Page 375 - A Course in Linear Algebra with Applications

9.4: Jordan Normal Form 359

Example 9.4.7
Put the matrix ^ 3 1 0
/
H 0 1 0
- 1
0
V
2
in Jordan normal form.
We follow the method of the proof of 9.4.5. The eigen-
values of A are 2, 2, 2, so define

1 1 0'
A' = A-2I=\-1 - 1 0
0 0 0

The column space C of A 1 is generated by the single vector

( l \
X = —1 . Note that AX = 2X, so X is a Jordan string
V 0/

of length 1 for A. Also the null space N of A' is generated by
X and the vector

Thus D — C D N — C is generated by X. The next step is to
write X in the form A'Y: in fact we can take

Y

0'
Thus the second basis element is Y. Finally, put Z = [ 0 | ,

so that {X, Z} is a basis for N. Then

A'X = 0, A'Y = X, A'Z = 0