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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...

Hyacinth
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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.

GReyes
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  • I'm probably just confused, and maybe this is just a notation issue, but since $M^$ seems to be defined only as homomorphisms (i.e. not necessarily bounded), is there an issue? It is definitely not true in infinite dimensions if instead $M^$ is the set of continuous linear functionals (I think $V=W=\ell^2$, and $A$ maps $(x_n)\mapsto (x_n/n)$ works). But here, $M^*$ seems to be defined just with respect to the vector space structure, not any topological structure, so I don't see why boundedness is the problem... – J.G Jan 16 '20 at 17:54
  • The problem I see is how you extend (in the infinite dimensional case) a functional (one-form) defined on a subspace to the whole space. Typically you add new basic vectors and you define your extended functional on the extended subspace, etc, but this process has to converge when the number of dimensions goes to infinity. I do not see how you can deal with this unless you assume boundedness and resort to Hahn-Banach. In a general vector space, you have a Hamel basis and you can definitely add one dimension at a time to extend the form, but can you repeat the process infinitely many times? – GReyes Jan 16 '20 at 19:09
  • Would the same extension argument in Hahn-Banach work, just without using the sublinear part of H-B? Namely define a partial order on linear extensions, show every chain has an upper bound in the usual way, so there is a maximal element, then show it must be defined on the entire vector space as otherwise you could extend another dimension? This clearly need not give any bounded thing in general, but would it not give a well-defined linear extension on the whole space? – J.G Jan 16 '20 at 20:26
  • Actually, I do not see any flaw in your argument. You get a purely algebraic result relating subspaces of the duals. Will keep thinking about it.. – GReyes Jan 17 '20 at 03:51