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6.2 De¯nitions and examples

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DEFINITION 6.2.1 (Eigenvalue, eigenvector)

Let A be a complex square matrix. Then if ¸ is a complex number and

X a non{zero complex column vector satisfying AX = ¸X, we call X an

eigenvector of A, while ¸ is called an eigenvalue of A. We also say that X

is an eigenvector corresponding to the eigenvalue ¸.

So in the above example P1 and P2 are eigenvectors corresponding to ¸1

and ¸2, respectively. We shall give an algorithm which starts from the

eigenvalues of A =

a h

h b

and constructs a rotation matrix P such that

PtAP is diagonal.

As noted above, if ¸ is an eigenvalue of an n £ n matrix A, with

corresponding eigenvector X, then (A ¡ ¸In)X = 0, with X 6= 0, so

det (A ¡ ¸In) = 0 and there are at most n distinct eigenvalues of A.

Conversely if det (A ¡ ¸In) = 0, then (A ¡ ¸In)X = 0 has a non{trivial

solution X and so ¸ is an eigenvalue of A with X a corresponding eigenvector.

DEFINITION 6.2.2 (Characteristic equation, polynomial)

The equation det (A ¡ ¸In) = 0 is called the characteristic equation of A,

while the polynomial det (A¡¸In) is called the characteristic polynomial of

A. The characteristic polynomial of A is often denoted by chA(¸).

Hence the eigenvalues of A are the roots of the characteristic polynomial

of A.

For a 2£2 matrix A =

a b

c d

, it is easily veri¯ed that the character-

istic polynomial is ¸2 ¡(traceA)¸+detA, where traceA = a+d is the sum

of the diagonal elements of A.

EXAMPLE 6.2.1 Find the eigenvalues of A =

2 1

1 2

and ¯nd all eigen-

vectors.

Solution. The characteristic equation of A is ¸2 ¡ 4¸ + 3 = 0, or

(¸ ¡ 1)(¸ ¡ 3) = 0:

Hence ¸ = 1 or 3. The eigenvector equation (A ¡ ¸In)X = 0 reduces to

2 ¡ ¸ 1

1 2 ¡ ¸

¸ ·

;

6.2. DEFINITIONS AND EXAMPLES 119

(2 ¡ ¸)x + y = 0

x + (2 ¡ ¸)y = 0:

Taking ¸ = 1 gives

x + y = 0

x + y = 0;

which has solution x = ¡y; y arbitrary. Consequently the eigenvectors

corresponding to ¸ = 1 are the vectors

¡y

, with y 6= 0.

Taking ¸ = 3 gives

¡x + y = 0

x ¡ y = 0;

which has solution x = y; y arbitrary. Consequently the eigenvectors corre-

sponding to ¸ = 3 are the vectors

, with y 6= 0.

Our next result has wide applicability:

THEOREM 6.2.1 Let A be a 2£2 matrix having distinct eigenvalues ¸1

and ¸2 and corresponding eigenvectors X1 and X2. Let P be the matrix

whose columns are X1 and X2, respectively. Then P is non{singular and

P¡1AP =

¸1 0

0 ¸2

Proof. Suppose AX1 = ¸1X1 and AX2 = ¸2X2. We show that the system

of homogeneous equations

xX1 + yX2 = 0

has only the trivial solution. Then by theorem 2.5.10 the matrix P =

[X1jX2] is non{singular. So assume

xX1 + yX2 = 0: (6.3)

Then A(xX1 + yX2) = A0 = 0, so x(AX1) + y(AX2) = 0. Hence

x¸1X1 + y¸2X2 = 0: (6.4)

120 CHAPTER 6. EIGENVALUES AND EIGENVECTORS

Multiplying equation 6.3 by ¸1 and subtracting from equation 6.4 gives

(¸2 ¡ ¸1)yX2 = 0:

Hence y = 0, as (¸2¡¸1) 6= 0 and X2 6= 0. Then from equation 6.3, xX1 = 0

and hence x = 0.

Then the equations AX1 = ¸1X1 and AX2 = ¸2X2 give

AP = A[X1jX2] = [AX1jAX2] = [¸1X1j¸2X2]

= [X1jX2]

¸1 0

0 ¸2

= P

¸1 0

0 ¸2

;

P¡1AP =

¸1 0

0 ¸2

EXAMPLE 6.2.2 Let A =

2 1

1 2

be the matrix of example 6.2.1. Then

X1 =

¡1

and X2 =

are eigenvectors corresponding to eigenvalues

1 and 3, respectively. Hence if P =

¡1 1

1 1

, we have

P¡1AP =

1 0

0 3

There are two immediate applications of theorem 6.2.1. The ¯rst is to the

calculation of An: If P¡1AP =diag (¸1; ¸2), then A = Pdiag (¸1; ¸2)P¡1

and

An =

¸1 0

0 ¸2

P¡1

¶n

= P

¸1 0

0 ¸2

¸n

P¡1 = P

¸n

1 0

0 ¸n

P¡1:

The second application is to solving a system of linear di®erential equations

= ax + by

= cx + dy;

where A =

a b

c d

is a matrix of real or complex numbers and x and y

are functions of t. The system can be written in matrix form as _X = AX,

where

X =

and _X =

· dx

6.2. DEFINITIONS AND EXAMPLES 121

We make the substitution X = PY , where Y =

. Then x1 and y1

are also functions of t and

= P _Y = AX = A(PY ); so _Y = (P¡1AP)Y =

¸1 0

0 ¸2

Hence x_1 = ¸1x1 and y_1 = ¸2y1.

These di®erential equations are well{known to have the solutions x1 =

x1(0)e¸1t and x2 = x2(0)e¸2t, where x1(0) is the value of x1 when t = 0.

[If dx

dt = kx, where k is a constant, then

e¡ktx

= ¡ke¡ktx + e¡kt dx

= ¡ke¡ktx + e¡ktkx = 0:

Hence e¡ktx is constant, so e¡ktx = e¡k0x(0) = x(0). Hence x = x(0)ekt.]

However

x1(0)

y1(0)

= P¡1

x(0)

y(0)

, so this determines x1(0) and y1(0) in

terms of x(0) and y(0). Hence ultimately x and y are determined as explicit

functions of t, using the equation X = PY .

EXAMPLE 6.2.3 Let A =

2 ¡3

4 ¡5

. Use the eigenvalue method to

derive an explicit formula for An and also solve the system of di®erential

equations

= 2x ¡ 3y

= 4x ¡ 5y;

given x = 7 and y = 13 when t = 0.

Solution. The characteristic polynomial of A is ¸2+3¸+2 which has distinct

roots ¸1 = ¡1 and ¸2 = ¡2. We ¯nd corresponding eigenvectors X1 =

and X2 =

. Hence if P =

1 3

1 4

, we have P¡1AP = diag (¡1; ¡2).

Hence

An =

Pdiag (¡1; ¡2)P¡1¢n

= Pdiag ((¡1)n; (¡2)n)P¡1

1 3

1 4

¸ ·

(¡1)n 0

0 (¡2)n

¸ ·

4 ¡3

¡1 1

122 CHAPTER 6. EIGENVALUES AND EIGENVECTORS

= (¡1)n

1 3

1 4

¸ ·

1 0

0 2n

¸ ·

4 ¡3

¡1 1

= (¡1)n

1 3 £ 2n

1 4 £ 2n

¸ ·

4 ¡3

¡1 1

= (¡1)n

4 ¡ 3 £ 2n ¡3 + 3 £ 2n

4 ¡ 4 £ 2n ¡3 + 4 £ 2n

To solve the di®erential equation system, make the substitution X =

PY . Then x = x1 + 3y1; y = x1 + 4y1. The system then becomes

x_ 1 = ¡x1

y_1 = ¡2y1;

so x1 = x1(0)e¡t; y1 = y1(0)e¡2t. Now

x1(0)

y1(0)

= P¡1

x(0)

y(0)

4 ¡3

¡1 1

¸ ·

¡11

;

so x1 = ¡11e¡t and y1 = 6e¡2t. Hence x = ¡11e¡t + 3(6e¡2t) = ¡11e¡t +

18e¡2t; y = ¡11e¡t + 4(6e¡2t) = ¡11e¡t + 24e¡2t:

For a more complicated example we solve a system of inhomogeneous

recurrence relations.

EXAMPLE 6.2.4 Solve the system of recurrence relations

xn+1 = 2xn ¡ yn ¡ 1

yn+1 = ¡xn + 2yn + 2;

given that x0 = 0 and y0 = ¡1.

Solution. The system can be written in matrix form as

Xn+1 = AXn + B;

where

A =

2 ¡1

¡1 2

and B =

¡1

It is then an easy induction to prove that

Xn = AnX0 + (An¡1 + ¢ ¢ ¢ + A + I2)B: (6.5)

6.2. DEFINITIONS AND EXAMPLES 123

Also it is easy to verify by the eigenvalue method that

An =

1 + 3n 1 ¡ 3n

1 ¡ 3n 1 + 3n

U +

where U =

1 1

and V =

1 ¡1

¡1 1

. Hence

An¡1 + ¢ ¢ ¢ + A + I2 =

U +

(3n¡1 + ¢ ¢ ¢ + 3 + 1)

U +

(3n¡1 ¡ 1)

Then equation 6.5 gives

Xn =

U +

¶·

¡1

U +

(3n¡1 ¡ 1)

¶·

¡1

;

which simpli¯es to

(2n + 1 ¡ 3n)=4

(2n ¡ 5 + 3n)=4

Hence xn = (2n ¡ 1 + 3n)=4 and yn = (2n ¡ 5 + 3n)=4.

REMARK 6.2.1 If (A ¡ I2)¡1 existed (that is, if det (A ¡ I2) 6= 0, or

equivalently, if 1 is not an eigenvalue of A), then we could have used the

formula

An¡1 + ¢ ¢ ¢ + A + I2 = (An ¡ I2)(A ¡ I2)¡1: (6.6)

However the eigenvalues of A are 1 and 3 in the above problem, so formula 6.6

cannot be used there.

Our discussion of eigenvalues and eigenvectors has been limited to 2 £ 2

matrices. The discussion is a more complicated for matrices of size greater

than two and is best left to a second course in linear algebra. Nevertheless

the following result is a useful generalization of theorem 6.2.1. The reader

is referred to [28, page 350] for a proof.

THEOREM 6.2.2 Let A be an n £ n matrix having distinct eigenvalues

¸1; : : : ; ¸n and corresponding eigenvectors X1; : : : ;Xn. Let P be the matrix

whose columns are respectively X1; : : : ;Xn. Then P is non{singular and

P¡1AP =

6664

¸1 0 ¢ ¢ ¢ 0

0 ¸2 ¢ ¢ ¢ 0

...

0 0 ¢ ¢ ¢ ¸n

7775

124 CHAPTER 6. EIGENVALUES AND EIGENVECTORS

Another useful result which covers the case where there are multiple eigen-

values is the following (The reader is referred to [28, pages 351{352] for a

proof):

THEOREM 6.2.3 Suppose the characteristic polynomial of A has the fac-

torization

det (¸In ¡ A) = (¸ ¡ c1)n1 ¢ ¢ ¢ (¸ ¡ ct)nt ;

where c1; : : : ; ct are the distinct eigenvalues of A. Suppose that for i =

1; : : : ; t, we have nullity (ciIn¡A) = ni. For each i, choose a basis Xi1; : : : ;Xini

for the eigenspace N(ciIn ¡ A). Then the matrix

P = [X11j ¢ ¢ ¢ jX1n1 j ¢ ¢ ¢ jXt1j ¢ ¢ ¢ jXtnt ]

is non{singular and P¡1AP is the following diagonal matrix

P¡1AP =

6664

c1In1 0 ¢ ¢ ¢ 0

0 c2In2 ¢ ¢ ¢ 0

...

0 0 ¢ ¢ ¢ ctInt

7775

(The notation means that on the diagonal there are n1 elements c1, followed

by n2 elements c2,. . . , nt elements ct.)

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