Suppose $V$ and $W$ are finite dimensional vector spaces over a field $F$. Let $T : V \to W$ be linear. Prove that $\operatorname{Im} ((T'))^0 \subseteq \phi(\operatorname{Ker}(T))$, where $\phi$ is the canonical isomorphism from $V \to V''$ and $T'$ is the adjoint of $T$.
Any idea on how to proceed here? Many thanks!
Let $v \in V$ such that $\phi(v) \in (\operatorname{im} T')^0$, that is, $\operatorname{im}T' \subseteq \ker\phi(v)$. Then, for any $f \in W'$ we have $$f(T(v)) = T'(f)(v) = \phi(v)(T'(f)) = 0 \tag{$*$}$$ meaning that$^1$ $T(v) = 0_W$. So, we are started with $v \in \phi^{-1}((\operatorname{im} T')^0)$ and we are showed that $v \in \ker T$; thus $\phi^{-1}((\operatorname{im} T')^0) \subseteq \ker T$ and then$^2$ $$(\operatorname{im} T')^0 = \phi(\phi^{-1}((\operatorname{im} T')^0)) \subseteq \phi(\ker T).$$
$^1$ If $T(v) \neq 0_W$, then (since $W$ is finite dimensional) we can choose a basis $\beta$ of $W$ whose first element is $T(v)$, and then define $f : W \to F$ by $f(T(v)) = 1$, $f(w) = 0$ for every $w \in \beta \setminus \{T(v)\}$, and extend it by linearity. This contradicts $(*)$.
$^2$ Here is where I used that $V$ is finite dimensional.