Proof that $\operatorname{rank}(SAT)= \operatorname{rank}(A)$

Question

Let A be $m\times n$ matrix, $S$ nonsingular $m\times m$ matrix and $T$ nonsingular $n\times n$ matrix. Prove that $\operatorname{rank}(SAT) = \operatorname{rank}(A)$.

I know

$\operatorname{rank}(SAT) \leq \min\{\operatorname{rank}(S), \operatorname{rank}(A), \operatorname{rank}(T)\}$

$\iff \operatorname{rank}(SAT) \leq \min\{m, \operatorname{rank}(A), n\}$

and since the rank of $A$ is at most $\min\{m,n\}$:

$\operatorname{rank}(SAT) \leq \operatorname{rank}(A)$.

But how do I prove that $\operatorname{rank}(SAT)$ is exactly $\operatorname{rank}(A)$?

It just seems intuitive and logical to me but I can't put it into actual proof.

Answer 1

Going the other way around, we have $$\operatorname{rank}(A) = \operatorname{rank}(S^{-1}SATT^{-1}) \leq \min\{\operatorname{rank}(S^{-1}),\operatorname{rank}(SAT),\operatorname{rank}(T^{-1})\}\leq \operatorname{rank}(SAT).$$