Annihilators and dual mapping

Question

Let $T'$ denote the dual map of $T$ where $T$ is a linear map from $V$ to $W$ and $V$, $W$ are both finite dimensional. Let $U^0$ denote the annihilator of the subspace $U$ of $V$.

I wanted to prove $(\ker T)^0=(\operatorname{Im}T')$ without using the fact that $\operatorname{Rank}(T)=\operatorname{Rank}(T')$. Right now, I am only stuck at the point of showing $(\ker T)^0\subset(\operatorname{Im}T').$ I saw this answer on Stackexchange earlier but it seems longer than expected. Since this was only a section of past paper question, I was wondering if there was a quicker way in showing so.

I thought about taking $f\notin \operatorname{Im}(T')$ and tried to show that $f\notin (\ker T)^0$ as well but I didn't get far with that one.

Answer 1

We have $T: V \to W$ and $T': W' \to V'$. Then: $$ \begin{aligned} f \in (\ker T)^0 & \iff f(v) = 0 \text{ for all } v \in \ker T \\ & \iff \ker T \subseteq \ker f \\ & \iff f = g \circ T \text{ for some } g \in W' \\ & \iff f \in \operatorname{im} T' \end{aligned}$$

The step that needs explaining is going between the conditions $\ker T \subseteq \ker f$ and $f = g \circ T \text{ for some } g \in W'$. Does this hint suffice?