What is the rank of the product of two full rank matrices?

Question

Let $S$ be an $m \times n$ matrix with $m<n$ and rank $m$. Also, let $A$ be a $n \times d$ matrix with $n > d$ and rank $d$. Given that $m > d$, is the matrix $SA$ full rank with rank $d$ ?

Answer 1

Hints: $\DeclareMathOperator{\rank}{rank}\rank(S)+\rank(A)-n \le \rank(SA) \le \min\{\rank(S),\rank(A)\}$.

Edit: $\text{karate_kid} $, no, $\rank(SA) $ may not be $d$, I give the hints to show you $\rank(SA)$ is not exactly $d$ always, as, $(m-n)+d \le \rank(SA) \le d $ , where $(m-n) \lt 0 $,,, so, $\rank(SA) $ runs in this $[d-(n-m), d ]$

For $\rank(SA)=0 \neq 1=d $ , take $S=\begin{pmatrix}1&0 &0\\ 0&1&0\end{pmatrix}$ $A=\begin{pmatrix} 0\\0\\1\end{pmatrix} $ $SA=\begin{pmatrix} 0\\0\end{pmatrix} $

For , $\rank(SA) =1= d $ ,take, $S=\begin{pmatrix}1&0 &0\\ 0&1&0\end{pmatrix}$ $A=\left(\begin{array} {c}1\\0\\1\end{array}\right)$ $SA=\left(\begin{array} {c}1\\0\end{array}\right)$

Answer 2

$S=\left(\begin{array}{ccc}1&0 &1\\ 0&1&1\end{array}\right)$

$A=\left(\begin{array} {c}1\\1\\-1\end{array}\right)$