I want to prove that: $\textrm{If}\ P\ \textrm{invertible}\implies\textrm{rank}(PA)=\textrm{rank}(A),$ given that $A\in M_{m\times n}(F)$, $P\in M_{m\times m}(F)$. My textbook omits the details and I need some check about the correctness of my proof.
$\dagger$ Given a function $T$, $R(T)$ means the range of $T$, and $\textrm{Dim}(R(T))$ means the dimension of the range of $T$.
Proof.
Assume $P$ invertible.
$\begin{align}{} R(L_{PA})=R(L_PL_A)=L_PL_A(\mathsf{F}^n)&=L_P(\ L_A(\mathsf{F}^n)\ )\\ &=L_P(\ R(L_A)\ ), \end{align}$
Since $\textrm{rank}(PA)=\textrm{Dim}(R(L_{PA}))=\textrm{Dim}(L_P(R(L_A)),$ as long as I can show that $R(L_A)$ is a subspace of $\mathsf{F}^m$ and with the assumption that $P$ is invertible, then
$$\textrm{Dim}(\ L_P(R(L_A))\ )=\textrm{Dim}(R(L_A))=\textrm{rank}(A),$$
by the fact that $L_P$ is an isomorphism on $\mathsf{F}^m,$ and this complete the proof. Since $L_A$ is linear, $0\in R(L_A)$. Given $a,b\in R(L_A),$ there exist $x,y\in \mathsf{F}^n$ s.t. $L_A(x)=a,L_A(y)=b,$ but this implies that given any $c\in\mathsf{F}, L_A(cx+y)=c\cdot L_A(x)+L_A(y)=ca+b\in R(L_A),$ so $R(L_A)$ does be a subspace of $\mathsf{F}^m.$
Preliminary: $P$ is invertible iff $\mathsf{L}_P$ is invertible.
So $\mathsf{L}_{P}:\mathsf{F}^m\to\mathsf{F}^m$ is an isomorphism, and clearly $R(\mathsf{L}_A)$ is a subspace of $\mathsf{F}^m,$ and the isomorphism keeps its dimension, i.e. $\textrm{Dim}(\ \mathsf{L}_P(R(\mathsf{L}_A))\ )=\textrm{Dim}(R(\mathsf{L}_A))$, which complete the proof.