Page 144 - Matrix Analysis & Applied Linear Algebra

138 Chapter 3 Matrix Algebra

Proof. To establish the validity of (3.9.8), observe that rank (A)= rank (B)
implies A ∼ N r and B ∼ N r . Therefore, A ∼ N r ∼ B. Conversely, if A ∼ B,
where rank (A)= r and rank (B)= s, then A ∼ N r and B ∼ N s , and
hence N r ∼ A ∼ B ∼ N s . Clearly, N r ∼ N s implies r = s. To prove (3.9.9),
row row row
suppose first that A ∼ B. Because B ∼ E B , it follows that A ∼ E B . Since
a matrix has a uniquely determined reduced echelon form, it must be the case
that E B = E A . Conversely, if E A = E B , then

row row row
A ∼ E A = E B ∼ B =⇒ A ∼ B.
The proof of (3.9.10) follows from (3.9.9) by considering transposes because

col T T
A ∼ B ⇐⇒ AQ = B ⇐⇒ (AQ) = B
T
T
T
⇐⇒ Q A = B ⇐⇒ A T row T
∼ B .
Example 3.9.4
Problem: Are the relationships that exist among the columns in A the same
as the column relationships in B, and are the row relationships in A the same
as the row relationships in B, where

   
1 1 1 −1 −1 −1
A =  −4 −3 −1  and B =  2 2 2  ?
2 1 −1 2 1 −1

Solution: Straightforward computation reveals that
 
10 −2
E A = E B =  01 3  ,
00 0

row
and hence A ∼ B. Therefore, the column relationships in A and B must be
identical, and they must be the same as those in E A . Examining E A reveals
that E ∗3 = −2E ∗1 +3E ∗2 , so it must be the case that

A ∗3 = −2A ∗1 +3A ∗2 and B ∗3 = −2B ∗1 +3B ∗2 .
The row relationships in A and B are different because E A T
= E B T .

On the surface, it may not seem plausible that a matrix and its transpose
should have the same rank. After all, if A is 3 × 100, then A can have as
many as 100 basic columns, but A T can have at most three. Nevertheless, we

can now demonstrate that rank (A)= rank A T .