A A problem in multilinear algebra

[steenis]

TL;DR

A one-to-one correspondence between bilinear non-degenerate maps and invertible linear maps

I have the following problem in multilinear algebra:
Let $W$ and $V$ be real finite-dimensional vector spaces, $V^*$ is the dual space of $V$
Let $L:W \times V \rightarrow \mathbb{R}$ be a non-degenerate bilinear map
Define $g:W \rightarrow V^*$ by $g(w)(v)=L(w,v)$
To prove: $g$ is an isomorphism

The map $g$ is obviously linear
$g$ is injective, because if $g(w)=0$ for $w \in W$, then for all $v \in V$ we have that $g(w)(v)=L(w,v)=0$, so, because $L$ is non-degenerate, we have $w=0$
Remains to prove that $g$ is surjective

If $dim W=dim V=n$, then we are ready, because in that case $dim W=dim V^*=n$ and $g$ is surjective if and only if $g$ is injective. On the other hand, if $g$ is bijective, then $g$ is an isomorphism and $dim W=dimV=dim V^*$. However, it is not a-priori given that $W$ and $V$ have the same dimension

Can anybody help me with this problem, can the statement be proven or is the statement not true?

 

[fresh_42]

You are correct. The difficulty is that we only can conclude $\dim W \leq \dim V$.
What happens if you consider $h\, : \,V\longrightarrow W^*$ with $h(v)(w):=L(w,v)$?

 

[steenis]

In the same way, $h$ is injective and therefore $dim V \leq dim W$. Now we have $dim W=dim V=dim V^*$, thus the map $g$ is surjective and therefore an isomorphism ($h$ is also an isomorphism)