14.1. THE POWER METHOD FOR EIGENVALUES 369

Now pick the eigenvalue λq which is closest to q. Then

|Ax− qx|2 =

n∑k=1

|ak|2 (λk − q)2 ≥ (λq − q)

2n∑

k=1

|ak|2 = (λq − q)2 |x|2

which implies 14.5. ■

Example 14.1.10 Consider the symmetric matrix A =

 1 2 3

2 2 1

3 1 4

 . Let x =(1, 1, 1)T.

How close is the Rayleigh quotient to some eigenvalue of A? Find the eigenvector and eigen-value to several decimal places.

Everything is real and so there is no need to worry about taking conjugates. Therefore,the Rayleigh quotient is

(1 1 1

) 1 2 3

2 2 1

3 1 4

 1

1

1

3

=19

3

According to the above theorem, there is some eigenvalue of this matrix λq such that

∣∣∣∣λq − 19

3

∣∣∣∣ ≤

∣∣∣∣∣∣∣ 1 2 3

2 2 1

3 1 4

 1

1

1

− 193

 1

1

1

∣∣∣∣∣∣∣

√3

=1√3

 − 13

− 43

53

=

√19 +

(43

)2+(53

)2√3

= 1. 247 2

Could you find this eigenvalue and associated eigenvector? Of course you could. This iswhat the shifted inverse power method is all about.

Solve  1 2 3

2 2 1

3 1 4

− 19

3

 1 0 0

0 1 0

0 0 1

 x

y

z

 =

 1

1

1

In other words solve  − 16

3 2 3

2 − 133 1

3 1 − 73

 x

y

z

 =

 1

1

1

and divide by the entry which is largest, 3. 870 7, to get

u2 =

 . 699 25

. 493 89

1.0

