# Thread: Rank of a product of two matrices

1. ## Rank of a product of two matrices

There is a remark my professor made in his notes that I simply can't wrap my head around. He even gave a proof but it made me even more confused. Here it is:

Two matrices:
$\displaystyle A_{mxn}, Q_{nxn}$
Q is invertible
AQ is mxn
$\displaystyle \therefore$ rank(AQ) = rank A

Note: I'm confused by this alone (I'll show the proof after if someone can help me understand this first). What is the relevance of Q's invertibility?

2. Originally Posted by Canadian0469 There is a remark my professor made in his notes that I simply can't wrap my head around. He even gave a proof but it made me even more confused. Here it is:

Two matrices:
$\displaystyle A_{mxn}, Q_{nxn}$
Q is invertible
AQ is mxn
$\displaystyle \therefore$ rank(AQ) = rank A

Note: I'm confused by this alone (I'll show the proof after if someone can help me understand this first). What is the relevance of Q's invertibility?
one way to see this is to look at the transformations that A and Q define: suppose V and W are two vector spaces with dimensions n and m respectively. let $\displaystyle T: V \longrightarrow V$ and $\displaystyle S: V \longrightarrow W$

be defined by $\displaystyle T(v)=Qv, \ S(v)=Av.$ then $\displaystyle ST(v)=AQv.$ now since Q is invertible, $\displaystyle T(V)=V.$ hence: $\displaystyle ST(V)=S(V),$ and thus: $\displaystyle \text{rank}(AQ)=\dim ST(V) = \dim S(V)= \text{rank} A. \ \ \Box$

matrices, product, rank 