Chapter 6 Eigenvalues. Example In a certain town, 30 percent of the married women get divorced each year and 20 percent of the single women get married.

Chapter 6

Eigenvalues

Example In a certain town, 30 percent of the married

women get divorced each year and 20 percent of the single

women get married each year. There are 8000 married

women and 2000 single women, and the total population

remains constant.

Let us investigate the long-range prospects if these

percentage of marriages and divorces continue indefinitely

into the future.

1 Eigenvalues and Eigenvectors

Definition

Let A be an n×n matrix. A scalar is said to be an eigenvalu

e or a characteristic value of A if there exists a nonzero vecto

r x such that . The vector x is said to be an eigenve

ctor or a characteristic vector belonging to .

xx A

Example Let

11

24A and

1

2x

The subspace N(A- I) is called the eigenspace corresponding

to the eigenvalue .

The polynomial is called the characteristic

polynomial, and equation is called the

characteristic equation for the matrix A.

)det()( IAp

0)det( IA

Let A be an n×n matrix and be a scalar. The following

statements are equivalent:

(a) is an eigenvalue of A.

(b) has a nontrivial solution.

(c)

(d) is singular.

(e)

0x)( IA

0)( IAN

IA

0)det( IA

Example Let

Example Find the eigenvalues and the corresponding

eigenvectors of the matrix

23

23A

231

121

132

A

Find the eigenvalues and the corresponding eigenspaces.

The Product and Sum of the Eigenvalues

nnnn

n

n

aaa

aaa

aaa

IAp

21

22221

11211

)det()(

Expanding along the first column, we get

)det()1()det()()det(2

11

11111

n

ii

ii MaMaIA

)det()0(21 Apn

n

iii

n

ii a

11

The sum of the diagonal elements of A is called the trace of A

and is denoted by tr(A).

Example If

11

185A

Some Properties of the Eigenvalues:

1. Let A be a nonsingular matrix and let be an eigenvalue

of A, then is an eigenvalue of A-1.

1

2. Let be an eigenvalue of A and let x be an eigenvector

belonging to , then is an eigenvalue of and x is

an eigenvector of belonging to for m=1, 2, ….

m mA

mA m

3. Let , and let be

an eigenvalue of A, then is an eigenvalue of .

1

0 1 1( ) m m

m mf x a x a x a x a

)(f )(Af

Example If the eigenvalues of matrix A are: 2, 1, -1, then

find the eigenvalues for the following matrices:

(a)

(b)

IAA 2

1AA

Similar Matrices

Theorem 6.1.1 Let A and B be n×n matrices. If B is similar to

A, then the two matrices both have the same characteristic

polynomial and consequently both have the same eigenvalues.

3 Diagonalization

Theorem 6.3.1 If are distinct eigenvalues of an

n×n matrix A with corresponding eigenvectors x1, x2, …,xk,

then x1, …, xk are linearly independent.

k ，，， 21

Definition

An n×n matrix A is said to be diagonalizable if there exists a

nonsingular matrix X and a diagonal matrix D such that

X-1AX=D

We say that X diagonalizes A.

Theorem 6.3.2 An n×n matrix A is diagonalizable if and

only if A has n linearly independent eigenvectors.

Remarks

1. If A is diagonalizable, then the column vectors of the

diagonalizing matrix X are eigenvectors of A, and the diagonal

elements of D are the corresponding eigenvalues of A.

2. The diagonalizing matrix X is not unique. Reordering the

columns of a given diagonalizing matrix X or multiplying them

by nonzero scalars will produce a new diagonalizing matrix.

3. If A is an n×n matrix and A has n distinct eigenvalues, then

A is diagonalizable. If the eigenvalues are not distinct, then A

may or may not be diagonalizable depending on whether A has

n linearly independent eigenvectors.

4. If A is diagonalizable, then A can be factored into a product

XDX-1.

12

1

1

XX

XXDA

kn

k

k

kk

Example Let

52

32A

Example Let

112

202

213

A

Determine whether the matrix is diagonalizable or not.

Determine whether the matrix is diagonalizable or not.

Definition

If an n×n matrix A has fewer than n linearly independent

eigenvectors, we say that A is defective.

Theorem 6.3.3

If A is an n×n matrix and are s distinct eigenvalues

for A, let be the basis of , where

, then

are linearly independent.

s ，，， 21

)()2()1( ,,, irniii

)( AIN i

)( AIrr ii ),,1( si ,,,,,,, )(2

)2(2

)1(2

)(1

)2(1

)1(1

21 rnrn ，)()2()1( ,,,, srn

sss

Example Let

Determine whether the two matrices are diagonalizable or not.

201

040

002

A and

263

041

002

B

Some Results for Real Symmetric Matrix:

2. If are distinct eigenvalues of an

n×n real symmetric matrix A with corresponding eigenvectors

x1, x2, …,xk, then x1, …, xk are orthogonal.

k ，，， 21

3. If A is a real symmetric matrix, then there is an orthogonal

matrix U that diagonalizes A, that is, U-1AU=UTAU=D, where

D is diagonal.

1. The eigenvalues of a real symmetric matrix are all real.

542452222

AExample Let

Find an orthogonal matrix U that diagonalizes A.

Example Let

Find an orthogonal matrix U that diagonalizes A.

320222

021A

6 Quadratic Forms

Definition

A quadratic equation in two variables x and y is an equation o

f the form

(1)

Equation (1) may be rewritten in the form

(2)

Let

The term

is called the quadratic form associated with (1).

02 22 feydxcybxyax

0

f

y

xed

y

x

cb

bayx

y

xx and

cb

baA

22 2xx cybxyaxAT

Conic Sections

The graph of an equation of the form (1) is called a conic section.

A conic section is said to be in standard position if its equation

can be put into one of these four standard forms:

222)1( ryx

1)2(2

2

2

2

yx

1)3(2

2

2

2

yx

or 12

2

2

2

xy

yx 2)4( or xy 2

(circle)

(ellipse)

(hyperbola)

(parabola)

Example Consider the conic section

08323 22 yxyx

This equation can be written in the form

831

13

y

xyx

The matrix

31

13has eigenvalues 2 and 4

with corresponding unit eigenvectors

2

12

1

and

2

12

1

Let

45cos45sin

45sin45cos

2

1

2

12

1

2

1

Q

and set

'

'

2

1

2

12

1

2

1

y

x

y

x

Thus

40

02AQQT

and the equation of the conic becomes

12

)(

4

)( 2'2'

yx

Quadratic Surfaces

2 2 2

2 2 2(1) 1

x y z

a b c

(ellipsoid) (cone)

(hyperboloid of one sheet)

2 2 2

2 2 2(2) 0

x y z

a b c

2 2 2

2 2 2(3) 1

x y z

a b c

2 2 2

2 2 2(4) 1

x y z

a b c

(hyperboloid of two sheets)

2 2

2 2(5) 2

x yz

a b

(elliptic paraboloid)

2 2

2 2(6) 2

x yz

a b

(hyperbolic paraboloid)

A quadratic form including n variables is： nnn xxaxxaxxaxaxxxf 1131132112

211121 222

nn xxaxxaxa 2232232222 22

2nnnxa

AXX

x

x

x

aaa

aaa

aaa

xxx T

nnnnn

n

n

n

2

1

21

22212

11211

21 )(

Theorem 6.3.5

For any quadratic form XTAX, we can find an orthogonal

transformation X=CY such that YTBY is in standard form.

Example Let

4342324131214321 222222 xxxxxxxxxxxxxxxxf

Find an orthogonal transformation X=CY such that YTBY is in

Standard form.

Example For the conic section

01222 323121 xxxxxx

Find an orthogonal transformation X=CY such that YTBY is

in standard form.

Definition

A quadratic form f(x)=xTAx is said to be definite if it takes o

n only one sign as x varies over all nonzero vectors in Rn.

The form is positive definite if xTAx>0 for all nonzero x in R

n and negative definite if xTAx<0 for all nonzero x in Rn.

A quadratic form is said to be indefinite if it takes on value

s that differ in sign.

If f(x)=xTAx ≥0 and assumes the value 0 for some x≠0, the

n f(x) is said to be positive semidefinite.

If f(x) ≤0 and assumes the value 0 for some x≠0, then f(x)

is said to be negative semidefinite.

Definition

A real symmetric matrix A is said to be

I. Positive definite if xTAx>0 for all nonzero x in Rn.

Ⅱ. Negative definite if xTAx<0 for all nonzero x in Rn.

III. Positive semidefinite if xTAx≥0 for all nonzero x in Rn.

IV. Negative semidefinite if xTAx≤0 for all nonzero x in Rn.

V. Indefinite if xTAx takes on values that differ in sign.

Theorem 6.6.2

Let A be a real symmetric n×n matrix. Then A is positive

Definite if and only if all its eigenvalues are positive.

Theorem 6.6.3

Let A be a real symmetric n×n matrix. Then A is positive

definite if the leading principal submatrices A1, A2, …, An of

A are all positive definite.

Example Determine whether the quadratic form2 2 2

1 2 3 1 2 3 1 2 2 3( , , ) 5 6 4 4 4f x x x x x x x x x x

is positive definite.