Putting m ¼ ν + ν
0
À 1, if i ! ν
0 , N
0i
¼ 0 from the supposition. If i < ν
0 , we have
m À i > m À ν
0
¼ ν + ν
0
À 1 À ν
0
¼ ν À 1; i.e., m À i ! ν. Therefore, N
m À i
¼ 0.
Consequently, we have N
m À i N
0i
¼ 0 with any i in (12.113). Thus, we get
(N Æ N
0
)
m
¼ 0, indicating that N Æ N
0 is nilpotent.
(iii) Let S be a semi-simple and nilpotent matrix. We describe S as
S $
α 1 Á Á Á 0
⋮ ⋱ ⋮
0 Á Á Á α n
0
B
@
1
C
A,
ð12:114Þ
where some of α i (1
i
n) may be identical. Since S is nilpotent, all
α i (1 i n) is zero. We have then S~0; i.e., S ¼ 0 accordingly.
Now, suppose that a matrix A is decomposed differently from (12.109). That is,
we have
A ¼ S þ N ¼ S
0
þ N
0
or S À S
0
¼ N
0
À N:
ð12:115Þ
From the assumption, S
0 and N
0 are commutable. Moreover, since S, S
0 , N, and N
0
are described by a polynomial of A, they are commutable with one another. Hence,
from (i) and (ii) along with the second equation of (12.115), S À S
0 and N
0
À N are
both semi-simple and nilpotent at once. Consequently, from (iii) S À S
0
¼ N
0
À N ¼ 0.
Thus, we finally get S ¼ S
0 and N ¼ N
0 . That is, the decomposition is unique.
These complete the proof.
∎
Theorem 12.6 implies that the matrix decomposition of (12.109) is unique. On the
basis of Theorem 12.6, we investigate Jordan canonical forms of matrices in the next
section.
12.6 Jordan Canonical Form
Once the vector space V
n has been decomposed to a direct sum of generalized
eigenspaces with the matrix reduced in parallel, we are able to deal with individual
eigenspaces e
W α i 1 i s
ð
Þand the corresponding A
(i) (1 i s) separately.
12.6.1 Canonical Form of Nilpotent Matrix
To avoid complication of notation, we think of a following example where we
assume a (n, n) matrix that operates on V
n : Let the nilpotent matrix be N: Suppose
that N
ν À 1
6 ¼ 0 and N
ν
¼ 0 (1 ν n). If ν ¼ 1, the nilpotent matrix N is a zero
488
12 Canonical Forms of Matrices
0
À 1, if i ! ν
0 , N
0i
¼ 0 from the supposition. If i < ν
0 , we have
m À i > m À ν
0
¼ ν + ν
0
À 1 À ν
0
¼ ν À 1; i.e., m À i ! ν. Therefore, N
m À i
¼ 0.
Consequently, we have N
m À i N
0i
¼ 0 with any i in (12.113). Thus, we get
(N Æ N
0
)
m
¼ 0, indicating that N Æ N
0 is nilpotent.
(iii) Let S be a semi-simple and nilpotent matrix. We describe S as
S $
α 1 Á Á Á 0
⋮ ⋱ ⋮
0 Á Á Á α n
0
B
@
1
C
A,
ð12:114Þ
where some of α i (1
i
n) may be identical. Since S is nilpotent, all
α i (1 i n) is zero. We have then S~0; i.e., S ¼ 0 accordingly.
Now, suppose that a matrix A is decomposed differently from (12.109). That is,
we have
A ¼ S þ N ¼ S
0
þ N
0
or S À S
0
¼ N
0
À N:
ð12:115Þ
From the assumption, S
0 and N
0 are commutable. Moreover, since S, S
0 , N, and N
0
are described by a polynomial of A, they are commutable with one another. Hence,
from (i) and (ii) along with the second equation of (12.115), S À S
0 and N
0
À N are
both semi-simple and nilpotent at once. Consequently, from (iii) S À S
0
¼ N
0
À N ¼ 0.
Thus, we finally get S ¼ S
0 and N ¼ N
0 . That is, the decomposition is unique.
These complete the proof.
∎
Theorem 12.6 implies that the matrix decomposition of (12.109) is unique. On the
basis of Theorem 12.6, we investigate Jordan canonical forms of matrices in the next
section.
12.6 Jordan Canonical Form
Once the vector space V
n has been decomposed to a direct sum of generalized
eigenspaces with the matrix reduced in parallel, we are able to deal with individual
eigenspaces e
W α i 1 i s
ð
Þand the corresponding A
(i) (1 i s) separately.
12.6.1 Canonical Form of Nilpotent Matrix
To avoid complication of notation, we think of a following example where we
assume a (n, n) matrix that operates on V
n : Let the nilpotent matrix be N: Suppose
that N
ν À 1
6 ¼ 0 and N
ν
¼ 0 (1 ν n). If ν ¼ 1, the nilpotent matrix N is a zero
488
12 Canonical Forms of Matrices