applied numerical methods lec13

Eigenvalue – Eigenvector Problem

EIGENVALUES & EIGENVECTORS

The eigenvalue problem is a problem of considerable

theoretical interest and wide-ranging application.

For example, this problem is crucial in solving systems of

differential equations, analyzing population growth models,

and calculating powers of matrices (in order to define the

exponential matrix (A100)).

Other areas such as physics, sociology, biology, economics

and statistics have focused considerable attention on

“eigenvalues” and “eigenvectors”-their applications and their

computations

Eigenvalue Problems(Mathematical Background)

0xaxaxa

0xaxaxa

0xaxa xλa

nnn2n21n1

n1n222121

n1n212111

0XA

The roots of polynomial D(λ) are the eigenvalues of the eigen system

A solution {X} to [A]{X} = λ{X} is an eigen vector

(homogeneous system)

0 XIA

IA tDeterminanD

(eigen system)

Example 1

CW

Example 2

15

The Power Method(An iterative approach for determining the largest eigenvalue)

Example (3):

Iteration 1: initialization [x1, x2, x3]T = [1 1 1]T

Iteration 2: A [1 0 1]T

Iteration 3: A [1 -1 1]T

Iteration 4: A [-0.75 1 -0.75]T

Iteration 4: A [-0.714 1 -0.714]T

(Exact solution = 6.070)

Class Work: Use the power method to find the dominant eigenvalue and

eigenvector for the matrix

Example 4

21

Power Method for Lowest Eigenvalue(An iterative approach for determining the lowest eigenvalue)

422028101410

281056202810

1410281042201

...

...

...

A

7510

1

7510

1241

8840

1241

8840

1

1

1

422028101410

281056202810

141028104220

.

.

.

.

.

.

...

...

...

Example (5):

Iteration 1:

Iteration 3:

Exact solution is 0.955 which is the reciprocal of the smallest eigenvalue,

1.0472 of [A].

Iteration 2:

7150

1

7150

9840

7040

9840

7040

7510

1

7510

422028101410

281056202810

141028104220

.

.

.

.

.

.

.

.

...

...

...

7090

1

7090

9640

6840

9640

6840

7150

1

7150

422028101410

281056202810

141028104220

.

.

.

.

.

.

.

.

...

...

...

Idea: The largest eigenvalue of [A]-1 is the reciprocal of the lowest

eigenvalue of [A]

5563

422077810

778155632810

077815563

7781 .

..

...

..

.A

Faddeev-Leverrier Method for Eigenvalues

Use Faddeev’s Method to find the eigenvalues of the following matrix

Eigenvalue Problems(Physical Background 1)

1212

1

2

1 xxkkxdt

xdm

tsinAx ii

Mass-spring system

Analytical solution

(vibration theory)

2122

22

2 kxxxkdt

xdm

k kAi = the amplitude of the vibration of mass i

and ω = the frequency of the vibration, which is

equal to ω = 2π/Tp, where Tp is the period.

1

2

3

tsin-Ax 2ii 4

To Find A1, A2, and :

Substitute equations 3 and 4 into 1 and 2

0A A

0A A

21

21

2

22

1

2

1

2

2

m

k

m

k

m

k

m

k

The eigenvalues to this system are the

correct frequency , and the eigenvectors

are the correct A1 and A2.

take 2 as λ

0A A

0A A

21

21

2

22

1

2

1

2

2

m

k

m

k

m

k

m

k

The eigenvalues to this system are the

correct frequency , and the

eigenvectors are the correct A1 and A2.

take 2 as λ

0xaxaxa

0xaxaxa

0xaxa xλa

nnn2n21n1

n1n222121

n1n212111

0XA (homogeneous system)

0 XIA (eigen system)

30

Eigenvalue Problems(An instance of mass-spring problem)

0A A

0A A

21

21

2

22

1

2

1

2

2

m

k

m

k

m

k

m

k

mode second s

mode first s-

-

1

1

2362

8733

.

.

k=200 N/m

First mode: A1 = -A2

Second mode: A1 = A2

0A A

0A A

21

21

2

2

105

510

m1=m2=40 kg

A unique set of values cannot be obtained

for the unknowns. However, their ratios can

be specified by substituting the eigenvalues

back into the equations. For example, for

the first mode (ω2 = 15 s−2), Al=−A2. For the

second mode (ω2 = 5 s−2), A1 = A2.

Note:

Eigenvalue Problems(Physical Background 2)

Eigenvalue Problems(Physical Background 3)

Special Problem 5

Home Work

1.

2.

3.