Applied numerical methods lec13
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The document discusses the eigenvalue-eigenvector problem, which has applications in solving differential equations, modeling population growth, and calculating matrix powers. It provides mathematical background on homogeneous systems of equations where the eigenvalues are the roots of the characteristic polynomial. Iterative methods like the power method are presented for finding the dominant or lowest eigenvalue of a matrix. Physical examples of mass-spring systems are given where the eigenvalues correspond to vibration frequencies and the eigenvectors to mode shapes.
![Eigenvalue Problems
(Mathematical Background)
0xaxaxa
0xaxaxa
0xaxaxλa
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)](https://image.slidesharecdn.com/appliednumericalmethodslec13-150507042423-lva1-app6892/85/Applied-numerical-methods-lec13-4-320.jpg)
![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)](https://image.slidesharecdn.com/appliednumericalmethodslec13-150507042423-lva1-app6892/85/Applied-numerical-methods-lec13-15-320.jpg)
![21
Power Method for Lowest Eigenvalue
(An iterative approach for determining the lowest eigenvalue)
422028101410
281056202810
141028104220
1
...
...
...
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](https://image.slidesharecdn.com/appliednumericalmethodslec13-150507042423-lva1-app6892/85/Applied-numerical-methods-lec13-21-320.jpg)
Applied numerical methods lec13
- 1. Eigenvalue – Eigenvector Problem
- 2. 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
- 4. Eigenvalue Problems (Mathematical Background) 0xaxaxa 0xaxaxa 0xaxaxλa 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)
- 6. Example 1
- 8. CW
- 9. Example 2
- 15. 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)
- 16. Class Work: Use the power method to find the dominant eigenvalue and eigenvector for the matrix
- 19. Example 4
- 21. 21 Power Method for Lowest Eigenvalue (An iterative approach for determining the lowest eigenvalue) 422028101410 281056202810 141028104220 1 ... ... ... 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
- 22. Faddeev-Leverrier Method for Eigenvalues
- 27. Use Faddeev’s Method to find the eigenvalues of the following matrix
- 28. Eigenvalue Problems (Physical Background 1) 1212 1 2 1 xxkkx dt xd m tsinAx ii Mass-spring system Analytical solution (vibration theory) 2122 2 2 2 kxxxk dt xd m k k Ai = 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 2 ii 4 To Find A1, A2, and : Substitute equations 3 and 4 into 1 and 2 0AA 0AA 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
- 29. 0AA 0AA 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 0xaxaxλa nnn2n21n1 n1n222121 n1n212111 0XA (homogeneous system) 0 XIA (eigen system)
- 30. 30 Eigenvalue Problems (An instance of mass-spring problem) 0AA 0AA 21 21 2 22 1 2 1 2 2 m k m k m k m k modeseconds modefirsts - - 1 1 2362 8733 . . k=200 N/m First mode: A1 = -A2 Second mode: A1 = A2 0AA 0AA 21 21 2 2 105 510 m1=m2=40 kg
- 31. 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: