Eigenvalues

Dominant Eigenvalues

&

The Power Method

Eigenvalues & Eigenvectors

In linear algebra we learned that a scalar is an eigenvalue for a square nn matrix A if there is a non-zero vector w such that Aw = w, we call the vector w an eigenvector for matrix A. The eigenvalue acts like scalar multiplication instead of matrix multiplication for the vector w. Eigenvalues are important for many applications in mathematics, physics, engineering and other disciplines.

The Dominant Eigenvalue

A nn matrix A will have n eigenvalues (some may be repeated). By the dominant eigenvalue we refer to the one that is biggest in terms of absolute value. This would include any eigenvalues that are complex.

200

140

731

A

The matrix A to the right has as its eigenvalues the set of numbers {1,-4,2} (i.e. it is upper triangular). In absolute value this set is {|1|,|-4|,|2|}={1,4,2}. Since -4 is the largest in absolute value we say that -4 is the dominant eigenvalue.

The problem we want to solve is that if we are given a matrix A can we estimate the dominant eigenvalue?

The Power Method

The Power Method works for matrices sort of like how the fixed point method works for functions. The iteration step is a bit different though. To explain how it works we need to introduce a bit of terminology. The dominant term of a vector v is the term that has the greatest absolute value (careful: it is the term itself not the absolute value of the term). If there are two terms that have the same absolute value you can pick either one for our purposes.

6

3

7

1v

dominant term=7

8

4

8

5

2v

dominant term=-8

10

63v

dominant term=-10

213

4 37

5

v

dominant term=6.5

The algorithm consists of the following steps. Start with an initial vector w0. Let the approximation for dominant eigenvalue be z0 the dominant term in w0. Use the iteration to the right:

zk+1 = dominate term in Awk

wk+1 = (1/zk+1) Awk

To get the next approximation for the dominant eigenvalue multiply the previous eigenvector by the matrix A and take that vectors dominant term. To get the next approximation for the eigenvector divide the product by it dominant term. The vector w0 given to the right is often used as the initial vector.

Example:

Apply the power method for 3 iterations to find z3 and w3 for the matrix A given to the right.

1

1

1

0 w

102

011

210

A

1

1

1

0w

10 z

3

2

3

1

1

1

102

011

210

0Aw

31 z

1

1

32

1w

31

1

102

011

210

35

38

32

1Aw

32 z

195

98

2w

925

913

923

95

98

2

1102

011

210

Aw

7.2925

3 z

12513

2523

3w

Power Method Convergence

The Power method will not converge for a real matrix A the power method will converge to the dominate eigenvalue if the dominant eigenvalue is a real number. If the dominant eigenvalue is a complex number an initial vector with complex entries would need to be used. If the dominant eigenvalue is repeated it will find it.

How this convergence can be seen is as follows.

Given a nn matrix A with n eigenvalues 1,2,3,…,n with 1>2>3>…>n (i.e. 1 is the dominant eigenvalue) find 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 (replace with eigenvalues)

=a11w1+a22w2+a33w3+…+annwn

Akw0 =a1(1)kw1+a2(2)kw2+…+an(n)kwn (repeat for powers of A)

Akw0/(1)k-1 =a1(1)k /(1)k-1 w1+ a2(2)k /(1)k-1 w2 +…+an(n)k /(1)k-1 wn

n

k

nnn

kk

k

k

aaaa wwwwwA

1

13

1

1

3332

1

1

2221111

1

0

For large values of k (i.e. 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

c

c

2

1

1w

nn

k

k

nn

k

k

ca

ca

ca

c

c

c

aand

ca

ca

ca

c

c

c

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.