Image of dual map is annihilator of kernel Suppose $T:V\to W$ and that $V$ is finite-dimensional.
I want to prove that $$\text{Im }T'=(\ker T)^0$$ where $T'$ is the dual/transpose map and $(\ker T)^0$ is the annihilator of the kernel.
I know that $\phi \in V'$ is an annihilator of $\ker T$ if and only if $$\phi(v)=0 \space \forall v\in \ker T$$ if and only if $$\phi(v)=0 \space \forall v \in V \text{ such that } T(v)=0$$
Now I also know that $\phi \in \text{Im }T'\subset V'$ if and only if $$\phi = f \circ T \text{ for some } f \in W'$$
I can see that if $v\in \ker T$, then this implies $\phi(v) = f(T(v)) = 0$, so $\phi \in (\ker T)^0.$
However, I'm having trouble showing the other way, that if $\phi \in (\ker T)^0$, then $\phi \in \text{Im }T'.$
Could anybody help?
 A: By the first isomorphism theorem the map $T$ factors over $V/\ker T$ as
$$V\ \stackrel{\pi}{\longrightarrow}\ V/\ker T\ \stackrel{\overline{T}}{\longrightarrow}\ \operatorname{im}T,$$
where $\pi$ is the canonical quotient map and $\overline{T}$ is an isomorphism. Note that $\pi=\overline{T}^{-1}\circ T$.
If $\phi\in(\ker T)^{\circ}$ then $\ker T\subseteq\ker\phi$. This means $\phi$ also factors over $V/\ker T$, as
$$V\ \stackrel{\pi}{\longrightarrow}\ V/\ker T\ \stackrel{\psi}{\longrightarrow}\ F,$$
where $F$ denotes the base field, and $\psi\in(V/\ker T)'$. It follows that
$$\phi=\psi\circ\pi=\psi\circ(\overline{T}^{-1}\circ T)=(\psi\circ\overline{T}^{-1})\circ T,$$
where of course $\overline{T}^{-1}$ is an isomorphism, hence $\psi\circ\overline{T}^{-1}\in(\operatorname{im}T)'$. Because $\operatorname{im}T$ is a linear subspace of $W$, the linear functional $\psi\circ\overline{T}^{-1}$ on $\operatorname{im}T$ extends to a linear functional $f$ on $W$, which then satisfies
$$\phi=f\circ T,$$
by construction, as desired. Note that $f$ is far from unique in general.
A: Because you already have that $(\ker T)^\circ\subseteq \text{im } T^*$ you can simply show that their dimensions must be equal and thus we know that they must be equal. We know that $$\dim V^*=\dim V=\dim\; \ker T+\dim\; \text{im }T$$ and that $$\dim V^*=\dim\; \text{im } T^*+\dim\; (\text{im } T^*)^\circ.$$ 
It is easy to show that $\text{rank } T^*=\text{rank }T.$ Thus we can re-write the equality and it follows immediately that $$\dim \ker T=\dim \; (\text{im } T^*)^\circ.$$ Therefore, the two must be equal. 
