Can every 2 form be represented as a linear combination of these specific two forms? 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.
 A: I think you are encountering an issue with your calculation due to a misreading of the problem.  The way I understand it, the question is, for any $2$ form $\omega$, does there exist a basis $\sigma_i$ such that
$$\omega(u, v) = \sigma_1\wedge\sigma_2(u, v)+\dots +\sigma_{2r-1}\wedge\sigma_{2r}(u, v)$$
In your example, it is fine to start with a basis $\sigma_i$ and consider the two form: $\omega = \sigma_1 \wedge \sigma_3$.  However, you are then allowed to choose another basis $\sigma_1, \sigma_3, \sigma_2, \sigma_4, \ldots$ such that the equation is satisfied.  In this case:
$$\omega(u, v) = \sigma_1\wedge\sigma_3(u, v)$$
where we choose $r = 1$.
A concrete way to proceed with the problem is to choose a basis $e_i$ and consider an arbitrary two form in that basis:
$$\omega = \sum_{i < j} \alpha_{ij} e_i \wedge e_j$$
and ask, do there exists linearly independent vectors of the form
$$\sigma_i = \sum_j \beta_{ij} e_j$$ 
such that
$$\omega(u, v) = \sigma_1\wedge\sigma_2(u, v)+\dots +\sigma_{2r-1}\wedge\sigma_{2r}(u, v)$$
for some $r$.
Examples: 
$$\alpha_{12}e_1 \wedge e_2 + \alpha_{13}e_1 \wedge e_3 = e_1 \wedge (\alpha_{12}e_2 + \alpha_{13}e_3)$$
$$e_1 \wedge e_2 + e_1 \wedge e_3 + e_2 \wedge e_3 = e_1 \wedge (e_2 + e_3) + e_2 \wedge (e_2 + e_3) = (e_1 + e_2) \wedge (e_2 + e_3)$$
