Let $A \in M_{m \times n}(\mathbb{F})$, and let $\beta$ and $\gamma$ ordinated basis for $\mathbb{F}^n$ and $\mathbb{F}^{m}$ respectively. Let $B=[L_{A}]_{\beta} ^{\gamma}$ . Show $B=P^{-1}AQ$ where $P \in M_{m \times m}$ is a matrix such their $j$-th column is the $j$-th vector of $\gamma$ and $Q \in M_{n \times n}$ is a matrix such their $j$-th column is the $j$-th vector of $\beta$.
I know $L_{A}:\mathbb{F}^n \to \mathbb{F}^m$ where $L_{A}(v)=Av$ for every $v \in \mathbb{F}^n$. Then having $\beta= \lbrace \beta_{1}, \beta_{2},...,\beta_{n} \rbrace$ where $\beta_{i} \in \mathbb{F}^{n}$ for each $1 \leq i \leq n$. We can think each vector $\beta_{i}$ as $\beta_{i}=(\beta_{i_{1}},\beta_{i_{2}},...,\beta_{i_{n}})$ for every $1 \leq i \leq n$.
Then as $Q=\begin{pmatrix} \beta_{1}, \beta_{2},....,\beta_{n} \end{pmatrix}$ thinking the vectors $\beta_{i}$ vertically. Calculating $AQ$ we have $AQ=(aq)_{ij}$ where $$aq_{ij}=a_{i_{1}}\beta_{j_{1}}+a_{i_{2}}\beta_{j_{2}}+...+a_{i_{n}}\beta_{j_{n}}$$ where $1 \leq i \leq m$ and $1 \leq j \leq n$ Now Im struggled to end up the proof. I wold aprecciate any help proving the statement.
Expanding in terms of $\beta$, we have $$v = \sum_{k=1}^n v_k\beta_k = Q\begin{bmatrix}v_1\\\vdots\\v_n\end{bmatrix} = Q[v]_\beta$$ so $$L_A(v) = Av = AQ[v]_\beta.$$ On the other hand, we can expand $L_A(v)$ in terms of $\gamma$: $$L_A(v) = \sum_{k=1}^m w_k\gamma_k = P\begin{bmatrix}w_1\\\vdots\\w_m\end{bmatrix} = P[L_A(v)]_\gamma.$$ Equating and multiplying both sides by $P^{-1}$ we get $$[L_A(v)]_\gamma = P^{-1}AQ[v]_\beta,$$ therefore $[L_A]_\beta^\gamma = P^{-1}AQ$.