I'm presented with the following question:
Suppose $T:V \to W$ is a linear map and $V$ is finite dimensional. Prove Im $T'$ = (Ker $T$)$^0$.
I have showed that Im $T'$ $\subseteq$ (Ker $T$)$^0$:
Let $f \in$Im $T'$. Then for $v \in $ Ker $T$.
$$\begin{align} f(v) &= T'(g)(v) \text{ for some } g\in W'\\ &=g(T(v))\\ &=g(0)\\ &=0\end{align}$$ and so $f\in$(Ker $T$)$^0$ $\Rightarrow$ Im $T'$ $\subseteq$ (Ker $T$)$^0$
But when it comes to showing Im $T'$ $\supseteq$ (Ker $T$)$^0$, I am not sure how to proceed. I'm quite sure I need to use the finite dimensionality of $V$ somewhere, but I'm not sure where.
One way to do things is by counting dimensions.
In particular, let $r$ be the dimension of the image of $T$, and let $n$ be the dimension of $V$. By the rank-nullity theorem, $\ker T$ has dimension $n - r$. Then, we note that $(\ker T)^0$ must have dimension $r$ (you may have to prove this yourself).
On the other hand, note that $T'$ has the same rank as $T$, which means that $\operatorname{Im}T'$ is $r$-dimensional.
Now, by your work, $\operatorname{Im}T'$ is an $r$-dimensional subspace of $(\ker T)^0$, which is also $r$-dimensional. It follows that $\operatorname{Im}T' = (\ker T)^0$, which was the desired conclusion.