Prove $\operatorname{rank}(A)=\operatorname{rank}(PAQ)$ , when $A$ is a $m\times n$ matrix and $P$ and $Q$ are invertible.

Question

Prove $\operatorname{rank}(A)=\operatorname{rank}(PAQ)$ , when $A$ is a $m\times n$ matrix and $P$ and $Q$ are invertible.

So I've figured I can solve this in two steps, proving that $\operatorname{rank}(A)=\operatorname{rank}(AQ)$ and then that $\operatorname{rank}(A)=\operatorname{rank}(PA)$ . In my linear algebra book they go the route of $$ R(L_{AQ})=R(L_AL_Q)=L_AL_Q(F^n)=L_A(L_Q(F^n))=L_A(F^n)=R(L_A) $$ Since $\operatorname{rank}(AQ)=\dim(R(L_{AQ})$ we have now proven the first half.

What I don't understand is how $L_Q(F^n) = F^n $ and also why $R(L_AL_Q)=L_AL_Q(F^n)$ .

My guess is that if you enter $F^n$ into the transformation that is $L_Q$ you will return $F^n$ no matter what $Q$ is. If that is the case then does $L_A(F^n)$ also produce $F^n$ ?

Answer 1

1. Since $Q$ is invertible, the mapping $L_Q: F^n \to F^n$ is bijective, hence $L_Q(F^n) = F^n$.

2. $R(L_AL_Q)=(L_AL_Q)(F^n)$

Answer 2

Note that, in general, $\text{rank}(AB) \leq \text{min}(\text{rank}(A), \text{rank}(B))$. Thus:

$\text{rank}(A) = \text{rank}((AQ)Q^{-1}) \leq \text{rank}(AQ) = \text{rank}(P^{-1}(PAQ)) \leq \text{rank}(PAQ) = \text{rank}((PA)Q) \leq \text{rank}(PA) \leq \text{rank}(A).$

We have showed that $\text{rank}(A) \leq \text{rank}(PAQ) \leq \text{rank}(A)$, i.e. $\text{rank}(A) = \text{rank}(PAQ)$.