Page 399 - Matrix Analysis & Applied Linear Algebra
P. 399
5.10 Range-Nullspace Decomposition 395
holds, and this forces n< dim N A n+1 , which is impossible. A similar
argument proves equality exists somewhere in the range chain.
Property 2. Once equality is attained, it is maintained throughout the rest of
both chains in (5.10.2). In other words,
0 k k+1 k+2
N A ⊂ N (A) ⊂· · · ⊂ N A = N A = N A = ···
(5.10.3)
0 k k+1 k+2
R A ⊃ R (A) ⊃ ··· ⊃ R A = R A = R A = ··· .
To prove this for the range chain, observe that if k is the smallest nonneg-
ative integer such that R A k = R A k+1 , then for all i ≥ 1,
i+k i k i k i k+1 i+k+1
R A = R A A = A R A = A R A = R A .
The nullspace chain stops growing at exactly the same place the ranges
stop shrinking because the rank plus nullity theorem (p. 199) insures that
p
p
dim N (A )= n − dim R (A ).
Property 3. If k is the value at which the ranges stop shrinking and the
k k
nullspaces stop growing in (5.10.3), then R A ∩ N A = 0.
k k k n
Proof. If x ∈ R A ∩ N A , then A y = x for some y ∈ , and
k
2k
k
A x = 0. Hence A y = A x = 0 ⇒ y ∈ N A 2k = N A k ⇒ x = 0.
Property 4. If k is the value at which the ranges stop shrinking and the
k k n
nullspaces stop growing in (5.10.3), then R A + N A = .
Proof. Use Property 3 along with (4.4.19), (4.4.15), and (4.4.6), to write
k k k k k k
dim R A + N A = dim R A + dim N A − dim R A ∩ N A
k k
= dim R A + dim N A = n
k k n
=⇒ R A + N A = .
Below is a summary of our observations concerning the index of a square matrix.
Index
The index of a square matrix A is the smallest nonnegative integer k
such that any one of the three following statements is true.
k k+1
• rank A = rank A .
k k+1 k
• R A = R A —i.e., the point where R A stops shrinking.
k k+1 k
• N A = N A —i.e., the point where N A stops growing.
For nonsingular matrices, index (A)=0. For singular matrices,
index (A)is the smallest positive integer k such that either of the fol-
lowing two statements is true.
k k
• R A ∩ N A = 0. (5.10.4)
k
n
• = R A k ⊕ N A .