14.1. THE POWER METHOD FOR EIGENVALUES 361

Proof: Let λk and xk be as described in the statement of the lemma. Then

(A− αI)xk = (λk − α)xk

and so1

λk − αxk = (A− αI)

−1xk.

Suppose 14.3. Then y = 1λ−α [Ay − αy] . Solving for Ay leads to Ay = λy. ■

Now assume α is closer to λ than to any other eigenvalue. Then the magnitude of 1λ−α

is greater than the magnitude of all the other eigenvalues of (A− αI)−1

. Therefore, the

power method applied to (A− αI)−1

will yield 1λ−α . You end up with sn+1 ≊ 1

λ−α andsolve for λ.

14.1.2 The Explicit Description of the Method

Here is how you use this method to find the eigenvalue closest to α and thecorresponding eigenvector.

1. Find (A− αI)−1.

2. Pick u1. If you are not phenomenally unlucky, the iterations will converge.

3. If uk has been obtained,

uk+1 =(A− αI)

−1uk

sk+1

where sk+1 is the entry of (A− αI)−1

uk which has largest absolute value.

4. When the scaling factors, sk are not changing much and the uk are not changing much,find the approximation to the eigenvalue by solving

sk+1 =1

λ− α

for λ. The eigenvector is approximated by uk+1.

5. Check your work by multiplying by the original matrix to see how well what you havefound works.

Thus this amounts to the power method for the matrix (A− αI)−1

but you are free topick α.

Example 14.1.4 Find the eigenvalue of A =

 5 −14 11

−4 4 −4

3 6 −3

 which is closest to −7.

Also find an eigenvector which goes with this eigenvalue.

In this case the eigenvalues are −6, 0, and 12 so the correct answer is −6 for theeigenvalue. Then from the above procedure, I will start with an initial vector, u1 =(

1 1 1)T

. Then I must solve the following equation. 5 −14 11

−4 4 −4

3 6 −3

+ 7

 1 0 0

0 1 0

0 0 1

 x

y

z

 =

 1

1

1

