Page 398 - Matrix Analysis & Applied Linear Algebra

394 Chapter 5 Norms, Inner Products, and Orthogonality
5.10 RANGE-NULLSPACE DECOMPOSITION

Since there are infinitely many different pairs of complementary subspaces in
n n 54
(or C ), is some pair more “natural” than the rest? Without reference to
anything else the question is hard to answer. But if we start with a given matrix
n
A n×n , then there is a very natural direct sum decomposition of defined
by fundamental subspaces associated with powers of A. The rank plus nullity
theorem on p. 199 says that dim R (A)+dim N (A)= n, so it’s reasonable to ask
about the possibility of R (A) and N (A)being complementary subspaces. If A
is nonsingular, then it’s trivially true that R (A) and N (A) are complementary,
but when A is singular, this need not be the case because R (A) and N (A)
need not be disjoint. For example,

01 1
A = =⇒ ∈ R (A) ∩ N (A).
00 0
But all is not lost if we are willing to consider powers of A.

Range-Nullspace Decomposition

Forevery singular matrix A n×n , there exists a positive integer k such
k k
that R A and N A are complementary subspaces. That is,

n
k
= R A k ⊕ N A . (5.10.1)
The smallest positive integer k for which (5.10.1) holds is called the
index of A. For nonsingular matrices we define index(A)=0.
Proof. First observe that as A is powered the nullspaces grow and the ranges
shrink—recall Exercise 4.2.12.
0 2 k k+1
N A ⊆ N (A) ⊆ N A ⊆· · · ⊆ N A ⊆ N A ⊆· · ·
(5.10.2)
0 2 k k+1
R A ⊇ R (A) ⊇ R A ⊇· · · ⊇ R A ⊇ R A ⊇· · · .
The proof of (5.10.1) is attained by combining the four following properties.
Property 1. There is equality at some point in each of the chains (5.10.2).
Proof. If there is strict containment at each link in the nullspace chain in
(5.10.2), then the sequence of inequalities
0 2 3
dim N A < dim N (A) < dim N A < dim N A < ···

54
n
All statements and arguments in this section are phrased in terms of , but everything we
n
say has a trivial extension to C .