CHAPTER2 Systems of Linear Equations EXERCISE SET 2.1 1.(a) and {c)are linear.(b)is not linear due to the x1x3 term.{d) is not' linear due to the x}2 term. 2.(a)a.nd(d)are linear.(b)isnot linearbecause of the xyzt erm.(c)isnotlinearbecauseof the 3/5 x1 term. 3.(a)islinear.(b)islinear if k# 0.(c)islinear only if k= 1. 4.(a)islinear.(b)is linearjf m=f:0.(c)is linear only if m=1. 5.(a),(d),and(c)are solutions;thesesets of values satisfyallthree equations.(b)and(c)are not solutions. 6.(b) ,(d),and(e)are solutions;these setsof valuessatisfyall three equations.(a)and(c)arenot solutions. 7.The tluee linesintersectat thepoint {1, 0)(see figure).Thevaluesx= 1,y= 0satisfyallthree equationsand this isthe unique solution of thesystem. 3.x-3y"' 3 :c The augmented matrixof the system is lr231~~~ ] -3! 3 Add-2 timesrow 1 to row2 andadd-3 timesrow1 to row3: [ ~=!!~ ] Multiply row2 by- jand add9timesthenew row2t.orow3: [ ~~~ ] Fromthe last rowwes ~thatt hesystem isredundant(reduces to only two equations).Fromthe secondrowwe see that y= 0 and, fromback substitution, it followst hat x= 1 - 2y = 1. 22 Exercise Set 2.1 8.The three lines do not intersectina. common point(see figure).This system has no solution. )' The augmentedmatrix of the systemis and the reduced row echelonformof this matrix(details omitted)is: The last rowcorrespondsto theequation 0 =1,sothe system isjnconsistent. 23 9.(a)The solution set of the equation 7x - 5y = 3 can be described parametrically by (for example) solvingthe equationforxintermsofyandthenmakingyinto aparameter.This leadsto x== 3'*:,5t,y=t ,where-oo ], 20.a(,ivc'? that5A- = 52 it folowst hat 5A= 5 :;::;. ._, and so= - =- .A 1 [-:.!.- lJT1 [-25) 5535-)3 (b)Given(I+ 2A)-1 = wehaveI+ ZA = = 113 r-: and soitfollowsthat 1( 1[-52][I OJ)1[-9 l] A:-::213-11- ol.=132-c' 21.Thematrix A=isinvertibleif and onlyif det(A) =c2 - c f.0,i.e.if and onlyif c :/:.0,1. 22.The matrix A=[- :)isinvertible if anJ only if dct(A) =-c2 + 10,i.e. if aud only if c =/=L 23.One such example isA=In genera),anymatrix of the form[:; ] . 323JeJc 24.One such example isA =

In general,any matrix of the form;]. 3-20-e- !0 25.LetX= [xiJJ.ThenAX= Iif and only if 3x3

0 '] ['" X}2xnl

"

lJ 10X21X22 X23 j 1 11X31X32X33 0 Exercise Set 3.2 i.e.if and onlyif the entries ofXsatisfy thefollowing system of equations; Xn + X:z t + X22 Xt3+ X23 xz1 -!- X3 1=} + X32= 0 + X:tJ=:0 =0 =1 =0 =0 = 0 + X33=1 69 Thissystemhastheuniquesolutionxu = :t12=xzz= x23= X31= X33::;;~andx 13= x21 = x;12= - !ThusAisinvertibleandA-1 =[-!~-!] '2-- 2'l 26.LetX=lx,iJ. Then AX = Iifanrlonlyif 3X3 [~~~ ][::: 20lXJJ X13][}0()] X23=010 x:n00l i.e. if and only i i xu = 1, :tu = 0, x13= 0, :t21= 0, xn =1,r23= 0,2x11+ X31=0,2x12 + X32= 0,and2x13+ X33=l. This isavery simple syst.emof equationswhicbhas thesolution Xll=- 1,X12= 0,X13= 0,X21=0,X22= l ,X23=0,X31=- 2,X32=0,X33=1 Thu& AisinvertibleandA- t=[~~~ ] - 20l 28.(a)Thematt.rueiff di- 0or1 for i=1, 2,ancl3.There area.tOtalof eight such matrices (3x3matriceswhose diagonal entries are either 0 or 1) (b)There are at otal of 2n such matrices (n xn. rnatrice.s whose diagonal entries are either0 or 1). If A=[d' 0],thenA2 + 5A + 6!2 =0if and onlyifd?+ 5'-.+ G = 0fori= 1 and2;i.e.if and 0dl onlyif d;= - 2or- 3for i= 1 and 2.There are at.ota.lof fonrsuchmatrices(auy 2x2diagonal matrixdiagonal entriesare either - 2 or- 3) . D7.IfAisboth symmetricandskewsymmetric,thenAT= AandAT =- A; thusA =-A andso A =0. 0 8.Inasymmetricmatrix,enlriesthataresymmetr icallypositionedacrossthemaindiagonalare equalto eachother.Thus asymmetricmatrix iscompletelydeterminedbythe entriesthat lieon orabovethemainrliagonal,andentriest hat.appearbelowthemaindiagonala.reduplicates of entries that appear above the mai n diagonal.Au nxnmatrix has nentriP.s onthe maindiagonal , n- lentries onthediagonaljust abovethemaiudiagonal,etc.Thusthereare atotal of n(n + 1) n+ (n- 1)+ + 2+ 1 = 2 entrieslhatlieonorabovethemaindiagonal.Fora;;ymmet.ricmatrix,t hisisthemaximum numberof distinct entriesthematrix canhave. Inaskew-symmetricmatrix,the diagonalentriesare0andentriest hataresymmetricallyposi-tioned ac:rossthe main diagonalarc the negatives of each other.The maximum nwnber of distinct entries can be attained by selecting distinct positive entries for thep,.,sit.ions above the 1.0ain diagonalThe entriesin then(n2-11 positionsbelowthe maindiagonalwillthen automaticallybe distinct fromeach otherand fromthe entries on or above themain diagonal.Thus the maximum number of distinct entriesin askew-symmetric matrix is n(n- 1}n(n- I ) --'--:---.....:.++ 1 = n(n- 1) + 1 22 09.If D== thenAD = ldt a1d2a2Jwherea1 anda2arelherolumnsofA.ThusAD = I= lete2](wheree1and arethestandardunitvectorsinR2)ifandonlyifdlaJ=e1 anddza2=e2.Butthisistrueifandonlyifd1i- 0,d2:/;0,a1 = -J;e1,anda2 = :, Thus 102 Chapter 3 [_!.0] A=1 whered1,d2 =1- 0.Althoughdescribedhereforthecasen= 2,it should.beclear d2' that the same argument can be appliedto asquare matrix of anysize.Thus, if .1D = 1,then the f..!.00 dt lo.l... .o d2 tliagonalentri es d1, dz , ... ,d.nof Dmust benonzero,andA=::. . .;. 00.....1. d,.. DlO.(a)False.If Aisnot squarethenAisnotinvertible;it doesntmatterwhetherAAT(wh.ich isalwayssquare)isinvertibleor not.(ButifAissquareandAATisinvertible,then Ais invertiblebyTheorem 3.6.5.] (b)False.Forexampleif A =G and B=!] , thenA + B= G is( c)True.IfAisbothsymmetricandtriangular,thenAmustbeadiagonalmatrix.Thus A = :,andsop(A) = :]isalsoa.diagonalmatrix(both 00Un00p(d., ) symmetric andtriangular). (d)True.Forexample,inthe3x3 case,wehave (e)True.If Ax= 0has only thetrivial solution,thenAisinvertible.But if Aisinvertible then soisAT {Theorem3.2.11);thusAr x=a hasonlythe trivialsol ution. Dll.(a)False.For = (b)True.If A is invertiblethenAkisinvertible,andthusAk=1- 0,forevery k= 1, 2, 3, .. . . This showsthat aninvertiblematrix cannotbe nilpotent; equivalently, anilpotent matrix cannot be inverti ble. (c)True(assumingAi=0).If A3 = A,then.46 = A,A9 =A,A12 =A, . . . .Thusisitnot possible to haveA,..::::0 foranypositiveintegerk , since this wouldimplythat=0 forall j?.k. (d)1'rue.See Theorem 3.2.11. (e)False.For example,Iisinvert)blebut I-1 = 0 isnotinvertible. WORKING WITHPROOFS Pl.If Aand1Jare symmetric,thenA 1'= AandBT = B.It followsthat (AT)T = AT (A+ B)TAT + BT = A+ B (A- B)T = AT - BT =A- B (kA)T =kAT= kA th11st.hcmatricesAl' .A+ B,A - B , and kAarealso symmetric. EXERCISE SET 3.7 P 2.Our proof isby inductionon the exponent k . [d0J:o:]=[d0I: Step 1.We haveD1 =D=...., 0] ;thusthe statement istruefork= I. rl l n 103 Step 2 st ep).Suppose the statement is t rue fork=j, wherejis an integer1.'rhen and sothe statement isalso true fork=j+ l. Thesetwo stepscomplete the proof by induction.

;2 :::1=P 3.If d, d2, ... ,d11 arenonzero,then 00d.,Q0I00

} ,0.. .01 0};0 isinver tiblewithn-' =On the otherhandif auyoneof 00:i;;- . D=

:, :l 00 dn thediagonalentries iszero,thenDhas arow of zerosandthus isnot.invertible. P 4.Wewill showt hat if Aissymmetric(i.e.if AT==A),then( An)T=A 11 foreachpositiveinteger . n.Ourproofisby induction onthe exponent n. StepLSinceAissymmetric,wehave(A1) TA =A1;thust hestatementistruefor n=1. Step 2(induction step).Suppose the statement istruefor n=j, where jis Aninteger1.Then (Ai+l f= (AAi)T =(Ai )T A1'=AJ A= AH1 and sothe statement isalsotrueforn=j+ I. Thesetwo steps completetheproof by induction. P5.If Aisinvertible,then Theorem 3.2.11impliesAT is !::1vertible;thust heproductsAATandAT A areinvertibleaswell.Ontheotherhand,ifeitherAATorAT Aisinvertible,thenTheorem 3.3.8impliesthatAisinvertihle.It followsthatA,AAT,andAT Aareallinvert ibleor all singular. EXERCISE SET 3.7 1.Wewill solveAx =bbyfirst soiving Ly =bfory,and t hen solvingUx = yforx. Thesystem Ly = bis 3yl= 0 -2yl + Y2= 1 from which,by forwardsubstitution,we obtain Yl=0,Y2= 1. 104 The systemU x= y1s x1- 2x2= 0 X2=1 Chapter 3 fromwhich,bybacksubstitution, we obtainx1 = 2,X2=1.It is easyto check that this is in fact thesolutionof Ax =b . 2.The solutionof Ly = bis y1 =3,Y2= ThesolutionofUx =y(a.ndofAx= b )isx1 =X - l 2- '1 3.We willsolveAx =bbyfirstsolving Ly =bfor y , andthen solving Ux= yforx. The systemLy = bis 3yx=-3 2yl + 4y2=-22 - 4yl- Y2+ 2y3 =3 fromwhich, byforwardsubstitution,obta.inYl=-1, Y2=-5,YJ=- 3. The systemU x=yis XJ- 2X2X3-=-1 x2+ 2x3=-5 X3=-3 fromwhich,byback substitution,weobtain:x:1 = -2, x2 =1,x3 =- 3.Itiseasyto checkt ha.t thisisinfactthe solutionof Ax =b . 4 .T hesolutionof Ly =bis Yt= 1,Y2= 5,Y3= 11.The solutionof Ux= y{andof Ax =b )is X- ill X- 37X_11 l- 14'2- i4)3- i4. 5.T hematrixAcanbereducedto rowechelonformbythe followingoperations: The mult.ipliers associated wit hthese operations 1,andA=LU =[2 0][14] -1301 isanLU-factorization ofA. To solvethesystemAx= bwhereb= wefirstsolveLy = bfory ,and then solveUx =y forx : The systemLy ;::bis fromwhichweobtainy1 =:=.- 1, Y2:::::- 1. The system Ux= yis 2yl=- 2 -yl + 3y2= - 2 X 1+ 4X2= - 1 X2=- 1 fromwhichweobtainx1 = 3,x2 =- 1.Itiseasytocheckthatthi!isinfactthesolutionof Ax. =:::.b. EXERCISE SET 3.7105 6.AnLU-decomposition of thematrixA= [ - : isA= LU = (-: Thesolutionof Ly = b,;hereb= isYl=2, = - 1.Thesolutionof Ux = y(and of Ax =b)isx1 =4,x2= ..-1. 7.The matrixA canbere= I, Yl= 0.The solution or U x= y(and or Ax = b) is Xt= -1, xz = 1,x3= 0. 106 9. Chapter 3 ThematrixAcan be reducedto row echelonformbythe following sequence of operations: 01 0- 1 0 - 1

3- 23- 230 A= 2 - 1 -120 -1-1 2 0010101

0-1

0-1 ;10-1

10l010 -12 0201 010101 ."!.:.

0-1 01[1 0- 1

10 2J0 l0 011001 004000 The multipliersassociatedwiththese operationsare-1,-2,0(forthethirdrow),0, 1,0,-1,and thus [-1 00

0-1

A-LU-3010 - 1201 0100 isanLU-decomposition ofA. TosolvethesystemAx= bwhereh= [-;],we firstsolveLy - blory, and then solve Ux - y forx: The systemLy = bis -yl==5 2yl+3yz= - 1 Y2+ 2y3==3 Y3+ 4y4=7 fromwhichweobtain Y1=- 5,Y2=3, Y3= 3,Y4==1. The sy.:>tcm Ux = yis x1 - X3= - 5 xz+ 2x4=3 X3+X4=3 X4=1 fromwhich weobtain x1 =-3, x2 =1,x3 =2,x4 = 1.It iseasyto checkthatthisisinfactthe solution of Ax =b. EXERCISESET3.7 107 10. ['-' 00][' 00 j - 20 0] An LU-decomposition of !1 = 12 is A= LU = 40130 2 0201 . The 0-1-4- 50-1-100 ro!utiono[Ly= b,wh"e b= [!lisy,= 4, Y>= - l, y,Y bwhere p-> P [!:].Usingthe given decomposition,the systemp - l A;x = p - l bcan bewritten as 18. [10 LUx=1 3-5 WesolvethisbyfirstsolvingLy =p-lb fory,andthensolvingUx =yforx. Thesystem Ly =p - I bis Y1=1 Y2=2 3yl- 5y2+= 5 fromwhichwecbt..ainy1 = 1,'!/2=2,y3 = 12.Finally,t he system Ux =yis x1+ 2x2 +2x3 =1 X2+4X3=2 17x3= 12 fromwhichweobtaiuXt=x2:::::- = g, Thisisthe solution of Ax= b. ThesystemAx hiseq uivolenttop- 'Ax P -' bwhe>ep- 'P [i decomposition,the syst.emp-l Ax =p-lb canbe wri tten M (\0] o1.Usingthe given 10 Thesolution ofLy = p - l bisy1 =3, y2 = 0, Y3=0.The solution of Ux =y(and ofAx= b )is X1=X2= 0,X3= 0. 19.If weinterchangerows2 and 3of A,thenthe resultingmatrix can be reducedt orowechelonform withoutany furtherrow!nterchanges.This isequivalent to firstmultiplyingAon the leftbythe correspondingpermutationmatrix P: [3-1 02 3-1 -1 -1 2 :1 J 1103 Thereduct ion ofP Ato rowechelonform proceeds asfollows: [3-1 PA =02 3- 1 Thiscorresponds to t heLU-decomposition [3-1 PA =02 3-1 0][30 0][1 1=0200 13010 of t.hematrix PA, ortothefollowingP LU-decompositionof A [1 0 0][30 0][1A =00102001 01030100

=p -1LU Notethat, sincep-1 decompositionca.nalsobe as A =P LU. The systemAx =b=! is equivalentto PAx= Pb=and,using theLU-decornposjtion obtainedabove,thiscan bewrittenas [3 o olll-4ol[x1][-2] PAx ==DUx ==02001!x 24= Pb 30}00}X31 Finally,thesolutjonof Ly = Pb isYt= y2 =2,y3 = 3,andthe solution of Ux =y(and of Ax =b)isx 1 =-x2= X3= 3. 20.If weinterchangerows1 and2 of A,thenthe resulti ng matrixcanbe reducedtorow echelonform wi thoul anyfurt her rowinterchanges.This isequi valent to first multiplyingAontheleftbythe correspondingpermut at ionmatrixP: The LU-decomposit ionoft hematrixPAis P A = LU = -;] 20-3001 a.ndthecorresponding FLU-decomposit ionofAis SinceP 1 =P ,this.2 + 2>. - 3 = (..\ -1)(>. + 3}.Thusdet(A) =0if and onlyif/\= 1 or >.=- 3. 14.>.= 1, ,\= 3,or ,.\ =- 2. 15.det(A) =(>. - 1)(>. + l }.Thusdct(A}=0if and only if A""1or.A= - 1. 16.A = 2or ,\ = 5. 17.We have 1; 1-=_1xl= x( l- x) + 3 = -x2 + x + 3,and 18. 19. 20. 21. 10-3 2x-6= ((x(x - 5) + 0- 18) - ( -3x- 18 + 0)= x2 - 2x ]3X- 5 Thus t he gi venequation isif and only if -:t2 + :-r+ 3 =.x2 - 2x, i.e.if 2x2 - 3x- 3 =0.The roots ofthis quadro.tic equationare x= 3*{33. y=3 0 000

10 0 (a)0= (1)(-1)(1) = - 1(b) 12 - 1 0 3 =0 00 0 '1 0 1238 127- 3 (c) 0I-41 =(1)(1)(2) (3)= 6 0 () 27 000a 111 200 (a)020= 23 =8(b) 022 = (1 )(2)(3)( 4)=24 002 003 0004 000 {c) l200 =( - 3)(2)( -1)(3) =18 4010- 1u 100200-- 233 Mn = 1; -11 4 = 29. c11=29 Ml2 =-11 4 = 2l,Ct2 =- 21M13= 21,= 27 ,-2 M21=1 !I = G21=11 M22= !I= 13,c22=13 M23= -5, C23= s 1-2 M3t=731_1 =- 19, C31=- 19 M3231_1 ===-19, C32==19M33==19, c33= 19 120 22. Mll = 6, cn =6 M12=12,012 = -12M1s= 3, C1a=3 M21= 2, C21= - 2 M22= 4, Cn = 4M2a= l , Oza=-1 M31 21) 6 =O,Ca1= 0Ma2=O,C32=0Ms3=0, Cas=0 003 23.(a)M13 =4114= (0 + 0 +12) - (12 + 0 +0)=0013 = 0 412 4-16 (b)M23= 4114= (8 - 56+24)-(24+56-8)=-96023= 96 412 4l6 (c)Mn=4014=(0+ 56 +72)-(0+8+168) = -48022= - 48 432 - 116 (ct)M21=1o14= {O+I4+18)-(0+2 - 42) = 72c21= - n 132 24.(a)M3z=- 30, C32= 30 (c).M41= - l,C41= l {b)M44=13, 044= 13 (d)M24=0,024 =0 25.(a)det(A)= (l) Cu+ (- 2)C12+ (3)013 = (1)(29) + (- 2)(-21) + (3)(27)= 152 (b)det(A)= ( I)Cll +(6)C21+(-3)C31 =(1){29) +(6)(11) + (-3)(-19) =152 (c)det(A)==(6)C21+ (7)C22+(-1)023 = (6)(11)+(7)(13)+(-1)(5)= 152 (d)det.(A)= (-2)C12 +(7)C22+(1)032 = (-2)(- 21) +(7)(13) +{1){19)==152 (e)det(A)=( -3)CJI+(l)Ca2 +( 4)Caa =( -3)( - 19) +( 1)(19) +(4)( 19) =152 (f)det(A) = (3)013 + (-l)Czs + (4)033 =(3)(27} + ( - 1)(5) + (4)(19)=152 26.(a)det(A)= (l)Cll +(l)Ct2 + (2)013 =(1)(6) +(1)(- 12)+(2)(3)=0 (b)det(A)= (l)011 + {6) C21+ (3)Ca1 =(1)(6) + (3)(-2) + (0)(0) = 0 (c)det(A)= (3)021+ (3)022 + (6)023=(3){-2) + (3)(4)+ (6)(-1} = 0 (d)det(A) = (l)Ou +(3)022 + (I)Cs2 = (1)(-12) + {3)(4)+ (1)(0)= 0 (e)det(A) =(O)C31+ (l)C32 + (4)C33 = {0)(0)+ (1)(0) + (4)(0) = 0 (f) = (2)013+( G)02J +{4)033= (2)(3}+ (6)(- 1) +(4)(0) = 0 27.Using column2:det{A}= ;j = (5){-15+ 7)= - 40 28.Usingrow2:det(A)= (-4)=(1)( -18) + (-4)(12) = - 66 Chapter 4 DISCUSSIONAND DISCOVERY 30.Jet( A):::::k3- 8k2- IOk + 95 3353 :n.Usingcolumn 3:det(A)=( -3) 22-2- (3)2 21024 32.det(A) =0 33.By expandingalong the t hird column,wehave 35 2- 2= (- 3)(128) - (3)( -48) = -240 0 sinO -cosO cos B 0 I sinBcosO' sin80= (1) 8 .=sin2fJ+cos2B=1 -cossmB s inO- cosOsinO+ cosB1 forallvalues of 0. 121 34.AB = bf)a.ndBA =

bd ce]. ThusAB =BA if andonJyifae + bf= bd + ce ,and thisisequi valenttot he condition that lb 0- cl =b(.--=-83 11 = (,X-+ 1} =0. Thus..X= 3 and>.=- 1 are eigenvalues of A;hasalgebraicmultiplicity 1. The characteristic equation is I.X=410 )..:21 =(>. - lO)(..l. + 2) + 36 =(>. - 4)2= 0.Thus ,X =4 isthe onlyit has algebraicmultiplicity 2. The characteristicequation is 1.>.-2 , 0 I= (..l.- 2)2 = 0.Thus ,X= 2 isthe onlyit -1"'-2 has algebraic multiplicity 2. The characteristicequationis.X2- 16 = 0.Thus,X =4 areeigenvalues;eachhasaJgebra.ic multiplicity1. Thecharacteristicequationis>..2 = 0.Thus,\=0istheonlyeigenvalue;it hasaJgebra.ic multiplicity2. The characteristic equationis(>..= 0.Thus>..= 1isthe only eigen.value;it fi8salgebraic multiplicity 2. .>. - 10- 1 The characteristic l!quat io!l is2.>.- 1o=).3 - 6).2 + 11 .A- 6 =(.A - 1)(>. - 2){>..- 3)= 20A - 1 0.Thus.A=1,>..=2,and>.=3are eigenvalues; eachhas .algebraic multiplicity1. .A-455 (b)The characteristic equation is .X-11= A3- 4>.2 + 4A=..\(A - 2)2 = 0.Thus A= 0 -t3). + 1.. and).=2areeigenvalues;>.=0hasalgebraicmultiplicity 1,and).= 2hFlSmultiplicity 2. )..-3-41 (c)The characteristic equation is.x+ 2- 1= ).3 - >..2 - 8>.+ 12 =(,X+ 3)(A- 2)2 = 0.Thus - 3-9). A =- 3and,\= 2 areeigenvalues; A= - 3has multiplicity1,and>..= 2has multiplicity 2. 8.{a)Thecharacteristicequationis>.3 + 2..\2 + A=>.(>.+ 1)2 =0.Thus>.=0isan eigenvalueof multiplicity1,and)..=-1 isan eigenvalueof multiplicity2. (b)The characteristic tlltuationis).3- 6)..2 + 12)., - 8= (A- 2)3 =D;thus,\ =2 is an eigenvalue of multiplicity3. (c)Thechara.ctcrist.icequationis>..3- 2A2- 15>. + 36 =(,\ + 4)(A - 3)2 = 0;t husA= -4 isan eigenvalueof multiplicity1, and ,\ =3isan eigenvalueof multiplicity2. 9.(a)Theeigenspacecorrespondingto>.= 3isfoundbysolvingthe system[ (;]=This yields the general solutionx= t, y= 2t;t.husthe eigenspace consists of allvectors of the form [:J=Geometrically,this is the line y= 2xinthe xy-plane. The eigenspaceto A =- 1isfoundby solvingthe system[ =:(:] = This yields solutionx= 0,y = t;thusthe eigenspace consistsof allvectors of the form = tGeometrica.lly,this is the linex=0(y-a.x.is). (b)The eigenspacecorrespondingto.A= 4isfoundbysolvingthe system[=: !] [:] = .This yields the generalsolutionx= 3t, y= 2t;thus the eigenspace consist of a llvectors of the form [:J=Geometrically,thisisthe line y= 3x-E.XERCISE SET 4.4141 (c)Theeigenspace correspondingto >.=2isfoundby solvingthesystem[ 00)(x]=(0].This -10y0. yieldsthegeneral solutionx= 0,y= t;thustheeigenspaceconsistsofallvectorsof theform [:] = Geometrically,t hisis the linex= 0. 11).(a)Theeigenspacecorrespondingto>. =4consist sof allvectorsofthe for m(:)= thisis theliney= Theeigenspacecorrespondingto >.=- 4consists ofallvectorsof theforrri [:] = tthisist he liney=-11. (b)The eigenspace corresponding to>.= 0consists of allvectors ofthe form= + this is the entire xy-plane. (c)Theeigenspacecorrespondingto A.= 1consistsof allvectorsof the form(:J= t thisisthe line x= 0. (a) (b) (c) Theeigenspacecorrespondingto>.= 1isobtainedhysolving" [-[oool yieldsthe general solution x==0,y= t,z= 0;thusthe eigenspaceconsistsof allvectorsof the [t [:] ;thls conespondtoaHnethmughtheoigin(the y-axis)inR3Slmilady, theeigenspaceto!.2 consistsofallvecto" of thefo-2- 8), + 15)(,\2 - 9)=(>-- 5)(.), - 3)2(-X+ 3) Thustheeigenvaluesare 5,),=3(withmultiplicity2),and), = -3. 16.Using theblocktriangular structure,the characteristic polynomial ofthegiven matrix is IA011 ).+ 20I, p(>-)=I= >- 2 ().+ 2)(>.- 1) - 1).0>- - 1 Thus the eigenvaluesofBare).=0(withmuitiplicity2),,\= - 2,and),= l. 17 .The characteristicpolynomial of Ais >. +12 p( A)= det( AI - A)=-1).- 2 1 2 -1=(.-\+ 1)(>. -1)2 ). EXERCISESET 4.4143 thustheeigenvaluesare.X =-1 and..\=1(withmultiplicity2}.TheeigenspacecorrespontJingto .>.= - 1 is obtainedby sol.-ing the system whkh yield$[:]l[ - :]. Similarly,the eigenspace co"espondingto !.=Iis obtainedbysol viug H -:-:1m =m whichhas the generalsolution' =t,y"-t - ' '= ' ' or(invedorform)= ' [-:] + t [-1] Theeigenvalues of A.!5 are). =( - 1)25 =-1 and >.=( 1)25 =1.Correspcnding eigenvectorsthe same asabove. 18.Thoeigenvalues ofAare !.=J ,!.=j, !.=0,end!.= 2 ngeigcnveclo" are[:j. r:J.. rll and[r"'peotively.TheeigenvaluesofA9 areA=(1)' 1,A =(!)' =,:, . :0)' =0 0. and>.=={2)!}==512.Corresponding eigenvectorsarcthe sameasabove. l9.ThecharacteristicpolynomialofAisp()..)= >.3 -- >.2- 5).. - 3=(..>.- 3)().. + 1)2;thusthecigeo- are.>.1 = 3,.A2=-1,>..3= -l.Webnvedet( A)= 3andtr(A) =1.Thusdet(A)= 3:; (3)(- 1)(-1} = .A1.A2>..3andtr(A) = I= (3) + (- 1) + ( -1)=.At+ .X2+ .>.3. W.The characteristicpolynomial of Ais p(>..)=>.3- 6.:\2 I12.>- - 8 =(.>.- 2)3;thus the eigenvalues are At =2,-\2= 2,A3= 2.Wehavedet (A)= 8andtr(A)= ().Thusdet(A)= 8=(2)(2)(2)= ..\.1.\2.:\3 andtr(A}= 6 =(2)+(2)+(2)= )q+.:\2+.\3. !1.The eigenvalues are .\ = 0and.>.= 5, with associa.t.edeigenvectors ( annGJrespectively.Thusthe eigenspacescorrespondtot he perpendicular li nesy= -!x andy .:::::2x. :2.Theeigenvaluesare..\= 2and>.=-1,withassociatedeigenvec-tors[ v;]and[ respectively.Thus the eigenspace::; correspond totheperpendicular lines y""' andy= - J2x. 144Chapter 4 23.Theinv..2- (b+ l )A + (b- 6a), so Ahas the stated eigenvalues if and onlyif p( 4)= p( - 3) =0.Thisleadstotheequations fromwhichweconcludethata = 2l b.,.,0. 6a - 4b = 12 6a+ 3b=12 25.T he characteristic polynomial of A i:; p( A)= >.2 - (b + 3)A + (3b - 2a),so A has the s tatedeigenvalW!S if l.l.lldonl yif p(2)= p(5)= 0.Thisleadstotheequations - 2a + b = 2 a+ b=5 fromwhi chweconcludet.hat a =1 andb = 4. 26.The chl'l.r!'l.ctcristicpolynomial of Ais p(>. ) = (A - 3)(A2- 2Ax + x2 - 4).Note thatthe second factor inthispolynomial cannot.have adouble root (forany value of x) since( -2x)2- 4(x2 - 4)=16 #0. Thust heonlypossiblerepeatedeigenvalueof Ais.>- =3,andthisoccursifandonlyif .\ =3isa rootoft.hesecondfactor of p(A),i.e.ifandonlyif 9-- 6x + x2 - 4= 0.The roots ofthis quadratic equationarex=1 and :t - 5.Fort hese values ofx,A=3isaneigenvalueof multiplicity2. 27.lf A'l=.f. thenA(xl- tlx)= Ax+ A2x:-:- Ax+ x= x +Ax; thusy=x+Ax is an eigenvectorof A correspondingtoA =1.Sirni lady,z= x- Ax isa.neigenvectorof AcorrespondingtoA = -1. 28.Accorrlingto Theorern 4.4.8,the characteristicpolynomial of Acanbe expressedas where>.1,>.2, ... ,AkarcthedistincteigenvaluesofAandm1 + m2 + + mk= n.Theconstant t.crrniuthispolynomialisp(O) .Onthe otherhar.J,p(O)==det( - A)=(-l)ndet(A). 29.(a)lJsinf:!Formulu.(22),the d1atacteristic equation of Ais.\2 + (a+ d)>..+(ad - be)= 0.This isn qundratric equationwithdiscriminant (a+ d)2 - 4(ad - be)=a2 + 2ad + d2- 4ad + 4bc=(a - d)2 + 4bc ThustheeigenvaluesofA aregivenby.A=d) J(a- d}'l+ 4bcj. {b)If (ud)2 + 4/>c>0then,from(a) ,the characteristic equation hastwo distinctrealroots. (c)If (a- df + 4bc:-:o0then, from(a), il:lone real eigenval ue {of multiplicity2). {d)If {n.- dV+ 4bc< ()then,from(a), t herea renorealeigenvalues. EXERCISE SET 4.4145 30.lf (a- d)Z+ 4bc> 0, we have two distinct real eigenvalues A1 and Az.The corresponding eigenvectors are obtainedby solvingthe homogeneoussystem a):tt- bx2= 0 - cx1+ (Ai- d)x2= 0 SinceA;isan eigenval uethis systemisredundant,and(usingthefirstequation)ageneral sol ution isgivenby x1 = t, Finally,settingt= -b. we ;,;eethat[a =b>.Jisaneigenvectorcorresponding to >.=..\i n.If thecharacteristicpolynomial of Aisp(.A)= A2 + 3>.- 4 = (>.- l){A + 4)then the eigenvaluesof Aare>.1=1 andA2=- 4. (a)FromExercise P3below,A-1 haseigenvalues.-\1 = 1and ,\2 =(b)From(a),togetherwithTheorem4.4.6,itfollowsthat.A-3 haseigenvaluesA1 "'""(1 )3 =1and >-z-=- ( -- - o\. (c)PromP4below,A - 41haseigenvalues AJ=1 --4 = -3 and,\2= - 4 - 4 = - 8. (d)FromP 5below,5Ahas eigenvalues)"= 5and ..\2;:::-20. ( e)From P2( a.)below, t he eigenvalues of .(x x) ==.AIIxll2 andsoA=(i1:1?t. '3.(a)Thecharacteristicpolynomialof the matrix Cis >..0 () 0co - 1A00Ct p( ,\}= det ( >. J- C) "" 0-I ,\ 0Cz 000- 1 A+ Cn- 1 Add.>.l ime.->therowto the firstrow,then expandbycofactorsa.longthe firstcolumn: 0 A2 00Co+ CtA )..'2 00co+Ci A - 1>.00Ct ..\0 .>. -1 Cz p(A)= 0 -1 0cz = 000- 1 ..\+C:n - 1 00-1 ..\+ Cn- 1 AddA2 timest.hesecvnd rowto the firstrow,then expand by cofactors alongthe firstcolumn. 0 Az 0co+ Ct >.+cz>..2 ,\2 0co + C)A+ C2A2 - 1 ). 0C2 p(.>.)== 00 - 1..\ + c,._l 0 -1 A+ Cn-l 146Chapter4 Continuinginthis fashion for n- 2 steps,we obtain = 1,\n-lco +Ct A + c2.\2 + + Cn.-2>-.n-21 -1>-+Cn-t = Co+ CtA+ C2 A2 + + Cn-2A11-2 + .isaneigenva lueofA,then.-\2 isaneigenvalueof A2;thus (.>.2I - A1)x= 0has uontrivial sol utions. False.If A =0 isan eigenvalueofA, thenthesystemAx = 0ha.snontrivial solutions;thus Aisnotinvertibleanrlsotherowvectorsand columnvectors ofAarelinearlydependent. !The statementbecomes trueif"independent"isreplacedby"dependent" .J False.Forexample,A= G haseigenva.lues).= 1and.\ = 2.!Butit istruethata :;ymmetricmat rixhasrealeigenvalucs.J flOJ.[1OJ False.For example,thereducedrowechelonformofA= h 2 l SI= 01 . True.WehavP.A(x1 + x2)=A1x1 + .>.2x2and,if ..X1f. >.2it canbe shown(sine"'x1andX2 mustbelinearlyindep(:ndent)thal .)qxl+ A2Xzf. {J(xl + X2)forany valueof a. WORKING WITH PROOFS147 (c)'!'rue.Thecharacteristic polynomial of Ais a.cubicpolynomial, o.ndeverycubicpolynomial has at least onerealroot. (d)1Tue.If r(A) = .>."+ 1,thendet(A) = (-l }np(D) = 1 =I0;thusAisinvertible. WORKING WITH PROOFS Pl.If A=[acb], t henA2 = [02 +beab + bdl2 andtr(A)A = (a+ d) [ab]= [02 +doab + d!]; thus dca + decb+ d2jcdac+dcad+ a-A2 _tr (A)A = [be -0 adcb _oad]=(- de0t(A)0] _det.(A)=- det(A)J nodso p(A)....:A'l- tr(A)A + det(A)J = 0. P 2.(a)UsingpreviouslyestabJishedproperties,we have 111')= det(Afr - AT)= dct((M- A)r) = det(M - 11) ThusAa11dAThavet.hc*lamecharacteristicpolynomial. (b )The e;gcnvaluesare2 and3 ineach case.The eigenspace of Aconesponrlingto).= 2is ob-tai nedby solvingthe systemr:J t.he eigenspaceof AT corresponding ':' t o>..=2 isobto.inf:dby solving [:] Thus the eigenspace of A corresponds to the liney =2x,whereaseigenspaceof A'rcorrespondsto 11= 0.Simtlarly,forA 3,the eigf'nspacc of Acorrespondsto x= 0;whereasthe eigeuspaceof AT corresponds toy= P3.SupposethatAx =.Xxwherex i- 0andAisinvert.ible.Thenx= A-1 Ax =A-1>.x=AA-1x and,s1nce,\:f;0(because.1isinvertible),itfollowsthatA-1x= }x.Thuslisan eigenvalue of .t\ - tandxiscormsponding P4.Supposetlt;HAx.\.xwhcr.x)= (s>..)x.Thus s.>.isaaeigen-valueof sAandxisa.corres ponciingeigenvector. P 6.If the matrixA =[::]issymmP.tric,then c :=:b a.ndso (a- d)2 + 4b and thenmaking bzand o3 into parameters. (c)The augmentedmatrixof the systemAx ::;0canbe row reducedto

(l 0 1 0 1L: 0] 1-1l 0 00:0 Thus thekernel of TA(i .e.the solution s pace of Ax = 0) consistsof allvectorsof theform X= [-s- tl[-1][-1] -s + t-11 = s+ t s10 t01 182Chapter 6 DISCUSSION AND DISCOVERY DL(a)True.If Tisone-to-one,thenTx = 0ifandonlyifx= 0;thusT( u - v) = 0implies u - v= 0andu= v. (b)True.lf T: Rn-+ Rnisonto,then (fromTheorem 6.3.14)it is one-to-oneand so the argu-ment given inpart(a)applies. (c)'ITue.See Theorem 6.3.15. (d)'ltue.IfTAisnotone-to-one,thenthehomogeneouslinearsystemAx =0hasinfinitely many nontrivialsolutions. (e)True.The standard matrix of a shear operator Tis of the form A=[or A=.In either case, wehavedet(A) =:=1'f:.0 andso T =TAisone-to-one. D2.No.The transformation isnot one-to-one since T(v)= ax v=0forall vectors vtb.at:a.reparallel to a. D3.The transformation TA: R"-+Rm.isonto. D4.No (assumingvoisnotascalar multipleof v).Thelinex=vo +tvdoesnotpassthroughthe originandthusisnotasubspace of nn . It.followsfromTheorem 6.3.7thatthis line Calmotbe flQ ualt orangeof alinear operator. WORKING WITHPROOFS P l.Jf Bx =0,then(11B)x= A(Bx)=AO=0;thusxisint he nullspace of AB. EXERCISE SET 6.4

- 3 !][: - 2 0][ 5 -1 2!] 1.!Tn

= BA =01-3=10- 8 2445 25 [: -2 [: - 3 :1][-8- 3I] [TA oToJ =A E3=101=-5- 15- 8 246744- 1145 [4 0

3 -1][40 0 20] 2.[Ts oTAJ= BA= - 1501=12- 918 2-3- 3638-1843 fTAoTaf=AB =3 -1][4 0 .Jr 18 22] 01-152=10- 316. - 362- 3831 58 3.(a)iT!) = [12)={b)[72c T1]== [!ITt oTz) ==_:] (c)T2(Tt (x1, x2)) =(3xt+ Jx2, 6x1- 2x2)T1 (T2(x1t x2))= (Sx1+ 4x2, x1- 4x2} Exercise Set 6.4183 0 2 4.{a)!T1]=1[T2]= () - 1 -30- 1

2 0][ 4 0

u 2

{b)[12o T1]:=0- 1-2l=3 0-1- 1-33 {T1o T3][- 0

2 0] [ 4 8 0] 10-1 = -2-4-1 -1 -30-1- 1-23 (c)T2{T,(x1,x2,x3))=(2x2,:r1+3x2,17:r1+3x2) T1 (T2(x1, x2, x3))=( 4:rt + 8x2, -2x1 - 4x2 - x3, -xi - 2x2 + 3x3) 5.(a)The st andard matrix for the rotat1onisA1 = [- andthe standardmatrix forthe tion isA2=[Thust hestandardmatrixfortherotation followedbythe reflectionis [0 1][0-1][lOJ A2A1 =1010=0- 1 (b)Thest andardmatrixfortheprojectionfollowedbythe contractionis (c)Thestandardmatrixforthereflectionfollowedbythe dilation is .[:lOJ[1OJ[301 .ilz A1== 030-10-3 6.(a)The st andardmatrixfor ther.ompositionis A,A, A,[_J] ( b)Thestandardmatrixforthe compositioni:s (c)The composition correspondstoacounterclockwise rotation of 180.The standard matrix is [-10] R zChf'lonformofthe m11.trix.4is 100 1 -!] 4 010 1 R= 25 001 t3 ;j 5 00000 Thusthevectorsf )=(1, 0, 0,n.r 2= (0, 1, 0, 4. a.ndTJ= (0, 0, 1, },formabasisforthe rowof A\VealsoconcludefromnninspectionofRthatthenullspace ofA(solutions of Ax -:::O)consists of vectors of the form Thusthevect orsn1 =(1, 0),n2 =:( -0, 1)formabasisforthenulls pace ofA. It iseasyto cherocolumnvectors andA =uvr, thenAx =(uvT)x= u(vrx) = (v. x) u .Thus Ax= 0if Bndonlyif v x= 0.i.e.if and onlyif xisorthogonalto v .Thisshowsthat ker (T) =v.l. . Similar!_\,I herange ofT conststs of allvectors of the formAx= (v x }u,and so ran(T) =span{ u}. 23.(a)If B1. xfor some nonzero vector x, then A2x= >.2x.On the other hand, since A2 = (u v )A, wehaveA2x= (u v )Ax= (u v)>.x .Thus >.2 =(u v)>.,a.ndif>. :f nit. followsthat>.= u v . DISCUSSION ANDDISCOVERY217 (c)If A isany square matrix, thenI - A failsto be jnvertible if and only if thereis anonzero .vector xsuchtha.t(I - A)x =0; this isequivalentto saying t hat>. =1isan eigenvalue ofA.Thusif A :::::uvT isanmk 1rtatrixforwhich I- Aisinvertible,wemusthave u v=/=1 and.42 :fA. DISCUSSION AND DISCOVERY Dl.(a)True.For example,if Aism xnwherem> n(morerowsthan columns)thenthe rowsof Aformaset of mvectorsinRnandmustthereforebe linearlydependent.Ont.heother hand, if m< nthen the columns of Amust be linearly dependent. (b)False.If theadditional rowisali nearcombinationof theexisting rows,thentherank will notbe increased. (c)False.Forexample, if m=1then rank( A)= 1and nullity( A) =n - 1. (d)True.such amatrix musthave rank lessthan n; thusnullity(A)::::::n - rank(A)==L (e)False.UAx= bisinconsistentforsomebthenAisnotinvertibleandsoAx=0hM nontrivialsolutions;thusnulHty(A);:::1. (f)11-uP..Wemusthaverank(A) + nullity(A)=3;thusit isnotpossibletohaverank( A)and nullity(A)bothC' qualt.o1. D2.If Ais mxn,then ATis nxmand so,by Theorem 7.4.1,wehaverank(AT) + nullity( AT)= m. D3.If Aisa3x5matri."l- 2bt.Thusthe system Ax = b 00 1 b1- 2i>l + b3 ist! itherinconsistent(ifbt- 2b2+ b3 =I=0),orhasexactlyonesolution(if b1 - 2b2 + 63 =0).The latter incl udest he case b1 = b2=b3 = 0;thusthesystemAx = 0hasonlythetrivial sdl'i.ition. IfA= [ thenA rA[ 612].Itisclearfrominspectiont hnttherowsofAand.4. T A - ) 2 I1224 aremultiplesofthe singlevectoru= (1,2).Thusrow(A)= row(Ar A)isthe!-dimensionalspace consistingofallscalarmultiplesofu .Similarly,null (A) = null(AT A)isthe!-dimensionalspace consisting of allvectorsvinR2 whichare orthogonaltou ,i eallV('ctorsof the formv=s( -2, 1). The reduced row echelonfromof A=[111]is[0107]andthereducedrow echelonformof 23-4I-6 ATA =[Thusrow(A)=row(ATA)isthe2-dimensionalspaceconsisting -7- 1117000 of allli near comb1ne.tions of the vect orsu1 =(1, 0, 7)andu2 ={0, I , - 6).Thusnull(A)=null( AT A) 1sthe1-dimtlnsional space consisting of aJIvectorsvinR3 whichn.reorthogonaltobothu1 andu 2, i e, all\'CClorof theformv= s( - 7,6, 1) 17.augmentedma.trixof the systemAx =bcan berel!\lo.:edto -3bt 01 b2- bl 00b3- lb2 + 3bl ) 00b4+ 2b, 00b5- 8b2+ 7bt thusthe systemwillbeinconsistentunless(b1, b2, b3,b4, b5)theequationsb3=-3bt -14b2, b,.=2bt- bs=-?b1 + whereb1 canassume anyvalues. 18.The augmentedmatrix of the systemAx= bcanbe 1educedto [123-1: 0- 7-85! 0000: thus the system isconsistent if and onlyif b3 = b2 + b1 and, inthis case, therewillbe in!'lnitely many solutions. WORKJNG WITH PROOFS223 DISCUSSION AND DISCOVERY D l.If A is a 7X5 matrix wit h rank 3, t hen A'ralso has rank 3; t hus dim( row( AT)) = dim( col( AT))= 3 and dim(nuli(AT))= 7 - 3=4. 02.If Ahasrankkthen,fromTheorems7.5.2and7.5.9,wehavedim(row(AT A))= rank(A1'A) = rank(A1')= rank(A)= k anddim(row(AAT})=rank(AAT)=rank( A) =k. D3.If AT x= 0has onlythe trivial solution then, fromTheorem 7.5.11, Ahas full rowrank.Thus,if Ais m>pectionthatPis symmetric.It isalso apparent thatPhasrank1 si nce each ofits rowsisasc,\larmult1plcuf aFinally,itiseasytocheckthat P2 = Pand soPisidl.mpot enl. [ 3?.].16'l LetM=- 1uTheni\f1 M=[ 91;]and.fromTheorem7 7 5,thestandardmatrixforthe I:l orthogonalprOJCcltonof !1.3 ontoW= span {a1, a2}isgivenby P= AI(MTM)-IMT --=[-! [13-9][3-4 13257- 92620 [ 113-84 1]= -1- -84208 32579656 96] 56 193 Wenot efrominspectiont.hatlhematrixPis symmetric.The reducedrow echelon formof Pis ! andfromthisweconcludethat Phasrank2.Finally,it iseasytocheckthat P' = [,m2!1[,m = 2!1[ =P 193 and so Pic::idl'mpotent EXERCISE SET 7.7231 16.LetM[ - : ThenMT M:]andthe tandW"dmatrix forthe orthogonal of R3 onto W= span{ a1, a2}is givenby -23] [1-2 304- 2 -102305 From inspection we seethatPis symmetr ic.The reducedrow echelonform of P is

0 1 0 and fromthisweconcludethat Phas ranK.2.Finally,it iseasy to checkthat P2 = P. 17.ThestandardmatrixfortheorthogonalprojectionofR3 ontothe xz-planeisP=This 001 agceeswiththefollowingcomputationrn;ingPo, mula(2nLetM[:; l ThenMT M, andM(MT M) -0 MT [: [;:1 :]= :J. 18.Thestandardmatrixfort heorthogonalprojectionof R3 ontothe yz-p]aneisP=This 00I 19.WeproceeriasiuExample6.Thegeneral ::oolution of the equationx + y + z=0canLewrit ten as andsothetwocolumnvectorsontherightformabasisfortheplane.IfMisthe3x2matrix thesevectorsas Jtscolumns,thenMT M= andthe standardmatrix of the orthogonal projectionontothe planeis [ 2 -l] [-1 1 0]= 2-12- 1013 lJ-1- 12 Theorthogonalprojectionof thevector von the plane isPv =-1 - J-1] [ 2][ 1] 2-14= 17. - 12- 1- 8 232Chapter 7 20.Thegeneralsolutionof theequation2x - y + 3z =0can be written as andsothetwocolumnvectorsontherightformabasisfortheplane.1f Misthe3x2matrix havingthese vectors as its columns,then NfT M=[!

and the standard matnx of the projection ontotheplane is P = M(MT M)-1 MT =23_!_[ 10 -6][120]=_!_2133 [10][ 102-6] 0 14-65031146 1- 35 [ 10 Theorthogonalprojection of thevectorvon the plane isPv =]_2 14-6 2 -6] [ 2]1 [34] 1334=1453 1li- 1-5 21.LetAbethematrix havingthe givenvectorsasits columns.The reducedrowechel. is nn eigenvalue of A with corresponding eigenvectorx(xf0).Then A2x=A(Ax) = A(.h) = ..\2x.On theotherhand, sinceA2 =A, we haveA2x= Ax =.>.x.Since x=/=0, it follows 'that >.?=>.andso >.=0or 1. DlO.Using calculus:The reducedrow echelonformof lA I bj isthus the general solution of 00010 Ax =bis x=(7- 3t, 3 - t , t} where-oo < t< oo.Wehave ll xll2 = {7- 3t)2 + (3 - t)2 + t2 =58 - 48t + 1lt2 andsothesolut.ionvectorofsmallestlengthcorrespondstoft lllxii2J =-48 + 22t =0,i.e.,to t :.=:Weconcludethat Xrow= (7=-- ii, 3- H, =( 15p

Usinganorthogonalprojecti on:The solutionXrowisequaltothe orthogonalprojectionof a.ny solutionofAx = b,e. g.,x= (7, 3, 0},onto t herow space of A.Fromtherowreductionalluded toabove,we sec that thevectorsv1 =(1, 0, 3)andv2 ==(0, 1, J)formabasis forthe row space of .4.Let.Bbethe3x2matrix havingthese ver,toras its columns.ThenBT B=

andthe standardmatrix forthe orthogonalprojection ofR3 onto W= row(A}isgivenby = 1\ 3110 Finally,in agreement withthe calculus solution,wehave 1 Xrow=Px =-11

[;]=f [ !] 31100 1 24 238 Chapter 7 Dll.The rowsof RCormabasisforthe rowspace of A, and G= RThasthese vectors asits columns. Thus,fromTheorem7.7.5,G(GTG)-1GTisthe standard matrixforthe orthogonal projection of RnontolV = row{A). WORKINGWITHPROOFS Pl.If xanJ yarevectors inRnand if aand /3arescalars,thena (ax+ /3y)=a(a x) + /3(a y). Thus a (ax + /3y)(a x)(a y) T(ax + /3y)=llall2 a =a:lja112a += aT(x) + /3T(y) whichshowsthat Tislinear. P 2.If b=ta,thenbTb=b b=(ta) (ta) = t2a a =t2aTaand(similarly)b bT =t2aaT;thus 1T12T1T bTb bb=t2aTa ta aaT a aa P3.Lf'lPhl'asymmetricnxnmatrixthatisidempotent.andhasrankk.ThenW-=col(?)isa k-dimensonalsubspaceof R".WewillshowthatPisthestandardmatrixfortheorthogonal projecl ionof Rnonto W,i.e, thatPx =projwxforallxinRn .Tothis end,wefirstnotethat Px belongstoWandthat x=Px + (x- Px) =Px t(I- P)x ToshowthatPx = projwxitsuffices(fromTheorem7.7.4)toshowthat(I- P)xbelongsto \1' .1ndsinceW- col{ A)= ran(P).this IS equtvalemto showmg thatPy (J- P)x =0forall ymR"Finally,sincepT = P= P2 (Pis symmetricandidempotent),wehaveP(J- P) = P- P2 - P - P= 0and so forevcrvxandymR"Thiscompletesthe proof. EXERCISE SET 7.8 J .Firstwenotethatthe columnsof Aare linearlyindependent sincetheyarenot scalarmultiples of eachother;thusAhasfullcolumn rank.It followsfromTheorem7.8.3(b)thatthe systemAx= b has aunique least squares solution given by The least squares error vectoris b- Ax =

= 545 11811 15 andit is easyto chP"kthat thisvector isin factorthogonal to each of the columns of A.For example, (b- Ax) Ct(A)= 1\!(-6){1) + {-27}(2) + (15)(4)]= 0 EXERCISESEl' 7.8239 2.The colunms of A are linearlyindependent and soA has full columnrank.Thus the systemAx= b has aunique least squares solution givenby 0213- 1- 2..9 -l [[21 6]-2I I]lj- 21[-14] The least squareserror vector is anditiseasy ,to.check that this vector is orthogonal to eachof the columnsof A. 3.FromExercise1,the least squares solution of Ax= bisx=f1 Ax =23

= 16 [1 -1][28] 45 11811 40 On the other hand,the standardmatrix forthe orthogona.J projection of R3 onto col(A)is p_A(ATA.)-tAT::::.:[2125]-l[ 1 2 4]= _12535-135220 452090170 :).ndsowehave projcoi(A)b= Pb =

2090 [- = 111 = Ax 170f>40 4.FromExercise2,theleast squaressolutionof Ax =bisx= 2\[ -t:];thus [2-- 2][ tj6] Ax =1.l.!..[ 9]=..!:_- 5 3121-142113 On the other hand,thestandardmatrix of the orthogonalprojectionont.ocol(A)is andsowehave =211 =Ax 171.13 240Chapter 7 5.The least squares solutions of Ax =b are obtained by solving the associated normal system AT Ax= .4ThwhichIS Sincpthernntrix onthe leftisnonsingular,this systemhas the unique solution X=[J:1]= [248]-l[12]=2_[6 -8][12]=[to] X286880-8248l The errorvectoris andtheleast squares errorisli b - AxiJ= + + (0)2 = Ji = 6.The least. squares solutions of Ax= b are obtained by solving the normal systemAT Ax =ATb which IS [1442][Xl][ 4] 42126X2=12 This(redundant)systemhas infinJtelymany solutions givenby X=[::J= 3tl= [!J + t r n The errorve,.lorjc:; andthe least squares errorisli b - Axil= ++ = J* = :Lfl. 7.The least SCJUf\rf'Ssolut1ons of Ax= b are obtai;1edby solving the normal system ATAx =ATb which is Theaugmentedmatdx of thissystemreducesto[: solutionsgivenby 0 0 1 '7] '-a 1:i; thusthereareinfinitelymany 0'0 EXERCISE SET 7.8241 Theerrorvectoris andl hcleastsquareserrorisli b - Axil =J(!)2 + = =8.The least squares solutions of Ax= bare obtained by solvingAT Ax= ATbwhichis [1 [::]=173350X36 Tbeaugmentedmatrix of thissystemreducesto 0 1:-sl. 1:tf; thusthereareinfinitelymany 00I0 :;olutionsgivenby x =[:;][W1,=:]T heerrorvt.'Ctoris andt heleast squareserror isllb - Axll= V( !iY + {fr )2+ {- fi-)2=,f1j; = 2f[I . 9.Thehnc"' modelfo,the givendatabM v= ywh"cM [i il andy= [;] .Theleastsqua li b- projwb ll ; thusprojwbisthe onlybest approximationtobfromW. P5.If ao, a 1, a2, ... , amare scalars sucht hat aocdM) + a 1c2(M)ta2c3(M ) +.. + amCm+l(M) = 0, then ao + n1:ri++ + amxi = 0 foreacht=1, 2, ... , nThuseachx,isarootof thepolynomialP(x)= a0 + a1x+ + amxm. But suchapolynomial(if notidenticallyzero)can haveat most mdistinct roots.Thus, if n> m and if at.least.m + 1 of the numbt>rsx 1, x2,. , Xnare distinct,then ao= a 1= a2= =am= 0. This showsthatthe columnvectorsof Mare linearlyindependent P6.If atleastm+ Iof thenumbers:r1, :r2, ... :rnaredistinctthen,fr omExercise P5,thecolumn vect orsofMarelinearlyindependent,thusMhas fullcolumnrankandMT Misinvertible. EXERCISE SET 7.9 1.(a)v, v2 = (2)(3)-t(3)(2)=12-:/;0,thusthevectorsv1,v2 donotformanorthogonal set (b)Vtv2 = (1}(1)t(1}(1) = 0; thus the vectors v1, v 2forman orthogonal set.The correspond-ingOrthonormalSCtISql= u::u= = = (-32, (c)Wehnvev1 vz= V t v3 = v2 v3 = 0;thusthe vectors v1,v2,v3 forman orthogonal set.The correspondmgorthonormal set isql= = (7s,O.js), q3 = (d)Althoughv1 v2- Vtv3 = 0wehavev2 v3= (1)(4) + (2)(-3) + (5)(0)= - 2-:/;0;thusthe vectorsv1 ,v2,v3do not forman orthogonal set. 2.(a).,. v2 = 0;thusthe vectorsv1,v 2 forman orthogonal set.The corresponding orthonormal set isq l= ?tJ) (b)v1 v2 :1:0;thusthe vectorsv 1ov2 donot forman orthogonal set. (c)V tv2 -:/;0,thusthevectorsv1,v2,v3 donot forman orthogonal set . ( d )Wehavev1 v2 = v1 v3 = v2 v3 = 0;t husthevectorsv1,v2,v3 formanorthogonalseL. The corresponding orthonormal set isq1 = -j, = ! .-i), q3 = (!, 3.(a)These vectorsforman orthonormal set. (b)Thesevectorsdonotforman orthogonal setsincev2 v3= - '3;-:/;0. (c)Theseformanorthogonal setbut not orthonormnl set si ncell v3ll= .j3 :1:1. EXERCISE SET 7.9247 4.(a)Yes. ( b)No;ll v1 1lf. 1 ;,.ndll v zllf. 1. (c)No;Vtvzf. 0.vz vJf. 0,llv= {:3, I) .Then{v1, v2}isanorthogonalbasisforR2,andthe vectorsq1= II= ( jfo) and q2=

=(fro, ftc) forrnan or-thononnalhAS isforR2. 28.Let V J= Wt=(1, 0)andV2=W 'J- v 1=(3, -5) - n)(l ,0)=(0, - :>) . ThenQ1 = = (1,0)amiq 2= ft:!n= {0, - 1} formanorthonormalba:;is forR2 29.LetYJ= w, = (1, 1, 1),voz=w 2--=(-1, 1,0)-J. , 1)=(--1, 1, 0},and formu.northonormalhaslsforR3. 30.Let.v,,..WJ=(1, o. 0), Y2= w 2- u;;u! v l=(3. 7, -2)- (f)(l , O, 0)= (0, 7, -2). and Then {v 1 1 v2, v3}isan orthogonalbasis forR3 1 and the vect ors -s VJ q,= llvdl= (l, O, O), VJ(Q30105) q3 = llv3U=' ./119251 J uns form anorthonormal basis forR3. 250Chapter 7 31.LetV l=Wt= (0, 2, 1, 0), V2= w2-VJ= (1, -1, 0, 0) - (1, 0)= (1, -i, and Then {v 1,v2, v3, v 4}isan orthogonal basisfor andthevectors - YJ- {02VsJ50)- - ( v'3(i- VJo:.&.Q0) ql - llvdl- '"5 Sl,q2 - llv2116I30'ISI' - VJ- (.{IQ.@v'iO)-- (v'iS.illuTI.iil) q 3-11v3ll- 10 '10,- s,--s- q 4 151s, - tss form anorthonormalbasisforR4. 32.LetVJ=WJ= (1. 2.1. 0). V2= Wz- = (1,1, 2, 0)-2, 1, 0)= (0), and Then{v1, v2, v3,v 4}isan orthogonalbasisforR4,andthe vectors form an orthonormalbasisforR4 33.The vectorsw,= (0),w2 = ( 0),andw3 =(0, 0, 1)forman"rthonormalbasisfor R3. EXERCISE SET 7.9 251 34.Let A be the 2x4 ma.trixhavingthe vectors w1 =!) and w2= 7J,asrows. Then row(A) = spa.n{w1, w2},and uull{A)= spa.n{wt, w2}.1. .Abasisfornull(A)canbefoundby solvingthelinPnrsystemAx =0 .The reducedrowechelonformof the augmentedmatrix forthis system is 10 _l 2 l 2 4I .!.:0] _!I0 4I andooageneralootut;on;, by[m , [\ -1;]= [-!]+ [=;] the vootornw, = 4, 1, 0)and w4= ( 0, 1)formabasisforspan{w1, w2}-'-,andB= {wJ, w2, w3, w4}is abasisforR4.Notealsot ha t,inadrlitiont obeing orthogonaltow1a.ndW'J,thevectorsw3 and w4arasi::;{qt , Qz, Q:$,q 4}whereQ1= Wt(LQ2= Wz::::(-7.f,7:3 0),Q3= 11:;11= (-js ,-)5, -:76,0),andq4 = 11:!g = (--dis, - )r8, 7rn,o). 35.Notethat W3= w1 + Wz .Thus the subspace Wspannedbyt hegivenvectors is2-diinensiooal with basis{w1, w2} .Let v1=Wt=(0, 1, 2)and w2V1)( 2)()(21) v2= w2- ll vdl2v1 =(-1, 0,l -5 0,1 ,2;;- Then{v1. ,l}i:::an Ol lbogonalbasisfor\\' ,antithe vect-or!' V1l2) Ut= ll vdl= tO.-/S', formanorthonor malba.c;isfor 36.NoLethat w.1= w1 - W -zt- w3Thus the ::;uhspace\V spannedhythe givenvectors:3-dimensional with { w 1 , W ?, w3}.Letv 1 =w 1 =(-1, 2, 4, 7) .t\ndlet. v ...= \1: 2 -=(-304-2)- (2.)(-l24.., , )""' (- 1!_J. ..Jlv1jj 2''' ;o11 '-:,12 w3 v 1v 2,(9) llvdl2 v1- llv2jj2v2 = (2, 2, 7, -3)-70 (- 1, 2,-1,7)-( 31843)( - 41- 26 401 14'7)7'2 14 (9876376858913032) =2005 ) 2005 >2005 I- 200.'5 Then{vt . v2, VJ}isanorthogonalha.osisforw. andthevectorsUj= n::u= (j.ffl,,;!1.,.(- 41- 2s2-35)I(987637685891-3032) U2 =Uv211=JS'iit4 '/5614 1 v15iii4'J5614'an(UJ=llvlil=J 155630105'v'l55630t05 ',/155630105 1 J l55630105 forman orthonormalbasisfor W. 37.that u1 and u 2are orthonormal vectors.Thus the projectior. vf wonto the subspace Wspanned bythesetwovectors isgivenby WJ= projww= (w u t) u1 + (w u 2)u 2=(0, + (2)(0, 1,0)=2,andthe componentof worthogonaltoWis Wz= W-Wl =(1,2, 3)

252Chapter 7 38.First we findvDorthonorma.lbasis {Q], Q2}forw by applying the Gram-Schmidt process to {UtI u2} LetVt= U t= (-1,01 112)1v2= U2- =

-1(-1,0,1,2) = (j,l, -!,!},andlet ql==( --Js, 0,7s, fs), q2 ==( Jh. Ji2).Then {qt , Q2}isan orthonormal basisforH' . and sothe orthogonalprojection of w= (- 1, 2, 6, 0)onto Wisgiven by andthe component of worthogonalto Wis W2= W - Wt=( -1, 2,6, Q) - ( 1 = ( 1

1491 39.If w=(a.,b,c),thenthe vector isanorthonormal forlhe!-dimensional subspacelV spannedbyw.Thus,usingFormula (6), the sta.ndardmatrix forLheorthogonalprojection of R3 onto Wis P= u Tu = 2 :22 bIabc]= 2

2 abb2 be [a][a 2 abacl a1- + ca++ cb.'2 c.a.ccc-DISCUSSION ANDDISCOVERY D1.If aand b"arenonzero,thenu1 = (1, 0, a) and u2 = (0, 1, b)formabasis forthe plane z= ax + by, amiapplKHion of tlw Gram-Schm1dt process to these vectors yields'\n orthonormal basis {Q1, Q2} where 02.(a)span{ vt} =span{ wl}, span{ v1, v2}=span{w1, w2} 1 and spa.n{vJ , v2 , v 3}=span{ Wt, w2, w3 (b)v 3ISorthogonalto span{wt, w2}. D3.If the vectorsWt, w2, .. , are linearlydependent,t henat leastone of the vectorsinthe listis a linearcombination oftheprevious ones.If w1 is alinear combination of w1, w2, . ,w1_1 then, whenapplyingtheGram-Schmidt processat the jth step, thevector v iwillbe0. D4.If Ahasorthonormalcolumns,thenAATisthe standardmatrixfortheorthogotia.lprojection ontothe column space ofA. D5.(a)col(M) = col(P ) (b)Find an orthonormal basis forcol(P) andusethesevectorsas the columns of the matrixM. (c)No.Anyorthonormalbasisforcol(P )canbe usedt o formt hecolumnsof M . EXERCISESET 7.10 253 D 6.(a)True.Anyorthonormal set of vectors islinearlyiudeptmdent. (b )Pa.lse.Anorthogona:lsetmaycontain0.However,itist r uet hatanyorthogonal 'setof nonzerovect.ors islinea.rlyindependent. (c)False.Strictly speaking, the subspace {0} has no basis, hence no or thonormal basis.However , it true thatanynonzero subspacehasanorthonormalbasiS. (d )1'ruc.'Thevect.orq 3isort hogonalto the subspacespan{w1, w2}. WORKING WITH PROOFS Pl.If {v1, v2, .. . , isanorthogonalbasisforW,then{vt/llvl ll ,v2/llv2ll. . .. ,isa.n orthonor malbasis.Thus,usingpart(a),theorthogonalprojectionofavector xonWcanbe expressedas 0(Yt)V1(V2)Y2()Vk proJwX=x . llvtll +x.l!v2llll v1ll+ . .. +x.PZ.lf A.issymmet.ricandidempot ent,t henAisthes tandardmatrixof anorthogonalprojection operator;namelytheort hogonalprojectionofR"ontoW=col(A).ThusA= UlJTwhereU is any11xkmatrixwhose columnvectorsformanorthonormalbasisfortV . P3.Wernust.provethat v iEspan{w1, w2 ___, w,}foreach J- 1, 2, . . -- Theproof isby induction on J Step1.Since v1 =w1,we have v1 Espan{w1 } ;t.hllSt.he stat ementistrue forj=L Step2(inductionstep) .Suppose the st atemeot ist.r ne forintegerskwhichareless t han or toj,i.e.,fork =l , 2, . . . , j .Then andsincev 1C ,c;pan{ w d, v'/.Espan { Wt, w2), . .. 0 a.ndv;(7s pan{ Wt , w2,. .. ,Wj} ,it followsthat vJ+lCspan{ Wt ,w2, .. . , wi, wj 1.1}Thusifthes tatement.istruefor eachoftheintegersk= 1, 2, ... , jthenitisalsotruefork= j+ 1. T hesf:two s tepscompletetheproofbyinduction. EXERCISE SET 7.10 1 .T hecolumn vec.t,orsof the mat rix Aarew1 = g}andWz=_ Applicat ion of t heGram-Schmidt procE'SSto vect.oryields WehaveW1:={ w, q i)q 1 =J5q landw2 ={w2 q1 )q l+ (w2 Q2)Q2=-/5q l+ ../5q2.Thusap-plio.:ationof Form;.:!a(3)yieldsthe following QRdecompositionofA: A===fl l-i] =23v'5 [J5V5j= QR V50v'5j 254 Chapter 7 2.Applicationof the Gram-Schmidtprocesstothe column vectorsWtand w2 of Ayields WehaveWt= J2q1 andw2 = 3J2q1 + J3q2.This yieldsthe following QR-decomposition of A: [ l [11]1272-"J!J23v'2 A =01=013[l = QR 14110J3 7273 3.Applicationofthe Gram-Schmidtprocessto the columnvectors w1 andw 2of Ayields [ 8l aJ26 Q2=s0t JJ26 WehaveWJ= (wl Qt)CJ1- 3q,andw2= (w2 q, )q l+ (wz Q2)Q2- + "fq z_ thefollowing QR-decomposilion of A: [11]A= =-: 8l 37263.l

[o =QR 3726 1\ppliratton nfthPGrn111-Schrnidtprocesstothecolumn\'ectors w1, wz , w3ofAvields Thisyields \Vehavew1= J2q1,w2 ="lq1 + vl1q2,andw 3= .J2q1 - 1q2 + Q3This yields the follow-ing QR-dec01npositinnl.lfA: A [1 0 2] 011=073' 12QIl 7273' [v'2 7s0 --Ja0 v'2 v'3 0 J2] il - 3 h1 3 =QR 5.Application of theGram-Schmidtprocesstothe columnvectorsw11 w2, w3of Ayields

Q2 = 719 WehaveWt=J2q1,w2 = +andwa=J2ql + 3'(Pq2+Thisyieldsthe followingQR-decomposition of A: A=2 1]I1=72 3l0 1 738 1 -73'8 6 73'8 J2]=QR .i!! 19 0 EXERCISE SET 7.10255 6.Application of the Gram-Schmidtprocess to thecolumnvectors w1, w2, w3 of 11yields Wehavew1 = 2ql,w2= - q1+ qz,andw3 =+ + ./i-G3This yieldsthefollowi ngQR-decomposition of A: 2- 1 A= l

- 1 r :L [-;1 1oJ-22 0] [o1 100 J]=QR 72 h [-1 1] =[-3_: f3 7 .From3,wehave.4=21 3 :Jv"itl=R IJ ..rtuQ T husthennrmnlfor 2J27LO 3W'16 3 Ax =bcanbe expressedasRx = Qrb, which is:

,J,!l m=Lil Solvingthi.ssystemhyback substitutionyieldstheleastsquares solutionXz=x1 = -[102]- [-72- [v'2v"2 8.FromEx.erci$c1,wehaveA =011- 0,16 ov'3 120-jz73 -)60() systemforAx =bcanbe e.xprcssedasRx=Q1' b, whichis: ,.. v2

[::]= 0 WX3 3v 6 0 1 J3 2 v'G v'2] - :/f.""'QR.Thusthenormal 2..f6 :r Solvingthissystem byback substit utionyieldsX3= Xz=x,= 0.Note that,inthisexample, the syst.etnAx = bis consistent and this isits exact solution. [12 1] - kl [Vi 9.F'romExcrci11e5,we haveA =111= o 0 310'.;?S87190 41 0 J2] 3'(P=QR. Thus the normal .ill 19 systemforAx= bcanbe expressedas Rx = QTb ,whichis: w 2 1 v'2 .ill 2 l - v'38 3 0'J19 Solving this system by back substitutionyields X3=16,x2= -5, x,=-8.Note that, inthis exam-ple,the systemAx= bisconsistentand isits exact solution. 256Chapter 7 10.PromExercise6,wehaveA=! :] -H[:- Ill =QR.Thusthenormal system -110IlIO 272 for.-\x - bcanbe expressedasRx = Qrb,whichis

i] [=:]=[t 00X30 1 72 Solvingthis systembyback substitution yields x3 =-2, x2 = 121,x1 =11.Theplane2x - y + 3z= 0correspondstoa.l.wherea= (2, -1, 3).Thus,writingaas acolumn vector,thestandardmatrixforthe reflectionof R3 about the planeis 2T H= 1 - --aa= a r a

=[

0016-39-7 andthereflectiOnof I he vee lorb=(1, 2, 2)about that plane is given,incolumnform,by Hb= !2.lhevlane.x1-y- lzOcorrespondstoa1 wherea=(l,l,-4). Thus,writingaMacolumnvector, t.hestandardmatnxforthereflection of R3 about theplane tS [ 1-4][ -18I- 4:;-4- -l16:1.:_l 999 o.ndtheretlec:ti>nof thevectorb=(1,0,1)abouttltatplaneisgiven,incolumnform,by Hb H-:Ul

0 [-: - 1 -:] u 2 -!l 3 13. 2T 11 I H = l - - aa-= 3 aT a 0-1 2 3

0

2[ - : - 1 u I 3 14. 21 I1 2 H=1- --aa= = 3 aT a 62 4J 0-2 2 3

00 OJr 00

00 - 0] 2T 100201- 1 92 ITITII 15.Fi= 1- - aa 296 aT a 01 11 - 11 TIITIT 003-3 _... i 7 IIII -n EXERCISE SET7.10 257

= 1342 '] 00 0][ 1 -2-3T515515 21002-246 47_14. 16.H- l- -aar = 15H'5 -15 aT a 01o10-369 2412 5 -ii--s -g 00 1-123 2_.i_:.!13 1515 -5 15 17.(a)Leta=v-w=-=(3,,1)-(5,0)=(-2,4),

Then His the Householder matrix forthe reflectionabout a..l,and Hv =w. (b)Leta=v-w=(3,4)-(0,5)=(3,-1),

4 5 Then Histhe Householder matrix forthe reflectionabout a..l,and Hv =w. (c)Let a= v.- w= (3,4)- (,- {/) =(6-;12, S+l'l).Then the appropriate Householder ma-trix is: 2T[10]2 H=I- --aa=- ----;:::: aT a01&0- l7v'2 17-25/2]r .1__.. 2l'n =1 33+8../2--2- ../2 18.(a)Leta= v- w,., (1, 1)- (v'2,0) = (1- v'2, 1).Then the appropriate Householder matrix is: H=J .....[1OJ___2_[(1- v'2)2 1- v'2] a1'a011-2v'21-/2I = 222 (b)Letv- w=(1, 1)- {0,/2) =(l, 1- .J2).Thenthe appropriate Householdermatrix is: H= 1 __2aaT =[10]...__._2_ [ 11- v'2] aTa014-2J21- v2(1- /2)2 =[oto]_ [ lv'2 -2 =2- v'2J:i -2- 2 ( 1 ; 1 - ''22= ...Ir.:.t -'-,J2w _ill 22 0 .i _ _.& 22 00"' v'3 001 FromaUurd such construction,thefourt-h entry inthe thi rdcolumn of Q2Q1Acanbe zeroedout by multi plyingbythe orthogonalmatrix Q3 as indicatedbelow: sV2- :1 -2- 2 v'622= 0- ./3 0 \12 0 0 0 _ill 22 :&_:L 22=R 02 00 Finally,setting Q= Q[QfQT = Q1Q2Q3,weobtain thefollowingQR-dccomposition ofA: :1- .i1- :il 2626 ::111:il 26-26 0 1:il J-26 Ifl 22 __ill

A= 00 0 ...... 260Chapter 7 25.Since A = QR,the systemAx = b is equivalentto the uppertriangular systemRx = QTbwhlcbis: 26. [v'3- v'3-33][:ttl[73 0.../20:t2=0 00 4 X3.:i 7s3 Solvingt hissystembybacksubstit utionyields:t3 = 1, :t2=1, :t1 =1. (a) (b) Since aaTx = a (aTx) = (aTx)a, we have Hx =(I- af.aaT)x=/x - af.aaaTx=X- e:1T:)a. Using the formula in pa.rt (a), we have Hx = x- (2$:)a = (3,4, 1}-1:(1, 1, 1) =( -1, -Jj). Ontheotherha.nd,wehaveH= I - ;faaaT = j ""[-1-:andso 00l111 _1l a3_3 DISCUSSION ANDDISCOVERY Dl.The standard matrix for thereflectionof R3 about ef is(as shouldbe expected)

2[-001000001 and fort.heothers. D2.The standardmatnxforthereflectionof R2 aboutt heliney=mxis(takinga =(1, m))given by D3.If s = v's3,thenllwll= ll vllandtbeHouseholderreflect1onabout.(v- w).lmnpsvinto w D4.Sincellwll = llvll ,theHouseholderreflectionabout(v - w).l.mapsvintow ,Wehave v - w= (-8, 12),and so(v - w).l.istheline-8x + 12y= 0, ory =D5.Leta =v - w= ( 1, 2, 2)- (0, 0, 3) =(1, 2, -1).Thenthe reflectionofR3 abouta.l.mapsvinto w,andthe planea .lcorrespondsto x+ 2y- z = 0or z= x+ 2y. WORKING WITHPROOFS P2.Toshowthat H= I - af;aaT isorthogonalwemust showthat HT =H-1This followsfrom T(2T)(2TJ' T2T2T4TT H H=I - --aaI- - aa= I - - aa- --aa+ ---aa aa aTaaTaaTaaTa(aTa)2 =I- .2.. aaT - -2- aaT + -4- aaT =I aTaaTaaTa whereweha,eusedthe factthataaT aaT =a(aT a)aT =(aTa)naT. EXERCISE SET 7.11 261 P 3.One ofthe features of the Gram-Schmidt process is that span{ q1, q 2, . . .,q,} = span { w1, w2,,. ., Wj} forea.chj==1, 2, . .. , k.inthe expansion wemusthavew 3 CV-::j:.0,forotherwise w jwouldbe in span {Q1 , Q2, ... ,ClJ -1 }= span{w1 , w2 , ... , w;- l }whichwouldmeanthat { w1,w2, . . . , w1} isalinearlydependent set. P4.If A =QR isaQR-decompositionof A,then Q = AR-1.From thisit followsthat the columns of Q belongtothecolumn space of A.In particular,if R-1 = js,j],thenfromQ=AR-1 it follows that c; (Q) =Acj(R-1)= StjCI(A) + s2jc2(A) + + for each j=1, 2, ... , k.Finally,since dim(col(A))=k and t.hevectorsCt(Q), c2(Q), ... ,ck(Q) are linearly independent,it followsthat they formabasis forcol(A). EXERCISE SET 7.11 1.(a)Wehave w'""3v1 ...7v2; thus(w)B= (3, -7) and[w]B= (b)h 11I [ 2 a3)[cc2']-- [11)'and Tevect or equationc1 v1 + c2v2 = wisequiva. enttot1einearsystem - 4 the 1:.olut.ionof thissystemisc1 = 18,c2= 134.Thus(w ) B= {fa, f4)andlwl o =(2.(a)(w)B= ( -2, 5) (b)(w)n=(I , 1} 3.Thevectorequationc1 v1+ + c3v3 = wisequivalentto =Solvingthissys-o03C33 [ tembyback subi:ititut ionyir.ldsc.3 =l,c2 = -2,c1 = 3.Thus(w)a = (3,-2, 1)and[win= 4.Thevectorequationc1 v1 r.2v2 + c3 v3 =wisequiva.lentto - :._; ]= Solvingthis 3 69C:J3 system byrowreductionyif!k!sc1 ='-2, c2 =0,c3 =1.Thus(w)B =( - 2, 0, 1). 5.If (u)n =(7, - 2, 1),thenu= 7v1 - 2v2 + =7(1, 0, 0)- 2(2, 2, 0) + (3, 3, 3)= (6, - 1, 3). 6.If (u)B = (8, - 5, , 6) + 4(7, - 8, 9)=(56, -41, 30) . 7.(w}B = (w v 1 ,w v2)= = .. 8.(w)a =(w v1, w v 'l , w v3)::: (0, -2, 1) 9.(w)J=(W VtWv2 Wv3)=(....Q..._..2__hQy]) ''-/2',/3'v62'36 262Chapter 7 11.(a)Wehaveu= v1+ = v=- v1+4v2=1;). ( b)Using Theoretn 7.11.2:Uu ll=ll(u)all= y'(1)2 +(1)2 = /2, llvll= ll(v)sll=J(-1}2 +(4)2= m. andu. y= (u}s 0(v)a = (1)(-1) + (1)(4)= 3. Computing directlyllull= 4- = /s =/2, llvll= Je;p +e;p = !iii= Jl7. anduv= (t)(

+ (-!)( ;6)= = 3. 12.(a)Wehaveu= -2v1+v2+2v3= -i,v= 3vl+Ov2- 2v3=.!p,1). (b)ll u ll=ll(u)sll=j( -2)2 +(1)2 +(2)2= 3,ll v ll= ll(v)sll =J(3)2 +(0)2 +( -2)2 =Jf3, and u v= (u)s (v )s = {-2}(3) +(1)(0) +(2)(-2) = - 10. ll u ll= +(-D2 + = fj =3,llv ll==J =Jf3, andu v=-!) + (-i)( 13) + (=-99 =-10. 13.!lull= ll(u)Dil= J( - 1)2 1- (2)2 + (1)2 + (3)2= Jl5 llvll= li(v)sll =}(0)2 t(-3)2 +(1)2 +(5)2 = J35 Uwll- ll( w)sll = ,fi2)2 + ( -4)2 +(3)2 +(1)2=v'30 uv+w 11= 11 s+s 11= 11 .=8 (multiplicity 7).The eigenspace corresponding to >.""'-3 has dimension 1;the eigenspace correspondingto).=- Jhasdimension1,2,or 3;andtheeigenspacecorrespondingtoA = 8 havedimensionI,2,3,4,5,6,7,or8. 6.(a)Thematrixis5x5witheigenvaluesA= 0(multiplicity1),A=1(rm1ltiplicity1),.A= -2 (multiplicity 1), and .A= 3 (multiplicity 2).The eigenspaces corresponding to .A= 0,>.= 1; and >.=- 2 eachhavedimensionJ.The eigenspact!correspondingto A =3 has di mension1 or 2. (b)Thematrixis6x6witheigenvaluesA= 0(multiplicity2),A= 6(multiplici ty1) ,and.A= 2(multiplicity3).Theeigenspacecorresp(1ndingto>.= 6hasdimension1,theeigP.nspace corrcspondinr;toA = 0ha.dimension1or2;andthe eigenspacecorrespondingtoA =2has dimension1,2,(){3. 7.SinceAistriangular,itscharacteristic polynomial is p(.A)= (..\- l)(A- 1)(.\ - 2)=(.A -- 1)2(A- 2). Thus the eigelll'aluesof !I are .A=1 and>.= 2,with algebraicmultiplicities 2 a nd1 respectively.The eigenspacecorrespondingtoA=1 isthe solution spaceof the system(I- A)x = 0,whichis Thegeneralsolutionor thissystemisx=[i]=t [:] ;thustheeigenspais1-dime.'5ionaland so A = 1 hasgeometricmultiplicity1.Theeigenspace correspondingtoA = 2isthesolutionspaceof the system(2/- A)x =0whichis 278Chapter 8 Thesolutionspaceoftbis system;.,x~[':]= s [l thustheeigenspaceis1-dimensiotlaiandso ).= 2also hasgeometric multiplicity1. 8.Theeigenvaluesof Aare).=l ,>.=3,and>.= 5;eachwithalgebraicmultiplicity1and geometric rcultipliclty 1. 9.Thecharacteristic polynomial of Ais p(>.)=det(>J -A) = (.X- 5)2(>.- 3).Thus the eigenvaluesof A are>. =5 and >.= 3,with algebraic multiplicities 2 and 1 respectively.The eigenspace correspond-ingto).= 5isthe solution space of the system(51- A)x =0,whichis Tbegene>alsolutionof this systemisx~m~t [!l;thustheeigenspaceis1-.- 3)2.ThustheeigenvaluesofAare>.= - 1and >.= 3,withal gebraicmultiplici ties1 and2 respectively.The eigenspacecorrespondingto>.=-1JS 1-dirnensional,u.ndso >.= -1 hasgeometncmultiplicity1.Theeigenspace correspondingto>.= 3 isLhesolutionspace of the system(31- A)x =0whichis [ _ ~- ~ ~ ][:;] =[ ~ ] [-1"'[0] The general solution of Lhissystem is x~s~ J+ t~; thusthe eigenspace is2-dimensional, and so ).= 3geometricmultiplicity2. ll.ThecharacteristicpolynomialvfAisp(>.)= >.3 + 3>.2 = >.2(>.+ 3);thustheeigenvaluesare>.=0 and..\=- 3,withalgebraicmultiplicities2and1respectively.The rank of the matrix OJ - A =-A = [~1= ~ ] - 1-11 EXERCISE SET 6.2279 is dearly 1 since each of the rows is a scalar multiple of the lst row.Thus nullity(OJ - A):=:3- _1= 2, andthisistbe geometricmultiplicity of ..\= 0.Onthe other hand,the matrix -31 --A =-

[- 21-1] 1-2-1 - 1- 1-2

0 1] 11.Thus nullity( -31- A) ==3 - 2 =1,and 00 hasrank 2since its reducedrowechelonformis this is t hegeometric multiplicityof>.=- 3. 12.The characteris ticpolynomial ofAis(>.- l)(A2 - 2A + 2);thus >. =1isthe onlyrealeigenvalue of A.The reducedrow echelon form ofthe matrixli - A=is - thusthe rank - l4- 60(I0 of 11- Ais2and the geomet ricmulti plicityof )..=1isnulli ty( ll - A)= 3 - 2""1. 13.The characteristic polynomia.l of Ais p(>-.)=.>.3- 11>.2 + 39A - 45 = (A- 5)(>.- 3)2;thus the eigen-values are>.= 5and,\ =3,with algebraicmul tipli. - l)2;thustheeigenvaluesare,\= - 2andA= 1, withalgeb-2 - 3>. + 2=(>.- l) (A- 2);thusAho.stwodistinct eigenvalues,>. =1 and>.= 2.The eigenspace corresponding to A = 1 is obtained by solving the system (/ - A)x = 0.whichis= Thegeneralsolutionof thls systemisx=Thus, I 280Chapter 8 taking t= 5,weseethat ?l= [:]isaneigenvectorfor>.= 1.Similarly,P2 = [!]isan eigenvector forA= 2.Finally,thematrix P =[p1 p2]= [!!}hasthepropertythat p-1 AP = [4 -3][-1412][43]=[l0] -54-20175402 16.The characteristic polynomial of Ais(.>.- 1)(.>. + 1); thus Aha.stwodistinct eigenvalues,.>.1 =1 and .>.2=-1.Correspondingeigenvectorsare P1= andP2= andthematrixP=!P1 P2Jhas thepropertythat p -l AP = = 17.Thecharacteristic polynomialof Aisp(>.)=.>.(.>.- 1)(,\- 2);thusAhas threedistinct eigenvalues, >.= 0,>.=1,and>. = 2.The eigenspacecorresponding to>..= 0isobtainedbysolving the system (OJ- A)x 0,whichisn-:=!] [::] The solutionofthissystemisx+:J Similarly,the genml solution of (I- A)x 0is x=+l, and the gene.)= >.3 - 6>.2 + 11>..- 6 =(..>..- 1)(>. - 2)(.>.- 3);thus Ahas Lhreedistincteigenvalues>.1 =1,.>.2= 2and.>.3 = 3.Correspondingeigenvectorsarev1 = v,m, andv, [:]- ThusAisdiagonalizableandP[:::]hasthepmpe.)= >..3- -t >.2 + 5>.- 2= (>.- 2)(>.- 1)2;thusAbastwo :::: -179f7)0 whichshowsI hattheetgenspacPcorrespondingto>.= 2hasdunension1Simi larly,llwgeneral solutionof (I - A )x 0isx- s [ il whichshowsthattheeigenspaceconespondingtnA =talso hasdimension1.ItfollowsthatthematrixAisnotdiagonalizablesinceithas onlytwolinearly independenteigenvectors. 21.The characteristicpolynomialof Aisp(>.)= (>. - 5)3;thusAhas one eigenvalue,>.= 5,whichhas algebraicmultipUcity3.Theeigenspacecorrespondingto>..= 5isobtainedbysolvingthe system (51- A)x = 0,whichisH:[:;] = mThe genmlsolutionofthissystemisX- t which showsthatt ht:'eigenspacP has dimension 1,i.P,Lheeigenvalue hasgeometric mulliplicity1lL followsthat.Aisnnt. sincethn sumof t.hrgcomelrifmultiplicil.it!Svf itsis lessthan 3 22.The characteristic polynomtaJnf Ais>.2(>.- 1};thus the eigenvalues ofAare\= 0,and>.=1The eigenspacecorrespondingtoA0hMdimension2.andthevecto'5v,=[v,[:]fotma hMislotthisspace.The"'""' ' '= mfotmsaba.risttcpolynomtalofAisp( .\)=(>.+ 2)2t.\- 3)2thusAhastwoetgcnwtlues,.-\=-2 and.>- =3,eachofwhichhasn.lgebraicrnulLiplicity2.TheeigewJpacecorrespondingL(l..\= -2 is obtainedbysolvingthP.system( -2/- A)x =0,whi chis The gonctnl solution of this system ts x=r + swhteh shows that the eigenspace has dimenSton 2,i.e.,thattheeigenvalue>.=-2 hasgeomeLricmultiplicity2.Ontheotherhand,thegeneral solution of (31 - A )x- 0isx=Ind so =3hasgeomet de mu I liphcityIItfollowsthatAis not diagonalizable sincethe sum of thegeometricof itseigenvaluesislt>..ssthan4 282Chapter8 Thecharactensticpolynorrualof Aisp(>.)=(>.+ 2)2(>.- 3)2;thusAhastwoeigenvaluesA= -2 au.l3, eachof algebrakmultiplici 0,thonp(>.)has two distinctreo.! roots;thusAis since it has twodistinctPigenvalues. (b)Tf( o - d)2 + tbcn, t h"n r(>.)has noroots;t hue:Ahas no rc:1lt'lgf"nvaluesand is thenoldiagonahzablP. DISCUSSION AND DISCOVERY 01. arenot s1mtlar smcerank(A) = 1 andrank(8) = 2 02.(a)'ITue.WehavPA =p-I AP whereP=I (b)1Tue.If A1ss tnHiartoBandBissimilartoC.thentherearcmvert1blematricesP1 nndP2surhthatA- P1-1 BP1andB=F2-1 (' P2 .ltfollowsthatA=P1 1 (P2 1 C P2 )P1 = (P2P1) 1C(e'JJ>I ),thusAissinular to C (c)True.IfA - P 1BP,thenA-1=(P-1BP) -1= P -1B-1(P 1) 1 p-1e- 'P. ( cl)False.This statementdoesnotguaranteethatthereare enoughlinearlyindependent eigen-vectors.Forexample,thematrixA=hasonlyone(real)eigenvalue,>.=1, 0-10 wh1chhas mull1pltcity1,but Aisnot dJagonalizable 03.(a)False.Forexamplr,1- isdiagonalizahle (b)False.Forexc.Lmple,1fp -l AP isadiagonalmatrix thensoISQ 1 AQwhereQ=2P.The diagonalizingmatrix(if it exists)isnoturuque! (;:)True.Vectors fromdifferent eigenspaces correspond to djfferent eigenvalues and are therdore linearlyindependent.In the situationdescribed{ v1,v2,v3}isalinearlyindependent set. (d )True.If aninvert1blematrixAissimilartoadiagtlnalmatrixD,thenD must alsobem-vertible,thus D hasnonzero diagonal entries and o-1 is the diagoncJmatnxwhose diagonal entries are the reciprocals of the corresponding entries of D.Finally, 1f PIS an mvert1ble ma-trix suchthat p -I AP = D,wehavep- A-1 P= (P-1 AP) -1 =o-1 andsoA-1 issintilar too-l 284 (e) \ 04.(a) (b) (c) (d) 05.(a) (b) (c) Chapter 8 1'ru_yThevectorsin abasisarelinearlyindependent;thusAbasnlinearindependent

Ais a6 x6matrix. Thecigenspucecorrespondingto>.=1hasdimension1The eigt:nspacecorrespondingto >.= 3hasdimensionl or2.The eigenspace correspondingto >.= 4 has dtmension1,2 or3. IfAisdiagonalizable,thenthe eigenspaces correspondingto>.=1,>.= 3,and>.= 4have dimensions1,2,and3 respectively. These vectorsmust correspondtothe eigenvalue>.= 4. If >.1 hasgeometricmultiplicity2and>.2hasgeometricmultiplicity3,then>.3musthave multiplicityl.Thus the sum of the geometricmultiplicitiesis6andso Ais diagonalizable. Inthis casethematrix isnot diagonalizable sincethe sum of the geometricmultiplicities of t he e1genvaluesislessthan 6. The matrix may ormay notbe diagonalizable.The geomet.ric multiplicity of >.3.IIWStbe 1 or 2.If the geometricmultiplicity of >.3is2,thenthe matrix isdiagonalizable.If the geometric mllll.iplicitvof>-1Is1, thenthe matrix isnot dia.gonaliza.ble. WORKING WITHPROOFS Pl.If Aand Bare similar,then there is an invertible matnx P such that A =p - t BP.ThusPA = BP and so,usira,vtheresult of t he ci tedExercise, we have rank( A)=rank(PA)= rank(BP) =rank(B) andnullity(A)=nullity(PA)nullity(BP)=nnllity( B). P2.If Aand13aresumlarlhent heretsaninvert iblematnxPsuchthat A =p - BPThus,using part (e) of Theorem3 2 12,wehave tr(A)=tr(P -1 BP) ==tr(P 1 ( IJP))o..:tr(( BP)P -1)= tr(B). P3.If Xf.0andAx =>.xt.hen, sincepis invertible andcp-l = p - l A,wehave C' P-1x = p - I Ax =p-1(,\x) = >.P-1x wit hp - x-:f0T husp -xisan eigenvector of Ccorrespondingto >.. P4.If AandBure :m111lar,lhenthereisan invertible matrixP sucht ha tA= p - lBP.Wewillprove, byinducti on,tha tAirP 1 B" P(thus AkandBJrare similnr)foreverypositive integerk . SteplThefactthat, \1 = .1=p -1BP = p-l B1Pisgtven Step2.If Ak= p - l BI.)= ,\2- 5,\= >.(>.- 5).Thus the eigenvalues of Aare).= 0 and,\ = 5,andeach of the cigcnspaces has climension1. 2.The characteristic polynomial of A1s p(,\)==>.3 - 27...\-54== (,\- 6){>. + 3)2 Thus the eigenvalues of A are,\ = 6 and.>,= -3.Thf>t>igenspace corresponrling to,\ = C hl\.o:;dimension 1,anrllhe e1genspace correspondingto,\=- -3 ha.srlimPnsion2. 3.Thecharacteristicpolynom1a.lofAisp(>.)=>.3- 3).2 =..\2(.X- 3).Thust heeigenvaluesofAare A= 0and,\3.The eigensptLCt:correspondingto>.=0hasdillleu.sion2,andthe eigenspacecorre-spondingto >.- 3has dimcllSIOn1 4 .The charactensticpolynomialof Aisp(>.)= ,\,3- 9...\2 + 15).- 7- (,\- 7)(>-- 1)2 Thustheeigen-valuesof Aare>.= 7and>.- 1.The eigenspacecor respondingto,\ = 7hasdimensionI,andt he eigenspacecorrPspondmgto >.=lhasdimension2. 5.The gen"al solution or the system(0/- A)x0isxr + s [ -:], thus the ectorsv1 [andv3[-:]form a bastsforthe eigenspace corresponding to =0.Similarl y the vector v,= m forms abasisfort heeigenspace correspondingto>.= 3.Since v 3isor thogonal to bothv1 andv2 it followsthat thetwoeigenspnces are 286Chapter 8 6.Thegeneralwlutionor(7f>i )x0is T[i} thusthevcctocv,[:]romsabasisfoethe eigenspa.)= .>.3 + 6>.2- 32 = (>.- 2j(>. + 4)2;thus the eigenvalues of A 2 and.\=- 4.The g'ncral solu.)=\3 - 2>.2 = ).2(>,- 2);t.husthe eigf'nvalues of A are>.= 0 and> = 2.ThegenoraloluUonor(OJ- .4)x = 0isX= Tm+ s [ -:]' andthegeneral soluUonor ( 2/- A )x = 0isxI[ l] Thus the vedors v,= mandv, = [-:lformabasisorthe eigenspace ""respondintoI0.o>nd>hevectorv, =[l]mmsabasosortheeigenspaceeonespondingto >.= 2.These VPctorsare mutually orthogonal, andt.heorthogonalmatrix [ Vt F =llvdl 1 -72 I 72 0 ~ ] 72 0 hn.-,thepropt>rtv1hnl [ n - ~ 72 ~~ ][ ~~~ ][ ~ ~00001 1 l [l V2I)00 72=000=D 0002 12.The.4 - 1250..\2 + 390625= (>.- 25)2(..\+ 25)2;thusthe eigenvaluesof Aace 25,andA= -25.The vecto"v1= v3fo.= -1and).= 2,withcorrespondingeigenvectors 1 and Thusthematr ixP = [- :- hasthe propertythatp- t AP =lJ= [-Itfollowsthat -3][1OJ[ 2 201024- 1 3][30703069] -1- 2046-2045 290 Chapter 8 20.The marixAhas eigenvalues>.= 2and>. =-2, with corresponding eigenvectors Thus the matr;.,P = ;)hasthe property that p-l AP = D =It, followsthat AlO=PDIOp- 1 = [1-1][10240][-74]= [10240] 27010242- 101024 21.ThematrixAhaselgenv&luesand>.1.The ve.0,>.1,and>.- 1,withconesponding eigenvedorn[:].[:] , and[!]ThuP[:;!]hasthepcope,tythat p- AP D !andit followthat A1ooo= potooop-1 = [ 6:]23.(a)The characteristic polynomial ofAis p(>.)= >.3- 6>.2 + 12>.- 8Computing successivepowers of A,wehaveA2 ..,.[: :1andA3 =thus 12-241636-7244 -24 - 40 -72 6[: 4412-24 4][3-2 8+ 122-2 163-6 8/ which showsthat Asatisfiesits characteristic equation,i.e.,that p(A} = 0. {b )SinceA3 =-6A2- 12A + 8/, wehave A4 = 6A3 -l2A2 + 8A =24A2- 64A ..._4I (c)Since A3- 6A2 -112A - 81 = 0,wehave A(A2- 6A + 12/) = 81 and A-1 =

6A+ 12/). EXERCISE SET 8.3 291 24.(a)The characteristicpolynomialofAisp(>.)=>.3 - >.2 - >.+ 1.Computmg successive powers of A, wehaveA2 = =I andA3 = = A,thus 001-405 [-50 6][10 0][-5 0 6] .A2 - A=-313- 010- -3l3= -405001-405 which showsthat.Asatisfies its characteristic t>quat1on,i.e.,that p(A) = 0. (b)Since A3 =A, wehave A4 = AA3 = AA =A2 = I . (c)SinceA2 = I,wehaveA-1 =A. l 25 .FromExercise7wehnvepT AP =[t [== D.ThusA =PDPTand _PetD pT = [*) [e2'0][-?z]= 1 [te.4'-l''lt+ - -k-L0.:11 -h.J,. 2-e:!-Tt'11 ":!'+elt' "2.,!}"2"'2 27PromExercise9wehave [I I n

76 J;;1 pT AP ==f l -3 72 276 I 2 n'l ---rr.. v'3vJ../3\(> ThusA= PDP'l'nnd [I 1 [''' 0 76 -/2 e'A Pe'o J>r= I j e.,, 76 ..;2 -730 -36 0 0 () ....a [ e" + se-" e2'_e-4t e2t_e-H e21 +5e-41 () 2e2t- 2e-4t 2e2t- 2e-"' 28 .PromExercise10wehavP PTAP =_!00 [ 0 10][ -2 i 0i-36 ThusA= PDPTand 4 -g 0 3 s I

0

-72 1 - I =D 72../3 = 0 1 0- I 7J ][I I 07ti 76 0-?z I J2 e-4tI -73 2e21 - 2e-"] 2e2'- 2e-4t 4e2t+ 2e- 4t = D 0-50 292 Chapter 8 = 4 -5 0 3 5 [" e'" + -2-e-'" 2525 0 _12 e25t+ 12 e-sot 2525 0 -+ " e-'"]2525 e-31 0 0 ..!!. e25t+ !! e50t 2525 [sln(21f)00l [00OJ 29.Notethato0 sin( -0 4rr)0=oo0 o0 Thus,proceeding as inExercise 27: sin( - 47.xS10ceF2 =1 1tfollowslhat ..\2x==Jx= x,thus >.2 =1 andso >.= 1 DISCUSSION AND DISCOVERY Dl.(a)TrueThe malrixAAr issymmetricandhenceisorthogonallydiagonalizable. (b)FalseIf Asdtagonahzablebutnot symmetric(therefme not orthogonallydaagonalizable), thenthereasabasisforR"(but notanorthogonalbasis)cons1slmgof eJgenvectorsof A. WORKJNG WITH PROOFS (c)False.An orthogonalmatrix neednot be symmetric;forex\\mpleA= . 293 {d)'!TueIfAisaninvertible orthogonallydiagonalizablcmntnX1 thenthereis anorthogonal mat rixPsuchthatpT AP=DwhereDisadiagonalmatnxwithnonzeroentries{lhe eigenvaluesofA)onthe main diagonal.It followsthatpr A- lp - (PT AP)-1 =D-1 and t.hatD-1 isadtagonalmatnxwithnonzeroentries(thereciprocalsof theon themaindiagonalThus the matrixA -l is orthogonallydiagonalizable. (e)1TueIf Ais orthogonally cliagonalizable1 t henA is symmetnc andthus hasr eal eigenvalues. D2.(a)A = P DPT = [ ;; ] 00070 t 72 0 1 7z -72][3 0 0] 0=034 I043 72 (b)No.'l'hevectorsv 2andv3correspondtodifferenteigenvalues,butarenotorthogonal. T here foret hey cannotbe eigenvectorsofasymmet ricmat rix. 03.Yes.Since Ais cll!lgonn.lizable andthe eigenspaces are mutually orthogonlll1 there 1san orthonormal bastsforR" cons1stmgole1genvectorsof A.ThusAtsorthogonallyd1agouahzablcandtherefore mu11tbe symmetric WORKING WITH PROOFS Pl.\'\'efirstshowthat if AandCare orthogonallysmular,thenthere orthonormalha.c:eswith rc.. J tlv hkb tht:)rt'pn:scnlthe same l1ut:ar '-per ll JrFur1!11:.lJUrpose,letT the v1eratvr definedbyT(x) =AxfhenA=[T],i.e.,Aisthematnxof Trelativetothestandardbasis B={ e,, e211 en}SmceAandCareorthogonallysimilar,there1sanorthogonalmatrL RnISalinearoperator andB,fJ'n.reh1\SCSfornu.IfP=PD'-+8t.henPisanort,hogonnlmatrixandC=[T]a= pTITlnPpTA P.ThusAandCare orthogonally si1ni ln.r. P2.Suppose A - c1Ut U{Ic2u 2uf+ + Cn ll nu;,where {u1, u2,... , un}is an orthonormal basisfor Rn.Sincf>(u1 uJ)T = u J'7' u J= u1ufit follov.;sthat A'T=A;thusAis symmetnc.Furthermore, sinceu{' u 1 =u, u1 - o,, 1 \"ehave T\ Au1={c1u1uf +c2u2uf- 2: c, u,u?' u1 =c1u1 I foreach j=11 2,, nThus c1, c2 . 1 Cnare e1genvalues of A P3.T hes pect r aldecompositionA= >q u1uf+ A2 u 2ui+ -r tsequivalent.toA= PDPT whereP =!u tI u 2lI unJandD =A21.. , >-n);thus /(A)= P /(D)PT=Pdiag(f( At). /(>2), ... , /(An))PT =/(AI) ulUJf+ j(>.2) u2uf+ +294Chapter 8 P4.(a)SupposeAisasymmetricmatnx,and>.oisaneigenvalueofAhavinggeometricmulti-plicityJ..Letl\'betheeigenspacecorrespondingto>.0.Chooseanorthonormalbasis { u1, u2,., uk}for W, extend it to an orthonormal basisB= {u1, u2, ... ,Uk+J, .. . , un} forR", andletPbe the orthogonalmatrix having thevectors-:>fBas its columns.Then,as ..;howninExercise P6(b) of Section 8.2,the productAP canbewritten asAP= Since PISorthogonal,wehave andsincepT AP isa symmetric matrix,itfollowsthatX= 0. (b)SinceA issimilar to C =hasthesame characteristic polynomialasC, namely (>.- >.o)"det(>.In-k- Y)= (>.- >.o)kpy(>.)where py(>.) isthe characteristicpofynomialof Y .Wewillnowprovethatpy (>.o):f.0andthusthatthea lgebraicmultiplicityof >.0 is exactlyk.The proof isby contradiction: Supposepy (Ao)- 0,i.e.,that>.oisaneigenvo.lueofthematrixY.Thenthereisa nonzerovectoryinR"-k suchthat Yy = ..\oYLet x= [;]be LhevectorinR"whosefirst kcomponents are 0andwhose lastn- kcomponents arethose of yThen and sox ISaneigenvector ofCcorrespondingto >.oSinceAP = PC,itfollowsthat Px is an eigenvectorof A correspondingto >.oBut.note that e1,. , are also eigenvectors of C to>.o,andthat{e1.., ekx}ISalinearlyindependent setIt followsthat {Pet, .. , Px}1salinearlymdependent.setof eigenvectorsofAcorrespondingto>.0. But thisimpliesthat the geometricmuJtiplkity of >.0 ISgreaterthank,acontradiction! (c)It follows frompart (b)that the sum of the d1mensions of the eigenspaces of Aisequal ton; thusAisdJagonalizableF\u-thermore, smceAISsymmetric,the e1genspacescorresponrling todifferenteigenvaluPsare orthogonal.Thuswecanforman orthonormalbas1sforR"by choosinganorthonormalbasisforeachof tht>e1genspacesand JOiningthemtogether.Since the surnofthe dimensionsisn,thiswillbeanorthonormalbns1sconsistingof ejgenvectors uf A. ThusAISorthogonallydiagonalizable. EXERCISE SET 8.4 1.(a)3x?+ -!xJx2] ][ 4 -3][X'] X2 -3-9X2 EXERCISE SET8.4 295 5.The quadraticform Q = 2xr1- - 2x1x2 canbe expressedinmatrix notation as Thematrix. \hasds!n:alucs,\1 =1and,\2 = 3,wilhcorrespondingdgt:m'l(tor:>v1 - GJam.l v2= [-:]respective))Thus matrix P= orthogonallydiagona.IJZcs A, and the change of \'&riable[:21)= x- Py[tz- [Y1]eliminatesthe cross producttermsinQ 7z72Y2 Notethatthe in verst:relationshipbetweenxaadyis = y=pTx=[-J'i [r1] ,. 7iJ:'l - I G.rho v"'""""'' ,.;,!om' ""' " inLt ,,,notationMQ -1 A XwhcccA[:] ThematrixAhnsrigcnvahJe(o, m canbeexp2. -1-1Jc 0 3 0 .T hequadrat 1rformcanbeexpressedinmatrixnotationasQ = xr AxwhereA= u 1Jc.-tl k2 30-1 Jc=5- 3k2 2 The del.lt foUowsthat s; = xT Ax where l 1I n-n(n-1)- n{n-1) __1_ .!. I n(n-1)n- n(n-1) A= 11 l -n{n-1)- n(n-1)n DISCUSSION AND DISCOVERY301 (b)We havesi= ":11(x,- i)2 + (xz - i)2 -t- + (xn- :t)2J;:::0,and0if a ndonly if x1 = i, Xz=i, ... ,:en= :i:,i.e.,if and onlyif x1 = X2= :::::Xn.Thuss;isposit1ve semJdefinite. 35.(a)ThequadraticformQ = jx2 + + + + + canbeexpressedinmatrixtationasQx' Axwhe.j. The vectorn v, [-!]andv,formabas;. forthe eigenspace correspondingj, andv3 [:]formsahasisfortheeigenspacecorrespondingto>.!Applicationofthe Gram-Schmidt processto{ v1, v2, v3}produces orthonormal eigenvec tors{p 1, P2, PJ},andt he matrix PJ= [-i ' 76 P= Pt I P2-"JS 1 -76

73 orthogonally diagonahz.esA.Thus Langeof variablex=Px' converts Qinto aquadratic formmthevariablesx'- (x',y',z')w1thout crossproduct terms. FromthiswPcnnrludcI hatthrequationQ1 toanellipsoidwitha.xislengths 2j"i =J6inthPx' andy'directions,and2.jf = thez'direction {b)The matnxAmustbP.positivedefinite. DISCUSSIONAND DI SCOVERY Dl .(a) (b) (c) (d) (e) (f ) 0 2.(a) (b) (c) (d ) (e) False.ForcxmnplethematrixA=hn.sr.igcnvaluesland3;tlmiiA1smdefi nite. False.The term4x1.z:'2x3isnot quadratic tnthevariables x1,x2, XJ. True.Whencxpuudcd, eachof t.heterms of lhe res uiLingexpressionis (of degree 2)int hevariables True.The eigenvalues of apos1tJVedefinitematnxAare strictlypositive,mparticular,0Is not an eigenvalueof AandsoAisinvert.1ble FalseFor examplr th!C'matnxA= ispos1t1vesem1-definite True.rrtheeigenvalues of. \are posttive,tlu>nthe eigenvaluesof - Aarcnegat1ve True.Whenwr itten inmatrix form,wehave x x= xT Ax whereA =I . True.TfA haspos1t1veeigenvalues,then so doesA-1. True.See Theorem 8.113(a) True.Both of thE'principal S'lbmatrices ofAwi llhaveapositivedeterminant False.If A= (:] , thenxT= x' + y' 0the graph is an ellipse.If c < 0the graph isempty. D3.The eigenvaluesof Amustbepos1lheand equalto eachother;inotherwords,Amusthavea positive eigenvalue of muiLlplicily2. WORKING WITH PROOFS Pl.Rotatingthe coordinate axesthroughanangle8 correspondstothechange of va.ria.blex= Px! whereP= [:: -:::], i.e.,x=x' cosO- y' sinOandy= x' sin 8 + y' cosO.Substitutingthese expressionsintothequadraticformax2 + 2bxy + cy2 leadstoAx'2 + Bx'y' + Cy'2,wherethe coefficientof thecrossproduct term is B=-2acos0sin 8 + 2b(cos2 0- sin2 0) + 2ccos OsinO = (-a+ c) sin 20 + 2bc'os20 Thus the resulting quadratic formin the variables x' and y'hasno cross productter):nP.ifand only if (-a+ c) si n 28 + 2b cos 20= 0, or (equivalently)cot 28= P2.FromthePrint1paJAx1sTheorem(8.4. 1),thereisanorthogonalchangeof varial>lex= Pyfor wh1chxr Ax= yl'Dy - >-.t!Jf1-where)qand>.2arethe eigenvalues of A.Since ,\1 and>.2 are nonnegat1ve,it followsthat xT Ax 0 forevery vectorxinRn. EXERCISE SET 8.5 l.{a)Thefirstpartialderivativesot farefz(x , y)= 4y- 4x3 andf11(x, y)= 4x - 4y3 To findthe cnt1calpomtswesl'll:randfvequalto zero.Thisyieldstheequationsy = x3andx= y3. 1-\omLh1sweconduJeLhaty = y9 andsoy=0ory= 1Sincex= y3 thecorresponding values of xare x=0andx=1respectivelyThus there are three critical points:(0, 0),(1, 1), and(1,-1). (b)TheJl c:-.sianmatrixISH(x.y) = [ln((x.y))/xy((z,y)]= r-l2:r2 4 2]Evaluatingthismatrix ryfvvz.y)4- 12y 2.(a) {b) atthP.cnticf\1pointsoff yaelds /l(O,O) [0l] 4cI[-124] H(l,l) =4-12' [- 12 H( - 1,-1) = 4 TheeigenvaluesofareA= 4;thusthematrixH(O, 0)isindefiniteandso1hasa saddlepoinlat(0, 0) .The eigenvaluesof

are>.= -8 and>.=- 16;thusLhematrix H(l,l) = H( - 1, - 1)isnegativedefinHeandsofhasarelativemaximumat{1,1)andat (-1,-1) Thefirstpartial derivativesoff aref%(x, y)= 3x2- 6yand fv(x, y)= -6x- 3y2.Tofindthe points we set1%and f11 equal to zero.This yields the equationsy=4x2 and x=-

FromLhisweconcludethaty= h4 and so y= 0ory= 2.The corresponding valuesof xare x=0and 4- 2respectavdy.Thus there aretwo criticalpoints:(0, 0) and( -2, 2). TheHessianmatrixisH(x,y)= [/u((x,y)) 11zv((:z:,y))]= (6x-6].The eigenvaluesof H(O,O}= fvz:z;, Y1111%, Y-6-611 [are>.=6;thismatrixisindefiniteandsofhasasaddlepointat(0, 0) .The eigenvaluesof H( -2, 2)=r-12 -6]are >.=-6 and >.=-18,thismatnx isnegativedefinite -6-12 andsofhas arelnLivemaximum at( -2, 2) EXERCISE SET 8.5305 13.The constraint equation 4x2 + 8y2 =16 can be rewritten as ( )2 + ( )2 =l. Thus, with the change of variable (x, y)=(2x', v'2y'),the problem is to findthe exvreme values of z =xy =2v'2x'y' subject lo xn + y'2 = 1Notethat.z= xy =2v'2x'y'canbe expressedasz=xrT Ax' whereA= [The of Aare >.1 '=v'2 and>.,= -J2, with corresponding (normalized) eigenvectors v1 = l and v2= [ Thus the constrainedmaximumisz = J2 occurring at (x', y')= (x,y)= (J2, 1).Similarly,the constrainedminimumisz= -v'2 occurdngat(x',y')= (-72, or(x,y)=(-J2,1) 14.The constraint :r2 + 3y2 =16 can be rewritten as( )2 + ( 4y)2 =1.Thus, setting (x, y)= (4:r',lheproblem isLufindthe extremevaluesof z=x2 + xy12y2 = 16x'2 t- + v'2 subjectto x'2 + y'2 = 1.Notethatz=l6x'2 + t- y'2 canbeexpressedasz = xfT Ax'whereA = [ The eigenvalue:. of A are >.1= 8with correspondmg eigenvectors v,=[an:! v/=[Normalizedareu 1=u:: 11=[ _andll ?==[ Thusthe constrainedm;ucimumisz= 536 occurringat(x',y')= {4, !>or (x, y) = .J:(2J3, 1s>Similarly, the constrainedmmimumisz =a occurring at (x', y') =-4) or(x, y) = {2. -2) 15.Thelevelcurvecorrespondmgtotheconstramedmaximum1s thehyp.:!rbola5x2 - y2 = 5;ittouchestheunitci rcleat(x, y)= ( 1, 0).The curvecorre ponuingtotheconstrainedmini-mumISthehyperbola 5z-=-1,ittouchesthe unitCircleal (:r,y)= (0,1) 16.!'he ltv .. l cunp l'nrrPspondmgtoth"'constramedmaxirnum1sthe hype1bolaxy-1ttoucheslhr>un1t circle at (x, y)== 72). Thelevelcurvncorrespondingtotheconstrainedminimumis thehyperbolaxy - -!;ittouchestheunitcircleat.{x, y) = ( -J2) 'Y "_.!. 2 -4 I-2 .ry: 2 -4 xy=.!. 1 x.y= _l 1 17.Thearea of theinscribedrectangleisz=4xy.where(x, y)isthe cornerpomt thatliesinthefirst quadrant.Ourproblemistofindthe maximumvalue of z= 4xysubjectto theconstraktsx;::: 0, y;:::0,x2 += 25Theconstraintequat1oncanberewrittenasx'2 + y'2 =1wherex= Sx' andy=y'.Intermsof thevanablesx'andy',ourproblemIStofindthemaximumvalueofz = 20x'y' subject to x'2 + y'2 = l , x';:::0,y';::: 0.Note that z= x'T Ax' where 1A=

The largest eigenvalueofAIS>.= 10withc:orresponding(normalized)eigenvector[ Thusthemaximum areaISz = 10.andth1soccurswhl'n(z',y') = (-jz,(x,y) = 306Chapter8 18.Ourproblemistofindtheextremevaluesof z= 4x2 - 4xy + y2 subjecttox2 + y2 = 25.Setting x= 5x'and y=5y'thisis equivalentto findingthe extreme valuesof z= 100x'2 - lOOx'y' + 25y'2 subject to x'2 t- y'2 = 1Note that z=xff Ax'where.-\=[ The eigenvalues of Aare .-\1 = 125t'.nd >.2=0, wit h corresponding (normalized) eigenvectors v1 = [-andv2 = [Thus the maximum temperature encount ered by the ant is z =125, andth1s occurs at (x', y') = (--;h. 75) or (x, y) =(-2-.15, VS).The minimum temperature encountered is z= 0, and t his occurs at (x', y') = or {x,y) =(VS,2VS). DISCUSSION AND DISCOVERY Dl.(a)Wehavefz(x, y) = 4x3 and/,Ax, y ) =4y3;thusfhasacriticalpointat (0, 0).Simifarly, 9:z(x, y) =4x3 andg11(x, y)= - 4y3,andsoghasacriticalpointa.t{0, 0).TheHessian matricesforfandgareHJ(x,y) =[12;2 12Y2]andH9(x,y) = [12;2

respectively. Since H,(O,O)- H9(0.0)= secondderhntl\etestISmconclusivein both cases. (b)It isclearthat fhas arelativeminimum at{0, 0) since/{0, 0)= 0andf(x, y)= x4 + y4 is strictlypositive atallotherpoints(x,y).lncontrast,wehave g(O,O)=0,g(x,O)= x4 > 0 forx ::/;0,and g(O, y) =-y4 < 0fory f. 0.Thus g hasasaddle point at (0, 0). D2.The eigenvalues of Hare A= 6 and.,\= -2.Thus His indefinite and so the crit1cal points off (if any}nrcs'\dcile pmnts.Starting from fr.Axy\-(.ry}= 2 and /y:r(:ry) =/-ry(xy)=4 itfollows,usingpartialmlegrat.ion,that.the quadraticformf1sf(x, y)= :r2 + 4xy + y2.This function has one cril!calpoint(asaddle)whichislocatedatthe origin D3.lfx isauni tf>lgenvectorcorresponding to..\.thenq(x)=x1'Ax = xT(.-\x)=>.(x Tx )=>.(1)= >.. WORKING WITH PROOFS Pl.First.note that,asin03, wehaveur,Au m = mandur,AuM =MOn the otl.erhand, sinceum andliM are orthogonal, we have= =M(u;: u M)= M(O)= 0 and ur,Aum = 0.Itfollowsthatif Xrj .f1-_":r.,UM,then T(M- e) r(e-m ) 1 (M-e )(e-m) xcAXc=M-mumAum+O+O+J\f-mu MAUt.t=M -mm +M- mM=c EXERCISE SET 8.6 1.The ch.,.a.2(>. - 5);thus the eigenvalues ofAT AareAt=5andA2=0, and cr1 =v'5 isasingular value of A. 2.The eigenvaluesofAT A==

areA1= liJandA2 = 9;thusu1 =v'i6 = 4and u2=J9 =3are smgular values ofA. Exercise Set 8.6307 3.The eigenvalues of A7'A [_ = >.1 =5 and >.2 = 5(i.e., >. =5isan eigeJJ.value of mulliplicaty2);thusthe 'singular values of Aare crt=J5 and cr2.::JS. 4 .The eigemalucs of ATA= ('7, = [Jz v'2] 2 are >. 1 = 4and>.2 - 1;thusthe singular values of Aarecr1 =J4 - 2andcr2 =v'l = 1.---5.TheonlyeigenvalueofAT A= = is>.= 2(multiplicity2),andthe vectors VJ=andv2= formanorthonormalbasisfortheeigenspace(whichisallof R2).The singularvalues of Aare crt= ../2 ar:du2= .J2.We have Ut=;, Av1=(;= u2 ="\ Av2"" -= [ Thisres ultsint.hefollowingsingularvaluedecomposition ofA: A [1 -1]= [v2 [l OJ_UEVT 11720v20I 6.Theof AT A[-J 0][-30]= [090]are)q= 16and>.2 = 9,with corresponding 0-10-416 unitcagenvectorsv1 = andv2 =resperLivelyThesmgularvaluesofAareu1 = 4and a2= 3\VpIHl,eUt :1= .>3= 0The ve...!..J40:iS--oJS u,= [!lsothat { u, , u,, u,}Isan octhonocma\-basisroc R3.This cesults mtherollowmgsingular valuedecornposttton

I0 v5 I -::-s 0 0][80]['21] 1ll2.. /,_"J.= l/EVT 00ll..... $"5

6)[.J..-J.J[s UsingthesmgularvrdtH'decompositionA ==

0.+ 1)(.>.- 3)2;thus>. =- 1 isan eigenvalue cf multiplicity 1 and>.= 3isan Pig envalue of multiplicity 2.The vector v 1 = [-il forms abasis forthe eigenspace ingto>.- -1' andthevee toesv' =mandv'[il forman(orthogonal)basisfor the etgenspace ep"'udotnwse ofAts + _T_ 1T_1[26-321[70 5]_ [ ! A-(AA)A- goo-3274105- 5.(a)AA1-A =[1]=CJ=A (b)A+AA+= (ls/ll][ts = [1][I&is]= (tsfs)=A+ (c)AA+ = =issymmetnc;thus (AA+)T= AA+ . {d)A+ =Ill is svmmetnc;thus(.4+ A)T = A+ A. _!!.. ] IS I

0 0(e) Thee1genvaluesofAAT:; [ 912]are>q=25and =0,withcorrespondinguniteigen-1216 vectorsVt= rnandv2= [-:Jrespectively.The onlysingularvalueof ATis0'!= 5,and wehaveu 1 = o'.AT v1 =!(34] [i]= [1].This results inthe singular valuedecomposition ,1 T =[3J=II I [5OJ[-!ll =UEVT 314 6. Chapter 8 ThPcorrespondingreducedsingular valuedecompositionis AT = (34]= fll [5][l =UlEl vr andfromthisweobtam (AT>+ ="Ei"ur =[i]UJ111= =(A+>r (f)The eigenvalues of(A+)T A+=

are At= 215 and A2=0,withcorresponclingunit eigenvectorsv1 [i]andv2=respectively.The onlysingularvalueof A+is a1 = t andwehaveu 1 - ct1,A+v= 5[,\fs)[!] = [1).This results in the singular value decompo-sition (a) (b) (c) (d) (e) The correspondingreducedsingularvaluedecompositionis A+= [235 =[1)[t) = u.E. vt and fromthtsweobtam 4: 11- .4= .41A:\1 [l =

[ l =5 0 __]_ 30 2 5 . -30 2 5 -1] [l;] = r11[II]

=A [i [! 7

7 - 30- Jo 22 5s I -.t l 6 '2!1 3o I13 - yg 15 AA' = [;i]-1z

issymmetric,thus(AA+)T -=AA+ -1$E 1] [;i]is 'ymme.,ic;thus WA)TA. Theeigenvaluesof AAT =areA1 = 15,>.2 ='2,andAJ= 0,withcorresponding 3:] unoleognvocto.s v,},;; v,*' Hl and., =f,o nlThesingularvaluesof Exercise Set 8.7315 ATa" u,= Jj5 and =.Ji.Setting u,=:, AT v,=*' [: [':]=*' 1:1and u, = :/rv, =[::J[-!]= f,; [_:].we have kJ-u 't"'vr 4- 1 726 and fromthisit foUowst.hat [ i&-:15] (C)Theeignn-.;tluesof(A")rAi- =-1;40

arc,\1=ft . ..\2 ,\3=0,wtth 43-mns conespondmg uniteigenw>CIO