# Rank of a product of two matrices

• Nov 15th 2008, 05:07 PM
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?
• Nov 15th 2008, 06:47 PM
NonCommAlg
Quote:

$\displaystyle A_{mxn}, Q_{nxn}$
$\displaystyle \therefore$ rank(AQ) = rank A
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: $\displaystyle \text{rank}(AQ)=\dim ST(V) = \dim S(V)= \text{rank} A. \ \ \Box$