Proving that $\mathrm{rank}(T) = \mathrm{rank}(L_A)$ and $\mathrm{nullity}(T) = \mathrm{nullity}(L_A)$, where $A=[T]_\beta^\gamma$. 
Let $T:V \to W$ be a linear transformation from an $n$-dimensional vector space to an $m$-dimensional vector space $W$. Let $\beta$ and $\gamma$ be ordered bases for $V$ and $W$, respectively. Prove that $\operatorname{rank}(T) = \operatorname{rank}(L_A)$ and that $\operatorname{nullity}(T) = \operatorname{nullity}(L_A)$, where $A=[T]_\beta^\gamma$.

My attempt:
We know $T = \phi_\beta L_A \phi_\gamma^{-1}$. We are guaranteed $\phi_\gamma^{-1}$ exists since it is an isomorphism and, therefore, a bijection. Since both $\phi_{\beta,\gamma}$ are both isomorphisms their nullity is $0$.


*

*So from here the nullity of $\phi_\beta L_A \phi_\gamma^{-1}$ should be the nullity of $L_A$?

*If that's the case then the rank of $\phi_\beta L_A \phi_\gamma^{-1}$ is the rank of $L_A$? Since $\phi_\beta L_A \phi_\gamma^{-1} = T$ then $\operatorname{rank}(L_A)=\operatorname{rank}(T)$?


The details are not coming together here but I feel this might be the right direction.
Thanks!
 A: Lemma. Let $T:V\to W$ be a linear map of finite-dimensional vector spaces. Let 
\begin{align*}
\phi&:X\to V & \psi&:W\to Y
\end{align*}
be linear isomorphisms. Then $\DeclareMathOperator{rank}{rank}\rank(T\circ \phi)=\rank(T)$ and $\rank(\psi\circ T)=\rank(T)$.
Proof. One checks that the maps
\begin{align*}
\begin{array}{ccc}
\DeclareMathOperator{image}{image}\image(T)&\xrightarrow{\Phi}&\image(T\circ\phi)\\
T(v) & \mapsto & \phi^{-1}(v)
\end{array}&&
\begin{array}{ccc}
\DeclareMathOperator{image}{image}\image(T\circ \phi)&\xrightarrow{\Psi}&\image(T)\\
T(\phi(x)) & \mapsto & T(\phi(x))
\end{array}
\end{align*}
are well-defined linear isomorphisms. 
Recall that the rank of a linear map $S$ is defined as $\rank(S)=\dim\image(S)$. Also recall that dimension is invariant under isomorphism. It then follows that
$$
\rank(T\circ\phi)=\dim\image(T\circ\phi)=\dim\image(T)=\rank(T)
$$
The proof that $\rank(\psi\circ T)=\rank(T)$ is similar. $\Box$
That $\rank(T)=\rank(L_A)$ then follows from the lemma since $T\circ\phi_\gamma=\phi_\beta\circ L_A$. 
The rank-nullity theorem then implies that
$$
\DeclareMathOperator{nullity}{nullity}\nullity(T)=
\dim W-\rank(T)
= \dim W-\rank(L_A)
= \nullity(L_A)
$$
