Page 218 - Matrix Analysis & Applied Linear Algebra

4.5 More about Rank 213

T
Proof. First observe that N A ∩ R (A)= {0} because
T T
x ∈ N A ∩ R (A)=⇒ A x = 0 and x = Ay for some y

T T T 2
=⇒ x x = y A x =0 =⇒ x =0
i
=⇒ x = 0.
Formula (4.5.1) for the rank of a product now guarantees that

T T
rank A A = rank (A) − dim N A ∩ R (A)= rank (A),
which is half of (4.5.4)—the other half is obtained by reversing the roles of A
T
and A . To prove (4.5.5) and (4.5.6), use the facts R (AB) ⊆ R (A) and
T

N (B) ⊆ N (AB) (see Exercise 4.2.12) to write R A A ⊆ R A T and

T
N (A) ⊆ N A A . The first half of (4.5.5) and (4.5.6) now follows because
T T T T
dim R A A = rank A A = rank (A)= rank A = dim R A ,
T
T

dim N (A)= n − rank (A)= n − rank A A = dim N A A .
Reverse the roles of A and A T to get the second half of (4.5.5) and (4.5.6).
To see why (4.5.4)—(4.5.6) might be important, consider an m × n system
of equations Ax = b that may or may not be consistent. Multiplying on the
left-hand side by A T produces the n × n system
T
T
A Ax = A b
called the associated system of normal equations, which has some ex-
tremely interesting properties. First, notice that the normal equations are always
consistent, regardless of whether or not the original system is consistent because
T
T

(4.5.5) guarantees that A b ∈ R A T = R A A (i.e., the right-hand side is
in the range of the coefficient matrix), so (4.2.3) insures consistency. However, if
T
T
Ax = b happens to be consistent, then Ax = b and A Ax = A b have the
same solution set because if p is a particular solution of the original system,
T
T
then Ap = b implies A Ap = A b (i.e., p is also a particular solution of
the normal equations), so the general solution of Ax = b is S = p + N (A),
T
T
and the general solution of A Ax = A b is
T
p + N A A = p + N (A)= S.
Furthermore, if Ax = b is consistent and has a unique solution, then the same
T
T
is true for A Ax = A b, and the unique solution common to both systems is
−1
T T
x = A A A b. (4.5.7)