Chow T.L. Mathematical Methods for Physicists: A Concise Introduction

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Once the eigenvalues have been found, corresponding eigenvectors can be

found from the system (3.55). Since the system is homogeneous, if X is an

eigenvector of

A, then kX, where k is any constant (not zero), is also an eigen-

vector of

A corresponding to the same eigenvalue. It is very easy to show this.

Since

AX  X, multiplying by an arbitrary constant k will give k

AX  kX.

Now k

A 

Ak (every matrix commutes with a scalar), so we have

AkXkX; showing that kX is also an eigenvector of

A with the same

eigenvalue . But kX is linearly dependent on X , and if we were to count all

such eigenvectors separately, we would have an in®nite number of them. Such

eigenvectors are therefore not counted separately.

A matrix of order n does not necessarily have n linearly independent

eigenvectors; some of them may be repeated. (This will happen when the char-

acteristic polynomial has two or more identical roots.) If an eigenvalue occurs m

times, m is called the multiplicity of the eigenvalue. The matrix has at most m

linearly independent eigenvectors all corresponding to the same eigenvalue. Such

linearly independent eigenvectors having the same eigenvalue are said to be degen-

erate eigenvectors; in this case, m-fold degenerate. We will deal only with those

matrices that have n linearly independent eigenvectors and they are diagonal izable

matrices.

Example 3.12

Find (a) the eigenvalues and ( b ) the eigenvectors of the matrix

A 



Solution: (a) The eigenvalues: The characteristic equation is

det

A ÿ 

I

5 ÿ  4

12ÿ 

 

ÿ 7  6  0

which has two roots



 6 and 

 1:

(b) The eigenvectors: For   

the system (3.55) assumes the form

ÿx

 4x

 0;

ÿ 4x

 0:

Thus x

 4x

, and





126

MATRIX ALGEBRA

is an eigenvector of

A corresponding to 

 6. In the same way we ®nd the

eigenvector corresponding to 

 1:



ÿ1



Example 3.13

A is a non-singular matrix, show that the eigenvalues of

ÿ1

are the reciprocals

of those of

A and every eigenvec tor of

A is also an eigenvector of

ÿ1

Solution: Let  be an eigenvalue of

A corresponding to the eigenvector X,so

that

AX  X:

Since

ÿ1

exists, multiply the above equation from the left by

ÿ1

AX 

ÿ1

X ) X  

ÿ1

Since

A is non-singular,  must be non-zero. Now dividing the above equation by

, we have

ÿ1

X 1=X:

Since this is true for every value of

A, the results follows.

Example 3.14

Show that all the eigenvalues of a unitary matrix have unit magnitude.

Solution: Let

U be a unitary matrix and X an eigenvector of

U with the eigen-

value , so that

UX  X:

Taking the hermitian conjugate of both sides, we have

 *X

Multiplying the ®rst equation from the left by the second equation, we obtain

UX  *X

Since

U is unitary,

I, so that the last equation reduces to

Xjj

ÿ 10:

Now X

X is the square of the norm of X and hence cannot vanish unless X is a

null vector and so we must have jj

 1orjj1; proving the desired result.

127

THE MATRIX EIGENVALUE PROBLEM

Example 3.15

Show that similar matrices have the same characteristic polynomial and hence the

same eigenvalues. (Another way of stating this is to say that the eigenvalues of a

matrix are invariant under similarity transformations.)

Solution: Let

A and

B be similar matrices. Thus there exists a third matrix

such that

B 

ÿ1

S. Substituting this into the characteristic polynomi al of

matrix

B which is j

B ÿ

Ij , we obtain

B ÿ Ijj

ÿ1

S ÿ 

Ijj

ÿ1



A ÿ 

I

Sj:

Using the properties of determinants, we have

ÿ1



A ÿ 

I

Sjj

ÿ1

A ÿ 

Ijj

Sj:

Then it follows that

B ÿ 

Ijj

ÿ1



A ÿ 

I

Sjj

ÿ1

A ÿ 

Ijj

Sjj

A ÿ 

Ij ;

which shows that the characteristic polynomials of

A and

B are the same; their

eigenvalues will also be identical.

Eigenvalues and eigenvectors of hermitian matrices

In quantum mechanics complex variables are unavoidabl e because of the form of

the Schro

dinger equation. And all quantum observables are represented by her-

mitian operators. So physicists are almost always deali ng with adjoint matrices,

hermitian matrices, and unitary matrices. Why are physicists interested in hermi-

tian matrices? Because they have the following properties: (1) the eigenvalues of a

hermitian matrix are real, and (2) its eigenvectors correspondi ng to distinct eigen-

values are orthogonal, so they can be used as basis vectors. We now proceed to

prove these important properties.

(1) the eigenvalues of a hermitian matrix are real.

Let

H be a hermi tian matrix and X a non-trivial eigenvector corresponding to the

eigenvalue , so that

HX  X: 3:62

Taking the hermitian conjugate and note that



H, we have

H  *X

: 3:63

Multiplying (3.62) from the left by X

, and (3.63) from the right by X

, and then

subtracting, we get

 ÿ *X

X  0: 3:64

Now, since X

X cannot be zero, it follows that   *, or that  is real.

128

MATRIX ALGEBRA

(2) The eigenvectors corresponding to distinct eigenvalues are orthogonal.

Let X

and X

be eigenvectors of

H corresponding to the distinct eigenvalues 

and 

, respectively, so that

 

; 3:65

 

: 3:66

Taking the hermitian conjugate of (3.66) and noting that *  , we have

H  

: 3:67

Multiplying (3.65) from the left by X

and (3.67) from the right by X

, then

subtracting, we obtain



ÿ 

X

 X

 0: 3:68

Since 

 

, it follows that X

 0 or that X

and X

are orthogonal.

If X is an eigenvector of

H, any multiple of X, X, is also an eigenvector of

Thus we can normalize the eigenvector X with a properly chosen scalar . This

means that the eigenvectors of

H corresponding to distinct eigenvalues are ortho-

normal. Just as the three orthogonal unit coordinate vectors

;

; and

form

the basis of a three-dimensional vector space, the orthonormal eigenvectors of

may serve as a basis for a function space.

Diagonalization of a matrix

Let

A a

 be a square matrix of order n, which has n linearly independent

eigenvectors X

with the corresponding eigenvalues 

 

. If we denote

the eigenvectors X

by column vectors with elements x

; x

; ...; x

, then the

eigenvalue equation can be written in matrix form:

 a

 

: 3:69

From the above matrix equation we obtain

k1

 

: 3:69b

Now we want to diagonalize

A. To this purpose, we can follow these steps. We

®rst form a matrix

S of order n  n whose columns are the vector X

, that is,

129

DIAGONALIZATION OF A MATRIX

S 

 x

; 

S

 x

: 3:70

Since the vectors X

are linear independent,

S is non-singular and

ÿ1

exists. We

then form a matrix

ÿ1

S; this is a diagonal matrix whose diagonal elements are

the eigenvalues of

To show this, we ®rst de®ne a diagonal matrix

B whose diagonal elements are 

i  1; 2; ...; n:

B 



; 3:71

and we then demonstrate that

ÿ1

S 

B: 3:72a

Eq. (3.72a) can be rewritten by multiplying it from the left by

S as

S 

B: 3:72b

Consider the left ha nd side ®rst. Taking the jith element, we obtain



S



k1



A



S



k1

: 3 :73a

Similarly, the jith element of the right hand side is



B



k1



S



B



k1





 

: 3:73b

Eqs. (3.73a) and (3.73b) clearly show the validity of Eq. (3.72a).

It is important to note that the matrix

S that is able to diagonalize matrix

A is

not unique. This is because we could arrange the eigenvectors X

; X

; ...; X

any order to construct

We summarize the procedure for diagonalizing a diagonalizable n  n matrix

Step 1. Find n linearly independent eigenvectors of

A; X

; X

; ...; X

Step 2. Form the matrix

S having X

; X

; ...; X

as its column vectors.

Step 3. Find the inverse of

ÿ1

Step 4. The matrix

ÿ1

S will then be diagonal with 

;

; ...;

as its succes-

sive diagonal elements, where 

is the eigenvalue corresponding to X

130

MATRIX ALGEBRA

Example 3.16

Find a matrix

S that diagonalizes

A 

3 ÿ20

ÿ230

005

Solution: We have ®rst to ®nd the eigenvalues and the corresponding eigen-

vectors of matrix

A. The characteristic equation of

A is

3 ÿ  ÿ20

ÿ23ÿ  0

005ÿ 

 ÿ 1 ÿ 5

 0;

so that the eigenvalues of

A are   1 and   5.

By de®nition

X 

is an eigenvector of

A corresponding to  if and only if

X is a non-trivial solution

of (

I ÿ

A

X  0, that is, of

 ÿ 32 0

2  ÿ 30

00 ÿ 5



If   5 the above equation becomes

220

000



 2x

 0x

 2x

 0x



Solving this system yields

ÿs; x

 s; x

 t;

where s and t are arbitrary values. Thus the eigenvectors of

A corresponding to

  5 are the non-zero vectors of the form

X 

ÿs



ÿs



 s

ÿ1

 t

131

DIAGONALIZATION OF A MATRIX

Since

ÿ1

and

are linearly independent, they are the eigenvectors corresponding to   5.

For   1, we have

ÿ22 0

2 ÿ20

00ÿ4



ÿ2x

 2x

 0x

ÿ 2x

 0x

ÿ 4x



Solving this system yields

 t; x

 0;

where t is arbitrary. Thus the eigenvectors corresponding to   1 are non-zero

vectors of the form

X 

 t

It is easy to check that the three eigenvectors



ÿ1

;



;



;

are linearly independent. We now form the matrix

S that has

,and

as its

column vectors:

S 

ÿ101

101

010

The matrix

ÿ1

S is diagonal:

ÿ1

S 

ÿ1=21=20

001

1=21=20

3 ÿ20

ÿ230

005

ÿ101

101

010



500

050

001

There is no preferred order for the columns of

S. If had we written

S 

ÿ110

110

001

132

MATRIX ALGEBRA

then we would have obtained (verify)

ÿ1

S 

500

010

001

Example 3.17

Show that the matrix

A 

ÿ32

ÿ21



is not diagonalizable.

Solution: The characteristic equ ation of

A is

  3 ÿ2

2  ÿ 1

  1

 0:

Thus  ÿ1 the only eigenvalue of

A; the eigenvectors corresponding to  ÿ1

are the solutions of

  3 ÿ2

2  ÿ 1



ý!





)

2 ÿ2



ý!





from which we have

ÿ 2x

 0;

ÿ 2x

 0:

The solutions to this system are x

 t; x

 t; hence the eigenvectors are of the

form



 t



A doe s not have two linearly independent eigenvectors, and is therefore not

diagonalizable.

Eigenvectors of commuting matrices

There is a theorem on eigenvectors of commuting matrices that is of great impor-

tance in matrix algebra as well as in quantum mechanics. This theorem states that:

Two commuting matrices possess a common set of eigenvectors.

133

EIGENVECTORS OF COMMUTING MATRICES

We now proceed to prove it. Let

A and

B be two square matrices, each of order

n, which commute with each other, that is,

B ÿ

A 

B0:

First, let  be an eigenvalue of

A with multiplicity 1, corresponding to the eigen-

vector X, so that

AX  X: 3:74

Multiplying both sides from the left by

AX  

BX:

Because

A 

B, we have

A

BX

BX:

Now

B is an n  n matrix and X is an n  1 vector; hence

BX is also an n  1

vector. The above equation shows that

BX is also an eigenvector of

A with the

eigenvalue . Now X is a non-degenerate eigenvector of

A, any other vector which

is an eigenvector of

A with the same eigenvalue as that of X must be multiple of X.

Accordingly

BX  X;

where  is a scalar. Thus we have proved that:

If two matrices commute, every non-degenerate eigenvector of

one is also an eigenvector of the other, and vice versa.

Next, let  be an eigenvalue of

A with multiplicity k.So

A has k linearly inde-

pendent eigenvectors, say X

; X

; ...; X

, each corresponding to :

 X

; 1  i  k:

Multiplying both sides from the left by

B, we obtain

A



;

which shows again that

BX is also an eigenvector of

A with the same eigenvalue .

Cayley±Hamilton theorem

The Cayley±H amilton theorem is useful in evaluating the inverse of a square

matrix. We now introduce it here. As given by Eq. (3.57), the characteristic

equation associated with a square matrix

A of order n may be written as a poly-

nomial

f 

i0



nÿi

 0;

134

MATRIX ALGEBRA

where  are the eigenvalues given by the characteristic determinant (3.56). If we

replace  in f  by the matrix

A so that

f 

A

i0

nÿi

The Cayley±Hamilton theorem says that

f 

A0or

i0

nÿi

 0; 3:75

that is, the matrix

A satis®es its characteristic equatio n.

We now formally multiply Eq. (3.75) by

ÿ1

so that we obtain

ÿ1

f 

Ac

nÿ1

 c

nÿ2

c

nÿ1

I  c

ÿ1

 0:

Solving for

ÿ1

gives

ÿ1

ÿ

nÿ1

i0

nÿ1ÿi

; 3:76

we can use this to ®nd

ÿ1

(Problem 3.28).

Moment of inertia matrix

We shall see that physically diagonalization amounts to a simpli®cation of the

problem by a better choice of variable or coordinate system. As an illustrative

example, we consider the moment of inertia matrix

I of a rotating rigid body (see

Fig. 3.4). A rigid body can be considered to be a many-particle system, with the

135

MOMENT OF INERTIA MATRIX

Figure 3.4. A rotating rigid body.