This question is Question 2 from Ilka's book on page 8.

The first part is to prove that every $\omega^2\in \Lambda^2(V^{\ast})$ can be represented as \begin{equation*}\tag{1} \omega^2=\sigma_1\wedge\sigma_2+\dots +\sigma_{2r-1}\wedge\sigma_{2r} \end{equation*}for a certain basis $\sigma_1,\dots,\sigma_n$ of $V^{\ast}$.

I am having some difficulty with this. I tried to see how this might work for the two form: \begin{equation*} \sigma_1\wedge\sigma_3: V\times V\rightarrow\mathbb{R}. \end{equation*}Given $(u, v)\in V\times V$, we have on the one hand \begin{equation*} \sigma_1\wedge\sigma_3(u, v)=\sigma_1(u)\sigma_3(v)-\sigma_1(v)\sigma_3(u), \end{equation*}and if $(1)$ is to be satisfied we require that the RHS of this equation equals \begin{equation*} \sigma_1\wedge\sigma_2(u, v)+\dots +\sigma_{2r-1}\wedge\sigma_{2r}(u, v) \end{equation*} for some $r$. Now since the highest index on the left hand side is 3, I thought this would imply that $r=2$ would be a good candidate. However, this implies that \begin{equation*} \sigma_1(u)\sigma_3(v)-\sigma_1(v)\sigma_3(u)=\sigma_1(u)\sigma_2(v)-\sigma_1(v)\sigma_2(u)+\sigma_3(u)\sigma_4(v)-\sigma_3(v)\sigma_4(u). \end{equation*}I'm not sure how to proceed further without defining one of the forms as a linear combination of the others. This would be wrong because as basis vectors they must be linearly independent.