I have come across the following claim:
Assume a skew-symmetric bilinear form $\omega: V \times V \rightarrow \mathbb{R}$ is symplectic if, by definition, $\forall v \in V\setminus \left\lbrace 0_V \right\rbrace: \exists u \in V: \omega(u,v) \neq 0$ (non-degenerate). Then:
$\omega$ is symplectic $\Longleftrightarrow$ $\omega^{\flat}: V \rightarrow V^* $ is an isomorphism, where $\omega^{\flat}(v)(u)=\omega(v,u)$.
I am stuck at proving the $"\Longrightarrow"$ implication, even though it seems trivial. Linearity is obvious from bilinearity of $\omega$. I have tried reasoning as such:
If $\omega^\flat(v_1)=\omega^\flat(v_2)$, then $\omega(v_1-v_2, u)=0 \ \forall u \in V$, and, by non-degeneracy, $v_1=v_2$, hence injectivity.
For surjectivity, let $\left\lbrace e_1, \ldots, e_m \right\rbrace$ be a basis in $V$ and $\left\lbrace e^1, \ldots, e^m \right\rbrace$ its' dual basis. It is enough to show that $\forall i \in \left\lbrace 1, \ldots, m \right\rbrace: \exists v_i \in V \mathrm{s.t.} \ \ \omega^\flat(v_i)=e^i$. Hence I want vectors $v_i$ with the property that $\omega(v_i, e_j)=\delta_{ij}$. By nondegeneracy, I can pick a $v_i'$ with $\omega(v_i', e_i) \neq 0$ and then set $v_i = \left( \omega(v_i', e_i) \right)^{-1} v_i'$ and hence $\omega(v_i, e_i)=1$.
However, and this is where I am stuck, I can't seem to find a way to pick the $v_i$'s such that $\omega(v_i,e_j)=0$ for any $j \neq i$. I am sorry if it's trivial and I am missing something obvious. I'm aware that I haven't yet used that $\left\lbrace e_1, \ldots, e_m \right\rbrace$ is a basis, but I can't see how.
Motivation for the question is that it appears in some proofs of the normal basis for any symplectic form - I was studying the one in Bryant's Lie Groups and Symplectic geometry and the claim is also made here http://pi.math.cornell.edu/~web6630/S186630Lec1.pdf. I also found a proof which doesn't use this claim here but the claim still seems useful.
Thank you for your time