A Ax À x
ð
Þ¼0:
Then, we have
Ax = x or Ax = 0:
ð12:74Þ
From Theorem 11.3 (dimension theorem), we have
V
n
¼ A V
n
ð Þ È Ker A,
where
A V
n
ð Þ ¼ x; x 2 V
n , Ax = x
f
g , Ker A ¼ x; x 2 V
n , Ax = 0
f
g :
ð12:75Þ
Thus, we find that A decomposes V
n into a direct sum of A(V
n ) and Ker A.
Conversely, we can readily verify that if there exists an operator A that satisfies
(12.75), such A must be an idempotent operator. The verification is left for readers.
Meanwhile, we have
E 2 A
ð
Þ
2 ¼ E À 2A þ A
2
¼ E À 2A þ A ¼ E À A:
Hence, E À A is an idempotent matrix as well. Moreover, we have
A E À A
ð
Þ¼ E À A
ð
ÞA ¼ 0:
Putting E À A B and following a procedure similar to the above
Bx À x ¼ E À A
ð
Þx À x ¼ x À Ax À x ¼ ÀAx:
Therefore, B(V
n ) ¼ {x; x 2 V
n , Bx = x} is identical to Ker A. Writing
W A ¼ x; x 2 V
n , Ax = x
f
g , W A ¼ x; x 2 V
n , Ax = 0
f
g ,
ð12:76Þ
we get
W A ¼ AV
n , W A ¼ E À A
ð
ÞV
n
¼ BV
n
¼ V
n
À AV
n
¼ Ker A:
That is,
V
n
¼ W A þ W A :
Suppose that
∃ u 2 W A \ W A . Then, from (12.76) Au = u and Au = 0, namely
u = 0, and so V is a direct sum of W A and W A . That is,
12.4 Idempotent Matrices and Generalized Eigenspaces
479