Exercise 2.4.20

With the notation in Figure 2.2 we can prove first that ${\varphi }_{\gamma }(R(T))=L_A({𝔽}^n)$. Since ${\varphi }_{\beta }$ is surjective we have that

Since $R(T)$ is a subspace of $W$ and ${\varphi }_{\gamma }$ is an isomorphism, we have that rank$(T)=$rank$(L_A)$ by Exercise 2.4.17.

On the other hand, we may prove that ${\varphi }_{\beta }(N(T))=N(L_A)$. If $y\in {\varphi }_{\beta }(N(T))$, then we have that $y={\varphi }_{\beta }(x)$ for some $x\in N(T)$ and hence

Conversely, if $y\in N(L_A)$, then we have that $L_A(y)=0$. Since ${\varphi }_{\beta }$ is surjective, we have $y={\varphi }_{\beta }(x)$ for some $x\in V$. But we also have that

and $T(x)=0$ since ${\varphi }_{\gamma }$ is injective. So similarly by Exercise 2.4.17 we can conclude that nullity$(T)=$nullity$(L_A)$.