1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your definition of "rank" is? $\endgroup$ – Vim Oct 8 '17 at 16:26
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.