For $T: X \to Y$ with $T$ having finite dimensional image I'm trying to show: $$(\ker(T))^{\circ} = \operatorname{Im}(T')$$ Where $T'$ is the dual operator.
I've shown that if we take an element in $\operatorname{Im}(T')$ this is zero on the kernel but I'm struggling to show the other inclusion. I know that I'm going to have to use finite dimensionality of the image of $T$ and I'm looking for a $g$ so that $f = g \circ T$ whenever $f \in (\ker(T))^{\circ}$.
Thanks for any help