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!