Post on 18-Jan-2018
description
Scientific Computing
The Power Method for FindingDominant Eigenvalue
• The eigenvectors of a matrix are vectors that satisfy Ax = λx Or, (A – λI)x = 0 So, λ is an eigenvalue iff det(A – λI) = 0
Example:
Eigenvalues-Eigenvectors
2/10064/30541
A
)2/1)(4/3)(1(2/10064/30541
)det(
IA
2/1,4/3,1
• Eigenvalues are used in the solution of Engineering Problems involving vibrations, elasticity, oscillating systems, etc.
• Eigenvalues are also important for the analysis of Markov Chains in statistics.
• The next set of slides are from the course “Computer Applications in Engineering and Construction” at Texas A&M (Fall 2008).
Eigenvalues-Eigenvectors
Mass-Spring SystemMass-Spring System
21222
2
2
12121
2
1
kxxxkdt
xdm
xxkkxdt
xdm
)(
)(
Equilibrium positions
Homogeneous system
Find the eigenvalues from det[ ] = 0
Mass-Spring SystemMass-Spring System
0Xm
k2Xmk
0XmkX
mk2
0x2xkdt
xdm
0xx2kdt
xdm
T2 tXx tXx let
22
21
2
21
12
1
2122
2
2
1221
2
1
p2211
)(
)(
;sin,sin
00
XX
mk2mkmkmk2
2
12
22
22
1
////
m1 = m2 = 40 kg, k = 200 N/m
Characteristic equation det[ ] = 0
Two eigenvalues = 3.873s1 or 2.236 s 1 Period Tp = 2/ = 1.62 s or 2.81 s
Polynomial MethodPolynomial Method
0515 07520
105510
mk2mk
mkmk2
2224
2
2
222
22
1
))((
)/(//)/(
Principal Modes of VibrationPrincipal Modes of Vibration
Tp = 1.62 s
X1 = X2
Tp = 2.81 s
X1 = X2
Power method for finding eigenvalues
1. Start with an initial guess for x2. Calculate w = Ax 3. Largest value (magnitude) in w is the
estimate of eigenvalue 4. Get next x by rescaling w (to avoid
the computation of very large matrix An )
5. Continue until converged
Power MethodPower Method
• Start with initial guess z = x0
Power MethodPower Method
Azwz
w Azw
Azwz
w Azw
)2(k
)2(
)2(k
)2()3(
)2(k
)2(k
)2()2(
)1(k
)1(
)1(k
)1()2(
)1(k
)1(k
)1()1(
kn
k3
k2
k1
n321 then , If
rescaling
k is the dominant
eigenvalue
Power MethodPower Method
(1)0
(1) (1) (2)
(2) (2) (3)
( ) ( ) ( 1)
1. 1,1,...,1
2. ;
3. ;
...
T
k
k
k k k
normalize
n
Initial guess z x
Calculate Az w z z by biggest w
Calculate Az w z z by biggest w
Calculate
ormalize
Az w z
• For large number of iterations, should converge to the largest eigenvalue
• The normalization make the right hand side converge to , rather than n
Example: Power MethodExample: Power Method
111
xz
7410438
1082A
0)1(
017143095240
21211520
111
7410438
1082Az 1
...
)(
Start with
Assume all eigenvalues are equally important, since we don’t know which one is dominant
Consider
Eigenvalue estimate Eigenvector
ExampleExampleCurrent estimate for largest eigenvalue is 21 Rescale w by eigenvalue to get new x
017143095240
211520
211
wabswx
...
))(max(
Check Convergence (Norm < tol?)
13433618812382138122xAx
618812382138122
017143095240
2101
7143095240
7410438
1082xAx
222 .).().().(
.
.
.
...
*...
Norm
Update the estimated eigenvector and repeat
New estimate for largest eigenvalue is 19.381 Rescale w by eigenvalue to get new x
381197621361917
017143095240
7410438
1082Az 2
.
.
.
...
)(
017101090910
381197621361917
381191
wabswx
...
.
.
.
.))(max(
58800449603594012030xAx
449603594012030
017101090910
3811901
7101090910
7410438
1082xAx
222 .).().().(
.
.
.
...
*....
Norm
ExampleExample
One more iteration
017080092430
93118931184031349917
017101090910
7410438
1082Az 3
...
....
...
)(
18510144001153001470xAx
144001153001470
017080092430
9311801
7080092430
7410438
1082xAx
222 .).().().(
.
.
.
...
*....
Convergence criterion -- Norm (or relative error) < tol
Norm
Example: Power MethodExample: Power Method
0.17085.09200.0
030.19030.19482.13507.17
0.17085.09196.0
7410438
1082Az
0.17085.09196.0
040.19040.19490.13508.17
0.17084.09206.0
7410438
1082Az
0.17084.09206.0
016.19016.19471.13506.17
0.17087.09181.0
7410438
1082Az
0.17087.09181.0
075.19075.19519.13513.17
0.17080.09243.0
7410438
1082Az
)7(
)6(
)5(
)4(
Script file: Power_eig.mScript file: Power_eig.m
xAxNorm
» A=[2 8 10; 8 3 4; 10 4 7]A = 2 8 10 8 3 4 10 4 7» [z,m] = Power_eig(A,100,0.001);
it m z(1) z(2) z(3) z(4) z(5) 1.0000 21.0000 0.9524 0.7143 1.0000 2.0000 19.3810 0.9091 0.7101 1.0000 3.0000 18.9312 0.9243 0.7080 1.0000 4.0000 19.0753 0.9181 0.7087 1.0000 5.0000 19.0155 0.9206 0.7084 1.0000 6.0000 19.0396 0.9196 0.7085 1.0000 7.0000 19.0299 0.9200 0.7085 1.0000 8.0000 19.0338 0.9198 0.7085 1.0000 9.0000 19.0322 0.9199 0.7085 1.0000error = 8.3175e-004» zz = 0.9199 0.7085 1.0000» mm = 19.0322
» x=eig(A)x = -7.7013 0.6686 19.0327
MATLAB MATLAB Example:Example:
Power Power MethodMethod
eigenvector
eigenvalue
MATLAB function
MATLAB’s MethodsMATLAB’s Methods
• e = eig(A) gives eigenvalues of A
• [V, D] = eig(A) eigenvectors in V(:,k) eigenvalues = Dii (diagonal matrix D)
• [V, D] = eig(A, B) (more general eigenvalue problems) (Ax = Bx)
AV = BVD
Theorem: If A has a complete set of eigenvectors, then the Power method converges to the dominate eigenvalue of the matrix A.
Proof: A has n eigenvalues 1,2,3,…,n with 1>2>3>…>n with a corresponding basis of eigenvectors w1,w2,w3,…,wn. Let the initial vector w0 be a linear combination of the vectors w1,w2,w3,…,wn.
w0 = a1w1+a2w2+a3w3+…+anwn
Aw0 = A(a1w1+a2w2+a3w3+…+anwn)
=a1Aw1+a2Aw2+a3Aw3+…+anAwn
=a11w1+a22w2+a33w3+…+annwn
Akw0 =a1(1)kw1+a2(2)kw2+…+an(n)kwn
Akw0/(1)k-1 =a1(1)k /(1)k-1 w1 +…+an(n)k /(1)k-1 wn
n
k
nnn
kk
k
k
aaaa wwwwwA1
13
1
1
3332
1
1
2221111
1
0
For large values of k (as k goes to infinity) we get the following:
1,,1,1:since11
3
1
21111
1
0
n
k
k
a wwA
At each stage of the process we divide by the dominant term of the vector. If we write w1 as shown to the right and consider what happens between two consecutive estimates we get the following.
nc
cc
2
1
1w
nn
k
k
nn
k
k
ca
caca
c
cc
aand
ca
caca
c
cc
a
211
2211
1211
2
1
211
1
01
11
211
111
2
1
1111
0
wAwA
Dividing by the dominant term gives something that is approximately 1.