06. Eigenvalue Problem (1)

8/18/2019 06. Eigenvalue Problem (1)

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The Eigenvalue Problem


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Homogeneous Systems

• A homogeneous linear system

A x = 0• If det( A)!0, the unique solution is the trivial

solution x = 0.• If det( A)=0, there exist nontrivial solutions.


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Example 1

Find the nontrivial solutions to the

homogeneous system

045

02

02

321

321

321

=++

=++

=!

+

x x x

x x x

x x x


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Solution 1

The system reduces to:

Letting x3=t , the solution can be expressed as

x 1+ x

2 = 0

!

x 2 + x

3 = 0

x 1+ 2 x

2 ! x

3 = 0

2 x 1

+ x 2

+ x 3

= 0

5 x 1+ 4 x

2 + x

3 = 0

t x

t x

t x

=

=

!

=

3

2

1

!

"

!#

$

!

%

!&

'(=

1

1

1

t X


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Applications

•

Quantum mechanics, geology, mathematics,image processing, etc.

•

MechanicsPrincipal stresses and the orientation of theprincipal planes

•

Structural Dynamics

Natural frequencies and mode shapes ofstructures


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The Eigenvalue Problem

Let A be a square matrix of dimension n x n, and

let x be a vector of dimension n. The problem is

to find scalars ! for which there exists a

nonzero vector x such that

Eigenvalue

(characteristic value)

Eigenvector

(characteristic vector)

x x A ! =


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Solution Methods

• Analytical Solution

– Solution of the characteristic equation

• Numerical Solution

– Power method

– Inverse power method

– Jacobi’s method, QR method


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Analytical Method

•

The eigenvalues of A are the solutions of the characteristic equation

• When the determinant is expanded, it becomesa polynomial of degree n, called the

characteristic polynomial .

( ) 0 x I A =! "

( ) 0det

21

22221

11211

=

!

!

!

=!

"

"

"

"

nnnn

n

n

aaa

aaa

aaa

!

"#""

!

!

I A


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Eigenvalues

If A is an n x n real matrix, then its n eigenvalues! 1, ! 2, !, ! n. are the real and complex roots of

the characteristic polynomial

Properties:

( ) ( ) I A p ! ! "= det

( ) !!==

==

n

k

k

n

i

iiatrace

11

" A ( ) !=

=

n

k

k

1

det " A

( ) ( ) ( )( ) ( )nn p ! ! ! ! ! ! !

""""

=

!211


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Eigenvector

If ! is an eigenvalue of A and the nonzero vector v

has the property that

Then v is called the eigenvector of A

corresponding to the eigenvalue ! . v is obtainedby solving the homogeneous system ( A - ! I ) v =

0.

vv A ! =


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Analytical Method

Hand computations used to solve the eigenvalue

problem when the dimension n is small.

1.

Find the coefficients of the characteristicpolynomial p(! )= det( A - ! I )

2.

Find its roots (eigenvalues).

3. Find the nonzero solutions (eigenvectors) of

the homogeneous linear system ( A - ! I ) v = 0.


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Example 2

Find the eigenpairs ! j , v j for the matrix

!

!!

"

#

$

$$

%

&

'

'''

=

310

121

013

A


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Solution 2

Eigenvalues. The characteristic equation det( A - ! I )=0 is

Therefore, the three eigenvalues are

! 1 = 1, ! 2 = 3, and ! 3 = 4.

012198

310

121

013

23 =+!

+!

=

!!

!!!

!!

" " "

"

"

"

( )( )( ) 0431 =!!!! " " "


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Solution 2

Eigenvectors.

For ! 1 = 1,

For ! 2 = 3,

!"

!#

$

!%

!&

'

=

!"

!#

$

!%

!&

'

(((

)

*

+++

,

-

...

.

0

0

0

210

111

012

3

2

1

x

x

x

!"

!#

$

!%

!&

'

=

1

2

1

1 av

02

02

32

21

=+!

=!

x x

x x

a x

a x

a x

=

=

=

3

1

2 2

!"

!#

$

!%

!&

'

=

!"

!#

$

!%

!&

'

(((

)

*

+++

,

-

.

...

.

0

0

0

010

111

010

3

2

1

x

x

x

!"

!#

$

!%

!&

'

(

=

1

0

1

2 bv

0

0

2

31

=

=+

x

x x

b x

b x

x

!=

=

=

3

1

2 0


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Solution 2

Eigenvectors.

For ! 3 = 4,

!"

!#

$

!%

!&

'

=

!"

!#

$

!%

!&

'

(((

)

*

+++

,

-

..

...

..

0

0

0

110

121

011

3

2

1

x

x

x

!"

!#

$

!%

!&

'

(=

1

1

1

2 cv

0

0

32

21

=+

=+

x x

x x

c x

c x

c x

=

!=

=

1

2

3


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Numerical Methods

1. Power Method

2.

Shifted-inverse Power Method


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Power Method

Assume that the n x n matrix A has n distinct

eigenvalues ! 1, ! 2, !, ! n and that they are

ordered in decreasing magnitude; that is,

|! 1| > |! 2| " |! 3| " ! " |! n|

eigenvector v1 corresponding to ! 1 is called adominant eigenvector .


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Power Method

If x 0 is chosen appropriately, then the sequences{ x k = [ x1

(k ) x2(k ) ! xn

(k ) ]’} and {ck } generated

recursively by

yk = A x kand

x k +1 = (1/ck +1) yk ,

where

ck +1 = max{| xi

(k )|

will converge to the dominant eigenvector v1 andeigenvalue ! 1, respectively.


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Example 3

Use the power method to find the dominant

eigenpair for the matrix

!!!

"

#

$$$

%

&

''

'''

=

10264

7172

5110

A


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Solution 3

Start with x 0 = {1 1 1}T and use to generate the

sequence of vectors { x k } and constants {c

k }.

The first iteration produces

113

2

2

1

1

12

12

8

6

1

1

1

10264

7172

5110

x c=

!"

!#

$

!%

!&

'

=

!"

!#

$

!%

!&

'

=

!"

!#

$

!%

!&

'

(

((

)

*

+

++

,

-

..

..

.


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Solution 3

The second iteration produces

Iteration generates the sequence

The sequence converges to v = {2/5 3/5 1}T, ! =4.

228

5

16

7

316

3

10

3

7

3

2

2

1

13

16

110264

7172

5110

x c=

!"

!#

$

!%

!&

'

=

!"

!#

$

!%

!&

'

=

!"

!#

$

!%

!&

'

(((

)

*

+++

,

-

..

..

.

!,

119

78,

19

38,

12

9,

13

16,

1

1278

47

52

21

38

23

76

31

18

11

12

5

8

5

16

7

3

2

2

1

!"

!#

$

!%

!&

'

!"

!#

$

!%

!&

'

!"

!#

$

!%

!&

'

!"

!#

$

!%

!&

'

!"

!#

$

!%

!&

'


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C-Implementation

Vector PowerMethod(

Matrix A, /* nxn matrix A */

Vector x0, /* nx1 starting vector */

double e, /* tolerance */

int N, /* maximum iterations */

double * ev /* eigenvalue */

);

Returns: The dominant eigenvector


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Shifting Eigenvalues. Suppose that ! , v is an

eigenpair of A. If " is any constant, then ! - #, v

is an eigenpair of the matrix A - " I.

Inverse Eigenvalues. Suppose that ! , v is an

eigenpair of A. If ! !0, then 1 / ! , v is an

eigenpair of the matrix A-1.

Suppose that ! , v is an eigenpair of A. If " !! , then

1 / (! - #), v is an eigenpair of the matrix ( A-# I )-1.


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A constant " can be chosen so that µ 1 = 1/(! i - #)is the dominant eigenvalue of ( A - # I )-1.

Furthermore, if x 0 is chosen appropriately, then

the sequences { x k = [ x

1

(k ) x2

(k ) ! xn

(k ) ]’} and {ck }

generated recursively by

yk = ( A - " I )-1 x

k

and

x k +1 = (1/ck +1) yk ,

where

ck +1

= max{| xi(k )|}


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will converge to the dominant eigenpair of the

matrix ( A - # I )-1. Finally, the corresponding

eigenvalue for the matrix A is given by the

calculation

! µ

" +=

1

1

j


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Example 4

Employ the shifted-inverse power method to find theeigenpairs of the matrix

Note: The eigenvalues of A are ! 1 = 4,! 2 = 2 and ! 3 = 1,

and select and appropriate " and starting vector for

each case

!!!

"

#

$$$

%

&

''

''

'

=

10264

7172

5110

A


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Solution 4

Case (1): for the eigenvalue ! 1 = 4, we select" = 4.2 and the starting vector x

0 = {1 1 1}T. First

form the matrix ( A – 4.2 I )-1, iteration produces

0

1

1

1

2.14264

78.122

5112.4

0 x y ==

!!

!!!!

"#$

%&'

()

*+,

-

11

1

6078431373.0

4117647059.0

18181818.23

18181818.23

09090909.14

545454545.9

0 x y c=!=

!

!

!

=

"#$

%&'

"#$

%&'


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Solution 4


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Solution 4

the sequence converges to µ1 = -5 which is

the dominant eigenvalue of ( A – 4.2 I )-1 ,

and xk converges to v1 = [2/5, 3/5,1]

T.

the eigenvalue ! 1 of A is given by

! 1 = (1/ µ

1 ) + " = 1/(-5) + 4.2 = -0.2+4.2 = 4


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Solution 4

Case (2): for the eigenvalue ! 2 = 2, we select " =

2.1 and the starting vector x 0 = {1 1 1}T then form

the matrix (A – 2.1I)-1.

the sequence converges to µ1 = -10 and the

eigenpair of the matrix is

! 2 = (1/ µ2 ) + " = (1/-10) + 2.1 = -0.1+2.1 =2

v1 = [1/4, 1/2,1]T.


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Solution 4


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Solution 4

Case (3): for the eigenvalue ! 3 = 1, we select " =

0.875 and the starting vector x 0 = {0 1 1}T then

form the matrix (A – 0.875I)-1.

the sequence converges to µ1 = 8 and the

eigenpair of the matrix is

! 2 = (1/ µ1 ) + " = (1/8) + 0.875 = 0.125+0.875 = 1

v3 = [1/2, 1/2,1]T.


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Solution 4

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06. Eigenvalue Problem (1)