R
0
¼
0 0 À1 0
0 0 1
1
0 1 0
0
1 1 À1 0
0
B
B
B
@
1
C
C
C
A
:
ð12:177Þ
In this case, we also get the same Jordan canonical form as before. That is,
R
0 À1 AR
0
¼
4 0 0 0
0 2 0 0
0 0 2 1
0 0 0 2
0
B
B
B
@
1
C
C
C
A
:
Suppose that we choose R
00 such that
e 1 e 2 e 3 e 4
ð
Þ R
00
¼ e 1 e 2 e 3 e 4
ð
Þ
0 1
0
0
0 À1 À1 0
1 0
0
0
1 1
0
1
0
B
B
B
@
1
C
C
C
A
¼ e
2
ð Þ
1 e
2
ð Þ
2 g
2
ð Þ
2 e
4
ð Þ
1
:
In this case, we have
R
00 À1 AR
00
¼
2 0 0 0
0 2 1 0
0 0 2 0
0 0 0 4
0
B
B
B
@
1
C
C
C
A
:
Note that we get a different disposition of the matrix elements from that of
(12.172).
Next, we decompose A into a semi-simple matrix and a nilpotent matrix. In
(12.172), we had
R
À1 AR ¼
4 0 0 0
0 2 0 0
0 0 2 1
0 0 0 2
0
B
B
B
@
1
C
C
C
A
:
Defining
12.6 Jordan Canonical Form
509