Let $\omega\in\bigwedge^2(V)$ where $V$ is a real vector space with $\text{dim }V=n=2m$. There exists a basis $\{e_1,f_1,e_2,f_2,\dots,e_m,f_m\}$ of $V$ for which $\omega(e_i,e_j)=0=\omega(f_i,f_j)$ and $\omega(e_i,f_j)=\delta_{ij}$.
Suppose $\omega$ is non-degenerate; that is, the map $V\to V^\ast$ sending $X$ to $i_X\omega$ is an isomorphism (where $i_X$ is the so-called interior multiplication map). Since $\{e_1,f_1,e_2,f_2,\dots,e_m,f_m\}$ is a basis of $V$, the collection $\{\varepsilon^1,\delta^1,\varepsilon^2,\delta^2,\dots,\varepsilon^m,\delta^m\}$ is a basis of $V^\ast$ where $\varepsilon^i=i_{e_i}\omega$ and $\delta^i=i_{f_i}\omega$ as the map $V\to V^\ast$ sending $X$ to $i_X\omega$ is an isomorphism. How can I show $$ \omega=\varepsilon^1\wedge\delta^1+\varepsilon^2\wedge\delta^2+\cdots+\varepsilon^m\wedge\delta^m? $$
This is my attempt at proving the following:
For a non-degenerate $\omega\in\bigwedge^2(\omega)$ there exists exists a basis $\{\varepsilon^1,\delta^1,\varepsilon^2,\delta^2,\dots,\varepsilon^m,\delta^m\}$ of $V^\ast$ such that $$ \omega=\varepsilon^1\wedge\delta^1+\varepsilon^2\wedge\delta^2+\cdots+\varepsilon^m\wedge\delta^m. $$