Page 215 - Matrix Analysis & Applied Linear Algebra

210 Chapter 4 Vector Spaces
4.5 MORE ABOUT RANK

Since equivalent matrices have the same rank, it follows that if P and Q are
nonsingular matrices such that the product PAQ is defined, then
rank (A)= rank (PAQ)= rank (PA)= rank (AQ).
In other words, rank is invariant under multiplication by a nonsingular matrix.
However, multiplication by rectangular or singular matrices can alter the rank,
and the following formula shows exactly how much alteration occurs.

Rank of a Product
If A is m × n and B is n × p, then

rank (AB)= rank (B) − dim N (A) ∩ R (B). (4.5.1)

Proof. Start with a basis S = {x 1 , x 2 ,..., x s } for N (A) ∩ R (B), and no-
tice N (A) ∩ R (B) ⊆ R (B). If dim R (B)= s + t, then, as discussed in
Example 4.4.5, there exists an extension set S ext = {z 1 , z 2 ,..., z t } such that
B = {x 1 ,..., x s , z 1 ,..., z t } is a basis for R (B). The goal is to prove that
dim R (AB)= t, and this is done by showing T = {Az 1 , Az 2 ,..., Az t } is a
basis for R (AB). T spans R (AB) because if b ∈ R (AB), then b = ABy
s t
ξ
for some y, but By ∈ R (B) implies By = i=1 i x i + i=1 η i z i , so

s t s t t

b = A ξ i x i + η i z i = ξ i Ax i + η i Az i = η i Az i .
i=1 i=1 i=1 i=1 i=1
t t

T is linearly independent because if 0 = α i Az i = A α i z i , then
i=1 i=1
t
i=1 α i z i ∈ N (A) ∩ R (B), so there are scalars β j such that
t s t s

α i z i = β j x j or, equivalently, α i z i − β j x j = 0,
i=1 j=1 i=1 j=1
and hence the only solution for the α i ’s and β i ’s is the trivial solution because
B is an independent set. Thus T is a basis for R (AB), so t = dim R (AB)=
rank (AB), and hence
rank (B) = dim R (B)= s + t = dim N (A) ∩ R (B)+ rank (AB).
It’s sometimes necessary to determine an explicit basis for N (A) ∩ R (B).
In particular, such a basis is needed to construct the Jordan chains that are
associated with the Jordan form that is discussed on pp. 582 and 594. The
following example outlines a procedure for finding such a basis.