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Let $V$ be a finite dimensional vector space over $F$ and $T:V \rightarrow W$ linear transformation. Prove that $\text{im}(T^t)=(\ker T)^0$ where $T^t$ is the dual/transpose of $T$ and $(\ker T)^0$ is the annihilator of the kernel.

I have so far proved that $\text{im}(T^t) \subseteq (\ker T)^0$ but I have difficulty proving the other direction. Given that $f \in (\ker T)^0$, how do I construct $g \in W^*$ such that $f=T^t(g)$?

Here is a link to an exactly same question: Image of dual map is annihilator of kernel. I don't quite understand the answer using the first isomorphism theorem. Is there another way to prove it? How should I make sense of the answer in the link? Thanks!

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user112358
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For any subspace $U$ of finite-dimensional $V$, $\dim U^0=\dim V-\dim U$. We also know that the ranks of $T$ and $T^t$ are equal, so $$\dim\operatorname{im}T^t=\dim\operatorname{im}T=\dim V-\dim\ker T=\dim(\ker T)^0,$$ and you’ve already shown that $\operatorname{im}T^t\subseteq(\ker T)^0$, hence $\operatorname{im}T^t=(\ker T)^0$.

amd
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  • I'm having trouble understanding why having the dimension be equal to each other and one being a subset of the other gives the result we seek. – OrangeApple3 Feb 01 '16 at 13:02
  • @aznluster That’s a basic result for vector spaces: if $U$ is a subspace of $V$ and $\dim U=\dim V$, then $U=V$. Otherwise, we could find a linearly independent set of $\dim V+1$ vectors in $V$, which would be a contradiction. – amd Feb 01 '16 at 18:26