Page 402 - Matrix Analysis & Applied Linear Algebra

398 Chapter 5 Norms, Inner Products, and Orthogonality

and write

k U k
C 0 = Q −1 A Q = A X | Y = UA X 0 .
k
k

k
0 N k V VA X 0
C k 0
k
k
Therefore, N = 0 and Q −1 A Q = . Since C k is r × r and r =
0 0
k −1 k k k
rank A = rank Q A Q = rank C , it must be the case that C is
nonsingular, and hence C is nonsingular. Finally, notice that index(N)= k
because if index(N) = k, then N k−1 = 0, so
k−1 k−1
C 0 C 0
k−1 −1 k−1
rank A =rank Q A Q =rank k−1 =rank
0 N 0 0
k−1 k
= rank C = r = rank A ,
which is impossible because index(A)= k is the smallest integer for which there
is equality in ranks of powers.
Example 5.10.3
k
Problem: Let A n×n have index k with rank A = r, and let

0 U r×n

Q −1 AQ = C r×r with Q = X n×r | Y and Q −1 =
0 N V
be the core-nilpotent decomposition described in (5.10.5). Explain why

0
I r −1 k k
Q Q = XU = the projector onto R A along N A
0 0
and

0 0 −1 k k
Q Q = YV = the projector onto N A along R A .
0I n−r
k k
Solution: Because R A and N A are complementary subspaces, and
because the columns of X and Y constitute respective bases for these spaces,
it follows from the discussion concerning projectors on p. 386 that

I 0 −1 0
I r −1
P = X | Y X | Y = Q Q = XU
00 0 0
k k
must be the projector onto R A along N A , and

00 −1 0 0
−1
I − P = X | Y X | Y = Q Q = YV
0 I 0I n−r
k k
is the complementary projector onto N A along R A .