Suppose $V$ and $W$ are finite-dimensional, $T\in\mathscr L(V,W)$, and there exists $\phi\in V'$ such that range $T'$ = span $(\phi)$. Prove that null $T$ = null $\phi$.
I just need some assistance cleaning up my proof and then clarifying what to do at the end.
Let $v\in$ null $T$. Then $T(v)=0$. Since range $T'$ = span $(\phi)$, there exists some $\theta \in W'$ such that $T'(\theta)=\theta T=\phi$. Now, $\phi(v)=\theta T(v)=\theta(0)=0$. Thus, $v\in$ null$\phi$. This implies null $T \subset$ null $\phi$.
I am not sure what to do after this step to finally say that null $T$ = null $\phi$.
The direction that you are missing:
Assume that $v\in null(\phi)$.
Let $w'\in W'$. Then $w'(T(v))=T'(w')(v)$, by definition of $T'$.
Since $\phi$ generates the range of $T'$ we can write $T'(w')$ as $r\phi$.
Therefore, $w'(T(v))=T'(w')(v)=(r\phi)(v)=r\phi(v)=0$.
Since $w'$ was arbitrary, it follows that $T(v)=0$. $\leftarrow$ This simple way to prove that a vector is zero is extremely useful in linear algebra. Keep it always in your head.
This is, $v\in null(T)$.