dimT'(X) = rank(QT) dim Z = dim W — dim X.
Solution to 7.4.4: We have A(B(ker(AB))) = (0) implying that B(ker(AB)) CkerA, so dim B(ker(AB) dim ker A. Using the Rank—NullityTheorem [HK6I,p. 71], we have
dimker(AB) = dim B(ker(AB)) + dim ker (BIkel.(AB))
dimkerA + dim kerB.
Solution to 7.4.5: Let vo be a vector in V such that Avo 0. By the secondcondition there is a scalar a such that By0 = aAvo.
Suppose v is any vector in V such that Av0 and Av are linearly independent.By the second condition there are scalars and y such that By = andB(VO+V)=YA(VO+V).ThCfl
aAvo + f3Av, = yAvo + yAv,
which by the linear independence of Av0 and Av implies that a = = y. HenceBy = aAv.
Fix a vector in V such that Av0 and Avi are linearly independent. Supposev is a vector in V such that Av0 and Av are linearly dependent, but Av 0.Then Avi and Av are linearly independent, and the reasoning above shows thatBy = aAv.
Suppose finally that v is a nonzero vector in V such that Av = 0. Then A(v +vo) and Av1 are linearly independent, so the reasoning above gives B(v + =aA(v + aAv0, implying that By = 0 (since By0 = aAv0). Thus By =aAv for all v in V, i.e., B = aA.
Solution to 7.4.6: Let ..., be a basis for V such that yj, ..., is a basisfor W. Then the matrix for L in terms of this basis has the form where Misa k x k matrix and N is k x (n — k). It follows that M is the matrix of withrespect to the basis ..., vk. As the matrix of 1 — tL is it followsthat det(l — tL) = det(l — tM) = det(l —
Solution to 7.4.7: The trace of f equals the trace of the restriction 11w plus thetrace of the induced endomorphism g of V/ W. Since f maps V into W, g is zeroand has trace zero.
Solution to 7.4.8: Suppose W is a k-dimensional invariant subspace for T. Anybasis {eI, ... , of W extends to a basis of V, (ei, . .., en). The matrix ofTwith respect to this basis has the form where B is the matrix of TIw. Itfollows that the characteristic polynomial of T Iw is a factor of the characteristicpolynomial of T, of degree k.
Conversely, suppose det(a — A!) = f(t)g(t) where f, g E F[t] are both non-constant. Then f(A)g(A) = 0, by the Cayley—Hamilton Theorem [HK61, p.194], so at least one of f(A), g(A), is singular. Assume f(A) is singular, andchoose a nonzero vector w E kerf(A). Then the subspace W spanned by
w, Tw, ..., Tk_Iw,
where k = deg f, is invariant under T, nonzero, and has dimension at mostk<n.
Solution to 7.4.9: Let f(x) E REx] denote the characteristic polynomial of T. Iff(x) has a nonreal root A, let v E C be a nonzero eigenvector with eigenvalueA; then the 2-dimensional subspace M spanned by the real vectors v + 15 and
(15 — v)/i will be mapped into itself by T.
Otherwise all roots of the degree-n polynomial f(x) are real. if f(x) has two
distinct real roots TI, then there exist nonzero eigenvectors vi and V2 with theseeigenvalues, and we may take M to be the span of vi and v2.
We are left with the case in which f(x) = (x — r E IL Let VI be a
nonzero eigenvector. Then the characteristic polynomial of the transformation T'of R'1/(Rv1) induced by T is (x — and there exists a nonzero eigenvector
1)2 E of T', since n > 1. Choose V2 E representing Then Tmaps the span M of and into itself.
Solution to 7.4.10: Let V1 = {v E V I x(L)(v) = 0), for i = 1, 2. Clearly,each V, is a subspace of V with X(L)Vi = 0. To show that V is the directsum of V1 and V2, choose polynomials a and b over F for which aXI+bX2=1.Tbena(L)X1(L)+b(L)X2(L)= 1.Ifv E V1flV2,thenv= 1•v
b(L)X2(L)v = a(L)0 + b(L)0 = 0, so Vi fl V2 = (0). If V E V, then bythe Cayley—Hamilton Theorem [11K61, p. 194], we have X(L)v = 0. Hence,Vi = (L)v is annihilated by X2(L) and, therefore, belongs to V2. Like-wise, v2 = belongs to V1. Since v = VI + V2, this shows thatV=v1+v2.
Solution to 7.4.11: It is sufficient to prove that y (x) and z g (x, y). Supposethat y = Ax for some A E Q. Then z = Ty = Ày = A2x, and x + y = Tz =
= A3x. Hence x + Ax = 13x, (A3 — A — 1)x = 0 and A3 — A — 1 = 0 sincex 0. This is a contradiction, since A3 — A — 1 = 0 has no rational solutions.Therefore y g (x).
Suppose that z = ax + fly for some a, fi E Q. Then
x+y = Tz = aTx+flTy =ay+flz = ay+fl(ax+ fly) = aflx+(a+fl2)y.By the independence of x, y we obtain
lafl, 1=a+fl2.Eliminating fi yields a3 —a2+ 1 =0. This is a contradiction since a3—a2 + I = 0has no rational solutions.
Hence x 0, y (x) and z g (x, y), so x, z are linearly independent.
Solution to 7.4.14: Since the linear transformation f has rank n — 1, we know thatis an n — 1—dimensional subspace of R?Z. Hence, there exist real constants
Aj, .. . , not all zero, such that
>Ajfj(v)=0
for all v E The 's are unique up to constant multiples. Further, this equationdetermines the subspace: If w E R'1 satisfies it, then w E f(]Rm).
Now suppose that the 'S all have the same sign, or, without loss of generality,that they are all nonnegative. Then if there existed v E ]Rtm with f, (v) > 0 for alli, we would have f, (v) > 0, a contradiction. Hence, there can be no such v.
Conversely, suppose that two of the Aj's, say Aj and A2, have different signs.Letx3 =X4 ••• = 1,andchoosex1 > Osufficiently largeso that
>0.i5/2
Then there is a real number X2 > 0 such that
= 0.
But then we know that there exists v E such that f(v) — (xi, . .. ,Since each of the x, 's is positive, we have found the desired point v.
Solution to 7.4.15: Let (, ) denote the ordinary inner product. From d(s, t)2 =d(s, 0)2 + d(t, 0)2 — 2(s, t) and the hypothesis, it follows that
(ço(s), ço(t)) = (s, t) for all s, t E S.
Let V C R'3 denote the subspace spanned by S, and choose a subset T C S thatis a basis of V. Clearly, there is a unique linear map f: V —÷ V that agrees withço on T. Then one has ff(t), f(t')) = (t, t') for all t and t' E T. By bilinearity, itfollows that (f(v), f(v')) = (v, v') for all v and v' E V. Taking v = v', one findsthat f(v) 0 for v 0, so f is injective, and f(V) = V. Taking v s E S andv'=t E T,onefindsthat
(f(s), f(t)) = (s, t) = ((p(s), ço(t)) = (ç2(s), f(t)),
so f(s) — tp(s) is orthogonal to f(t) for all t E T, and hence to all of f(V) = V.That is, for all s e S, one has f(s) — ço(s) E V-1-; but also f(s) — ço(s) E V, sof(s) — ço(s) = 0. This shows that f agrees with ço on S. It now suffices to extendf to a linear map R'1 —÷ Ri?, which one can do by supplementing T to a basis forR'1 and defining f arbitrarily on the new basis vectors.
Solution to 7.4.16: Define U (x) = T (x) — T (0), then II U (x) — U (y) ii = lix — yfor all x and y and, in particular, since 0 is a fixed point for U, 1IU(x)ll = lix ii.
Squaring the first equality yields
(U(x) — U(y), U(x) — U(y)) = (x — y, x — y),
or
flU(x)fl2 — 2(U(x), U(y)) + OU(y)112 = 11x02 — 2(x, y) + 11y112.
Canceling equal terms gives (U(x), U'(y)) = (x,
y).
Expanding the inner prod-
uct (U(x+y)—U(x)—U(y), U(x+y)— U(x)—U(y)) and applying this equalityimmediately gives that U (x + y) — U (x) — U (y) = 0. A similar calculation shows
that U(ax) — aU(x) = 0. Hence, U is lineai and (U(x), U(y)) = (x, y), so Uis in fact orthogonal. Letting a = T(0) gives the desired decomposition of T.
Solution 2. An isometry T of R2 is a bijection R2 -÷ R2 which preserves dis-tance. Such a map also preserves the inner product, and hence preserves angles.Euclidean geometry shows that a point in the plane is specified by its distancesfrom any three non-collinear points. Thus an isometry is determined by its imagesof any three non-collinear points.
The orthogonal linear maps of ]R2 correspond to rotations about 0, and re-flections in lines through 0. The rotation of through the angle 0 is the mapR: z F-+ The reflection on a line with an angle a with the x-axis is givenby composing three maps: rotation through —a, reflection in the x-axis, rotationthrough a. These are the maps z '—* zeia, z i-+ and z respectively.Thus reflection in that line is the map S: z i-÷
Let T(0) = a. The translation W(x) = x—a is an isometry, and U = WT is anisometry that satisfies U(0) = 0. Now d(U(1), 0) = d(U(1), U(0)) = d(1, 0) =1 and also d(U(i), 0) = 1. Further if U(i) = e'9, then, by preservation of angles,we see that U(i) = ±.ie'0.
First the case U(i) = iU(1). As R : z has the same effect as U onthe non-collinear points 0, 1, i, we deduce that U is that rotation. Next the caseU(i) = —iU(1). Let S: z i-+ be the reflection in line of slope a = 0/2. Wesee that that U agrees with S at 0, 1, i. Thus U is reflection in the line with slope0/2. In either case, U is an orthogonal map.
Finally, as U = WT we have T(x) = W1U(x) = a + U(x), where U is anorthogonal linear transformation.
Solution to 7.4.17: We use Complete Induction [MH93, p. 32] on the dimensionofV.IfdimV = 1,then V hasasinglebasisvectorf1 0,sothereisx1 E Xsuch that fi(xi) 0. Hence, the map f i-+ f(xi) is the desired isomorphism.
Now suppose the result is true for dimensions less than n and let dim V = n.Fix a basis (fit 12...., of V. Then, by the induction hypothesis, there arepOiflts X1, ... , such that the map f (f(x1), ..., 0) is anisomorphism of the subspace of V spanned by (ft, ..., In—I) onto cIn particular, the vector (fn(xi), ..., 0)is a linear combination of thebasis vectors ((f1(x1), ... , 0), 1 i n — 1), so there exists a uniqueset of A1's, 1 i n, such that
>J>jfj(xj)=0,
Suppose there is no point x E X such that the given map is an isomorphism fromV onto R'2. This implies that the Set ..., f(x)), 1 i n)is linearly dependent for all x. But because of the uniqueness of the 's, this, inturn, implies that for all x,
= 0.
Hence, the ft 's are linearly dependent in V, a contradiction. Therefore, such an xexists and we are done.
Solution to 7.4.18: We argue by induction on the dimension of V.If dim V = 1 let be a basis of V. g is not the zero function, therefore we
have g(xi) 0 for some xj E X. Thus fi (x) = gI(x)/gI(xi) satisfies the givencondition.
Suppose now that for any given subspace W of the vector space of continuousreal valued functions on X with dim W = n there exists a basis (fi. ... , forWand ..., E X such that f,(x3) =
Let V besuchasubspacewithdimV = n + 1. Let {gi,for V, and let W be the subspace spanned by (gi, ... , Then there exists abasis for W, (h1, . . . , h,j, and points xI, ... , E X with h- (xj) =
As each h1 is a linear combination of elements of the basis . .. , gn}, the set(h1, .. . , hn, is linerly independent, therefore a basis for V. Let
—
then (h1, . . . , is linearly independent and, for each 1 j n, we have
= gn÷1(xj)—
= — gn÷i(xj) 0.
As is nonzero, there is E X with (Xn+i) 0. We have
for 1 i n since = 0. Let =h sothat = 0.The set
independent and satisfies ft =
Solution to 7.4.19: Since the formula holds, irrespective of the values of Ck, forthe polynomials it suffices, by linearity, to restrict to the vector spaceof polynomials of degree, at most, 2n. This vector space has dimension 2n + Iand the map —* given by p i-÷ (p(—n), p(—n + 1),..., p(n)) is anisomorphism. As the integral is a linear function on there exist unique realnumbers ... , such that
1'1 n
p(x)dx = Ck p (k) for all p EJ—i k=—n
We have
i'l ,1 n
p(x)dx = / p(—x)dx = Ckp(k) for all p EJi J1 k=—n
so Ck = C—k by uniqueness of the Ck, and, therefore,
p1n
j p(x)dx = cop(O) + Ck (p(k) + p(—k)) for all p Ek=I
Setting p = 1, we find that
so, upon eliminating co,
p(x)dx = 2p(O) + cj (p(k) + p(—k) — 2p(O)).
Solution to 7.4.20: Let {vj, V2, ..., be a basis for consisting of eigenvec-
tors of T, say T = Let (u i, U2, ... , be the orthonormal basis one
obtains from (vi, v2, ..., by the Gram—Schmidt Procedure [HK6I, p. 280].
Then, for each index k, the vector Uk is a linear combination of vi, say
Also, each vk is a linear combination of ..., uj. (This is guaranteed by theGram—Schmidt Procedure; in fact, fuj, ..., ujj is an orthonormal basis for thesubspace generated by (vi, ..., vk).) We have
Tuk =ckiTvi= Cki AiVt + Ck2A2V2 + + CkkAkVk.
In view of the preceding remark, it follows that Tuk is a linear combination ofUi, ..., and, thus, T has upper-triangular matrix in the basis (ui, U2, ... ,
Solution to 7.4.21: Let dim V = k. Relative to a basis of V, each linear operatorf on V is represented by a matrix A over F. The correspondence f a—k A is aring isomorphism from L(V, V) onto Mk(F). Further, f is invertible if, and onlyif, A is invertible. As F is finite, the multiplicative group of invertible matricesGLk(F) is a finite group, so each element in GLk(F) has finite order. Thus if Pis invertible, there is n such that =
Solution to 7.4.23: 1. The characteristic polynomial of T has degree 3 so it hasat least one real root. The space generated by the eigenvector associated with thiseigenvalue is invariant under T.2. The linear transformation T — A! has rank 0, 1, or 2. If the rank is 0 thenT = Al and all subspaces are invariant, if it is 1 then ker(T — Al) will do andif it is 2 the image of (T — Al) is the desired subspace. This is equivalent to theJordan Canonical Form [HK61, p. 247] of T being either a diagonal matrix withthree I x I blocks or with one 1 x I and one 2 x 2 block, in both cases there is a2-dimensional invariant subspace.
Solution to 7.4.24: Since A is symmetric and orthogonal, we have A2 = I, there-fore its minimal polynomial, ILA, divides x2 — 1, which leaves three possibilities:
• = x — 1. In this case A = I.
•
• = — 1. Here we consider two cases. If dim ker(A — I) = 2, there is a2-dimensional subspace where every vector remains fixed (ker(A — I)), andthe complement of ker(A—I), ker(A+I), has dimension 1, is perpendicularto ker(A — I), since A is orthogonal. In this case we have a reflection aboutker(A — 1).
If dim ker(A — I) = 1, there is an axis (ker(A — I)) that remains fixed andker(A + I) has dimension 2. Moreover, since A is orthogonal, ker(A + I) -Lker(A — I), hence for x E ker(A + 1) we have A(x) = —x, that is, A is arotation by 1800 about the axis ker(A — 1).
Solution to 7.4.25: Clearly, both R and S are rotations and so have rank 3. There-fore, T, their composition, is a rank 3 operator. In particular, it must have trivialkernel. Since T is an operator on R3, its characteristic polynomial is of degree 3,and so it has a real root. This root is an eigenvalue, which must be nontrivial sinceT has trivial kernel. Hence, the associated eigenspace must contain a line whichis fixed by T.
Solution to 7.4.26: Let x = (XI, X2, x3) in the standard basis of R3. The line join-ing the points x and Tx intersects the line containing e at the pointf = (e, x)e and is perpendicular to it. We then have Tx = 2(f —x)+x 2f—x,or, in the standard basis, Tx = (2(e, x)a —xi, 2(e, x)b —x2, 2(e, x)c —x3). Withrespect to the standard basis for R3, the columns of the matrix of T are Tei, Te2,and Te3. Applying our formula and noting that (e, eI) = a, (e, e2) = b, and(e, e3) = c, we get that the matrix for T is
2ab 2ac2ab 2b2—1 2bc
\ 2ac 2bc 2c2—1
Solution to 7.4.27: Since the minimal polynomial divides the characteristic poly-nomial and this last one has degree 3, it follows that the characteristic polynomialof T is (r2 + 1)(t — 10) and the eigenvalues ±i and 10.
Now T (1, 1, 1) A (1, 1, 1) implies that A = 10 because 10 is the unique realeigenvalue of T.
The plane perpendicular to (1, 1, 1) is generated by (1, —1,0) and —1)
since these are perpendicular to each other and to (1, 1, 1).Let
1' = (1,1,1)12 =
=
we have Tfi = 1011 and, for ±i to be the other eigenvalues of T, Tf2 = and
Tf3=-f2.ThematrixofT in thebasis (fit f21 f3} = isthen
110 0 0[T]p=( 0 0 1
—1 0
The matrix that transforms the coordinates relative to the basis into the coordi-
nates relative to the canonical basis is
/1p=(i
ki 0
and a calculation gives
Therefore, the matrix of T in the canonical basis is2
Solutionto7.4.28:AA = IsolAl2 = landindeedlAl = 1.ThusalsolAtj= 1.
We have
IA—1I=I(A—!)IIA1I=l(A—I)AtI=I!—Atl=l(!—At)tt= I! — Al = (—1)31A — Il;
hence IA — = 0 and there is a nonzero v such that Av = v.
Solution to 7.4.29: For n = 1,2, ..., let be the space of polynomials whosedegrees are, at most, n. The subspaces are invariant under E, they increasewith n, and their union is P. To prove E is invertible (i.e., one-to-one and onto),it will suffice to prove that each restriction is invertible. The subspace isof dimension n + 1, it has the basis 1, x, x2, . . ., with respect to which thematrix of is
p—'=(/ 1/3 1/3
2/3vZ
1/30 •
[T}=P[TIpP 1_—
T10
10
T
I ...0...0• • • 0
110001200013
000 0 in001
In particular, the matrix is upper-triangular, with I at every diagonal entry, soits determinant is 1. Thus, is invertible, as desired. Alternatively, since degEf = deg f, the kernel of E is trivia!, so its restriction to any finite dimensionalinvariant subspace is invertible.
Solution 2. We can describe E to be I + D, where I is the identity operator andD is the derivative operator, on the vector space of all real polynomials P. Forany element f of P. there exists n such that D'2(f) = 0; namely n = deg p + 1.Thus,the inverse ofEcan be descri bed asl—D+D2—D3+.•.
Specifically, writing elements of P as polynomials in x, we have E'(l) = 1,
E'(x2)=x2—2x+2,etc.
Solution to 7.4.30: Given the polynomial n(x), there are constants a andr > 0 and a polynomial ço(x) such that ir(x) = x'ço(x) + a. If ço(x) 0,then = a!, it follows that the minimal polynomial of the operator n(D)is x — a. If ço(x) is not zero, then for any polynomial f E by the defini-tion of D, (ir(D) — aI)(f(x)) = g(x), where g(x) is some suchthat deg g = max(deg f — r, 0). Hence, letting E = — al, we have
(f) = 0 for all f (Ln/rJ denotes the greatest integer less thanor equal to n/r..) The polynomial f(x) = shows that +1 is the minimaldegree such that this is true. It follows from this that the minimal polynomial of
is (x —
Solutionto7.4.31: 1.Considerthebasis (I, x, x2, ... , x10)forthisspaceofpoly-nomials of dimension 11. On this basis the operator takes each element into a mul-tiple of a previous one, of one degree less, so the diagonal on this basis consist ofzeroes and trD =0.2. Since D decreases the degree of the polynomial by one, the equation
Dp(x) = ap(x)
has no non-trivial solutions for a, and D has only 0 as eigenvalue, the eigenvectorsbeing the constant polynomials.
Now since the 1 of any of the polynomials in this space is zero,the expression for the exponential of D becomes a finite sum
D2 D10
10!'
then any eigenvector p(x) of degree k of would have to satisfy
'F (10)
=ap
since p is the only polynomial of degree k in this sum, the eigenvalue a = I and
the summation can be reduced to'I (10)
10!=0
now p' is the only polynomial of degree k —1 in the sum so its leading coefficientk.ak is zero and the same argument repeated over the lower degrees shows that allof them are zero except for the constant one, making it again the only eigenvector.
Solution to 7.4.32: Let 1',, be the vector space of polynomials with real coeffi-cients and degrees at most n. It is a real vector space of dimension n + 1. Definethe map V : -÷ by
Vp = (p(xo), p(xi), ..., p(xn))
This map is linear, and its null space is trivial because a nonzero polynomialin cannot vanish at n + I distinct points. Hence V is invertible. Define thehomomorphism q : —÷ R by
=Jo
Being a linear functional q.' is induced by a unique vector a = (ao, ai, ... , an) in
The numbers at,), ai, ... , have the required property.
Solution to 7.4.33: 1. If p satisfies the equations then it does not have a constantsign on j = 1,..., n, sop has at least n zeros. Ifdegp <n this implies thatp is the zero polynomial.2. Let denote the vector space of real polynomials of degree at most n. Itsdimension is n + I. Each defines a functional in the dual space of by
=J Ii
The dual space of has dimension n + 1, so the functionals . .. , do notspan it. Hence there is a nonzero p in such that (p) = 0 for each j, in otherwords, all equations hold.
7.5 Eigenvalues and Eigenvectors
Solution to 7.5.1: 1. The minimal polynomial of M divides
—1 =(x — l)(x2+x +1).
Since M 1, the minimal polynomial (and characteristic as well) is
(x—
and the only possible real eigenvalue is 1.2.
0 0
I0
C05-f
Solution to 7.5.2: Suppose such X exists; then the characteristic polynomial forX is Xx(t) = t", but this is a contradiction since 0 and 2n — 1 >
Solution to 7.53: Since A is a root of the polynomialp(x) = xm. By the definition of the minimal polynomial, so
= tk for somek n then A?Z = A' =
Solution to 73.4: Let be a basis for the kernel of B, and let y be anothervector so that (a, 8, y) is.a basis for the three dimensional space. The representa-tion of B with respect to this basis is
fOOabob0 c
and the characteristic polynomial is: x (A) = A2(c — A), from which the two first
assertions follow.The third is false since if we take a 1, b = c = 0 above we get a matrix all of
whose eigenvalues vanish, but which is not diagonalizable since it is not similarto the zero matrix.
Solution to 733: The condition on the minimal polynomial implies that T is notthe zero operator and that its square, T2, is. Therefore, dim ker T is at most six,and dim ker T2 is 7. Since the nullity of the square of any linear operator is atmost twice the nullity of the operator, the nullity of T is at least 4.
Each of the three remaining integers, 4, 5, 6, is possible, as is illustrated by theexamples below, where (eI, . . . e7) is a basis for the space.
• dim ker T = 6. Define T by Te7 = and Te, = 0 for all i <7.
• dim ker T = 5. Let Te7 = e5, Te6 = and Te, =0 for all i <6.
• dimkerT = 4. Define T by Te7 = Te6 = Te5 = and Te1 = 0
for all 1 <5.
Solution to 7.5.6: 1. Being real and symmetric, L has an orthonormal basis ofeigenvectors ej, ... , Let A1, ... be the associated eigenvalues. We canassume that A1 0 and Ofori> 1. Write the two vectors v = and
x = x1 in terms of its coordinates, then the equation Lx + e = v becomes
+ ex, = for each i, which has the unique solution X' = v,/(A1 + e),provided that 0 <62. Writing ex in e coordinates
cx= =
as e —÷ 0, all terms in the summation on the right tend to 0, except the first which
approaches viei = (v, ej)el.
Solution to 757: As M divides — I =(t — l)(tP' + t + 1). Since M fixes no nontrivial vector, 1 is notan eigenvalue of M, so it cannot be a root of the minimal polynomial. Therefore,/LM(t)I(t"' •+t + 1). Since p is prime, the polynomial on the right isirreducible, so must equal it. The minimal and characteristic polynomialsof M have the same irreducible factors, so XM(t) = p,M(t)1 for some k 1.
Therefore,
dim V = deg XM(t) = k(p — 1)
and we are done.
Solution to 7.5.8:1. Let d = deg j.t. Since 0, the vector is linearlydependent on the vectors v, Tv,..., v. Hence, v is linearly dependenton T'1v, ..., and so, by the Induction Principle [MH93, p. 7],on v, Tv, ..., (n = 1,2, . . .). Thus, v, Tv, ..., span Vi, so dimIfj
On the other hand, the minimal polynomial of v1 must divide (since/2(T y1) = 0), so it equals because ji is irreducible. Thus, dim V1 d. Thedesired equality, dim Vi = d, now follows.2. In the case Vi V, let T1 be the linear transformation on the quotient spaceV/V1 induced by T. (It is well defined because Vi is T—invariant.) Clearly,
= 0, so the minimal polynomial of T1 divides p,, hence equals p. There-fore, by Part 1, V/V1 has a Ti—invariant subspace of dimension d, whose inverseimage under the quotient map is a T—invariant subspace V2 of V of dimension 2d.In the case V2 V. we can repeat the argument to show that V has a T—invariantsubspace of dimension 3d, and so on. After finitely many repetitions, we find dimV = kd for some integers k.
Solution to 7.59: Let v be an eigenvector in C?Z for M with nonzero eigenvaluea. Then,
MA'v = =
a2 is also an eigenvalue. Thus a2k is an eigenvalue for all nonnegative integersk. Since the set of eigenvalues is finite there exist 0 k < m Z such thata =a .Thusaisarootofunity.
Solution to 7.5.10: Since the matrix is real and symmetric, its eigenvalues arereal. As the trace of the matrix is 0, and equal to the sum of its eigenvalues, it hasat least one positive and one negative eigenvalue.
The matrix is invertible because the span of its columns has dimension 4. Infact, the space of the first and last columns contains all columns of the form
/01*1*\0
The span of all four columns thus contains
and
which together span all columns of the form
1*10(0
Since the matrix is invertible it does not have 0 as an eigenvalue. There are nowonly three possibilities:
• three positive and one negative eigenvalues;
• two positive and two negative eigenvalues;
• one positive and three negative eigenvalues.
A calculation shows that the determinant is positive. Since it equals the productof the eigenvalues, we can only have two positives and two negatives, completingthe proof.
Solution to 7.5.11: A calculation shows that the characteristic polynomial of thegiven matrix is
—x(x2 — 3x — 2(1.000012 — 1))
therefore one of the eigenvalues is 0 and the product of the other two is
_2(l.000012 — 1)) <0
so one is negative and the other is positive.
Solution to 7.5.12: Denote the matrix by A. A calculation shows that A is a rootof the polynomial p(t) = t3 —ct2 —bt —a. In fact, this is the minimal polynomialof A. To prove this, it suffices to find a vector x E F3 such that A2x, Ax, and x are
linearly independent. Let x = (1, 0, 0). Then Ax = (0, 1, 0) and A2x = (0, 0, 1);these three vectors are linearly independent, so we are done.
Solution to 73.13: The matrix A + I is diagonalizable, being unitary, so A also isdiagonalizable. It will therefore suffice to show that A has no nonzero eigenvalues.
Suppose A is a nonzero eigenvalue of A. Then +1 is an eigenvalue of + I(J = 1, 2,3) and so has unit modulus. Thus A, A2, all lie on the circle Jz +1J =1, so each has an argument strictly between and But if argA
then arg A2 2ir, a contradiction. And if < arg A < then
arg A3 < (mod 2ir), again a contradiction. The case where A is in thethird quadrant can be handled similarly.
Solution to 7.5.14: The first direction is obvious, lets assume now that everynonzero vector is an eigenvector of A, then in particular, the vectors of the stan-dard basis, e are are eigenvectors and A is diagonal. Lets assume the diagonalentries are A11 = A1. If A1 then A(e1 + = + is not a multiplescalar of + contradicting the hypothesis that e1 + is an eigenvector of A.Then all diagonal entries are equal, finishing the proof.
Solution to 7.5.15: 1. This is false. Consider A = and B =AB = and has (1, 1) as an eigenvector, which is clearly not an eigenvector
of BA = The condition that AB and BA have a common eigenvector isalgebraic in the entries of A and B; therefore it is satisfied either by all of A andB or by a subset of codimension at least one, so in dimensions two and higheralmost every pair of matrices would be a counterexample.2. This is true. Let x be an eigenvector associated with the eigenvalue A of AB.We have
BA(Bx) = B(ABx) = B(Ax) = ABx
so A is an eigenvalue of BA.
Solution to 73.16: Regard A and B as linear transformations on C Then A hasan eigenvalue A. Let be the corresponding eigenspace. For v in Sj., we have
A(Bv) = B(Av) = B(Av) = ABv,
showing that S,. is invariant under B. The linear transformation BISA has an eigen-value v is a corresponding eigenvector we have At' = At' and By =
Solution to 7.5.17: We use the Induction Principle [MH93, p. 7]. As the space(C) is finite dimensional, we may assume that S is finite. If S has one element,
the result is trivial. Now suppose any commuting set of n elements has a commoneigenvector, and let Shave n+l elements A1, ... , By induction hypothesis,the matrices A1, ..., have a common eigenvector v. Let E be the vector spacespanned by the common eigenvectors of A1, ..., A,,. If v E E, A,,+i v =A,,+j v = v for all i, so v E. Hence, fixes E. Let B be therestriction of A,,+i to E. The minimal polynomial of B splits into linear factors
(since we are dealing with complex matrices), so B has an eigenvector in E, whichmust be an eigenvector of by the definition of B, and an eigenvector for eachof the other A' 's by the definition of E.
Solution to 7.5.20: 1. For (ai, a3,...) to be an eigenvector associated withthe eigenvalue A, we must have
S((al,a2a3, ...))=A(a2,a3,a4,...)
which is equivalent to
a2=AaI, a3=Aa2,...,
so the eigenvectors are of the form ai (:1, A, A2,•
2. Let x = (xi, x2,...) E W. Thenx is completely determined by the first twocomponents and X2. Therefore, the dimension of W is, at most, two. If anelement of W is an eigenvector, it must be associated with an eigenvalue satisfyingA2 = A + 1, which gives the two possible eigenvalues
____
-12
and= 2
A basis for W is then
{
602, •..), . .
which is clearly invariant under S.3. To express the Fibonacci sequence in the basis above, we have just to find the
constants k1 and k2 that satisfy
J 1 =
k1 = = —k2. We then have, for the Fibonacci numbers,
1
Solution to 7.5.21: If T(v) = Av then T'2(v) = A'1v for all n 1, by induction.If f = ao + aix + + then f(T) = aol + aiT + + andf(T)(v) = aov+aiT(v)+. . = f(A)v so that f(A) is an eigenvalue
of f(T).
If A is an eigenvalue of f(T) then Jf(T) — All = 0. Factorize f(x) — A =—ar). .. (x (The case of constant f(x) is trivial.) Then f(T) —Al =— au). • (T — Taking determinants we see that there is i such that
IT — a,II = 0. Therefore is an eigenvalue of T with f(a,) = A.
Solution to 7.5.22:
flA—uu—I where
and I is the identity matrix. If Ax Ax, where x 0, then
— x = (ux)u — x = Ax
so x is either perpendicular or parallel to a. In the latter case, we can supposewithout loss of generality that x = u, so Wuu — u = Au and A = n — 1. Thisgives a 1-dimensional eigenspace spanned by u with eigenvalue n — 1. In theformer case x lies in a (n — 1)—dimensional eigenspace which is the null space ofthe rank one matrix uu', so
Ax = (uut — l)x = —lx = —x
and the eigenvalue associated with this eigenspace is —1, with multiplicity n — 1.
Since the determinant is the product of the eigenvalues, we havedet(A) = (n — 1).
Solution to 7.5.23: Since A is positive definite, there is an invertible Hermitianmatrix C such that C2 = A. Thus, we have C1(AB)C = C1C2BC = CBC.By taking adjoints, we see that CBC is Hermitian, so it has real eigenvalues.Since similar matrices have the same eigenvalues, AB has real eigenvalues.
Solution to 7.5.24: Let v be the column vector defined by ( a b c d )t,then
the operator given, that we will call T, is literally the product of the two matricesvv'. and the operator T applied to a vector x is:
Tx = (vvt)x = v(vtx) = v(v, x)
where (, ) is the standard euclidean inner product inLet u be the vector of norm 1 in the direction of v, that is, u = v/IvI, V the
uni-dimensional space generated by v, and its orthogonal complement inThen any vector in x E 1R4 can be written as x = + Xj where
x11 = (u,x)uXj =x
In this decomposition Tu = 1v12 u u u an eigenvector ofT witheigenvalue 1v12 and multiplicity 1. For any vector in the V1 space
Txj = IvI2u (u, xjj =0
so taking any basis for this 3-dimensional space, the matrix representing T willbe
a2+b2+c2+d2 0 0 0o oooo oooo ooo
Solution to 7.5.25: J 2: As A is symmetric, it is similar to a diagonal matrix,A2 A3). We have trA = Ai +A2 +A3. The eigenvalues of A are the zeros
of the polynomial (A — A1)(A — A2)(A — A3) and the conclusion follows.2 3: Let (Cl, C2, c3) be the ordered basis of R3 with respect to which therepresentation of A is diag(Ai, A2, A3). For W E S we have Wt = — W, therefore,using the standard inner product in R3, we obtain, for i = 1,2,3,
= (C,, = — (C1, = — ( Wc1, C,),
therefore Wc, ..LL is clearly a linear self map of 5, so it suffices to show that it is injective.Suppose L(W) = 0. Then AW = —WA, so we get, for i j,
== (ACt, WCJ)
= WCJ)
and
== cj)
= —AJ(CI, WCJ)
as A, + 0, we must have I c,. We conclude then that Wc, = 0 for
i = 1,2,3,therefore W =0.3 1: Suppose tr A is an eigenvalue. According to what we saw previously inthe Part 1 2, there is no loss in generality in assuming A1 + = 0. Let W
be defined on the vectors of the basis (Cl, C2, c3} by WC1 —C2, Wc2 = —cj,
Wc3 =0. Clearly W E S, and W 0. We have,
(AW + WA)cj = + WAc1
= —Ac2 +A1WC1
= —A2c2 — A1C2
=0
so L is not an isomorphism.
Solution to 73.26: The characteristic polynomial of A is
XA(t) = — (a + d)t + (ad — bc).
which has roots
is positive, so A has real eigenvalues. Let = (a + d + and let
v = (x, y) be an eigenvector associated with this eigenvalue with x > 0.
Expanding the first entry of Av, we get
ax+by =
or
Since b > 0, to see that y > 0 it suffices to show that d — a + > 0, or
a — d. But this is immediate from the definition of and we are done.
Solution to 7.5.27: It suffices to show that A is positive definite. Letx = (xl, ... , we have
(Ax, x) = 2x? — X2 — xIx2 + — X2X3 — -- - — i +
= + (xl — X2)2 + (x2 — X3)2 + .. + i — +
Thus, for all nonzero x, (Ax, x) 0. In fact, it is strictly positive, since one of thecenter terms is greater than 0, or = X2 = ... = and all the x, 's are nonzero,so x? > 0. Hence, A is positive definite and we are done.
Solution 2. Since A is symmetric, all eigenvalues are real. Let x = (xj)? be an
eigenvector with eigenvalue A. Since x 0, we have maxj lxii > 0. Let k bethe least i with ix1 I maximum. Replacing x by —x, if necessary, we may assume
Xk > 0. We have
AXk = Xk—I + 2Xk — Xk+l
where nonexistent terms are taken to be zero. By the choice of we have
<Xk and so we get AXk > 0 and A> 0.
Solution to 7.5.28: Let A0 be the largest eigenvalue of A. We have
Ao=max{(Ax,x)IxER, ilxiI=1),
and the maximum it attained precisely when x is an eigenvector of A with eigen-
value A0. Suppose v is a unit vector for which the maximum is attained, and let
u be the vector whose coordinates are the absolute values of the coordinates of v.Since the entries of A are nonnegative, we have
(Au, u) (Av, v) = Ao,
implying that (Au, u) = A0 and so that u is an eigenvector of A for the eigenvalueAo.
Solution to 7.5.29: Let be the characteristic polynomial of We have= Z, = — 1, and, for n ? 2, (z) — Thus, by
induction, contains only even powers of z for even n, and only odd powers of zfor odd n. It follows that the zeros of are symmetric with respect to the origin.Solution 2. Let be the diagonal n x n matrix with alternating l's and —I'son the main diagonal. Then S1AS = —A, from which the desired conclusionfollows.
Solution to 7330: Let A be an eigenvalue of A and x = (Xl, . . - ,
x, x whose absolute value is greatest.We have
= >2 ajxf
lxjl>2a,j lxii.
Hence, IAI 1.
Solution to 7.531: 1. Regard S as a linear transformation on Since 5 isorthogonal it preserves norms, ilSxli lix Ii for all x. Hence all eigenvalues havemodulo 1. The characteristic polynomial of S has real coefficients, so the nonrealeigenvalues occur in conjugate pairs. There is thus an even number of nonrealeigenvalues, counting multiplicities, and the product of all of them is 1. Assumingn is odd, there must be an odd number of real eigenvalues, which can only equal 1or —1. The product of all the eigenvalues is det(S) = 1. Hence —1 must occur asan eigenvalue with even multiplicity, so that 1 occurs with an odd, hence nonzero,multiplicity.2. If n is even then is a special orthogonal matrix, yet it does not have 1 as aneigenvalue.
Solution to 7332: Since A is Hermitian, by Rayleigh's Theorem [ND88, p.418], we have
(x, Ax) _Amjn "max(x,x)
for x (Cm, x 0, where Amin and Amax are its smallest and largest eigenvalues,
respectively. Therefore,(x,Ax)(x,x)
Similarly for B:L (x,Bx) -k'
(x,x)Hence,
(x,(A+B)x)(x,x)
However, A + B is Hermitian, since A and B are, so the middle term aboveis bounded above and below by the largest and smallest eigenvalues of A + B.But, again by Rayleigh's Theorem, we know these bounds are sharp, so all theeigenvalues of A + B must lie in [a + b, a' + b'].
Solution to 7.5.33: Let v = (1, 1,0, . .., 0). A calculation shows thatAv =(k+1,k+1,1,0,...,0),SO
(Av,v) =k+1.(v, v)
Similarly,foru = (1,—1,0,...,O),we have Au = (k—i, I —k,—1,0,...,0)and so
(Au,u) =k—1.(u,u)
By Rayleigh's Theorem [ND88, p. 418], we know that
(Av, v)Amin Amax.
(v,v)
for all nonzero vectors v, and the desired conclusion follows.
Solution to 7.5.34: As B is positive definite, there is an invertible matrix C suchthat B = C'C, so
(Ax,x) (Ax,x) — (Ax,x)(Bx,x) — (C'Cx,x) — (Cx,Cx)
Let Cx = y. The right-hand side equals
(AChy, C1y) y)
(y,y) (y,y)
Since the matrix (c')' AC—' is symmetric, by Rayleigh's Theorem [ND88, p.418], the right-hand side is bounded by X, where A is the largest eigenvalue of(c-')' AC1. Further, the maximum is attained at the associated eigenvector.Let yo be such an eigenvector. Then G(x) attains its maximum at x = C'yo,which is an eigenvector of the matrix (C—' ) A.
Solution to 7.5.35: Let y 0 in A is real symmetric, so there is an orthogonalmatrix, P. such that B = P' AP is diagonal. Since P is invertible, there is anonzero vector z such that y = Pz. Therefore,
— Pz) — (ptAm+IpZ,Z) — (Bm+1Z,Z)
(Atmy, y) — (Atm Pz, Pz) — (P'A"' Pz, z) — (Bmz, z)
Since A is positive definite, we may assume without loss of generality that B hasthe form
Aj 0 •.. 0o A2 0
o
where A1 A2 •.. > 0. Let z = (zi, ..., Zn) 0, and i n be suchthat Zj is the first nonzero coordinate of z. Then
(Bm+lZ,Z) —
=+ + • • + (X/A.)m+1Z2
\ + +• +(m —* oo).
7.6 Canonical Forms
Solution to 7.6.1: The minimal polynomial of A divides xk — 1 so it has nomultiple roots, which implies that A is diagonalizable.
Solution to 7.6.2: Assume Atm is diagonalizable. Then its minimal polynomial,(x), has no repeated roots, that is,
(x) = (x — a (x — ak)
where a, fori j.The matrix A"' satisfies the equation
(Atm — -(A"' — akl) = 0
so A is a root of the polynomial (X"' — al)-. (X"' — ak), therefore, (x) di-vides this polynomial. To show that A is diagonalizable, it is enough to show thispolynomial has no repeated roots, which is clear, because the roots of the factors
— a1 are different, and different factors have different roots.This proves more than what was asked; it shows that if A is an invertible linear
transformation on a finite dimensional vector space over a field F of characteristicnot dividing n, the characteristic polynomial of A factors completely over F, andif A'2 is diagonalizable, then A is diagonalizable.
On this footing, we can rewrite the above proof as follows: We may sup-pose that the vector space V has positive dimension m. Let A be an eigenvalueof A. Then A 0. We may replace V by the largest subspace of V on whichA — Al is nilpotent, so that we may suppose the characteristic polynomial of A is
(x — A)"'. Since A" is diagonalizable, we must have A" = A" I since A'1 is theonly eigenvalue of A'1. Thus, A satisfies the equation x" — A" = 0. Since the onlycommon factor of x" — A" and (x — A)" is x — A, and as the characteristic of Fdoes not divide n, A = Xl and, hence, is diagonal.
Solution to 7.6.3: The characteristic polynomial of A is XA (x) — 3, soA2 = 31, and multiplying both sides by A1, we have
A1 =
Solution to 7.6.4: 1. Subtracting the second line from the other three and expand-ing along the first line, we have
x 1 1 1 1 1
1—x x—1 0 +(x—1) 0 x—1 01—x 0 x—1 0 0 x—1
= (x — l)3(x+3).
2. Suppose now that x I and —3. Then is invertible and the characteristicpolynomial is given by:
t—x —1 —1 —1 x—t 1 1 1
—1 t—x —1 —1 1 x—t 1 1
—1 —1 t—x —1 = 1 1 1
—1 —l —1 t—x 1 1 1 x—t=(x—z — 1)3(x —t+3)
Now an easy substitution shows that the minimal polynomial is
= (x—t— 1)(x —t—3)
so substituting t by we have
((x — =00
multiplying both sides by A;1,
= —A_x_2
A1 = —(x — 1)1(x +
Solution to 7.6.5: The characteristic polynomial of the matrix A isXA (t) = — + 20t — 16 = (t — 4)(t — 2)2 and the minimal polynomial
is I.LA (t) = (t — 2)(t — 4). By the Euclidean Algorithm [Her75, p. 155], there is apolynomial p(t) and constants a and b such that
t10 = + at + b.
Substituting t = 2 and t =4 and solving for a and b yields a = 29(210 — 1) andb = _2h1 — 1). Therefore, since A is a root of its minimal polynomial,
f3a+b a aA'°=aA+bI=I 2a 4a+b 2a
—a —a a+b
Solution to 7.6.6: The characteristic polynomial of A is XA(t) = t2 — 2t + I =(t — 1)2. By the Euclidean Algorithm [Her75, p. 155], there is a polynomial q(t)and constants a and b such that = q(t)(t — 1)2 + at + b. Differentiatingboth sides of this equation, we get = q'(t)(t — 1)2 + 2q(t)(t — 1) + a.Substituting t = 1 into each equation and solving for a and b, we get a = 100
and b = —99. Therefore, since A satisfies its characteristic equation, substitutingit into the first equation yields = 100A — 99!, or
Aboo_(51 50
'¼
An identical calculation shows that A7 = 7A — 6!, so
7/2— '¼—7/2 —5/2
From this it follows immediately that
A7 — (—5/2 —7/27/2 9/2
Solution to 7.6.7: Counterexample: Let
(0 1'\ B2(a b'\(a b'\ (a2+bc ab+bdo) '¼c d)'¼c d)'¼ca+dc cb+d2
Equating entries, we find that c(a + d) = 0 and b(a + d) = 1, so b 0 and
a + d 0. Thus, c = 0. The vanishing of the diagonal entries of B2 then implies
that a2 = d2 = 0 and, thus, a + d = 0. This contradiction proves that no such B
can exist, so A has no square root.
Solution 2. Let(0 1
A_'¼o 0
Any square root B of A must have zero eigenvalues, and since it cannot be the
zero matrix, it must have Jordan Canonical Form [HK61, p. 247] JBJ1 = A.
But then B2 = J1 = 0 since A2 = 0, so no such B can exist.
Solution to 7.6.8: 1. Let(a b
d
thenA2_(a+1)c (a+d)b
bc+d2
Therefore, A2 = —I is equivalent to the system
a2+bc —1
(a+d)b = 0(a+d)c = 0
bc+d2 = —1
if a + d 0, the second equation above gives b = 0, and from the fourth, weobtain d2 = —1, which is absurd. We must then have a = —d and the resultfollows.2. The system that we get in this case is
a2+bc = —1
(a+d)b = 0(a+d)c = 0bc+d2 = —1—s
As above, we cannot have a —d. But combining a = —d with the first andfourth equations of the system, we get s = 0, a contradiction. Therefore, no suchmatrix exists.
Solution to 7.6.9: Suppose such a matrix A exists. One of the eingenvalues of Awould be w and the other (1 + e)1120w where w is a twentieth root of —1. Fromthe fact that A is real we can see that both eigenvalues are real or form a complexconjugate pair, but neither can occur because none the twentieth root of —1 arereal and the fact that
Iwl = 1 (1 +e)"2°
make it impossible for them to be a conjugate pair, so no such a matrix exist.
Solution to 7.6.10: = I implies that the minimal polynomial of A, jz(x)7L[x], satisfies — 1). Let ..., be the distinct roots of x'2 — 1 inC.We will separate the two possible cases for the degree of
• deg p. = I. We have /2(x) = x — I and A = I, or p.(x) = x + I andA=—J, A2=1.
• deg = 2. In this case, and are roots of for some i j, inwhich case we have = = say, since ji. has real coefficients. Thus,p(x) = (x — (x — = x2 — + 1. In particular, E Z, sothe possibilities are = 0, ±1/2, and ±1. We cannot have 91(r) = ±1because the corresponding polynomials, (x—1)2 and (x+1)2, have repeatedroots, so they are not divisors of x" — 1.
= 0. We have ji(x) = x2 + I and A2 = —1, = 1.
= 1/2. In this case p(x) = x2 — x + 1. is a primitive sixthroot of unity,so A6 = I.
= = x2+x +1. isaprimitivethirdrootof unity, so A3 = I.
From the above, we see that if = I for some n Z÷, then one of the followingholds:
A = I, A2 = I, A3 = I, A4 = I, A6 = I.
Further, for each n = 2, 3, 4, and 6 there is a matrix A such that A'1 = I butA1' forO <k <n:
• n=2.(—1 0
1
•n=3. (0 I
—1
• n = 4. (0 i0
• n=6. (0 1
1
Solution to 7.6.11: Since A is upper-triangular, its eigenvalues are its diagonalentries, that is, 1, 4, and 9. It can, thus, be diagonalized, and in, fact, we will have
S1AS =
where S is a matrix whose columns are eigenvectors of A for the respective eigen-
values 1,4, and 9. The matrix
fioo\B=SI0 2\o03j
will then be a square root of A.Carrying out the computations, one obtains
/1 1 i\ fi —1 0s=(o I and 1 —1
o iJ \o 0
givingfi 1 —1\
2 1 j.\oo 3/
The number of square roots of A is the same as the number of square roots ofits diagonalization, D = S1AS. Any matrix commuting with D preserves itseigenspaces and so is diagonal. In particular, any square root of D is diagonal.Hence, D has exactly eight square roots, namely
±1 0 0o ±2 0o 0 ±3
Solution to 7.6.12: n = 1. There is the solution X = A.n =2. A is similar to the matrix
1001010 0 0 0Io 0 0 0
0 0 0
under the transformation that interchanges the third and fourth basis vectors andleaves the first and second basis vectors fixed. The latter matrix is the square of
fO 1 0 010010\0000
Hence, A is the square of 010000010000•0000n =3. The Jordan matrix [HK6I, p. 247J
/0 100x_I0 0 1 0
is a solution.n 4. If X A the characteristic polyno-mial of X divides x4, so that X4 = 0 and, a fortiori, = 0 for n 4. There is,thus, no solution for n 4.
Solution to 7.6.13: Suppose such a matrix A exists. Its minimal polynomialmust divide t2 + 2t + 5. However, this polynomial is irreducible over IR, soI.LA(t) = t2 + 2t + 5. Since the characteristic and minimal polynomials havethe same irreducible factors, XA (0 = p..A (tY. Therefore, deg XA (t) n must beeven.
Conversely, a calculation shows that the 2 x 2 real matrix
10 —5—2
is a root of this polynomial. Therefore, any 2n x 2n block diagonal matrix whichhas n copies of Ao on the diagonal will satisfy this equation as well.
Solution to 7.6.14: Let p(t) = t5+t3 +t —3. As p(A) = 0, we haveHowever, since A is Hermitian, its minimal polynomial has only real roots. Takingthe derivative of p. we see that p'(t) = 5t4 + 3t2 + 1 > 0 for all t, so p(t)has exactly one real root. A calculation shows that p(1) = 0, but p'(l) 0.Therefore, p(t) = (t — l)q(t), where q(t) has only nonreal complex roots. Itfollows that — 1). Since t — I is irreducible, /LA(t) = t — 1 and A = I.
Solution to 7.6.16: Note that
f2 0 0A=fO 2 0
—1 1
can be decomposed into the two blocks (2) and since the space spanned
by (1 Ø0)t is invariant. We will find a 2 x 2 matrix C such that C4 = =say.
The eigenvalues of the matrix D are 2 and 1, and the corresponding La-grange Polynomials [MH93, p. 286] are = (x — 2)/(1 —2) = 2— x andp2(x) = (x — I )/(2 — 1) = x — 1. Therefore, the spectral projection of D can begiven by
(2 (io\ (001
11 o\ (2 o\ (1 00
We have
D=(? ?)+2(L g).
As P1 . P2 = P2. P1 0 and = = p2, letting C = =1P1 P2, we get
C4 P2)4 = P1 + 2P2 D.
Thenf
1
and B is
1 0 0
1 0 0
0 I
Solution to 7.6.17: It suffices to show that every element w E W is a sum ofeigenvectors of T in W. Let ai, . .., be the distinct eigenvalues of T. We may
write
W = Vi +" +V is an eigenvector of T with eigenvalue a,. Then
fl(T —a1)w = fl(a, —aj)vj.
i3L:j i#i
This element lies in W since W is T invariant. Hence, v, W for all i and theresult follows.
Solution 2. To see this in a matrix form, take an ordered basis of W and extend itto a basis of V; on this basis, a matrix representing T will have the block form
IACB
because of the invariance of the subspace W with respect to T.Using the block structure of T, we can see that the characteristic and minimal
polynomials of A divide the ones for T. For the characteristic polynomial, it isimmediate from the fact that
det(x/ — [T]8) = det(xl — A) det(xI — B)
For the minimal polynomial, observe that
Tk_(Ak CkI Bk
where Ck is some r x (ii — r) matrix. Therefore, any polynomial that annihilates[T} also annihilates A and B; so the minimal polynomial of A divides the one for[T].
Now, since T is diagonalizable, the minimal polynomial factors out in differentlinear terms and so does the one for A, proving the result.
Solution to 7.6.19: Let A be an eigenvalue of A and v a vector in the associatedeigenspace, Then A(Bv) = BAy = B(Av) = X(Bv), so By E AA. Now fixan eigenvalue A and let C be the linear transformation obtained by restricting Bto Take any v E Then, since C is the restriction of B,
(C)v = !2B (B)v =0,
so C is a root of (t). It follows from this that (t)I/.tB (t). But B was diago-nalizable, so (t) splits into distinct linear factors. Therefore, must splitinto distinct linear factors as well and so has a basis of eigenvectors of C. AsA is diagonalizable, V can be written as the direct sum of the eigenspaces of A.However, each of these eigenspaces has a basis which consists of vectors whichare simultaneously eigenvectors of A and of B. Therefore, V itself must have sucha basis, and this is the basis which simultaneously diagonalizes A and B.
Solution 2. (This one, in fact, shows much more; it proves that a set of n x ndiagonalizable matrices over a field F which commute with each other are allsimultaneously diagonalizable.) Let S be a set of n x n diagonalizable matricesover a field F which commute with each other. Let V = Suppose T is amaximal subset of S such that there exists a decomposition of
V=®,V1
where is a nonzero eigenspace for each element of T such that for i j,there exists an element of T with distinct eigenvalues on V1 and V3. We claim thatT = S. If not, there exists an N S — T. Since N commutes with all the elementsof T, NV, c V,. Indeed, there exists a function a, : T -÷ F such that v E ifand onlyif Mv = a,(M)v foraliM E T. Nowifv E V1 and ME T,
MNv = NMv = Na,(M) = a,(M)Nv
so Nv V1. Since N is diagonalizable on V. it is diagonalizable on V1. (SeeProblem 7.6.17; it satisfies a polynomial with distinct roots in K.) This meanswe can decompose each V, into eigenspaces for N with distinct eigenvalues.Hence, we have a decomposition of the right sort for T U N,
Hence, T = S. We may now make a basis for V by choosing a basis for V1 andtaking the union. Then A will be the change of basis matrix.
Solution to 7.6.20: 1. Take
fl),
they are both diagonalizable since the characteristic polynomial factors in linearterms, but A + B = is not diagonalizable.
2. If A = and B = then A and B have distinct eigenvalues and are
diagonalizable. However AB = is not diagonalizable.
3. If A2 = A then P-.A dividesx2 — x, whence ILA =x orx — 1 orx(x — 1). Itfollows that A is diagonalizable, as is a product of distinct linear factors.4. If A2 is diagonalizable its minimal polynomial is the product of distinct linearfactors: P#A2 = (x — •• (x — ?.k), where E C. Therefore
(A2 —A11)•••(A2—AkI) =0.
It follows that (A — .JXjI)(A + 44/XjI). . (A — ./X1)(A + ./X1) = 0. Theminimum polynomial of A divides the polynomial
If A is invertible then none of the is zero, and thus the factors of f are distinct.It follows that /.LA is a product of distinct linear factors and A is diagonalizable.
Solution to 7.6.21: The characteristic polynomial of A is
x—7 —152 x+4 =(x—1)(x—2)
so A is diagonalizable and a short calculation shows that eigenvectors associatedwith the eigenvalues I and 2 are (5, —2)' and (3, —1)', so the matrix B is
Indeed, in thiscase, B1AB =
Solution to 7.6.23: The characteristic polynomial is XA(t) = (t — 1)(t — 4)2•
Since the minimal polynomial and the characteristic polynomial share the sameirreducible factors, another calculation shows that j.LA (t) = (t — 1)(t _4)2•fore, the Jordan Canonical Form [HK6 1, p. 247] of A must have one Jordan blockof order 2 associated with 4 and one Jordan block of order 1 associated with 1.
Hence, the Jordan form of A is
/1001041.\O 0 4
Solution to 7.6.24: We have
2—A 1 1 1—A 1 0A—AJl= 1 2—A 1 = —1+A 2—A —1-j-A
1 1 2—A 0 1 1—A
—1 1 0=(A—1)2 I 2—A 1 =—(A—1)3,
0 1 —1
using column operations C1 — C2 and C3 — C2. The single eigenvalue of A isA = 1 with algebraic multiplicity 3.
The eigenvectors of A are the solutions to the equation (A — 1)x = 0. That is,xl + x2 + X3 = 0, or x (—x2 — X3, x2, x3). The eigenvectors corresponding totheeigenvalueA= 1 arex2(—1, l,0)+x3(—l,0, 1), (x2,x3 EF3) (includingthe zero vector).
The characteristic polynomial of A is (x — Now
/1 1 i\A—I= Ii 1 1 (A—I)2=0.\iiij
The minimal polynomial of A is (x — 1)2. The Jordan form has an elementaryJordan block of size 2. The Jordan form of A is therefore
/1 1 o\10 1 01.\ooiJ
Solution to 7.6.25: Combining the equations, we get = j.t(x)(x —i)(x2 +1)and, thus, = (x — i)2(x + i). So the Jordan blocks of the Jordan CanonicalForm [HK61, p. 247] JA, correspond to the eigenvalues ±i. There is at least oneblock of size 2 corresponding to the eigenvalue i and no larger block correspond-ing to i. Similarly, there is at least one block of size 1 corresponding to —i. Wehave x(x) = (x — i)3(x + i), so n = deg x = 4, and the remaining block is ablock of size 1 corresponding to i, since the total dimension of the eigenspace isthe degree with which the factor appears in the characteristic polynomial. There-fore,
fi 1 0 010 i 0 0JA=1Ø 0\o 0 o —i
Solution to 7.6.26: 1. As all the rows of M are equal, M must have rank 1,so its nullity is n — 1. It is easy to see that M2 = nM, or M(M — nI) = 0,
so the minimal polynomial is /2M = x(x — n), since the null space associatedwith the characteristic value 0 is n — 1, then the characteristic polynomial isXM —_x"'(x —n)2. If charF = 0 or if charF = p and p does not dividen, then 0 andn are the twodistinct eigenvalues, and since the minimal polynomial does not have repeatedroots, M is diagonalizable.
If char F = p. pin, then n is identified with 0 in F. Therefore, the minimalpolynomial of M is (x) x2 and M is not diagonalizable.
3. In the first case, since the null space has dimension n — 1, the Jordan form
[HK61, p. 247] is no...0o 0 •.. 0
IfcharF = p, pin, then all the eigenvalues of M are 0, and there is one 2-block
and n — 1 1-blocks in the Jordan form:
01...oo 0 •.. 0
Solution to 7.6.27: A computation gives
T= (—x21 Xfl —X22
\X21 X22J 0 X21
In particular, for the basis elements
Ii o\ to i\ 10 o\ 10 0o)' o)' o)' i
we haveTEI__(00 TE2=O,
TE3=(01 ?)=_EI +E4, TE4=
The matrix for T with respect to the basis {E1, E2, E3, E4) is then
100—101100—i0 0
0
A calculation shows that the characteristic polynomial of S is Thus, S is nilpo-tent. Moreover, the index of nilpotency is 3, since we have
T2EI = T2E2 T2E4 = 0, T2E3 = —2E2.
The only 4 x 4 nilpotent Jordan matrix [HK61, p. 247] with index of nilpotency3 is foooo
1001010 0 0 1
\o 0 0
which is, therefore, the Jordan Canonical Form of T. A basis in which T is repre-sented by the preceding matrix is
Solution to 7.6.28: It will suffice to prove that ker(TA), the subspace of matricesthat commute with A, has dimension at least n. We can assume without loss ofgenerality that A is in Jordan form. A k x k Jordan block commutes with thek x k identity and with its own powers up to the (k — 1), hence with k linearlyindependent matrices. Thus, the commutant of each Jordan block has a dimensionat least as large as the size of the block, in fact, equality holds. By taking directsums of matrices commuting with the separate Jordan blocks of A, we thereforeobtain a subspace of matrices commuting with A of dimension at least n and, infact, equal to n.
Solution to 7.6.29: The answer is 0 and 1. If A is diagonal with diagonal entries1, 2, . . . , n and g(t) = (t—1)(t—2)... (t — (n—i)), then g(A) has (n — 1)! in thelower right corner and zeros elsewhere, so g (A) has rank 1. If A is the zero matrixand g(t) = t"', then g(A) has rank 0. We will show that these are the onlypossibilities for the rank of g (A), even if A is allowed to have complex entries.
Since the ground field is now algebraically closed, we may conjugate A to putit in Jordan canonical form, without affecting the rank of g(A). Let A1, ..., Arbe the Jordan blocks of A, and let f, (t) E C [t] be the characteristic polynomialof A1 for each i. Since g(t) divides fl 41(t) and degg n — 1, we can factorg(t) as fl g,(t) where gj = for all i except one. Without loss of generality theexception is i = 1. Then for some k I and A E C, A1 is a k x k block
AI0...0OAI...0A1= : : : .
o o 0 ... 1
o 0 0 ... A
= (t — A)k, and = c(t — A)k_l, where c 0. Then g(A) is formed ofthe blocks ... , g(A4 (not necessarily full Jordan blocks), and g(A,) = 0
for i 2, since the characteristic polynomial of A1 divides g. On the other handg (A1) is some matrix times
/0 1 0 ... 0k—i
0 0 0 ... c
0 0 1 ... 0 0 0 0 ... 0
gi(Ai)=c ...o 0 0 ... 1 0 0 0 ... 0
0 0 0 ... 0 0 0 0 ... 0
which is of rank 1, so the rank of g(Ai) is at most 1, and the rank of g(A) is atmost 1.
Solution to 7.630: A direct calculation shows that (A — J)3 = 0 and this is theleast positive exponent for which this is true. Hence, the minimal polynomial of Ais pA(t) = (t — Thus, its characteristic polynomial must be XA(t) = (t — 1)6.
Therefore, the Jordan Canonical Form [HK61, p. 247] of A must contain one3 x 3 Jordan block associated with 1. The number of blocks is the dimension ofthe eigenspace associated with 1. Letting x = (xi, . .. , xe)' and solving Ax = x,we get the two equationsxi = Oandx2 +x3 +X4 +x5 = 0. Sincex6 is notdetermined, these give four degrees of freedom, so the eigenspace has dimension4. Therefore, the Jordan Canonical Form of A must contain four Jordan blocksand so it must be 11000001100000100000010000001,0000001Solution to 7.631: Since AB has rank 3, both A and B must also have rank3, from which it follows that BA has rank 3. Hence the null space of BA hasdimension 2; let v1, v2 form a basis for it.
Let eI, be the standard basis vectors for C We have
ABei=ei, ABe2=eI+e2, ABe3=—e3,
so
BABe1 = Bej, BABe2 =
B has rank 3, no nontrivial linear combination of Bei, Be1 + Be2, —Be3can vanish, implying that Be1, Be2, Be3 are linearly independent modulo the nullspace of BA. Hence v2, Be1, Be2, Be3 form a basis for C Relative to hisbasis, the transformation induced by BA has the matrix
0000 00000 00011 00001 0000 0—1
which is the Jordan form of BA.
Solution to 7633: The diagonal element of AAt is and the same ele-ment on A'A is Comparing these expressions successively for
i = nn — l,...l, we conclude that all with j > i are zero, that is, A isdiagonal.
Solution to 7.6.38: Since A is nonsingular, AA is positive definite. LetB = Consider P = BA—1. Then PA = B, so it suffices to show that Pis orthogonal, for in that case, Q = P' = will be orthogonal and A = QB.We have
ptp (A'Y'B'BA' = (A)'B2A' = (A'Y'A'AA3 =1.
Suppose that we had a second factorization A = Qi B3. Then
B2 = AA = = B?.
Since a positive matrix has a unique positive square root, it follows that B = B1.As A is invertible, B is invertible, and canceling gives Q Qi.
Solution to 7.6.39: Conjugating A changes neither the convergence nor the eigen-values, so we may assume A in Jordan canonical form, A = or A = j.In the first case A" = and > A" converges when the eigenvalues, a and b,have absolute value less than 1, since the entries of the sum are geometric series.
In the second case write A = aI+N, with N the nilpotent matrix N = i,). Inthis case N2 =0, and A" = d'I+nd' N. If I+A+A2 +•• converges, then thediagonal entries, a", of the terms A" must converge to 0, so lal < 1. Conversely,if lal < 1, then >2a" and >na"' converge, by the Ratio Test [Rud87, p. 66],therefore, A" converges.
Solution to 7.6.40: An easy calculation shows that A has eigenvalues 0, 1, and 3,so A is similar to the diagonal matrix with entries 0, 1, and 3. Since clearly theproblem does not change when A is replaced by a similar matrix, we may replaceA by that diagonal matrix. Then the condition on a is that each of the sequences(0"), (a"), and ((3aY') has a limit, and that at least one of these limits is nonzero.This occurs if and only if a = 1/3.
Solution to 7.6.41: Let g be an element of the group. Consider the Jordan Canon-ical Form [HK61, p. 247] of the matrix g in a quadratic extension ofThe characteristic polynomial has degree 2 and is either irreducible in F,, andthe canonical form is diagonal with two conjugate entries in the extension or re-ducible with the Jordan Canonical Form having the same diagonal elements and a
I in the upper right-hand corner. In the first case, we can see that gP21 = I andin the second gP(P1) = I.
7.7 Similarity
Solution to 7.7.1: A simple calculation shows that A and B have the same char-acteristic polynomial, namely (x — 1)2(x — 2). However,
/0 oo\ /010A—I= —1 0 1 , B—1= 10 0 0
0 iJ \0 0 1
Since A — I has rank 1 and B — I has rank 2, these two matrices are not similar,and therefore, neither are A and B.
Solution to 7.7.3: A calculation gives
det(B—zI)=z2(z—1)(z+1)=det(A—zI).
The matrix A is in Jordan Canonical Form, and there are only two matrices withthe same characteristic polynomial as A, namely A and the diagonal matrix withdiagonal entries 1, —1, 0,0. Since B obviously has rank 3, its Jordan form mustbe A, i.e., B and A are similar.
Solution to 7.73: The eigenvalues of A an B are either ±1 and neither is Ior —I, since the equation AB + BA = 0 would force the other matrix to bezero. Therefore, A and B have distinct eigenvalues and are both diagonalizable.Let S be such that SAS' = 01). Multiplying on the left by S and on theright by S1 the relations above we see that C = SBS' satisfies C2 = I and
+(SBS')(SAS') = 0. We get
I/c)for
and taking D = we can easily see that T = DS satisfies
TBT'=(?
Solution to 7.7.7: Let x (x) = x2 — bx + d be the characteristic polynomial of A.Since d is the determinant of A, d" = 1, but the only roots of unity in Q are ±1sod = ±1. Let a and be the two complex roots oft. Over C A is similar to amatrix M of the form
Since the roots of the characteristic polynomial of A" are the the nth powers ofthe roots of it follows that c? = = 1. Since ±1 are the only real roots ofunity, if a is real, we get that a = = ±1 so in this case, since A is similar
toM, A2 = I.Now supposea is notreal. This means b2 —4d <0 andI and b = 0 or b = ±1. If b = 0, a and are fourth roots of unity so
M4 = I andtbus A4 = I. Ifb = ±l,a aresixtbrootsofunityso M6 andA6 equal I. Thus in all cases A12 = 1.
Solution 2. Alternatively, we can analyze the characteristic polynomial in the fol-lowing way: x(x) = — bx + d where d = det A = ±1 and b traceAis an integer. Since A has finite order, its eigenvalues are roots of unity, so havemodulus 1. Since b is the sum of the eigenvalues, Ibi 2.
Therefore the characteristic polynomial of A is one of x2 ± 1, x2 ± x + 1,x2 ± x — I, x2 ± 2x + 1, x2 ± 2x — 1. Note that x2 ± I divides x4 — 1, andx2 ± x + 1 divides x6 — 1, so any matrix with one of these polynomials has orderdividing 12. The zeros of x2 ± x — I and x2 ± 2x — 1 do not have modulus 1, sothese polynomials cannot occur.
A matrix A with polynomial x2 ±2x + 1 has repeated eigenvalues ± 1. Recallingthat A'2 = 1 for some n > 0, A is diagonalizable (for example, since it satisfies apolynomial equation with distinct linear factors), so A = ±1.
Solution 3. Since A'2 — I = 0 the minimal polynomial of A is a divisor of— 1. The irreducible factorization of x'2 — 1 is given by
X I flmm
where 4m = fl (x — &,), is the m-th cyclotomical polynomial, product of all
the factors (x — Wj) where are the primitive m-th roots of unity.=
I 1 r ( m, (r, m) = 1) and the numbers of elements isgiven by ç9(m), the Euler's totient function. Thus the irreducible factorizations of
!-4A and XA over Q have the form
mA = XA =
where r 1, mi, ... , m,. are distinct and divide n, and d1 1. It follows thatdiço(mi)+...+drç'(mr)=2.
There are two cases: solving ço(m 1) + (mz) = 2 gives (ml, m2) = (1, 1),
(1,2), (2,2) and solving ço(m) =2 gives m = 3,4,6.Take the case (1, 1), that is the minimal and characteristic polynomial of A are
given by
By the Cyclic Decomposition Theorem A "h- e a direct sum ofcompanion matrices.
The other cases are similar, that is, A is similar over Q to one of the six matrices
C(11) C(c11), C(42), (J) C(42), C(c13), C(c14), C(cló)
which are(1 o\ (1 0•'\ (—1 o\
—i)'
(0 —'\ (0 —i\ (0 —1\1)'
and these have orders 1, 2, 2, 3, 4, 6 respectively. (the order of is m). Inall cases A'2 = 1.
Solution to 7.7.8: 1. Let A be any element of the group:
(1) Every element in a finite group has finite order, so there is an n > 0 suchthat A" = I. Therefore, (det A)" = det(A") = 1. But A is an integermatrix, so det A must be ± 1.
(ii) If A is an eigenvalue of A, then A" = I, so each eigenvalue has modulo 1,and at the same time, A is a root of a second degree monic characteristicpolynomial XA (x) = x2 + ax + b for A. If = 1 then b = ±1 anda = 0, ±1, and ±2 since all roots are in the unit circle. Writing out all 10polynomials and eliminating the ones whose roots are not in the unit circle,we are left withx2 ± 1,x2 ±x + 1, andx2 ±2x + 1, and the possible rootsare A = ±1, ±i, and and the sixth roots of unity.
(iii) The Jordan Canonical Form [HK61, p. 247] of A, JA, must be diagonal,otherwise it would be of the form JA = and the subsequent powers
(JAYC which is never the identity matrix since 0(remember lxi = 1). So the Jordan Canonical Form of A is diagonal, withthe root above and the complex roots occurring in conjugate pairs only.
The Rational Canonical Form [HK61, p. 238] can be read off from thepossible polynomials.
(iv) A can only have order 1,2,3,4, or 6, depending on A.
Solution to 7.7.10: Let RA and RB be the Rational Canonical Forms [HK6I, p.238] of A and B, respectively, over R; that is, there are real invertible matrices Kand L such that
RA = KAK'R8 = LBL'.
Observe now that RA and RB are also the Rational Canonical Forms over C aswell, and by the uniqueness of the canonical form, they must be the same matrices.If = LBL' then A = , so K3L is a real matrixdefining the similarity over R. Observe that the proof works for any subfield; inparticular, two rational matrices that are similar over IR are similar over Q.
Solution 2. Let U K + iL where K and L are real and L 0 (otherwise weare done). Take real and imaginary parts of
A(K + iL) = AU = UB = (K + iL)B
and add them together after multiplying the imaginary part by z to get
A(K +zL)= (K +zL)B
for any complex z. Let p(z) = det(K + zL). Since p is a polynomial of degree n,not identically zero (p(i) 0), it has, at most, n roots. For real zo not one of theroots of p, V = K + is real and invertible and A = VBV'.
Solution to 7.7.11: The minimal polynomial of A divides (x — so I — A isnilpotent, say of order r. Thus, A is invertible with
A1
Suppose first that A is just a single Jordan block [HK61, p. 247], say with matrix
110••. 0011... 0
relative to the basis (vi, V2, ..., va). Then A—' has the same matrix relative to thearesimilar.
In the general case, by the theory of Jordan Canonical Form, the vector spacecan be written as a direct sum of A—invariant subspaces on each of which A actsas a single Jordan block. By the formula above for A', each subspace in thedecomposition is A
of A1. The general case thus reduces to the case where A is a single Jordanblock.
Solution to 7.7.12: The statement is true. First of all, A is similar to a Jordanmatrix [HK61, p. 247], A = S1JS, where S is invertible and J is a direct sumof Jordan blocks. Then A' = SJt(St)_l (since = that is, Nis similar to J'. Moreover, P is the direct sum of the transposes of the Jordanblocks whose direct sum is J. It will, thus, suffice to prove that each of theseJordan blocks is similar to its transpose. In other words, it will suffice to prove thestatement for the case where A is a Jordan block.
Let A be an n x n Jordan block:
X 1 0 ••• 0
A= OX!••. 0
0 0 ... A 1
Let Cl, . .. , be the standard basis vectors for C so that = + ej_iforj > 1 and Ae1 = Aej. Let the matrix S be defined by Se1 = Then
S = and
=
— J +en_j), f <fl
— 1 i = n
Ji<n
lAej, j=n
which shows that S1AS A'.
Solution to 7.7.14: Using the first condition the Jordan Canonical Form [HK61,
p. 247] of this matrix is a 6 x 6 matrix with five 1 's and one —1 on the diagonal.The blocks corresponding to the eigenvalue 1 are either 1 x 1 or 2 x 2, by thesecond condition, with at least one of them having dimension 2. Thus, there couldbe three 1-blocks and one 2-block (for the eigenvalue 1), or one 1-block and two2-blocks. In this way, we get the following two possibilities for the Jordan Form
of the matrix:
10000 0 10000 001000 0 01100 000100 0 00100 000011 0 00011 000001 0 00001 000000—1 00000—1
Solution to 7.7.15: Since A and B have the same characteristic polynomial, theyhave the same n distinct eigenvalues ... , Let X (x) = (x —ii )C1 (xbe the characteristic polynomial and let p(x) = (x — • • (x — be theminimal polynomial. Since a nondiagonal Jordan block [HK61, p. 247] must beat least 2 x 2, there can be, at most, one nondiagonal Jordan block for N 3.Hence, the Jordan Canonical Form is completely determined by /h(x) and x (x)for N 3. If = x (x), then each distinct eigenvalue corresponds to a singleJordan block of size equal to the multiplicity of the eigenvalue as a root of x (x),so the Jordan Canonical Form is completely determined by x (x), and A and Bmust then be similar.
Solution to 7.7.16: It will suffice to show that A is similar to a matrix of the formwhere B is (n — 1) x (ii — 1). For in that case B has the same trace as A,
hence trace 0, and the argument can be iterated.Assume, without loss of generality, that A is not the zero matrix. Since A has
trace 0, it is not a scalar multiple of the identity matrix. Hence, there is a nonzerovector v in C that is not an eigenvector of A. Take a basis for C in which thefirst basis vector is v and the second one is Av. The matrix with respect to such abasis of the linear transformation induced by A is similar to A and has the requiredform.
Solution to 7.7.17: Since M3 = ti, N3 = 0, so it suffices to show that N2 0.If N2 =0, then M2 = iF for some 3 x 3 matrix P with entries in R[r}. ThereforetPM = M3 = ii, hence PM = 1, implying that ti = M3 has an inverse P3with entries in IRE,'], a contradiction.
Solution to 7.7.19: Since the map X i-÷ X2 on is continuous, we have
2 /B = I lim A'2 I = lim = B.
\n-+oo / n—oo
The minimal polynomial of B divides x2 — x, so the eigenvalues of B are zerosand ones. Since the minimal polynomial is squarefree, the Jordan blocks of B areof size 1, i.e., B is similar to a diagonal matrix with its eigenvalues (zeros andones) along the main diagonal.
7.8 Bilinear, Quadratic Forms, and Inner ProductSpaces
Solutionto7.8.1: For allx E g(x, x) = f(2x)—2f(x) = 4f(x)—2f(x) =2f(x). Therefore f(x) = g(x, x)/2 is the quadratic form associated with thebilinear form g. For each x E IR", y i-÷ g(x, y) is linear, so there is exactly onevector A(x) E IR" such that g(x, y) = (A(x), y) for all y E Since g is linearin the first variable, A is a linear map.
Solution to 7.8.2: Let H be the Hermitian matrix that induces the quadratic form
/ABD\H=(B C El,\DEFJ
and let
G(AB
H is positive definite if and only if all of its eigenvalues are positive. Since theproduct of those eigenvalues is det H, thepositivityofdet H is a necessary condi-tion for H to be positive definite. Since the positive definiteness of H implies thepositive definiteness of G, whose associated quadratic form is Ax2 +2Bxy+Cy2,the positivity of det G is also a necessary condition. The positivity of A is simi-larly a necessary condition.
Assume that A > 0, det G > 0, and det H > 0, but that H is not positive def-inite. Then the product of the eigenvalues of H (= det H) is positive, but not allof the eigenvalues are positive. Hence H must have one positive eigenvalue andtwo negative ones, or a single negative one of multiplicity two. It follows that His negative definite on the two dimensional subspace spanned by the eigenvectors
corresponding to the negative eigenvalues. That subspace must have a nontrivialintersection with the subspace spanned by the first two basis vectors. Hence thereis a nonzero vector v in JR3 with its last coordinate 0 such that v' Hv <0. It fol-lows that G is not positive definite. But since A > 0 neither is G negative definite.Thus G must have one positive and one negative eigenvalue, in contradiction tothe assumption det G > 0.
Solution to 7,8.3: We have
/2 t 1\/xi+4+34+ 2tXIX2 + 2XjX3 = X3) t 1 0
\1 0 3J\x3
By the Solution to Problem 7.8.2 above, the form is positive definite if and onlyif the determinants
2t1 2tzlO>0 and 1>0103 t
thatis,when —1 +3(2—t2)=5—3t2 >Oand2—t2 >0.Both conditions hold if (tJ < For these values of t the form is positive
definite.
Solution to 7.8.4: Every vector in W is orthogonal to v = (a, b, c). Let Q bethe orthogonal projection of JR3 onto the space spanned by v, identified with itsmatrix. The columns of Q are I j 3, where the 's are the standardbasis vectors in JR3. But
Qej = (v, ej)v = (az, ab, ac)
Qe2 (v, e2)v = (ab, b2, bc)
Qe3 = (v, e3)v = (ac, bc, c2).
Therefore, the orthogonal projection onto W is given by
/1—a2 —abP_—J—Q= —ab 1—b2 —bc
—be I—c2
Solution to 7.83: 1. The monomials 1, t, t2, ..., form a basis for Ap-plying the Gram—Schmidt Procedure [HK6I, p. 280] to this basis gives us anorthonormal basis P0, i'i,..., The (k + vector in the latter basis, isalinear combination of 1, t, . • , the first k + 1 vectors in the former basis, witht" having a nonzero coefficient. (This is built into the Gram—Schmidt Procedure.)Hence, deg Pk = k.
2. Since has degree k — 1, it is a linear combination of ..., Pk— i, for
those functions form an orthonormal basis for Pk— 1. Since Pk is orthogonal toP0, it is orthogonal to
p is even, it is orthogonal on [—1, 1] to all odd polyno-mials. Hence a and b are determined by the two conditions
(I Illp(x)dx=O, j x2p(x)dx=O.
J—1 J—.J
Carrying out the integrations, one obtains the equations
2b 2 2a 2b 2
Solving these for a and b, one gets a = — b = therefore
3
7—x
Solution to 7.8.7: Let n = dim E, and choose a basis vi, ... , for E. Definethen x n matrix A = by ajk B(vk, The linear transformation TA onE induced by A is determined by the relations
TA v = >J B(v, (v E E). It follows that E1 = ker TA. Bysimilar reasoning, E2 = ker TA', where A' is the transpose of A. By the Rank—Nullity Theorem [HK6I, p. 71], dim E1 equals n minus the dimension of thecolumn space of A, and dim E2 equals n minus the dimension of the row space ofA. Since the row space and the column space of a matrix have the same dimension,the desired equality follows.
Solution to 7.8.8: We have, for any x E
(Ax,x) =(x,Ax) = (Atx,x)
hence, ((A + At)X, x) 0 for all x E R'1. As A + A' is symmetric, thereforediagonalizable, there exist an integer k n, positive numbers Xj, A2, Ak, and
a basis of W, {v1, ..., vj, such that (A + AT)v1 = A,v1 for I i k, and(A + AT)v, = 0 fork + 1 i n. The matrix representing A in this basis hasthe form
CD
where D is antisymmetric and
1x1 •.•: :
where B' is antisymmetric. We have
so x ker A' exactly when the first k components of x are zero and the last n — kform a vector in ker C fl ker D. As
IAIV)_(B c\' D)'
we conclude that a vector is in the kernel of A' if it is in the kernel of (AF)t.
Solution to 7.8.9: 1. Since A is positive definite, one can define a new innerproduct ()A on R'1 by
(x,y)A =(AX,y).
The linear operator A1 B is symmetric with respect to this inner product, that is,
(A1Bx, Y)A = (Bx, y) = (x, By) (x, By)
=(A1Ax,By) = (Ax,A'By) =(X,A'BY)A.
So there is a basis {vi, of R'1, orthonormal with respect to in whichthe matrix for A' B is diagonal. This is the basis we are looking for; in particular,
is an eigenvector for A' B, with eigenvalue A1 and
(vi, öIj
(By1, = Vj)A (A1v1, vj)A =
2. Let U be the matrix which takes the standard basis to (vi, above, that is,= v,. Since the e1 form an orthonormal basis, for any matrix M, we have
Mx =
in particular,
U'AUe, >(UtAUeI,
=
=
= >Sjiej = e1,
showing that UAU = 1.
Using the same decomposition for U' BU, we have
=>2(U'BUej,ej)ei
=
= >JAiöjjej =
so U'BU is diagonal.
Solution 2. Since A is positive definite A = W' W for some invertible matrix W.Setting N = W1 we get N'AN = W' WW1 = 1 and N'BN is stillpositive definite since (x, N' BNx) = (Nx, B(Nx)) 0. As N' BN is positivedefinite there is an orthogonal matrix 0 such that Ot(Nt BN)O = D, where D isa diagonal matrix. The entries of D are positive, since N'BN is positive definite.Let M = NO. Then M'BM = O'N'BNO = D and M'AM = O'N'ANO =0110 = 1, and we can easily see that M is invertible, since N and 0 are.
Solution to 7.8.11: From the Criteria proved in the Solution to Problem 7.8.2 wesee that the matrix is positive definite and V'AV = N(v) defines a norm in iiV.Now all norms in R3 are equivalent, see the Solution to Problem 2.1.6, so
aN(v) liv Ii j5N(v)
where a and are the minimum and the maximum of liv Ii on the set v' Av = 1.
We can use the Method of Lagrange Multipliers [MH93, p. 414] to find themaximum of the function
f (x, y, z) = + +
over the surface defined by y, z) = 6, where
= (13x2 + 13y2 + 10z2 — lOxy — 4xz — 4yz)
is the quadratic form defined by the matrix A. Setting up the equations we have:
=26x—IOy—4z=A-11 =2Axdx
=2Ayay
— =20z—4x---4y=A-— =2Azaz az
which is equivalent to the linear system of equations
/13—A —5 —2 \ fx\—5 13—A —2 IIyI=°
\ —2 —2 lO—A/ \zJtogether with the equation y, z) = 6. Now if the determinant of the systemis nonzero, there is only one solution to the system, the trivial one x = y = z = 0and that point is not in the surface (x, y, z) = 6, so let's consider the case wherethe determinant is zero. One can easily compute it and it is:
—A3 +36A2—396A+ 1296
and one can easily see that A = 18 is a root of the polynomial, because it rendersthe first two rows of the matrix the same, factoring it completely we get:
(18—A) (A—6) (A— 12)
Considering each one of the roots that will correspond to non-trivial solutions ofthe system of linear equations
• A = 18. In this case the system becomes:
f—5 —5 —2\ /x\—5 —211y1=0
\—2 —2 —8) \zJ
which reduces to { = g . Therefore z = 0 and y = -x.Substituting this back in the equation of the ellipsoid, we find x =and the solution points are:
—1,0)
• A =6. The system now is
17—5 —2\/x\7 —21(yI=O
\—2 —2 4/ \z/
and the solutions are: x y = z and the only points in this line on theellipsoid are:
1,1)
• A = 12. The system now is
fi —5 —2\fx\1 —211y1=0
\—2 —2 —2/ \z/and the solutions are: y — x and z — —2x the points in this line on theellipsoid are:
Computing the sizes of each (pair) of the vectors we obtain 1 andrespectively; so the least upper bound is 1.
Solution 2. A, being a symmetric matrix, can be diagonalized by an orthogonalmatrix S, so
D=SASwhere the matrix S is given by the eigenvalues of A. Following the computationof the previous solution, we get
fs=(0
but in our case we only need the matrix D which is the diagonal matrix with theeigenvalues of A, that is,
/18 0 0D=I0 60
\0 0 12
so after the change of basis given by the matrix S above, the form becomes
s, t) = 18r2 + 682 + 12t2 and on these variables the ellipsoid
21/3+8 + 1/2=1
has semi-axis 1 and and the least upper bound is 1.
Solution 3. Rotating around the z-axis by 450 on the positive direction is equiva-lent to the change of variables
x =
y=
which substituted in the equation of the ellipsoid
eliminates the mixed terms in xy and xz. After the substitution we end up with
18r2-F8s2+
1
Now we rotate around the r-axis by an angle a where cot 2a = =in order to eliminate the mixed term in sz. Using trigonometric identities we find
cos2a = = =
be written as
1/1-s=wcosa—tsina=—(wV2—t
iiz=wSifla+tCoSa =
and the final substitution renders the equation of the ellipsoid
18r2+6w2-I- 12t2= 6
whose largest semi-axis is 1.
Solution to 7.8.12: Suppose that A has a diagonalization with p strictly positiveand q strictly negative entries, and that B has one with p' strictly positive and q'strictly negative entries. Then contains a subspace V of dimension p such thatxtAx > 0 for all nonzero x E V, and a subspace W of dimension n — p' such thatXtBX 0 for all x E W. Since xtAx xBx for all x E R", x'Ax 0 for all
x E W. Therefore V fl W = (0), so dim V + dim W n, i.e., p + (ii — p') n
giving p p'. Similarly q' q, so p — q p' — q'.
Solution to 7.8.13: Let A = T*T. Then (x, y) =0 implies (Ax, y) =0. For anyx E C" let x1 = (y E C" I (x, y) = 0). Thus, (x, x1) = 0 so (Ax, x1) = 0,
implying that Ax E (x-'-)1, so Ax = Ax for some A E C. Since every vectoris an eigenvector of A, it follows that rI for some scalar r. The constant r is anonnegative real number since A is positive semidefinite. If r = 0 then A = 0,hence T = 0 and we may take k = 0, U = I. If r > 0 we take k = and setU = *T. Clearly k is real, and U is unitary because
_T*T_= -A=I.
MT =kU wearedone.
Solution to 7.8.15: Assume that such u, v exist. Then there is a 3 x 3 orthogonalmatrix Q whose three rows are (ui, U2, a), (Vi, v2, b), (Wi, W2, c). Every columnof Q is a unit vector. This implies that a2 + b2 1.
Conversely, assume that a2 + b2 I. Let c be a real number such that+ b2 + c2 = 1. The vector (a, b, c) can be extended to a basis of JR3 and by
the Gram—Schmidt Procedure there exist vectors (ui, vi, WI), (u2, V2, w2) thattogether with (a, b, c) form an orthonormal basis. Thus the matrix M given by
/ Ui U2 a
lvi\ WI W2 C
is an orthogonal, that is, M'M = I. Since a square matrix is a left-inverse if andonly if is a right-inverse we also have that MM' = I, and the first two rows ofthis matrix are the desired vectors.
Solution 2. Suppose we have such u and v. By Cauchy—Schwarz Inequality[MH93, p. 69], we have
(Ui
v = 0, (uivi +u2V2)2 = (ab)2; since huh = liv II = 1, 1—a2 =and 1 — b2 = + Combining these, we get
(ab)2 (I —a2)(l —b2)= I —a2—b2-F(ab)2,
which implies a2 + b2 1.
_____
Conversely, suppose that a2 + b2 1. Let u = (0, li — a2, a). IIu fl = I, andwe now find vi and v2 such that + + b2 = 1 and u2v2 + ab 0. If a = I,then b = 0, so we can take v = (0, 1,0). If a 1, solving the second equationfor v2, we get
—ab
%fl—a2Using this to solve for vi, we get
4/1 —a2—b2vi=
By our condition on a and b, both of these are real, so u and v = (vi, V2, b) are
the desired vectors.
7.9 General Theory of Matrices
Solution to 7.9.1: Let A be such an algebra, and assume A does not consist ofjust the scalar multiples of the identity. Take a matrix M in A that is not a scalar
multiple of the identity. We can assume without loss of generality that M is inJordan form, so there are two possibilities:
(i) M=
with A1 A2 (ii) M=
In the first case the matrices commuting with M are the 2x2 diagonal matrices,which form an algebra of dimension 2. Thus dim A 2.
In the second case M has a one-dimensional eigenspace spanned by the vector
(i), so any matrix commuting with M has as an eigenvector, i.e., it is uppertriangular. Also, a matrix commuting with M cannot have two distinct eigenval-ues, because then the corresponding eigenvectors would have to be eigenvectorsof M (and the only eigenvectors of M are the multiples of (i)). Thus, the matricescommuting with M are the matrices
(/Lv
The algebra of alJ such matrices has dimension 2, so dim A 2 in this case also.
Solution to 7.9.2: In what follows we will use frequently the facts that I +A is in-vertible, which can easily be concluded from the fact that —1 is not an eigenvalueof A, as we have ker(A + I) 0, as well as that A commutes with (I +which can be seen by factoring A+A2 in two different ways (I+A)A = A(I+A)and now multiplying on the right and left by (I + A)'.1.
((I — A)(I + A)1) = ((I +A)') (I — A)t
— At)
= (I + At)—' (i — At) since At = A'=(1+A—')' (i_A—i)
= (A—' (A + I))' (A—' (A — I))(A—I)
== — (I — A) (I +
on the last equality we used the fact that A commutes with (I + A)' and thisshows that the product is skew-symmetric.2. Now let's suppose S is skew-symmetric, then
(i—si+s-')' = ((I +S)_1)(I_S)t
=
= (I—'S)'(J+S)=
= ((1 — S) (1+ S)—I)
that is the product is orthogonal. Now to see that it doesn't have an eigenvalue 1,observe that
det((I —S)(I+S)' +i) = det((I—S+I +S)(I+Si')
= det (21 (I + si'):2hldet((J +5)_I)
3. Suppose the correspondence is not one-to-one, then
(I—A)(I+AY' =(I—B)(I+B)'
(I + A) (I —
A) = (I B) (I +
(1 —A) (l+B)=(I+A) (I—B)
simplifying we see that A = B.
Solution to 7.9.4: if x, y E IRTZ then xt Py is a scalar, so (x' Py)t = x' Py, thatis, yt Ptx Py. We conclude, from the hypothesis, that yt = —yt Pxfor all x, y. Therefore y'(P + P')x 0. Write A = P + P' = Then,aj = e = 0, where the e, 's are the standard basis vectors, proving thatA = 0, that is, P is skew-symmetric.
Solution to 7.9.5: We will use a powerful result on the structure of real normal op-erators, not commonly found in the literature. We provide also a second solution,not using the normal form, but which is inspired on it.
Lemma (Structure of Real Normal Operators): Given a normal operator A onthere exists an orthonormal basis in which the matrix of A has the form
(0.1 Ti
0.1
(cr2 r20.2
(crk Tk
ctk
A2k+1
An
where the numbers = + j = 1, ... , k and ... , are theeigenvalues of A.
The proof is obtained by embedding each component of IR'2 as the real slice ofeach component of C extending A to a normal operator on C and noticing thatthe new operator has the same real matrix (on the same basis) and over Cn hasbasis of eigenvectors. A change of basis, picking the new vectors as the real andimaginary parts of the eigenvectors associated with the imaginary eigenvalues,reduces it to the desired form. For details on the proof we refer the reader to[Shi77, p. 265-271] or [HS74, p. 117].
The matrix of an anti-symmetric operator A has the property
= =(ej,A*ej) =
Since anti-symmetric operators are normal, they have a basis of eigenvalues, thesesatisfy the above equality and are all pure imaginary.
Thus, in the standard decomposition described above all eigenvalues are pureimaginary, i.e., ai = ... = A2k÷] = = = 0 and the decompositionin this case is
(0 Ti
v—TI 0(0 T2
\—r2 0
(0 Tk
00
0
which obviously has even rank.
Solution 2. Consider A the complex ifi cation of A, that is, the linear operator fromC to C with the same matrix as A with respect to the standard basis. Since Ais skew-symmetric, all its eigenvalues are pure imaginary and from the fact thatthe characteristic polynomial has real coefficients, the non-real eigenvalues showup in conjugate pairs, therefore, the polynomial has the form
XA (t) =
where the p1 's are real, irreducible quadratics.From the diagonal form of A over C we can see that the minimal polynomial
has the factor in t with power 1, that is, of the form
/hA(t) = P.A(t) = tpi(t)m' .
Now consider the Rational Canonical Form [HK6I, p. 238] of A. It is a blockdiagonal matrix composed of blocks of even size and full rank, together with ablock of a zero matrix corresponding to the zero eigenvalues, showing that A haseven rank.
Solution to 7.9.6: Since A is real and symmetric, there is an orthogonal ma-trix U such that D = U'AU is diagonal, say with entries .. . , LetE = BU, then tr AB = tr DE = EL1 A and B are both posi-tive semi-definite, then so are D and E, therefore A, 0 and 0, givingtrAB
If A is not positive definite then A <0 for some i. Let E be a diagonal matrixwith e-1 = 1 and all other entries zero. Then B = U'EU is positive semi-definite and trAB <0.
Solution to 7.9.7: Since A is symmetric it can be diagonalized: Let
A = QDQ1
where D = diag(di, ... , 4,) and each d1 is nonnegative. Then
0 = Q'(AB + BA)Q = DC + CD
where C = Q'BQ. Individual entries of this equation read
0=(d,
so for each i and j we must have either cjj = 0 or d, = = 0. In either case,
=0
which is the same asDC = CD = 0.
Hence, AB = BA =0.
Example:11 o\ 0
A is symmetric, it is diagonalizable. Let v be an eigenvector of
A withAv =Av,then
A(Bv) = —BAy = —ABv
that is, By is an eigenvector of A with eigenvalue —).
Using one of the conditions we get (A By, By) 0 but on the other hand(A By, By) = —A(Bv, By) 0, so either X =0 or By =0. Writing A and B onthis basis, that diagonalizes A, ordered with the zero eigenvalues in a first block
we have
AkJ
B=
which implies that AB =0 and similarly that BA =0.
Solution to 7.9.8: Assuming first that A are B are invertible, we have
A(A + B)1B = += (B'(A += +
and the same reasoning shows that B(A + B)—' A = (B—' +> OthematricesA+XIandB—AJwillbe
invertible for 0 < IAI <cs. Then by what has already been established,
(A +XI)(A + BY'(B — )J) = (B — A1)(A + BY'(A +
Taking the limit as -÷ 0, we get the desired equality.
Solution to 7.9.9: 1. No, for example take A = ?).
A form an orthonormal set then A' A = I, transposingthe product AA' = (At A) = 1 and now the rows form an orthonormal set.
Solution to 7.9.10: The characteristic polynomial of M1 is z2 — 7z + 10, withroots 5, 2. Hence M1 is similar to so is bounded away from 0 butis not bounded.
The characteristic polynomial of M2 is z2 — z + 1, with roots (i ± /2,both of unit modulus. Hence M2 is similar to a unitary matrix, and isboth bounded and bounded away from 0.
The characteristic polynomial of M3 is z2 — z + 0.7, with roots (1 ± i4/ii)/2,each of modulus less than 1. Hence M2 is similar to a diagonal matrix with eachdiagonal entry of modulus less than 1, implying that is bounded but notbounded away from 0.
Solution to 7.9.13: 1. The matrix A—I has the property that the sum of the entriesin each column is equal to 0. The row operation R1 —+ + • + applied toA — I results in a matrix whose first row is zero. Hence IA — fl = 0. Thus 1 is aneigenvalue of A and there is an eigenvector x 0 such that Ax = x.2. Let A = with a, b, c, d > 0. The characteristic polynomial of A is
— (a + d)A + (ad — bc). The discriminant is (a + d)2 — 4(ad — bc) =(a — d)2 + 4bc> 0. Therefore, A has two distinct real eigenvalues A1, A2. NowAj + A2 = a + d > 0. Thus A has at least one positive eigenvalue A, and there isan eigenvector y 0 such that Ay = Ày.
Solution to 7.914: Let Y = AD — BC. We have
(A B\(D —B\ (AD—BC —AB+BA\ fY 0D) v—C A ) — DC —CB+DA) =
If Y is invertible, then so are and X.Assume now that X is invertible, and let v be vector in the kernel of
Y : (AD — BC)v 0. Then
(A B\(Dv'\_0_(A B\(—Bv
implying that Dv = Cv = By = Av = 0. But then = 0, so, by theinvertibility of X, v = 0, proving that Y is invertible.
Solution to 7.9.15: Let A E GL2(C) be a matrix representing a. Then A?Z = A!for some A E Ca". We may assume, without loss of generality, that = I. Sincethe polynomial x" — 1 has distinct roots, A is diagonalizable, and its eigenvaluesmust be the n-roots of 1. Now conjugating and dividing by the first root of 1,we may assume that A = (u). If for some m 1, = si withs Ci',then, comparing upper left hand corners we see that s = 1. Since the order of ais exactly n, the previous sentence implies that A has order exactly n, so is aprimitive n-root of 1.
In the same way we may represent b by a matrix that is conjugate to B =for some primitive n-root of 1, Then is a power of so B is a power A, andconsequently b is conjugate to a power of a.
Solution to 7.9.16: If det B 0, then det B 0 (mod 2). Hence, if we canshow that det B 0 over the field Z2, we are done. In the field Z2, I = —I, so
B is equal to the matrix with zeros along the diagonal and 1 's everywhere else.Since adding one row of a matrix to another row does not change the determinant,
we can replace B by the matrix obtained by adding the first nineteen rows of B tothe last row. Since each column of B contains exactly nineteen 1 's, B becomes
(0 1 1
Ii 0 1
iii111
•.. 1 1
... 1 1
•.. 0 1
•.. 1 1 IBy adding the last row of B to each of the other rows, B becomes
100010
11
•.. 0 0•.. 0 0
..: 1 0•.. 1 1
This is a lower-triangular matrix, so its determinant is the product of its diagonalelements. Hence, the determinant of B is equal to 1 over 7Z2, and we are done.
Solution 2. In the matrix modulo 2, the sum of all columns except column i is thestandard basis vector, so the span of the columns has dimension 20, and the
matrix is nonsingular.
Solution to 7.9.17: Let C be the set of real matrices that commute with It isclearly a vector space of dimension, at most, 4. The set {sI + tA I s, t R} is atwo-dimensional subspace of C, so it suffices to show that there are two linearlyindependent matrices which do not commute with A. A calculation show that thematrices and are such matrices.
Solution to 7.9.18: Let
if AX = XA, we have the three equations: bz = yc, ay + bwcx+dz=za+wc.
• b = c = 0. Since A is not a multiple of the identity, aequations reduce to ay = dy and dz = az, which, iny=z=0.Hence,
A_Ia bd and X=(X y
w
=xb+yd,and
d. The aboveturn, imply that
0" / a ")=Ix_ (x_w))I+( )A.
W a—d a—d
• b 0 or c 0. We can assume, without loss of generality, that b 0, asthe other case is identical. Then z = cy/b and w = x — y(a — d)/b. Hence,
— 1 (bx — ay + ay by (bx — ay\ Acy b )
Solution to 7.9.19: 1. Assume without loss of generality that X = 0 (otherwisereplace J by J — A!). Let ... , be the standard basis vectors for C sothat Jek = eu_i for k = 2, . .. , n, Jei = 0. Suppose AJ = JA. Then, fork = 1, . .., n — 1,
Aek = AJZ_ken = , (*)
(Cn \so A is completely determined by If = then
\cl /
Cj C3 ...o Cl C2 ... Cn—2
o o Cl ... Cn_2
o o 0 ... Cl C2
o o 0 ... 0
In other words, A has zero entries below the main diagonal, the entry Cl at eachposition on the main diagonal, the entry c2 at each position immediately above themain diagonal, the entry C3 at each position two slots above the main diagonal,etc. From (*) it follows that every matrix of this form commutes with J. The
commutant of J thus has dimension n.
2. Consider a 2n x 2n matrix in block form where A, B, C, D are n x n.Asimple computation shows that it commutes with J ® J if and only if A, B, C, Dall commute with J. The commutant of J , as a vector space, is thus the directsum of 4 vector spaces of dimension n, soit has dimension 4n.
Solution to 7.9.20: The characteristic polynomial
xA(x) = (2—x)((2—x)2 —2)
has distinct roots 2 — 2, 2 + therefore A is diagonalizable, and, as Bcommutes with A, B is diagonalizable over the same basis. Let T be the transfor-mation that realizes these diagonalizations,
TAT1 = diag (2,2+ 2— TBT1 = diag (a, y).
The linear system
always has a solution, as it is equivalent to the Vandermonde system
/1 2 22 \fa\ faIi\i \cJ
with nonzero determinant, see the Solution to Problem 7.2.11 or [HK6I, p. 125].Applying T1 on the left and T on the right to the above equation we get
B +bA+cA2.
Solution 2. The characteristic polynomial
XA(X) = (2—x)((2 _x)2 —2)
has distinct roots 2 — 2, 2 + that we will call A1, A2 and A3, and letA = diag(Ai, A2, A3), then if AB = BA and A = TAT' thenA(T1 BT) = (T1BT)A and since the are distinct, T1 BT must be di-agonal. Say T1 BT = diag(bi, b2, b3), choose a quadratic polynomial p suchthat p(A1) —b,,fori = 1,2,3,forexample
(A — ?.2)(A — A3) (A — Ai)(A — A3) (A — Ai)(A — A2)p(A) = b1
(A1 — A2)(A1 — A3) (A2 — Al)(A2 — A3)+b3
(A3 — A1)(A3 — A2)
Then p(A) = Tp(A)T' = Tdiag(bi, b2, b3)T' = B.
Solution 3. Letfafly
B=(8 e
Solving for AB = BA leads to a system of 9 equations in 9 unknows with asolution of the form
fa y
a
46
afi o o\ f2 —1 O\ /5 —4 1
a 0 1 0 +b( —1 2 —1 +c 1 —4 6 —4\ool) \o —12/ ki —45we obtain
a =a+2/3+3yb=—46---4y
c=y
thus,
B =(cr+2fl+3y)1+(—fi —4y)A+yA2.
Solution to 7.9.21: 1. Take A = and B = a multiplication shows thatA2 B2 =Obut (A -I-B)2=I,soA-l-Bis not nilpotent.2. If A is nhlpotent then neither ±1 are eigenvalues of A, otherwise I would beeigenvalue of A21' for all k, which is a contradiction, so ker(1 ± A) 0, that is,I ± A are invertible.3. Let kA and kB be exponents that take A and B into zero, respectively. Since Acommutes with B
kA+kB
(A +=
+ kB)
and one of the powers on the right is always zero, and so is the whole expression.
Solution to 7.9.23: Define the norm of a matrix X = (x,3) by I1X1I = Ej,, Ixi.jI.Notice that if Bk, k = 0, 1,..., are matrices such <oo, thenconverges, because the convergence of the norms clearly implies the absolute en-trywise convergence.
In our case, we have Bk = As'. The desired result follows from the fact thathA U/c/k! converges for any matrix A.
Solution to 7.9.24: Note that, if A is an invertible matrix, we have
ACM A' =
so we may assume that M is upper-triangular. Under this assumption eM is alsoupper-triangular, and if aj, . . . , are M's diagonal entries, then the diagonalentries of eM are
Dal'.
and we get
det (eM) = JJ = eEZ=1 a1 =
Solution to 7.9.26: It is easy to see that
(X, Y) = tr(XY*)
defines an inner product on the space of n x n complex matrices, in fact, it is the
standard inner product under the identification with . The inequality is thenthe Cauchy—Schwarz Inequality {MH93, p. 69] for the norm.
Solution to 7.9.27: 1. Let A and B be n x n real matrices and k a positive integer.
We have
(A + tB)k = Ak + BAk_l_i + 0(t2) (t —÷0)
where 0(t2) is composed of terms with a power oft higher than 1. Hence
urn ((A + tB)k — Ak) = AIBAk_11.t—÷Ot 1=0
2. We haved
(A + tB)i = urn! ((A + tB)k — A')dt :=0 I
Using the previous part of the problem we get
tr(A + tB)kI = tr
Solution to 7.9.28: 1. Let a = !tr(M), so that tr(M — a!) = 0. Let
A = S=
The desired conditions are then satisfied.2. We have
M2 =A2 +S2+a21+2aA+2aS+AS+SA.
The trace of M is the sum of the traces of the seven terms on the right. We havetr(A) = tr(B) =0. Also,
tr(AS) = tr((AS)') = tr (S'At) = tr(—SA) = —tr(SA),
so tr(AS + SA) 0 (in fact, tr(AS) 0 since tr(AS) = tr(SA)). The desiredequality now follows.
Solution to 7.9.29: We must show that 0 is the only eigenvalue of A. Suppose thecontrary. We may assume that A is in Jordan form, since the trace is preservedby similarity, in particular that A is upper-triangular. Let be the distinct
nonzero eigenvalues of A, each Xj repeated rn times in the diagonal of A. Thenthe nonzero entries in the diagonal of are ..., 4, repeated ml, times,respectively. We have
fork=1,...,n.
Therefore M(mi •• mp)t = 0 where the matrix M is
A1 A2 •.. A,,
A2 A2 A2
A!...
But1 1. ... 1
detM = Xi A2. . . det
1k ik"1 "p
which is nonzero since the X's are nonzero and the other factor is a Vandermondedeterminant. We got a contradiction, therefore A has no nonzero eigenvalues.
Solution to 7.9.30: Consider the function f defined on the Unit disc by1(z) = (1 + Z)11', and its Maclaurin expansion [MF187, p. 234]
1(z) = We have f(z)r = I + z. Let the matrix A be defined by
A =
a finite sum, because A is nilpotent. The computation of N involves the sameformal manipulations with power series as does the computation of f(z)'. If wereplace z by N everywhere in the latter computation, we arrive in the end at theequalityAT =I+N.
Solution to 7.9.31: We will prove the equivalent result that the kernel of A istrivial. Let x = (xl, ... , be a nonzero vector in ]R'7. We have
Axx= >Jajjxjxji,j=1
= +i=1
i=1 i#j
So (Ax, x) 0 and Ax 0, therefore, the kernel of A is trivial.
Solution to 7.932: Suppose x is in the kernel. Then
=
for each i. Let i be such that lxii maxk lxkl = M, say. Then
so
lauiM
Since the therm inside the parenthesis is strictly positive by assumption, we musthave M =0, so x = 0 and A is invertible.
Solution to 7.9.33: It will suffice to prove that ker(1 — A) is trivial.Let x = (xi, x2, .. ., )t be a nonzero vector in R", and let y = (I — A)x.Pick k such that jxk I = max{ lxi I). Then
lYki = Ixk — >Jakjxj I
) IxjIj=1
n
IXkI —>iakJi ixki
j=i
—
n
= IXkl (1 >0.
Hence, y 0, as desired.
Solution 2. Let a < 1 be a positive number such that, for all values of i, we haveI a. Then,
j,k j j
And so, inductively, the sum of the absolute values of the terms in one row of A"is bounded by a". Thus, the entries in the infinite sum
1+A+A2+A3+-•.
are all bounded by the geometric series 1 + a + a2 +•• , and so are absolutelyconvergent; thus, this sum exists and the product
(I—A)(I +A+A2+'•)_—I
is valid, so that the inverse of I — A is this infinite sum.
Solution to 7.9.34: 1. If A is symmetric then, by the Spectral Theorem [HK61,p. 335], [Str93, p. 235], there is an orthonormal basis {el, ... , for JR'S with
respect to which A is diagonal: = j = 1, . .. , n. Let x be any vector
Ax = Ai el + •• + Moreover,
11x112 = + •'• +IlAxU2
... , + . + = M211x 112,
which is the desired inequality.2. The matrix gives a counterexample with n = 2. Its only eigenvalue is 0,yet it is not the zero matrix.
Solution to 7.935: 1. Suppose As not an eigenvalue of A, so 0. Write
/2(z) = + (z — A)g(z)
with g a polynomial of degree k 1. Then
0= = j2(A)I + (A — AI)g(A),
so the polynomial = —g(z)//2(A) has the required property.
2. It will suffice to show that the polynomials ..., are linearly indepen-dent, for then they will form a basis for the vector space of polynomials of degree
k — 1, so some linear combination of them will equal the constant polyno-mial 1. Suppose al, ..., are complex numbers such that = 0. Then
— =0. Multiplying the last equality by fl(A — I), we find thatq(A) =0, where
q (z) = fl(z —
j=1 i5&j
a polynomial of degree less than k. Therefore q = 0. Hence, for each j,
= — A,) =0,
so = 0. This proves the desired linear independence.
Part Ill
Appendices
Appendix AHow to Get the Exams
A. 1 On-line
Open a browser of your choice on the URL
http://www.booksinbytes.com/bpmYou can then proceed to explore the set of exams or download them. You will alsofind new sets of problems and updates to the book. When this page changes, youcan do a search on the words Berkeley Problems in Mathematics andyou should be guided to a possible new location by one of the net search engines.
To suggest improvements, comments, or submit a solution, you can send e-mailto the authors. you are submitting a new solution, make sure you include yourfull name, so we can cite you, if you so wish.
A.2 Off-line, the Last Resort
Even if you do not have access to the net, you can reconstruct each of the examsusing the following tables. This method should be used only as a last resort, sincesome of the problems have been slightly altered to uniformize notation and/orcorrect some errors in the original text.
Spnng 77 Summer 77 Fall 77 Spring 78 Summer 78 FaIl 78
1.4.8 7.1.8 7.6.21 1.6.10 6.1.3 1.2.9
1.5.4 5.7.6 7.3.2 2.1.2 6.12.1 1.6.24
5.1.5 6.11.4 7.9.25 5.6.9 7.9.13 2.2.44
5.7.5 7.9.23 7.4.21 5.11.5 7.3.2 5.10.27
6.12.3 5.1.2 6.13.9 6.11.11 5.7.2 5.7.8
6.1.1 1.1.33 5.9.1 6.12.3 5.5.10 3.1.6
7.2.12 7.6.22 5.11.24 2.3.1 3.1.2,1.4.11 6.4.1
7.7.14 3.1.9 3.2.3 3.1.5 4.1.19 7.6.263.2.2 7.4.28 1.1.2 7.6.23 4.1.11 7.8.13
2.2.22 1.1.23 1.6.2 7.9.9 2.2.30 6.2.1
5,2.6 2.2.24 3.1.3 5.10.3 5.2.7 2.2.21
6.13.6 7.4.31 4.1.7 5.2.16 5.6.8 2.2.297.9.22 5.2.5 5.3.1 3.1.4 6.11.18 2.2.31
7.9.21 6.3.1 5.1.1 6.4.10 7.7.8 5.3.25.6.31 4.1.17 2.2.25 6.11.30 7.5.19 5.4.115.11.15 5.10.26 7.4.1 7.5.16 7.6.34 7.7.26.7.1 6.11.10 6.1.2 2.3.2 3.3.1 6.2.14
6.11.16 5.8.30 6.11.1 7.2.1 2.1.6 6.12.42.2.23 6.13.7 7.5.1 1.5.5 1.1.26 7.6.31.2.6 7.3.1 5.8.1 2.2.35 2.2.40 4.1.24
Spring 79 Summer 79 FaIl 79 Spring 80 Summer 80 Fall 80
2.2.27 7.5.10 5.8.2 1.7.1 7.4.27 1.1.357.8.14 6.11.20 5.4.13 1.6.2 7.6.15 7.7.16.5.5 2.2.20 6.2.8 2.2.19 7.4.12 5.6.34.2.5 6.2.6 6.11.13 5.10.8 6.2.5 6.4.2
3.1.11 1.4.4 7.9.5 5.4.6 3.4.3 5.11.215.8.16 5.7.8 7.5.2 6.4.13 5.10.10 6.9.1
5.1 1.16 5.6.35 7.5.20 6.13.8 5.3.5 1.7.67.7.13 2.1.3 1.3.13 7.6.20 4.1.8 6.5.1
7.7.4 7.4.23 1.1.34 7.9.17 1.3.21 2.2.176.11.19 3.4.2 3.4.2 3.1.8 2.2.11 2.2.426.8.11 7.7.6 4.3.1 2.3.6 6.8.18 5.9.2
6.4.11 6.13.10 4.1.20 7.2.18 2.2.18 6.8.33.4.1 7.1.3 1.6.12 6.5.11 6.11.13 1.3.11
2.2.45 6.6.4 3.2.6 7.6.16 7.7.5 7.7.97.7.2 5.2.10 5.5.5 7.1.11 7.8.7 3.4.85.2.9 5.2.17 5.11.21 2.2.8 1.6.20 7.9.75.10.5 2.2.32 7.6.35 6.5.6 4.3.3 1.6.12
1.6.32 7.4.11 7.9.3 3.4.19 5.8.6 5.8.37.9.4 1.6.25 6.10.2 5.4.4 5.8.32 4.2.61.4.25 3.2.4 6.7.3 5.3.4 3.1.9 6.1.4
Spring 81 Summer 81 FaIl 81 Spring 82 Summer 82 Fall 82
2.3.7 1.1.9 5.11.15 5.7.5 7.6.23 3.1.107.1.15 6.2.7 3.4.11 4.1.12 5.11.24 6.11.136.9.14 4.1.6 6.5.10 7.9.26 4.3.4 5.10.23.4.4 1.5.6 7.9.2 2.2.28 6.7.5 7.2.1
6.11.31 7.2.19 1.3.12 1.6.22 5.8.21 1.6.275.8.30 4.1.25 1.7.8 1.1.15 1.4.8 - 7.5.215.11.8 5.10.13 6.11.4 6.7.4 7.2.17 5.11.14
1.6.11 6.11.21 5.8.4 5.11.15 6.13.12 6.4.8
7.5.25 1.7.2 7.2.16 7.6.23 5.2.11 1.1.31
1.6.6 7.5.20 2.2.47 6.3.5 5.1.11 5.1.9
5.10.24 1.1.14 7.9.12 5.8.33 5.6.12 1.1.10
7.6.4 6.9.13 5.9.1 4.1.22 6.11.21 7.7.6
1.3.19 6.2.10 2.2.10 7.4.2 1.7.3 1.6.9
4.1.23 1.4.28 3.4.12 1.8.2 7.8.8 7.4.226.10.14 7.5.7 5.4.17 7.1.3 5.4.15 6.6.6
1.1.3 1.6.26 6.4.11 6.4.14 7.9.21 3.4.27.6.36 5.5.5 1.2.13 1.6.21 2.2.14 5.11.62.2.36 7.7.10 7.7.10 5.6.35 1.1.29 4.3.56.1.7 5.8.25 1.1.7 3.1.9 7.5.7 5.4.21.8.1 3.2.6 6.4.10 6.11.23 2.2.41 6.7.3
Spring 83 Summer 83 Fall 83 Spring 84 Summer 84 Fall 84
1.5.13 6.13.16 5.11.30 1.5.23 6.4.3 6.2.137.2.8 5.7.11 6.10.3 6.1.10 6.11.22 7.9.275.3.3 7.6.25 2.2.26 1.4.12 7.4.14 1.6.1
6.1.5 1.4.1 1.3.17 1.1.20 7.5.26 5.11.193.1.12 1.3.18 7.5.22 7.5.18 5.10.14 3.1.15
1.2.5 7.1.17 5.8.7 1.3.7 5.8.26 6.13.146.2.11 1.5.23 6.6.1 5.8.8 2.2.39 7.2.11
6.13.13 6.6.5 7.4.24 3.4.5 1.5.8 7.5.242.1.8 5.7.12 3.2.8 7.3.4 3.4.7 5.6.36
5.11.5 1.7.7 1.5.11 6.10.15 4.2.8 1.4.14
7.6.37 7.6.24 1.4.14 7.7.10 6.5.2 6.8.2
5.6.13 1.2.11 5.10.1 3.1.14 7.1.3 7.4.25
1.6.28 7.9.11 3.4.10 1.4.23 7.9.18 1.6.7
6.7.8 6.5.12 7.6.19 6.10.5 7.1.10 5.11.20,5.10.26
5.10.5 5.4.3 5.9.3 5.6.37 6.11.27 3.4.9
7.8.10 5.6.15 2.2.46 7.6.2 1.1.16 6.11.325.1 1.11 7.6.14 6.12.2 1.4.24 5.2.12 7.6.267.1.3 3.1.13 2.2.34 1.2.8 4.3.7 7.4.303.1.9 5.4.5 6.11.24 1.4.2 3.2.5 5.6.17
4.1.10 1.4.27 4.1.26 6.7.9 5.11.18 1.1.17
Spring 85 Summer 85 FaIl 85 Spring 86 FalL 86
1.4.3
6.1.115.1.35.2.146.11.287.5.32
5.11.313.2.131.6.297.5.126.12.7
5.11.215.8.181.5.246.11.31.6.301.4.137.5.236.12.85.4.21
7.6.81.1.257.6.386.4.4
6.11.25.1 1.2 1
1.5.71.6.131.4.6
5.8.166.3.47.2.36.11.66.10.35.4.81.2.105.8.313.4.21.3.1
6.12.10
5.11.25.8.17
6.11.296.12.45.8.127.5.333.1.161.5.227.8.111.6.46.5.45.7.15.4.147.5.151.5.9
6.12.124.3.86.1.127.8.91.5.10
7.4.261.7.4
5.10.151.4.186.9.37.4.133.4.135.8.116.4.96.9.4
5.11.11.8.3
7.5174.1.57.4.166.11.123.4.145.2.186.7.2
5. 10.16
1.6.165.1.42.3.45.8.227.6.273.1.18
6.12. 131.4.22
5.11.281.1.57.2.14.1.152.2.4
7.8.156.11.243.1.196.5.3
5.10.176.13.11.5.12
Spring 87
4.1.162.2.155.6.16
6. 11.361.6.3 1
5.4.223.2.95.8.137.3.4
6.8.131.4.101.5.152.2.266.8.45.6.294.2.103.3.37.6.196.12.5
5.10.28
FaIl 87Fall 88 Spring 89 Fall 89 Spring 90
1.1.1
1.6.31.5.162.2.377.6.66.6.7
7.5.346.12.15.10.295.4.187.6.177.7.106.2.176.10.25.6.235.8.85.9.53.1.14.3.21.1.30
Spring 88
1.4.196. 13.187.3.35.6.26.1.137.6.136.10.7
5.10.331.5.2 1
5.6.186.5.71.4.206.2.85.4.97.6.57.4.175.4.197.6.10
6.13. 191.7.5
6.9.55.6.303.4.167.6.301.5.181.2.146.8.55.8.277.5.326.8.125.7.136.11.347.1.126.7.62.2.125.10.4
5.11.264.3.6
1.3.257.5.31.2.155.8.166.4.167.1.133.4.205.4.176.4.77.9.146.9.6
5.10.186.5.132.2.9
7.9.165.8.346.10.82.1.4
6.7.101.4.117.6.335.4.226.4.174.1.137.6.181.5.23
6. 11.351.3.5
6.13.205.7.147.2.34.1.186.7.75.8.27.1.111.1.27
1.4.57.6.1
5.8.236.10.91.3.157.2.145.11.96.2.41.2.36.8.65.7.167.5.271.4.167.1.145.3.6
6.13.211.5.105.1.7
Fall 90 Spring 91 Fall 91 Spring 92 Fall 92 Spring 93
1.1.18 6.8.1 6.2.8 6.3.1 7.7.1 1.3.165.10.19 5.6.35 1.2.12 7.6.11 6.11.9 7.9.336.10.11 6.13.22 5.10.20 5.6.25 5.8.5 5.11.7
1.5.3 6.9.7 7.9.28 4.1.14 3.2.6 6.7.47.5.35 7.2.9 1.4.16 2.2.33 6.2.19 3.2.115.5.2 2.3.8 5.4.10 6.11.33 4.2.12 7.7.146.4.18 5.4.20 3.4.17 6.13.15 5.10.21 1.6.221.5.14 7.2.7 1.3.20 5.11.6 7.5.8 6.8.97.8.4 2.1.5 6.3.6 5.8.15 5.6.20 5.8.287.6.23 6.6.3 7.1.3 7.6.12 6.5.14 1.1.281.2.2 1.3.22 4.2.11 5.2.19 7.8.7 7.4.296.2.9 1.3.24 5.6.24 7.5.28 1.6.18 5.10.6
5.4.13 5.11.10 7.5.30 2.3.12 5.11.9 6.9.136.11.5 3.4.6 2.3.9 6.11.4 3.1.20 3.1.77.4.19 6.1.8 6.11.8 6.5.8 6.4.8 6.12.11.3.26 7.7.11 4.1.27 1.3.3 2.2.8 2.2.16.1.6 5.6.32 2.2.16 1.1.19 5.9.4 6.4.14
5.5.11 1.4.21 5.5.13 5.6.26 1.3.6 5.4.16
Fall 93 Spring 94 Fall 94 Spring 95 Fall 95 Spring 96
4.2.4 4.1.5 1.3.23 1.6.14 6.2.22 1.3.26.2.17 7.9.34 7.5.11 7.6.40 5.1.10 4.1.55.2.4 5.10.26 5.10.32 5.10.25 5.2.15 5.10.267.2.21 6.4.19 6.2.20 6.12.19 7.7.10 5.8.103.2.10 3.3.5 3.3.4 3.1.21 6.9.15 7.6.76.6.4 7.7.12 7.9.31 6.10.13 3.2.7 7.3.3
2.2.45 5.1.8 2.2.38 1.1.4 5.6.36 6.11.257.6.1 6.11.17 6.12.15 7.4.6 7.9.32 6.13.17
5.11.27 5.6.27 5.11.29 5.8.24 1.3.10 6.12.141.1.13 1.5.19 2.1.1 1.6.8 6.12.5 1.1.226.9.10 7.4.20 7.7.14 7.4.15 5.7.10 1.1.325.10.23 5.6.28 5.2.2 5.10.31 1.5.17 2.2.436.8.6 6.12.23 6.12.22 6.5.10 5.2.1 5.6.13.1.3 3.4.18 1.112 3.3.6 6.11.7 5.6.357.5.10 1.3.5 7.2.20 6.12.24 1.4.29 7.9.71.6.23 1.1.21 2.3.10 4.3.4 7.7.15 7.5.227.8.5 6.8.6 6.9.12 7.4.10 7.2.15 7.1.75.6.34 5.9.6 5.6.21 5.3.8 1.1.36 6.6.7
Fall 96 Spring 97 FaIl 97 Spring 98 Fall 98 Spring 99
1.6.19 1.3.21 1.3.4 5.8.19 4.1.5 4.1.11
1.2.10 4.2.7 1.4.4 5.8.20 3.3.2 1.6.5
5.11.22 1.5.14 5.5.4 4.3.8 5.11.21 1.4.9
5.5.12 5.5.8 5.11.24 1.5.2 5.4.12 5.1 1.21
7.4.23 5.11.33 5.4.17 6.10.1 7.8.1 5.6.197.7.5 7.4.18 7.1.16 6.3.2 7.4.4 7.1.1
7.6.41 7.5.15 7.8.12 7.5.16 7.9.6 7.5.56.13.11 6.8.11 6.4.12 6.13.4 6.9.11 7.7.176.13.14 6.10.4 6.8.8 7.9.10 6.8.14 6.1.91.1.24 1.2.14 1.1.8 1.1.12 1.5.1 1.1.11
1.1.31 3.2.1 1.1.17 7.8.2 2.2.2 1.4.265.8.8 5.11.5 2.1.7 2.3.3 2.3.5 5.8.295.8.1 5.6.30 5.3.7 5.7.3 5.5.7 5.2.85.4.3 7.9.24 5.11.13 1.4.7 5.8.35 7.6.327.6.9 7.2.4 7.1.13 7.8.6 7.5.15 7.7.18
7.9.20 6.12.25 7.7.7 5.6.33 7.5.4 7.9.366.11.14 7.1.5 6.7.6 7.7.16 7.1.3 6.12.176.8.18 6.4.20 6.3.7 7.9.30 6.12.16 6.6.2
Fall 99 Spring 00 Fall 00 Spring 01 FaIl 01 Spring 02
7.4.3 7.7.3 7.4.7 7.1.4 7.6.29 7.1.94.2.2 4.2.9 4.2.3 1.2.5 1.2.16 1.2.1
6.10.12 6.2.16 6.12.27 6.10.10 6.8.20 6.5.95.5.3 5.8.34 5.10.11 5.11.23 5.10.22 5.10.92.2.40 1.3.9 1.4.15 6.2.15 5.8.9 6.13.37.4.8 7.9.35 6.10.6 7.5.31 7.2.10 7.4.326.8.16 1.4.17 1.2.4 1.6.17 2.2.13 4.2.15.10.7 6.9.16 7.1.6 6.8.15 6.13.23 6.12.263.4.15 5.10.30 5.6.14 5.6.11 5.5.14 5.5.61.2.7 4.1.28 7.8.3 7.6.28 7.9.1 7.6.31
7.6.19 73.29 4.1.9 4.1.21 1.1.37 5.5.95.6.6 6.2.18 6.2.21 6.5.7 6.12.21 6.8.10
6.12.11 5.8.14 5.7.7 5.2.13 5.6.22 5.7.154.1.14 7.4.33 7.4.9 3.1.17 5.6.5 1.3.347.9.29 2.2.5 1.3.6 7.2.6 7.5.13 7.9.195.7.4 7.7.19 6.3.8 4.1.4 3.2.12 1.5.256.3.3 5.11.3 5.6.7 6.11.26 6.12.18 6.12.201.5.20 6.8.19 6.9.8 5.7.9 5.11.4 5.11.17
Appendix BPassing Scores
The passing scores and data presented here go back to the first PreliminaryExamination, which was held on January 1977.
Date Minimumpassing score
# of studentstaking
# of studentspassing
% passingthe exam
Fall 03 63/120 23 13 56.5%
Spring 03 63/120 20 15 75.0%
Fall 02 64/120 41 24 58.5%
Spring 02 67/120 28 17 60.1%Fall 01 65/120 41 22 53.6%
Spring 01 65/120 15 5 33.3%Fall 00 65/120 38 27 7 1.0%
Spring 00 64/120 11 5 45.4%
Fall 99 64/120 35 25 7 1.4%Spring 99 66/120 12 8 66.7%
Fall 98 66/120 29 16 55.1%
Spring 98 7 1/120 15 9 60.0%
Fall 97 64/120 41 28 68.3%
Spring 97 70/120 10 5 50.0%
Fall 96 80/120 24 17 70.8%
Spring 96 84/120 17 13 76.5%
Fall 95 64/120 41 19 46.3%
Spring 95 65/120 10 4 40.0%FaIl 94 71/120 28 16 57.1%
Spring 94 70/120 11 5 45.5%
Fall 93 79/120 40 28 70.0%
Spring 93 69/120 22 17 77.3%Fall 92 71/120 53 34 64.2%
Date Minimumpassing_score
# of studentstaking
# of studentspassing
% passingthe exam
Spring 92 58/120 27 13 48.1%Fall 91 66/120 66 42 63.6%Spring 91 60/120 43 21 48.8%Fall 90 71/120 89 50 56.2%Spring 90 68/120 47 24 5 1.0%
FaIl 89 71/120 56 18 32.1%Spring 89 66/120 31 16 51.6%Fall 88 70/1 20 44 22 50.0%Spring 88 74/140 25 16 64.0%Fall 87 81/140 29 21 72.4%Spring 87 73/140 46 31 67.4%Fall 86 93/140 24 13 54.2%Spring 86 75/140 37 25 67.6%Fall 85 89/140 23 16 69.6%Summer 85 90/140 16 11 68.8%Spring 85 97/140 19 12 63.2%Fall 84 99/140 26 18 69.2%Summer 84 82/140 10 6 60.0%Spring 84 82/140 35 24 69.0%Fall 83 92/140 16 11 68.8%Summer 83 86/140 14 7 50.0%Spring 83 8 1/140 26 17 65.4%Fall 82 84/140 15 8 53.0%Summer 82 88/140 23 17 73.9%Spring 82 90/140 15 12 80.0%Fall 81 82/140 15 10 66.7%Summer 81 65/140 21 14 67.0%Spring 81 106/140 32 24 75.0%FaIl 80 99/140 23 17 74.0%Summer 80 82/140 24 12 50.0%Spring 80 88/140 30 17 56.7%Fall 79 84/140 13 7 53.8%Summer 79 101/140 15 7 46.7%Spring 79 97/140 23 18 78.3%Fall 78 90/140 17 12 71.0%Summer 78 89/140 21 12 57.1%Spring 78 95/140 32 19 59.4%FaIl 77 809140 13 12 92.0%Summer 77 90/140 20 18 90.0%Spring 77 — 90/140 42 29 69.0%
Appendix CThe Syllabus
The syllabus is designed around a working knowledge and understanding of anhonors undergraduate mathematics major. Any student taking the examinationshould be familiar with the material outlined below.
Calculus
Basic first- and second-year calculus. Derivatives of maps from to gra-dient, chain rule; maxima and minima, Lagrange multipliers; line and surfaceintegrals of scalar and vector functions; Gauss', Green's and Stokes' theorems.Ordinary differential equations; explicit solutions of simple equations.
Classical Analysis
Point-set topology of and metric spaces; properties of continuous functions,compactness, connectedness, limit points; least upper bound property of IL Se-quences and series, Cauchy sequences, uniform convergence and its relation toderivatives and integrals; power series, radius of convergence, Weierstrass M-test;convergence of improper integrals. Compactness in function spaces. Inverse andImplicit Function Theorems and applications; the derivative as a linear map; ex-istence and uniqueness theorems for solutions of ordinary differential equations;elementary Fourier series. Texts: [Ros86], {MH93], {Bar76], [Rud87].
Abstract Algebra
Elementary set theory, e.g., uncountability of R. Groups, subgroups, normal sub-groups, homomorphisms, quotient groups, automorphisms, groups acting on sets,Sylow theorems and applications, finitely generated abelian groups. Examples:permutation groups, cyclic groups, dihedral groups, matrix groups. Basic prop-erties of rings, units, ideals, homomorphisms, quotient rings, prime and maximalideals, fields of fractions, Euclidean domains, principal ideal domains and uniquefactorization domains, polynomial rings. Elementary properties of finite field ex-tensions and roots of polynomials, finite fields. Texts: [Lan94}, [Hun96], [Her75].
Linear Algebra
Matrices, linear transformations, change of basis; nullity-rank theorem. Eigen-values and eigenvectors; determinants, characteristic and minimal polynomials,Cayley-}Iamilton Theorem; diagonalization and triangularization of operators;Jordan normal form, Rational Canonical Form; invariant subspaces and canonicalforms; inner product spaces, hermitian and unitary operators, adjoints. Quadraticforms. Texts: [ND88], [HK61), [Str93].
Complex Analysis
Basic properties of the complex number system. Analytic functions, conformal-ity, Cauchy-Riemann equations, elementary functions and their basic properties(rational functions, exponential function, logarithm function, trigonometric func-tions, roots, e.g., Cauchy's Theorem and Cauchy's integral formula, powerseries and Laurent series, isolation of zeros, classification of isolated singularities(including singularity at oo), analyticity of limit functions. Maximum Principle,Schwarz's Lemma, Liouville's Theorem, Morera's Theorem, Argument Principle,Rouché's Theorem. Basic properties of harmonic functions in the plane, connec-tion with analytic functions, harmonic conjugates, Mean Value Property, Maxi-mum Principle. Residue Theorem, evaluation of definite integrals. Mapping prop-erties of linear fractional transformations, conformal equivalences of the unit discwith itself and with the upper half-plane. Texts: [MH87}, [Ah179], [Con78].
References
[Ah179] L. V. Ahifors. Complex Analysis, an Introduction to the Theoiy ofAnalytic Functions of One Complex Variable. International Series inPure and Applied Mathematics. McGraw-Hill, New York, 1979.
[AM95] 0. L. Alexanderson and D. H. Mugler, editors. Lion Hunting & OtherMathematical Pursuits, A Collection of Mathematics, Verse, and Sto-ries by Ralph P Boas, Jr., volume 15 of The Dolciani MathematicalExpositions. Mathematical Association of America, Washington, DC,1995.
[Bar76} R. G. Bartle. The Elements of Real Analysis. John Wiley & Sons Inc.,New York, 1976. Also available in Spanish [13ar82].
[Bar82} R. 0. Bartle. Introducción al análisis matemdtico. Limusa, Mexico,1982. English original in [Bar76].
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Index
algebraically independentArgument PrincipleArzelà—Ascoli 's Theorem
22 1—224, 299
Baire Category Theorem ..... 291
Bertrand's Postulate 483Binomial Theorem •. . .. 158, 159Bolzano—Weierstrass Theorem . 161,
176, 177,291,292,298
Casorati—Weierstrass Theorem . 334,
Cauchy
integral formula 326, 333,345,347,348,356,378,403
integral formula for derivatives323,331,332,348
RootTest 313sequence . . . . . . 184, 249, 298Theoremof
315,339, 350, 354, 355,375, 377, 378, 382, 392—f%r% A A I API(t A '% I
334,336,337,
Cauchy—Schwarz
inequality . 157,203,207,224,240,247,249,365,553,563
Cayley's Theorem . . . . 454Cayley—Hamilton Theorem 280,424,
505,506Chain Rule 252,262Change of variables 259,264Chinese Remainder Theorem .. 486Comparison test 314, 315, 395, 396complexification 557Contraction Mapping Principle 302,
303critical point
nondegenerate 34
Deformation TheoremDescartes' Rule of SignsDirichlet
integralofproblemof
• ... 477• 363,366•... 215,
Cauchy—Riemannequations . 330,
341,372
342
•. 375,381364.
393386A
Euler
domainmetric
Theorem of . -
totient function.485,541
Extension Theorem
GaussLemma .
TheoremGauss—Lucas TheoremGaussian elimination
553
476245
298793
117• 33
10
170,248
469• 263359, 361,363
• 496
centralizerconjugacy classdihedralfixed point
Heine—Borel TheoremHurwitz Theorem
Identity Theorem 311,342
L'Hôpital's RuleLagrange
multiplierspolynomialsTheorem of
456,485Laplace expansionLaplacian,
eigenfunctionLaurent expansionLeibniz Criterion..Liouville's Theorem
99
• 99
105
105
176,293,297315
335, 339, 340,
196
246
• 531430,444,450,
256,257
• . . . . . . . . 41
339, 342, 356188,210
329—33 1, 333,
Eisenstein Criterion . 465,469,470,475
EuclideanAlgorithm . 465,467,472,476,
527
120,430,485
482, Implicit Function Theorem 252,273Induction Principle . 158, 171, 180,
285 181, 183, 184, 187, 190,
467,478,481, 198, 441, 449, 477, 492,494,497,498,508,516,518
12, 136,519Intermediate Value Property ... . 9
Intermediate Value Theorem •
180,300345
.. 161, 162,174,201,205,
236,275,326,363, 364
29Inverse Function Theorem 241—243,
254—256,268,341
Fermat Little Theorem485,486
Fibonacci numbersFixed Point TheoremFubini's Theoremfunction,
concaveconvexharinonic . . . .
orthonormalpolynomialproper . . . . . . . . . . • .
real analyticsemicontinuous
Fundamental Theorem
ofAlgebra357, 359
of Calculus
Jensen's Inequality . 239
JordanCanonical Form •.. 181,281,
493, 510, 528, 530, 534—536,538—540,542—545
Lemma of - 167,396—398,404
81, 112, 349—351,
429,
Gram—Schmidt Procedure 510, 546,
Green's Theorem •.. 260,264,372group
action, faithful . • 124
permutations, transitive .. 105
group,
center 97
334, 336, 342, 349, 373
Maclaurin expansion 166, 185, 189,206, 315, 332, 337, 343,
345,369,370,565
closedproperstrict contraction
matrix,axiscolumn rank ...commutant of afinite orderindexordern .
orthogonalpositive definiterowrankskew-symmetric . .
special orthogonaltridiagonalunitaryVandermonde
364
. I • • • • I • I
WeierstrassM-testTheorem
111
• 98,119163, 168, 191, 194,
428,442,
163, 166, 199,
184,220,298,
226,227
176,232
map Rank—Nullity Theorem . . 491,497,32 501,504,547
• ....... 32 Rational Canonical Form .... 54263 Rayleigh'sTheorem .. 288,523,524
regularvalue ..•.34133 Residue Theorem 348, 374,375, 377,125 379, 381—385, 387, 389—
•.. 125, 151 392, 397, 398, 400, 405,138 406,409,410,414—416,419
• . . . . . Rierriann surris . . 1 85
Lemma •. 230,
231ring
indexunit
Rolle's Theorem
148140152
• 148125
.. 149
.. 138
.. 126148
127,371,497,
498,500,562,565Maximum Modulus Principle
74, 323—327,329—331,349, 364, 372, 373
Mean ValueProperty ...... 321
Sub—Property .. 74Theorem .. 168, 173, 192, 194,
200,204,215,244,248,336for Integrals • . . . . 227, 232
Method of Undetermined Coefficients275
Morera's Theorem 315,345, 351,355
Nested Set Property .. 300norm . . . . . . . . 32
Nyquist diagram 362
Open Mapping Theorem .. 320,336
Parseval Identity ... .... 229,231Picard's Theorem 266—268,273—275,
277,284,287Pigeonhole Principle .... 457,480Poisson's formula . .. . . .. . . 386
Rank Theorem •.... 254
199
Rouché'sTheorem
323, 349,357,358,360,362—364,368,369,381
SchwarzLemma . 325—328
Reflection Principle .... 340Theorem 242
Second Isomorphism Theorem . 438Spectral Theorem 497,567Stokes' Theorem 262Stone—Weierstrass
Approximation Theorem .
.. 210,216,225,228,230,231,315, 344
Structure Theorem .. 431,449—451,458,474
Sylow's Theorems . . .
452—455,478
Taylor's Theorem 159,242,369
Triangle Inequality
458,477
Problem Books in Mathematics (conanued)
Algebraic Logicby S. G. Gindikin
Unsolved Problems in Number Theory (2nd ed.)by Richard K Guy
An Outline of Set Theoryby James M Henle
Demography Through Problemsby Nathan Keyfltz and John A. Beekman
Theorems and Problems in Functional AnalysisbyA.A. Kirillov andA.D. Gvis/ziani
Exercises in Classical Ring Theory (2nd ed.)byT.Y Lam
Problem-Solving Through Problemsby Loren C. Larson
Winning Solutionsby Edward Lozanslcy and Cecil Rosseau
A Problem Seminarby Donald J. Newman
Exercises in Number Theoryby D.P. Parent
Contests in Higher Mathematics:Miklós Schweitzer Competitions 1962—1991by GáborJ. Székely (editor)