Madam, @TeresaLisbon: I have one small question: Why is $\Phi$ surjective in answer to the question linked above?

@Koro Oh, I see you got an answer! It was a generalization of a question whose proof I knew, so I was trying to get that through but this came before, which is good to know.

So the point is to note that the image of $\Phi$ is definitely a subspace of $F^m$ because the image of a vector space must be a vector space under a linear map. Now, what can the dimension of that vector space be? Suppose that it's NOT equal to $m$. Then, using the inner product on $F^m$, there is at least one vector $x = (f_1,...,f_m)$ which is orthogonal to every element of the image subspace. In particular, for every $v$, we have $x \cdot \Phi(v) = 0$ in $F^m$.

However, this is the same as saying that $\sum f_i \phi_i(v) = 0$ in $F^m$, which implies that $\sum f_i\phi_i = 0$ as a linear map into $F^m$, which contradicts the fact that $\phi_i$ are linearly independent.

Hence, we are done.

Obviously we used a quite strong fact here : that every strict subspace has an orthogonal vector corresponding to it, but I can't think of something easier immediately.

Madam, I got an answer to that too:).

@Koro Oh I see, that's good to know. Nevertheless, I hope this reply will be helpful as well.

Inner product has not yet been covered by this chapter.

I see. I do think there will be a simpler solution, I was probably going too hard at it.

@Koro Not sure I can find a solution, in fact. What is the answer that you got? I had some rough ideas but none too clean.

The idea was: If U is a proper subspace of a finite dimensional vector space V, then there exists a linear functional on V whose nullspace is U.

@TeresaLisbon madam, so using the above if $\Phi (V) $ were a proper subspace of $F^m$ then we should be able to find a linear functional f on $F^m$ such that f is not identically zero.

But this can be shown to result in a contradiction by showing that all such f must be zero on $ F^m$.