370 CHAPTER 14. NUMERICAL METHODS, EIGENVALUES

Now solve  − 163 2 3

2 − 133 1

3 1 − 73

 x

y

z

 =

 . 699 25

. 493 89

1.0

and divide by the largest entry, 2. 997 9 to get

u3 =

 . 714 73

. 522 63

1. 0

Now solve  − 16

3 2 3

2 − 133 1

3 1 − 73

 x

y

z

 =

 . 714 73

. 522 63

1. 0

and divide by the largest entry, 3. 045 4, to get

u4 =

 . 713 7

. 520 56

1.0

Solve  − 16

3 2 3

2 − 133 1

3 1 − 73

 x

y

z

 =

 . 713 7

. 520 56

1.0

and divide by the largest entry, 3. 042 1 to get

u5 =

 . 713 78

. 520 73

1.0

You can see these scaling factors are not changing much. The predicted eigenvalue is thenabout

1

3. 042 1+

19

3= 6. 662 1.

How close is this?  1 2 3

2 2 1

3 1 4

 . 713 78

. 520 73

1.0

 =

 4. 755 2

3. 469

6. 662 1

while

6. 662 1

 . 713 78

. 520 73

1.0

 =

 4. 755 3

3. 469 2

6. 662 1

 .

You see that for practical purposes, this has found the eigenvalue and an eigenvector.