Annihilator of the Kernel is equal to image of the dual map

Question

I'm revising Linear Algebra and am stuck on this question.

Supposing that $T:V \rightarrow W$ is a linear map, and that V is finite dimensional, I want to prove that $Im(T')=(Ker(T))^0$.

I know that $T':W' \rightarrow V'$ so that $Im(T') \subseteq V'$ and that this map is defined by $f \mapsto f\circ T$ and furthermore that $(Ker(T))^0$= {$f \in V': f(x)=0, \forall x \in Ker(T)$} but I am not sure how to prove they are equal.

If I can prove one is a subset of the other, I can prove their dimensions are equal quite easily:

$dimIm(T')=dim(ImT) = dimV-dimKer(T) = dim((Ker(T))^0)$

So I'm just stuck about how to prove one is a subset of the other.

Any help appreciated-thanks!

Answer 1

Since the have the same dimension, you only need to prove that $Im(T')\subset Ker(T)^0$.

Let $g\in Im(T')=g=f\circ T$. For every $x\in Ker T$, $g(x)=f(T(x))=0$ implies that $g\in Ker(T)^0$, you deduce that $Im(T')\subset Ker(T)^0$ since $dim(Im(T'))=dim(Ker(T)^0)$, you deduce that $Im(T')=Ker(T)^0$.

Answer 2

Let $g\in\operatorname{im}(T')$ so that $g=f\circ T$ for some linear functional $f$ on $W$. Then for all $u\in\ker T$ you have $$g(u)=(f\circ T)(u)=f(T(u))=f(0)=0,$$ so $g\in(\ker(T))^{\circ}$. This shows that $\operatorname{im}(T')\subset(\ker(T))^{\circ}$.