January 4, 2014

Mathematical Methods with Boas

Physics 131. Mathematical Methods. Winter 2013

Instructor Chris FronsdalChapter 2

Lecture 5Why complex numbers?

The fact that we need complex numbers is very difficult to explain in one lecture; yetI have to try.

Real numbers are very inconvenient. First of all, we encounter equations to be solved,such as

t2 + 1 = 0.

There is no real number t that satisfies this equation. Consider

x2 + sin t = 0.

For some values of t the sine function is negative, and then the equation can be solved,then as the parameter t changes, and sin t turns positive, there is no solution for a while,until the sine turns negative again. We may feel unsatisfied here and we may ask whythere is no continuity in the problem.

Here is another difficulty, though we are borrowing from Chapter 3. Frequently weare in need to find eigenvalues of matrices, for example,

(1 02 3

)(0a

)= 3

(0a

).

1

Then one day we come up with another matrix

(0 11 0

)(ab

)=

(ba).

The vector is an eigenvector with eigenvalue if

(ba)=

(ab

)=

(ab

),

which requires that b = a = 2b; thus 2 = 1.It is very inconvenient to have to deal with equations that may or may not have

solutions, matrices that may or may not have eigenvectors.The difficulty presented by the equation x2 + 1 = 0 was recognized 2500 years ago,

but it was only around 500 years ago that the problem became acute. Someone managedto find an expression for solutions of the general cubic equation, *

t3 + at2 + bt+ c = 0.

Sometimes, for some values of a, b, c this has one (real) solution, and sometimes it has3. The formula for the solution was apparently general, but sometimes it involved squareroots of negative numbers, even in some cases when it was known that the solutions were allreal. At first sight it may seem that such a formula would be useless, but now it becameurgent to give some meaning to it, and that is what lead to the invention/discovery ofimaginary numbers.

Sections 1,2The rules of complex numbers

A complex number is actually a pair of ordinary (real) numbers.

z = (x, y),

like a vector with 2 components. Complex numbers can be added and multipled; theyform an algebra. To make it easy to remember the rules of this algebra one writes

z = x+ iy,

like a vector with components x and y and unit vectors 1 and i. The rules are

i2 = 1, (a+ ib)(c+ id) = ac+ (ad+ bc)i+ bdi2 = ac bd+ (ad+ bc)i.

The number i is called the imaginary unit and any number of the form z = xi, with xa real number, is said to be imaginary. If z = x+ iy then x is called the real part and y isthe imaginary part.

* To see the equation Google del Ferro

2

The fundamental theorem of algebra: Every polynomial P (t) of order n withcomplex coefficients can be factored,

P (t) = (t )Q(t),

where Q(t) is a polynomial of order n 1 and is a complex number. That means thatthe equation P (t) = 0 with complex coefficients can be solved by at least one complexvalue of t.

Example:tn 1 = (t 1)(1 + t+ t2 + ...+ tn1).

See Homework.You may notice that the vector (

1i

)

is an eigenvector of the matrix (0 11 0

),

with eigenvalue i.

Application of the fundamental theorem of algebra: Every matrix (of complex num-bers) has at least one eigenvector. The proof will come later.

3

Lecture 6An application

Consider the cubic equation

t3 + pt+ q = 0,

with p, q real.Depending on the values of p and q the following alternatives exist:(a) The equation has 3 real solutions or(b) The equation has 1 real solution and two complex solution.

Proof.First of all, a polynomial of third order can be factored:

P (t) = (t )Q(t) = (t )(t )R(t) = (t )(t )(t )k.

This says that there are 3 solutions, except that they are not necessarily distinct:

(1) Three distinct solutions.(2) Two solutions, one of them double: P (t) = (t )(t )2k.(3) One triple solution P (t) = (t )3k.

Suppose that the equation has a complex solution z; that is,

z3 + pz + q = 0, z = x+ iy.

The numberz = x iy

is called the complex conjugate of the number z = x+ iy. Complex conjugation has thefollowing properties: If A,B are complex numbers then

(AB) = AB, (A+B) = A +B, (A) = A.

Apply this operation to the last equation; since p and q are real, p = p, q = q, we get

z3 + pz + q = 0, z = x iy.

Therefore, if z is a solution, so is z. This narrows the possibilities:

(1) Three real solutions (or multiple solutions).(2) One real solution and a pair of complex conjugate solutions.

Both possibilities occur, depending on the values of p and q. Consider the plane withcoordinates p, q; to each point in this plane there is a pair of real numbers (p, q). The planecan then be divided into two regions, according to whether case (1) or case (2) prevails.

4

Now let us suppose first that, at the point (p, q) there are 3 real solutions, case (1).Let us change the values of p, q continuously. Eventually, we expect to get to values of pand q where case (2) prevails. But the change is continuous, so on the border between thetwo regions the two solutions are both complex conjugates and real; hence equal. That is,points on the border are characterized by the fact that

t3 + pt+ q = (t )(t )2k,

Now take the derivative with respect to t,

3t2 + p = (t )2k + 2(t )(t )k. ()

At the point t = both vanish, and we deduce that, on the border,

t3 + pt+ q = t3 +tp

3,

that is,

q = 2pt3, t = 3q

2p.

And thus by Eq.(*),

3(3q

2p)2 + p = 0.

Finally, these points make up the line

27q2 + 4p3 = 0.

The locus of these points is shown in the figure. See following page.

I shall examine the cubic equation from the point of view of the complex plane.

Section 3. The complex plane.Here I shall explain what was done in the application that I just finished. The idea

is this. To every complex number z = x+ iy there corresponds a point in the x, y plane.Addition of two complex numbers

z1 = x1 + iy1, z2 = x2 + iy2

is a vector addition of two 2-dimensional vectors,

z1 + z2 = (x1 + x2) + i(y1 + y2).

Multiplication of a complex number by a real number is just like multiplication of a vectorby a real number,

kz = kx+ i(ky).

5

But the multiplication of two complex numbers is not a vectorial operation,

z1z2 = (x1 + iy1)(x2 + iy2) = x1x2 y1y2 + i(x1y2 + x2y1). ()

It does not have a simple, geometrical interpretation. Or does it?In polar coordinates,

z = x+ iy = r cos+ ir sin = r(cos+ i sin).

Notice that r is the length of the vector and that is the angle that the vector makes withthe positive x axis. The length of the vector is denoted

|z| = r =x2 + y2 =

zz.

The angle is called the phase of z.If r = 1 we have a vector of length 1, on the unit circle

x2 + y2 = 1, z = cos+ i sin.

The combination cos+ i sin is important. If we expand in power series we have

z = (1)m

(2m)!2m + i

(1)m(2m+ 1)!

2m+1

Since (1)m = i2m this the same as

z = in

n!n = ei,

or Eulers equationei = cos+ i sin.

The cubic equation in the complex planeThe equation is

z3 + zt+ q = 0, z = x+ iy.

This equation requires that the real and the imaginary parts are separately zero,

x3 3xy2 + px+ q = 0, 3x2yy3 + py = 0

With the calculator we can show the loci of these two equations in the x, y plane. See nextpage.

6

ApplicationThe real and imaginary parts of z, in polar coordinates, are

x = r cos, y = r sin.

In circuit theory, if q is the charge

q(t) = q(0) sint = (q(0)eit),

the current is the derivative,

j(t) = q(t) = q(0) cost = (q(0)eit).

Calculation in circuit theory deals with the complex quantity

q(0)eit.

Only at the end of the calculations does one take the real part, to extract the physicalmeaning. What makes this advantageous is the simplified rule for derivatives,

d

dteit = ieit

Perhaps, this example explains why the phase of a complex number is called its phase.

Sections 4,5Complex numbers form an algebra. The inverse of a complex number is the complex

number

z1 =1

z=

z

|z|2 .

If z = x+ iy, then

z1 =x iyx2 + y2

.

If z = rei, thenz1 = r1ei.

Answer to a question, geometric interpretation of the product. In polar coordinates,

z1z2 = r1r2ei(1+2)

The phases add. Illustrate.

7

Homework due Wednesday Jan 23.1. Practice using the binomial expansion. The gravitational potential is a function

of position in 3-dimensional space. It depends only on the distance from the center of theearth, call this distance r, so we are talking about a function (r) that takes the form

(r) =k

r, k is a constant.

Suppose the radius of the earth (a perfect sphere) is R. Let z be a vertical coordinatemeasured upwards from the surface, then the gravitational potential at a height z is

(r) = (R + z).

Suppose that z

Sections 6,7. Complex infinite seriesThe theory of functions of a complex variable is studied in the course Physics 132.

But our book deals with the most elementary aspects of the theory and we shall take thetime to include this in our curriculum.

Since complex numbers can be multiplied and added, we can form polynomials in acomplex variable z = x+ iy,

a+ bz + cz2 + ...,

Nj=1

ajzj ,

and series:j=1

ajzj .

To deal with series we must define convergence for such series.This is done in a way that is a natural extension of what we did for real series. First

of all,j=1

ajzj = lim

N

Nj=1

ajzj .

But, as in the real case, we must define what we mean by a limit.Recall that for real numbers we said that a number S is the limit of the sequence

S1, S2, ..... if all but a finite set of the numbers in the sequence lie close to S, more preciselyif

|Sj S| < ,for all but a finite number of the Sj , no matter how small we choose the positive number. We make exactly the same definition for the complex case; the only novelty is that

|Sj S|

is the norm of the complex number Sj S. This has a nice geometrical interpretation. If

Sj = xj + iyj , S = x+ iy,

then

|Sj S| =(xj x)2 + (y yj)2

is the distance between the points Sj and S in the complex plane. Most of the points Sj ,all but a finite number, lie at a distance less than from the point S, and that is true nomatter how small the positive number is.

So now we understand how to verify convergence of a series of complex numbers. Butwe shall limit ourselves to absolute convergence. This is much simpler and much moreimportance. So consider

j

aj .

9

Absolute convergence means ordinary convergence of

j

|aj |,

and this is a sum of real numbers. For example, take

j

aj, aj =(1 + i)j

2j.

See Boas page 57. Here

|aj| = (2)j2j = (

12)j;

the sum is absolutely convergent. The most important example,

j

zj

converges absolutely for |z| < 1 and zj/j! converges for all z. Furthermore, these sumscan be found by exactly the same method as used for the real case.

Section 8 is for reading. Section 9 is on Eulers formula, discussed already, pleasereview.

Section 10This section presents examples designed to make you more familiar with complex

numbers. Complex numbers are represented in the complex plane, mostly with polarcoordinates:

z = rei.

One principle point is to interpret the formulas

zn = rnein and z1/n = r1/nei.

Here r is a positive real number and r1/n has the usual meaning of roots of real numbers.Perhaps the most interesting new fact is the following. In view of the geometric

interpretation, when we write z = ei we must have 0 < 2. If increases beyond 2we get

e2pi = 1, ei(2pi+d) = eid.

That is, this function is periodic with period 2.

Sections 11 and 12 is for reading; we shall probably not need this.

10

Section 13. LogarithmsThis material is very subtle and very important. It seems easy to explain, but the

implications are rather difficult to see. The book limits the discussion to the bare factsand I shall do the same. After all, we shall not have much use for it in this course, I think.

The point is, as we have seen, that the function ei of the real variable is periodic,

ei(+2pi) = ei.

This will not cause problems, except for one thing: When reaches and surpasses 2 thevalue drops from 2 to 0, see board. Hence this function is not continuous, and it doesnot have a derivative at this point. This simple fact has very interesting consequences, aswe shall see much later.

Given the value of ei we cannot immediately determine the value of . If a realnumber a fits: that is, if eia = ei, then a+2 and a+4 work as well. To avoid ambiguityone defines the principal value of the logarithm to be i times the unique value that liesin the interval 1 < 2.

Sections 14, 15 is for reading.

11

Section 16. ApplicationsI shall not present this material in class, and you shall not be responsible for it. But

we can talk about it in my office, or you can organize a meeting and invite me to give aprentation.

An alternating electric current flowing out of a plug in the wall to enter some applianceis time dependent,

J = J0 sint.

J0 is called the amplitude, is the frequency. It repeats itself every time t advances by1/(2) seconds. At the same time the voltage also oscillates,

V = V0 cost.

Circuit theory embraces a large number of simple differential equations, such as

J = Rd

dtQ,

where Q is the charge andd2

dt2J = 2J.

Circuit engineers manage these equation by using a symbol s for d/dt and treating s asa number. This means that they are thinking in terms of complex currents, voltages andcharges, for if

d

dtJ(t) = iJ(t),

thenJ(t) eit = cost + i sint.

In Boas, on page 77 is a circuit diagram that I shall draw on the board. In comprisesa voltage source (a generator of alternating current, for example), a resistance (a lamp ora heater), a capacitor and a coil.

Circuit theory is summarized by Kirkhoffs laws, giving the voltage change across eachof these three passive elements in terms of the current:

VR = RI,

VL = LdI

dT,

VC =1

CI.

All these quantities depend on the time, t.The voltage across the generator is V , it is usually periodic,

V (t) = V (0) cost = V (0)eit

12

and all the other voltages have factors cost or sint. Because the current is the sameeverywhere,

V = VR + VL + VC = (R+ Ld

dt+

1

C

dt)J(t).

We can use either a cosine or a sine to represent V , but it is much simpler to set

V (t) = V (0)eit.

The actual physical quantity is the real part, or the imaginary part; we shall take the realpart of our complex expressions at the end. Now the exponential factor cancels out andwe get

V (0) = (R+ iL+1

i)J(0).

What is given is V ; we wish to calculate the current. Evidently

J(0) = (R+ iL+1

i)1V (0).

We need to manipulate the factor (called the impedance of the circuit) to extract the realand the imaginary parts,

(R+ iL+1

i)1 = A+ iB.

Or we may prfefer to express it as a positive number times a phase factor,

(R+ iL+1

i)1 = Zeit

In the latter case,J(t) = ZV (0)ei(t+).

The factor Z gives the current as a real number; the phase factor produces a phase shift.taking the real part we get the actual physical current,

J(t) = Z cos(t+ ).

To calculate Z and t see homework.

13

Chapter 3. Linear algebraIntroduction to the chapter

Here we shall do some exercises independently of the book. Here is the plan:A. Practice vectors: what we do with them. Easy.B. Interpret these operations to learn what they mean. Not so easy,perhaps.C. Practice matrices: what one does with them. Easy.D. Interpret these operations, to study what they mean. More difficult.

References will be made to the book whenever useful.

A. Practice vectors: what we do with themSometimes we can think of vectors as a type of arrows, and make vectors that we

can pick up and manipulate; I am not sure that this helps, but we can try. So here wehave a couple of these arrows. They point in all directions of 3-dimensional space, but3-dimensional vectors are too difficult to manage with just a couple of hands, so we shallconfine them to the table top or the blackboard; this turns them into 2-dimensional vectors.

With vectors we can do precisely 2 things:1. We can add 2 vectors. We can add them by the parallellogram rule that I shall

illustrate on the blackboard.But this is not practical; we cannot send off to have these arrows made for every

occasion. So we have to make it easy by introducing the idea of components. But first,here is another kind of vectors:

(12

),

(ab

),

(xy

)

and xyz

.

Let us , at first, just agree to call these objects vectors. We define addition in this way:if

~u =

(12

), ~v =

(xy

)

then the sum ~u+ ~v is defined as follows:

~u+ ~v =

(12

)+

(xy

)=

(1 + x2 + y

).

The numbers that appear in these column vectors are called components; to add vectorswe just add their components.

The sum of two vectors is again a vector (of the same dimension), so we can add anynumber of vectors to produce any number of new vectors. The space of vectors is closedunder addition.

14

To add two vectors they must have the same number of components. This number iswhat we call the dimension of a vector: the number of components. Above we have several2-dimensional vectors and one 3-dimensional vector.

The zero vector is the unique vector ~0 that has the following property:

~0 + ~v = ~v +~0 = ~v,

this being true for any vector ~v. Evidently,

~0 =

(00

).

2. The only other operation that we can do with vectors is to multiply them bynumbers. If A is a number and ~v is the vector used above then we define the product A~vas

A~v =

(AxAy

).

Of course, we can write xA and yA in stead of Ax and Ay, and we also allow ourselves towrite ~vA, meaning the same as A~v. Furthermore we have some familiar rules of ordinaryalgebra that continue to hold:

A(B~v) = (AB)~v,

andA(~u+ ~v) = A~u+ A~v, (A+B)~u = A~u+B~u.

OK, now let us see what the book says about this. Nothing!Exercises.

B. Interpret these operations with vectors1. Collecting information. Suppose you wish to buy: a kg potatoes, b gallons of milk

and c pounds of cereal. Write the amount vector or shopping list

A = (a kg, b gallons, c pounds) or A = (a, b, c).

Suppose the price of 1 kg of potatoes is P , a gallon of milk is M and 1 pound of cereal isC. Write the price list, another type of vector.

B =

PM

C

.

I decide to use row vectors to indicate amounts and column vectors to indicate prices. Thetotal cost is

aP + bM + cC = (a, b, c)

PM

C

= AB.

15

This can be understood as follows. There is a function (a register) B (a column vector)that depends on your shopping list. The value B(A) is the total cost, it is calculated asshown.

The next day I buy amounts A = (a, b, c). Already you see that I have gained someeconomy of expression. The total amounts purchased over 2 days is another shopping list,

A+A = (a+ a, b+ b, c+ c)

and the total cost (if the prices did not change) is

(A+A)B = AB + AB.

But on the third day the prices went up by 5 percent and the cost is then

A(1.05 P, 1.05 M, 1.05 C) = A1.05B = 1.05AB.

Thus I have added vectors and I have multiplied a vector by a number. I have also definedthe product of a row by a column. Incidentally, I have shown that there are vectors thatshould not be added, A+B has no meaning.

The main lesson of this example is that vectors have whatever meaning you assign tothem. But we agree that, when we call them vectors then it implies that vector operationsare defined.

2. Displacements. This should be done with a map on a screen. Possibly on theboard. Compare Boas page 96 and so on.

Explain the map and use the streets to indicate points/addresses.Suppose that the map has streets running horizontally and avenues running vertically.

The intersections are locations, or addresses. Hollywood and Vine is an address. It is nota vector; (Westwood and Leconte) cannot be added to (Wilshire and Federal).

But suppose you proceed from (Westwood and Leconte) to (Sepulveda and Wilshire).This is a displacement: 1 mile south and 1 miles west:

~a = (1,1).

From that point we proceed to Sepulveda and Sunset, a displacement

~b = (0, 2)

and it certainly makes sense to add the 2 displacements,

~a+~b = (1,1) + (0, 2) = (1, 1).

Displacements are vectors and they are one of the most important kind of vectors that weshall have to do with.

Let us do this addition graphically, using the parallellogram rule. To do this we needa plane, with coordinates, and an origin. This is not a map, Westwood and Leconte is not

16

a point in this plane. It is the space of displacements. In this plane we can draw our twovectors, starting at the origin, and we can add them.

Confusion arises with the notion of radius vector. Return to the map. Insteadof street names, introduce coordinates. Coordinates are positive or negative numbers,including zero, so there must be a point (0,0) on the map. This point has no particularsignificance, but it may be convenient to have a reference point. Now if p is any point onthe map, we can draw an arrow from the origin to p and call this arrow the radius vectorof the point p. It is not clear that this is a vector, we can discuss it.

What is a vector, is the displacement that connects the origin to p. Well, I shall stopthis discussion here. I just wanted to warn you that the term radius vector is confusingand that you should give it some thought.

See Boas page 82.

C. Practice with matrices: what one does with themTo a printer, a matrix is a rectangular array. To us it is a rectangular array of numbers.

Examples on the blackboard.An m-by-n matrix has m rows of dimension n or n columns of dimension m.With matrices we have 3 operations.

1. Addition of any two matrices of the same dimension: examples on blackboard.2. Multiplication of a matrix by a number.3. Multiplication of two matrices A,B is defined only if the rows of A have the same

length as the columns of B. Examples on blackboard.See Boas page 114.

D. Interpret these operations, to study what they meanA couple has 3 children called A,B,C and 10 houses. The man owns 6 houses and

the wife owns 4 houses. When both die it is found that the children inherit as follows fromthe parents,

(4, 4, 2) = (4, 6)

(.5 .5 01/3 1/3 1/3

)

What this means that the mother divides her property 50/50 among the first two childrenand the father leaves 1/3 of his propertyto each. The matrix is the will; it convertsthe 2-vector of properties (wifes property, husbands prperty) to the inheritance of the 3children, a 3-vector.

Here we see the most fundamental matrix operation: a matrix acts on a vector toproduce another vector. We could just as well replace the rows by columns and write

442

=

.5 1/3.5 1/3

0 1/3

( 4

6

).

Compare Boas page 84. But skip the next pages for the time being. The concept of rowreduction may be taken up later, but not yet.

Boas, on page 83, offers 2 interpretations of a set of linear transformations. Actually,there can be other transformations as well, as we have seen.

17

E. Problems involving matricesThe most notorious problem involving matrices is the solution of linear equations.

Considering the last matrix equation the problem would be: if we know how much eachchild gets (the 3-vector) and we know the will (the matrix), can we calculate the amountthat each parent owned?

442

=

.5 1/3.5 1/3

0 1/3

(x

y

).

There are three equations here, each giving us some information about x and y. The firsttwo equations both involve the same combination of x and y, so the second line gives nonew information. But the third equation gives us new and independent information aboutx and y, so that should be enough. In general: among the three rows of the matrix at leasttwo must be linearly independent.

This problem is basic to matrix theory, the problem of solving an equation

~u = M~v

for the column vector ~v when the column vector u and the matrix M are known. The mostintuitive way of stating the main result is this:

We can solve, to find u when v andM are given, if the number of linearly independentrows in M is at least equal to the number of rows in v. If this is not intuitively clear thengo back to the last example. We shall now look at the same problem in general. Keepconsulting the book for discussions of the same problem.

Square matrices, the determinantThis subject is taken uo by Boas on page 89. Since the subject is difficult, you should

study that version as well.We want to solve the equation

~u = M~v.

Formally, the solution is~v = M1~u.

When the matrix M is square the matrix M1 is completely defined by one or the otherof the following two condtions,

MM1 = 1 that is, the unit matrix,

orM1M = 1.

Such a matrix exists if and only ifdetM 6= 0.

This is why the determinant is a very important concept, so we shall take some time todiscuss it.

18

Let us begin by defining the determinant:

detM =

()M i11 M i22 ....M inn ,

where i1, ..., in is a permutation of 1, ...n and the sign is + for an even permutation and -for an odd permutation. I shall now explain what this means by the simplest examples.

1. If the matrix is one-by-on then detM = M .

2. If it is 2-by-2 then

detM = M11M22 M21M12 .

About the notation. here M ji is the element in row i, column j,

M =

(M11 M

21

M12 M22

).

I shall discuss this notation a little later.You notice that, in the formula, the lower indices (the row indices) are in the usual

order, 1,2 in both terms. The upper indices appear in both orders, 1,2 and 2,1. We say that2,1 is a permutation of 1,2. This particular permutation is just one interchange, 1, 2 7 21.It is called an odd permutation.

Let us consider this formula for detM . . Notice that the numbers M11 ,M21 form a

column, denote it ~u, and so do M12 ,M22 , call it ~v. Then

detM = u1v2 u2v1.

this is antisymmetric under the interchange of u and v, therefore it is zero if the twovectors are parallel. In fact, this determinant is precisely the area of the parallelogramdefined by the two vectors; this are is zero if the two vectors are parallel.

In view of the foregoing discussion of an example, this suggests that the matrix hasan inverse precisely when the determinant is not zero. This means that all the rows arelinearly independent. Boas row reduction should be seen as a calculation that exploresthis concept.

3. If it is 3-by-3,

detM = M11 (M22M

33 M32M23 )M21 (M12M33 M32M13 ) +M31 (M12M23 M22M13 ).

If this case is completely understood you should be able to handle the next case with its24 terms.

Notice: the lower indices are always in the normal order. In the upper indices, thereis an odd permutation: 2, 3 7 3, 2 or 1, 2, 3 7 1, 3, 2. The fourth term also has an oddpermutation, 1, 2, 3 7 2, 1, 3. The 5th term has an even permutation, 1, 2, 3 7 1, 3, 2 7

19

3, 1, 2; that is why the sign is +. Finally, the 4th and 6th terms have signs becausethey are related to the preceding terms by a single exchange, an odd permutation.*

The even permutations are 123, 231, 312. the others are odd.The quantities P ji that I introduce next are the cofactors on page 90 of Boas. She

uses a different notation. What I call M ji and Pji appears as aij and Mij in Boas.

If we understand the determinant we can isolate the coefficient of each of the matrixelements M ji . Call them P

ij . Then consider the sums, first in the two-dimensional case

M11P11 +M

21P

12 , M

11P

21 +M

21P

22 .

Then you find by easy evaluation that the first number is equal to detM and the secondnumber is equal to zero. This result is valid in all cases, and we conclude that

j

M ji Pkj = detM, if i = k and 0, otherwise. ()

This is the same as saying, if the determinant is not zero, that

(M1)ji = Pji / detM.

Test it in the 2-dimensional case.We shall not insist on a proof of this case, but we can make it plausible. The proof is

not so difficult but it require some facility in working with so many indices.Boas does not obtain this result until page 119. The discussion on page 94 is very

important for the understanding. The rank of a square matrix is the number of linearlyindependent rows (or columns).

Recall what we noticed about the 2-dimensional case. the determinant is an antisym-metrical function of the columns that make up our matrix. Therefore, if two columns areequal, or just parallell, then the determinant is zero. Likewise, if we replace one of thecolumns by one of the others then we get zero. That is exactly what our Eq.(*) says.

This insight into the concept of the determinant, and the relation of the determinantto the existence of an inverse matrix, is one of the core facts of linear algebra.

Homework number 3. Due Wednesday Jan30 by 12:30AM1. Boas page 95, number 1.2. Boas page 95, number 2.3. Boas page 95, number 7. This problem may help you understand all the talk of

row reduction and what it is good for.4. Boas page 99, number 3.5. Boas page 105, number 11.6. Boas page 122, number 5.

This takes us to page 124 in Boas. There begins a more sophisticated view of linearalgebra. The concepts till this point may seem a little dry, but to manage them is necessarypreparation for the continuation.

* About placement of indices. You can place them all downstairs, as Boas does. I shalldiscuss this a little later.

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Here we are going to study linear algebra in general. It is going to get abstract. Thereason for that is that linear algebra has very many applications, of a very varied nature.If you do not see the fundamental structure of the theory you will not be able to use it ina nover situation. Still, we are not going to spend a lot of time on it. Let us introduce theideas and then see how they work in later chapters.

What I write here is a little more detailed than usual, because the subject is hard.The exposition begins just as Boas with

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Physics 131. Winter 2013. Midterm 1.Write your answers on the spaces provided. Prepare on another sheet. The amount

of space gives you an idea of how much work is provided.Continue on the back of the page if necessary, but call attention to it.Skip any difficult part and return to it at the end if there is time. Be sure that you

get through it all to catch any part that is easy.

1. Given any sequence, of real or complex numbers:

a0, a1, a2, ......an, an+1, .......,

1.a. what property makes it an arithmetic sequence?

Solution. It is an arithmetic sequence if an+1 an is the same number for all n; thatis, if an = a0 + nb, or

an+1 = an + b,

with a fixed number b, the same for all n,

1.b. and what property makes it a geometric sequence?

Solution. It is a geometric sequence if an+1/an is the same number for all n; that is,

an+1 = anb,

or an = bna0. with a fixed number b, the same for all n,

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2. Prove the following sum formula by induction. (Mathematical induction is themethod that was used several times in the first chapter of Boas. If you are not sure whatthat means, go ahead and prove the formula any way you can.)

Nn=1

xn =xN+1 1x 1 .

Solution. Proof by induction. Suppose the statement is true for some value M of N .Then check it for N = M + 1,

M+1n=1

xn =Mn=1

xn + xM+1.

Using the known result for the first term and manipulating the second term you get

M+1n=1

xn =xM+1 1x 1 +

xM+2 xM+1x 1 =

xM+1+1 1x 1 .

So the statement is verified for N = M + 1. It is true for the case that N = 1, thereforeit is true for N = 2, 3, ... and for all positive, integer values of N .

3. Give a completely convincing argument that the limit of the following sequence is1. Please think a little about what it means and how to make the argument convincing.

...., .99, .999, .9999, ......

Solution. Maybe the best way is this. The difference

1 .99999...(n nines) = .0000....01(n 1 zeros) = ( 110

)n.

This difference goes to zero in the limit as n.

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4. Consider the complex series

n=1

zn

n!, z = x+ iy,

where x, y are real numbers. Show, using the ratio test (the first and simplest ratiotest) that this series is absolutely convergent for all z. Begin by defining what absoluteconvergence means; then explain the ratio test, and finally use it.

Solution. Absolute convergence of the series means the the series with positive terms,

n=1

|zn

n!| =

n=1

rn

n!,

where r = |z|, is convergent. This is a series of the form0

anrn, an =

1

n!.

Now an+1/an = 1/(n+1), which tends to K = 0 as n grows. Hence Kx is always less thanone and the series is convergent. Therefore the original series is absolutely convergent.

5. Consider the complex function

f(z) =1

z2 + 2i.

Image expanding it as a complex Taylor series about z = 0.(a) Write the series, the nth term in terms of the nth derivative of f(z), but do not

calculate the derivatives.

Solution.

f(x) =n=0

f (n)(0)

n!

(b) We want to know the radius of absolute convergence of this series. You can ofcourse use the ratio test to determine the radius of convergence of the series, but there isnot enough time. Instead, you are asked to use your experience with other examples tomake an informed guess.

Solution. The function that your are expanding seems to be well defined for all zexcept when z2 = 2i. this is a distance 2 from the origin (the expansion point). Itfollows that the series is absolutely convergent when |z| < 2.

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6. Consider a function of the complex variable z. To make sure that you know whatthat means, consider the case of a polynomial

P (z) = a0 + a1z + a2z2, ...,+anz

n

with complex coefficients a0, a1, ..., an. Imagine expanding all the powers zn = (x+ iy)n,

but do not actually do so unless you really need to. The point of that is that you canwrite out all the terms and (using i2 = 1), write the whole thing as a sum of two parts,P (z) = u(x, y) + iv(x, y) where u and v are real functions of the real variables x and y.This is by way of explanation only. Please take care; do not think of u and v as functionsof z.

(a) Considering P (z) as a function of a function, express the derivative with respectto x in terms of dP/dz. Then express the derivative with repect to y in terms of dP/dz.

Solution. Write Px for P/x to save time. Now consider P as a function P (z(x)) orP (z(y)):

Px(x+ iy) = Pz(z), Py(x+ iy) = iPz(z).

Hence Py = iPx. This is because P depends on the combination x+ iy only.

(b) This should give you a relation between the four derivativesu/x, u/y, v/x, v/x.

Solution. (u+ iv)y = i(u+ iv)x.

(c) The real and the imaginary parts of this relation should give you two very simplerelations between the four derivatives. Write them here.

Solution. (u + iv)y = uy + ivy = i(u + iv)x = iux vx, and the real and imaginaryparts must each be equal:

uy = vx, vy = ux.

These relations are known as the Cauchy-Riemann relations. They are usually takenas the foundation of the theory of functions of a complex variable.

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Solutions to HW number 2.Homework due Wednesday Jan 23.

1. Practice using the binomial expansion. The gravitational potential is a functionof position in 3-dimensional space. It depends only on the distance from the center of theearth, call this distance r, so we are talking about a function (r) that takes the form

(r) =k

r, k is a constant.

Suppose the radius of the earth (a perfect sphere) is R. Let z be a vertical coordinatemeasured upwards from the surface, then the gravitational potential at a height z is

(r) = (R + z).

Suppose that z

Complex variables5. We have defined analytic function in these notes, page 12. In particular, any

polynomial in x is a function that is analytic for all x. Now consider a complex variablez = x + iy. The definition of analytic is actually the same, but let us consider onlypolynomials, for example

f(z) = a+ bz + cz2.

Suppose that a, b, c are real numbers. Write z = x + iy. Calculate the real and theimaginary parts of f(z). (They are polynomials in x and y.)

Solution. f(x, y) = (a+ bx+ c(x2 y2) + i(by + 2cxy).6. Optional, but please do it for me if you can. It will tell me something about what

you know. With the previous example in mind, let f(z) be an arbitrary poynomial andimagine having done the same thing to it:

f(z) = u(x, y) + iv(x, y)

where u and v are polynomials in x and y with real coefficients. Prove the followingtheorem properties,

ux = vy, uy = vx.Here ux = u/x and uy = u/y. verify the result for the case that f = z

2

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Linear vector spaceBoas page 142.

Definition. A linear vector space is a set on which the following operations are defined:addition and multiplication by a real (real vector space) or a complex number (complexvector space).

Comments.1. Elements of a linear vector space are usually called vectors. If we have to deal

with several different kinds of vectors we may use different names to distinguish. A veryimportant concept is a pair of vector spaces referred to as vectors and covectors.

2. The operations, ~u+ ~v, and multiplication by a number, k~u are supposed to satisfyprecisely the same rules that we are familiar with. In general there is no way to multiplya vector by a vector.

BasesA basis for a vector space V is a set (finite or countable; that is, it is a sequence),

such that every vector in V can be expressed as a linear combination of them, in just oneway. The number of elements in the basis is called the dimension of the space.

Familiar examples1. The most familiar, 2-dimensional vectors form a plane with a selected origin and

Cartesian coordinates. Addition is defined as earlier and multiplication by a real numberas well. This is a real vector space; you can think of it as a plane. A basis can be chosenin many ways; you must have precisely 2 of them, {~e1, ~e2}. (Demonstrated on the board.)

2. Row vectors ~u = {a1, a2}, possible bases

~e1 = {1, 0}, e2 = {0, 1}.

Notice that

~u = {u1, u2} = u1~e1 + u2~e2 =2i=1

ui~ei.

The numbers u1, u2 are the components of ~u in this basis.

IndicesThe upstairs indices are used to give names to the components of a row vector. The

downstairs indices in these formulas give names to the basis vectors. These are two verydifferent ways to use indices and that is one of the reasons to place them differently.

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Summation conventionsWe shall make it a rule that a repeated index is to be summed; we leave out the

summation sign. And it will turn out that all the summations satisfy the following rule:One index upstairs the other one downstairs, as in the example. This is mainly so thaterrors can be avoided. But it goes much deeper than that, as will be seen.

Important exampleWe shall consider Newtons equation for a free particle. It is a differential equation

for the function x(t) that gives the position of a particle that is moving without beingsubjected to any forces. The velocity is

v(t) = x = dx/dt

and the acceleration is

a(t) = v = dv/dt = d2x/dt2 = x =d2

dt2x(t).

If no force is acting then the acceleration is zero and thus

x(t) =d2

dt2x(t) = 0.

This is a linear differential equation.The general solution of this equation is

x(t) = a1 + a2t,

with a1, a2 constant. Incidentally, it is a Taylor series.

LinearFirst of all, the differential equation is a linear one, a linear condition on the function

x(t). In the present case, when the equation is (d2/dt2)x(t) = 0, that means that if x(t)and y(t) are two functions then

d2

dt2(x+ y) =

d2

dt2x+

d2

dt2y.

This has the following consequence: If x(t) and y(t) are two solutions of the differentialequation then so is the sum; and so is any linear combination.

Here we are adding functions; we shall consider them to be vectors. Multiplication bya constant defines another function: kx is the function

(kx)(t) = kx(t).

So functions form a vector space, but this space is too large so we shall confine ourselves toa subxpace: the solutions of Newtons equation. This is simply the space of all functions

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of the type x(t) = a1+a2t, and this space has precisely 2 linearly independent vectors, forexample the functions (the vectors)

~e1(t) = 1, ~e2(t) = t.

So the general solution is ai~ei and now the constants a1, a2 are seen to be the components

of this vector.

To sum up, the solutions of Newtons equation for a particle that is moving withoutbeing influenced by a force is a 2-dimensional linear vector space. This statement is thefoundation of the theory of linear differential equations that we shall take up later.

What does linear mean?The word begins to appear in Boas on page 124.

Linear combination If u, v are thigs that can be added and multiplied by numbersto form au+bv; in short, if u, v are vectors, then a linear combination of u and v is anothervector, of the form

au+ bv.

It should be clear that the word linear means nothing unless the subject has to do withvectors.

Linear (in)dependence. A set of vectors are linearly dependent if there is a linearcombination that is equal to zero, as in the example

~u = (1, 1), ~v = (3, 3), 3~u ~v = 0,

otherwise linearly independent. The vectors

(1, 0), (0, 1)

are linearly independent.The basis vectors of a linear space are required to be linearly independent. This is to

assure that the expression for a vector in terms of the basis should be unique. That is if

u =

ua~ea =

va~ea,

then the coefficients in both sums are precisely the same. That is, if

ua~ea

va~ea =

(ua va)~ea

is zero then ua = va = 0.The concept of linear dependence has already been discussed in connection with linear

equations and and determinants.

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Linear function. A linear function of a real variable is a function of the formf(x) = kx, k constant. It implies that f has the property

f(ax+ bv) = af(x) + bf(y).

It says that the summing first and evaluating the function afterwards is the same asevaluating the function and summing the results. See Homework.

Linear functions, operators, maps, mappings, transformations. These wordsare going to be thrown about constantly. They all mean exactly the same thing. To makesense we need two vector spaces U and V , although they are not necessarily distinct.

A vector space is a collection of objects that can be combined as vectors are com-bined. Any two elements of a vector space U can be combined in linear combinations.Thus column vectors must have a fixed dimension since we cannot add vectors of differentdimensions. That is; a vector space has a dimension. And of course we cannot add columnvectors to row vectors.

So in U we can make linear combinations and in V we can make linear combinations.Therefore the following makes sense: Ainear function, operator, map, mapping, transfor-mation from U to V is a machine f that takes any vector u from the space U and producesa vector v in V (we denote it this way: f : u 7 v = f(u)) such that

f(u+ u) = f(u) + f(u)

andf(ku) = kf(u),

where k is a number.

First and all-important example: the ordinary function

f(x) = ax+ b

is linear only of b = 0!!!!!!!

Matrices. Consider any m-by-n matrix M = (Mi ), where i runs from 1 to m and runs from 1 to n. I use letters from different alphabets to help me remember the difference.I can apply (from the left) the matrix M to any n-dimensional column vector. The resultof the calculation is an m-dimensional column vector. So a matrix is a linear operator (ormapping or whatever) from a space of dimension m to a space of dimension n. That iswhat a matrix IS!

Conversely, any linear operator (function, ) from a space U to a space V becomes amatrix if you have bases in both spaces. This procedure, to construct a matrix once anoperator, and bases, are given is extremely important. If you can master it you are wayahead.

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Let ~e1, ..., ~em be a basis for V and let ~f1, ..., ~fn be a basis for the space V . If A is alinear operator from U to V then each of the quantities

A(~ei), i = 1, ..., m,

belongs to V . Therefore it can be expanded in terms of the basis of that space:

A(ei) =

n=1

Ai f, i = 1, ..., m.

And there is your matrix:

A = (Ai ), = 1, ..., n, i = 1, ..., m.

If this looks difficult let us work out a concrete example, one that is already familiar:Let U be the 2-dimensional space of solutions of Newtons equations for a free particle,functions of the form

a+ bt = a~e1 + b~e2.

and let V be the same space with the same basis. Let A be the linear operator d/dt. Then

(d/dt)(a+ bt) = b,

orA~e1 = 0 =

j

Aj1ej

andA~e2 = 1 = ~e1 =

j

Aj2ej

The first says that the first row of the matrix has only zeros. The other that A12 = 1 andthat A22 = 0. So finally

A =

(0 10 0

).

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