Linear Algebra, Theory And Applications, 2012a

15.1. THE POWER METHOD FOR EIGENVALUES 375
Now begin with this one.
⎛
5 −14 11
⎞ ⎛
⎝ −4 4 −4 ⎠ ⎝
3 6 −3
Divide by 12 to get the next iterate.
⎛
⎝
⎞ ⎛
1.0
−0.5 ⎠ = ⎝
−1. 430 6 × 10 −6
⎞ ⎛
12. 000
−6. 000 0 ⎠ 1
4. 291 8 × 10 −6 12 = ⎝
1.0
⎞
−0.5 ⎠
12. 000
⎞
−6. 000 0 ⎠
Another iteration will reveal that the scaling factor is still 12. Thus this is an approximate
eigenvalue. In fact, it is the largest eigenvalue and the corresponding eigenvector is
⎛
⎝ 1.0
⎞
−0.5 ⎠
0
The process has worked very well.
15.1.1 The Shifted Inverse Power Method
This method can find various eigenvalues and eigenvectors. It is a significant generalization
of the above simple procedure and yields very good results. One can find complex eigenvalues
using this method. The situation is this: You have a number, α which is close to λ, some
eigenvalue of an n × n matrix A. You don’t know λ but you know that α is closer to λ
than to any other eigenvalue. Your problem is to find both λ and an eigenvector which goes
with λ. Another way to look at this is to start with α and seek the eigenvalue λ, which is
closest to α along with an eigenvector associated with λ. If α is an eigenvalue of A, then
you have what you want. Therefore, I will always assume α is not an eigenvalue of A and
so (A − αI) −1 exists. The method is based on the following lemma.
Lemma 15.1.3 Let {λ k } n k=1 be the eigenvalues of A. If x k is an eigenvector of A for the
eigenvalue λ k , then x k is an eigenvector for (A − αI) −1 corresponding to the eigenvalue
1
λ k −α . Conversely, if (A − αI) −1 y = 1
λ − α y (15.4)
and y ≠ 0, thenAy = λy.
Proof: Let λ k and x k be as described in the statement of the lemma. Then
(A − αI) x k =(λ k − α) x k
and so
1
λ k − α x k =(A − αI) −1 x k .
Suppose (15.4). Then y = 1
λ−α
[Ay − αy] . Solving for Ay leads to Ay = λy.
1
Now assume α is closer to λ than to any other eigenvalue. Then the magnitude of
λ−α
is greater than the magnitude of all the other eigenvalues of (A − αI) −1 . Therefore, the
power method applied to (A − αI) −1 1
will yield
λ−α . You end up with s n+1 ≈ 1
λ−α
and
solve for λ.