I came upon the following statement: let $V$ be a vector space of dimension $2m \geq 4$ and $\omega \in \Lambda^2(V^*)$ a non-degenerate (i.e. $\left( \omega(x,y)=0, \forall y \in V \right) \implies x=0$) $2-$form on $V$. Then the map $\omega \wedge (\cdot): \Lambda^1(V^*) \rightarrow \Lambda^{3}(V^*)$ is injective.
To prove this, I'm trying to use a result from symplectic geometry which says that $\displaystyle \omega = \sum\limits_{i=1}^m e^i \wedge f^i$, for a particular basis $e^1, \ldots, e^m, f^1, \ldots, f^m$ of $V^*$. Taking then $\displaystyle \alpha = \sum_j c_j e^j + \sum_k d_kf^k$, we have that $\omega \wedge \alpha = 0$ iff $$ \sum_{i,j} c_j e^i \wedge f^i \wedge e^j + \sum_{i,k} d_k e^i \wedge f^i \wedge f^k = 0. $$ I don't know where to take it from here; maybe it would work by painstakingly taking every possible combination such as $(e_r, e_s, e_p)$, then $(e_r, e_s, f_p)$ etc. of vectors corresponding to those in the dual, but is there a cleverer way to see that the coefficients of $\alpha$ are $0$? Is writing $\omega$ in this form even necessary? If you know of a quicker proof to be found somewhere, it would also be appreciated.