Page 217 - Matrix Analysis & Applied Linear Algebra

212 Chapter 4 Vector Spaces

Proof. In words, (4.5.2) says that the rank of a product cannot exceed the rank
of either factor. To prove rank (AB) ≤ rank (B), use (4.5.1) and write

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

This says that the rank of a product cannot exceed the rank of the right-hand
factor. To show that rank (AB) ≤ rank (A), remember that transposition does
not alter rank, and use the reverse order law for transposes together with the
previous statement to write

T T T T
rank (AB)= rank (AB) = rank B A ≤ rank A = rank (A).

To prove (4.5.3), notice that N (A)∩R (B) ⊆ N (A), and recall from (4.4.5) that
if M and N are spaces such that M⊆N, then dim M≤ dim N. Therefore,

dim N (A) ∩ R (B) ≤ dim N (A)= n − rank (A),

and the lower bound on rank (AB) is obtained from (4.5.1) by writing

rank (AB)= rank (B) − dim N (A) ∩ R (B) ≥ rank (B)+ rank (A) − n.

T
The products A A and AA T and their complex counterparts A A and
∗
AA ∗ deserve special attention because they naturally appear in a wide variety
of applications.

T T
Products A A and AA
m×n
For A ∈ , the following statements are true.
T T
• rank A A = rank (A)= rank AA . (4.5.4)
T T T
• R A A = R A and R AA = R (A). (4.5.5)
T T T
• N A A = N (A) and N AA = N A . (4.5.6)
m×n T
For A ∈C , the transpose operation (') must be replaced by the
conjugate transpose operation (') .
∗