I'm trying to show the following:
Let $M$ be a $R$ vector space, define $M^*=Hom_F(M,R)$. Let $V,W$ be finite dimension $R$ vector spaces, $A:V\to W$ be linear, define $A^*:W^*\to V^*$ to be $A^*(f)=f\circ A$. Let $N$ be a subspace of a vector space $M$, define $N^0$ to be $\{f\in M^*:\forall n\in N, f(n)=0\}$.
Show $Im(A^*)=(Ker(A))^0$.
My argument:
Note $Im(A^*)\subseteq (Ker(A))^0$ is trivial.
To show the other containment, let $f\in (Ker(A))^0$, then $\phi(v+Ker(A))=A(v)$ is an isomorphism. Thus $\phi^{-1}:Im(A)\to V/Ker(A)$ is well-defined. Note the kernal of A is a subset of the kernal of f, so $F:V/Ker(A)\to R, F(v+Ker(A))=f(v)$ is well-defined homomorphism. Thus, we have $A^*(F\circ \phi^{-1})(v)=f(v)$ for all $v\in V$ and so the annihilator of the kernal is a subset of Im(A^*).
This argument seems like did not use finite dimension, and my prof says it is wrong with infinite dimension, thus, I think I probably did something wrong in the proof, but where is it?
Remark: My solution is the same as the proof given in the link Image of dual map is annihilator of kernel. However, it did not use finite dimension as well...
You need to incorporate some topological considerations.
As DanielWainfleet mentioned, your one-form $F\circ\phi^{-1}$ is only defined on $Im(A)$, so you need to extend it. In order to extend it, you use Hahn-Banach in the infinite-dimensional case, and this requires your form to be bounded (you cannot extend merely linear forms to linear forms on all the space). For $\phi^{-1}$ to be bounded, you can assume, for example, that $Im(A)$ is closed and use the open mapping theorem. For $F\circ \phi^{-1}$ to be bounded you need $A$ to be a bounded operator. Then it extends to a bounded one-form on the whole space and your reasoning works. Observe that it follows, in particular, that in this case (that is, under the assumption that $Im(A)$ is closed and $A$ is bounded), $Im(A*)$ is also closed, being an orthogonal complement. If you do not assume $Im(A)$ to be closed, all you can say is that $$ \overline{Im(A^*)}=(Ker(A))^{\perp}. $$ which is an existence result (Fredholm alternative) not as strong as its finite-dimensional analog.
BTW for all this to work you need your spaces to be Banach spaces.