Prove that $\text{rank}(T) = \text{rank}([T]_{\beta}^{\gamma})$. Let $T:V\to W$ be a linear transformation between finite-dimensional vector spaces, and let $\beta$ and $\gamma$ be ordered basis for $V$ and $W$, respectively. Then $\text{rank}(T) = \text{rank}([T]_{\beta}^{\gamma})$.
MY ATTEMPT
Before presenting this theorem, the author defines $\text{rank}(A) = \text{rank}(L_{A})$, where $L_{A}:\textbf{F}^{n}\to\textbf{F}^{m}$ is the linear transformation defined by $L_{A}x = Ax$.
But we know that
$$\text{rank}(L_{A}) = \dim\text{span}\{L_{A}(e_{1}),L_{A}(e_{2}),\ldots,L_{A}(e_{n})\}$$
Moreover, $L_{A}(e_{j}) = a_{j}$, where $a_{j}$ is the $j$-th column of $A$. Consequently, $\text{rank}(A)$ equals the dimension of its column space.
Based on it, I do not know how to proceed. Am I on the right track?
This question has already been solved here, but the textbook has not presented the fact that invertible linear mappings preserve rank so far. Is there another way to solve it? Or should I insist a little bit more on it?
 A: Exercise. Let $V_1$ and $V_2$ two finite dimensional vector spaces over the same field. Then, if $f : V_1 \to V_2$ is a linear isomorphism, for any subspace $S$ of $V_1$, $f(S)$ is a subspace of $V_2$ such that $\dim f(S) = \dim S$.
Once you are proved this simple exercise, suppose $\beta = \{v_1,\dots,v_n\}$ and $\gamma = \{w_1,\dots,w_m\}$. Define $\phi_\beta : V \to \mathbf F^n$ by
$$\phi_\beta \Big( \sum_{i=1}^n a_iv_i \Big) = \sum_{i=1}^n a_ie_i = \begin{pmatrix}
a_1 \\ \vdots \\ a_n
\end{pmatrix},$$
and similarly for $\phi_\gamma : W \to \mathbf F^m$. It is easy to see that $\phi_\beta$ and $\phi_\gamma$ are linear isomorphisms and that the following diagram
$$
\require{AMScd}
\begin{CD}
V @>{T}>> W\\
@V{\phi_\beta}VV @VV{\phi_\gamma}V \\
\mathbf{F}^n @>>{L_A}> \mathbf{F}^m
\end{CD} \qquad \textrm{where } A := [T]_\beta^\gamma
$$
commutes, that is, $\phi_\gamma \circ T = L_A \circ \phi_\beta$. Then, if we take $V_1 = \mathbf F^m$, $V_2 = W$, $f = \phi_\gamma^{-1}$ and $S = L_A(\mathbf F^n)$ in the preceding exercise, $\phi_\gamma^{-1}(L_A(\mathbf F^n))$ is a subspace of $W$ with the same dimension of $L_A(\mathbf F^n)$, i.e., with dimension $\operatorname{rank}(A)$. But also, since $\phi_\beta$ is surjective, $\mathbf F^n = \phi_\beta(V)$, and then
$$\phi_\gamma^{-1}(L_A(\mathbf F^n)) = \phi_\gamma^{-1}(L_A(\phi_\beta(V))) = T(V).$$
