# 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!

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 $\rank(\psi\circ T)=\rank(T)$.
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 $$\DeclareMathOperator{nullity}{nullity}\nullity(T)= \dim W-\rank(T) = \dim W-\rank(L_A) = \nullity(L_A)$$