360 CHAPTER 15. NUMERICAL METHODS, EIGENVALUES
and divide by the entry with largest absolute value, 2.9979 to get
u3 =
.71473.52263
1.0
Now solve
− 163 2 3
2 − 133 1
3 1 − 73
x
yz
=
.71473.52263
1.0
and divide by the entry with largest absolute value, 3.0454, to get
u4 =
.7137.52056
1.0
Solve
− 163 2 3
2 − 133 1
3 1 − 73
x
yz
=
.7137.52056
1.0
and divide by the largest entry, 3.0421 to get
u5 =
.71378.52073
1.0
You can see these scaling factors are not changing much. The predicted eigenvalue isobtained by solving
1λ − 19
3
= 3.0421
to obtain λ = 6.6621. How close is this? 1 2 32 2 13 1 4
.71378
.520731.0
=
4.75523.469
6.6621
while
6.6621
.71378.52073
1.0
=
4.75533.46926.6621
.
You see that for practical purposes, this has found the eigenvalue and an eigenvector.