Basis for the Dual Space Involving Wedge Products I'm pretty much stuck on the following problem. 

Let $V$ be an $n$-dimensional vector space, and let $\omega\in\Lambda^{2}(V^{*})$. Show that there is a basis $\{e^{1},e^{2},\ldots,e^{n}\}$ of $V^{*}$ such that $$\omega=e^{1}\wedge e^{2}+e^{3}\wedge e^{4}+\cdots+e^{2r-1}\wedge e^{2r}$$ for some $r$. Furthermore, show that $\omega^{r}\neq 0$, but that $\omega^{r+1}=0$.

I know that $\omega$ is a skew-symmetric bilinear form, and we showed in my class that if $\{e_{i}\}$ is a basis for $V$ with a dual basis $\{e^{i}\}$, then $\{e^{i_{1}}\wedge e^{i_{2}}~:~i_{1}<i_{2}\}$ is a basis for $\Lambda^{2}(V^{*})$. 
I'm completely new to wedge products and pretty much anything involving the dual space, so I'm completely lost and I'm not even sure where to start. Thanks in advance for any suggestions!
 A: Maybe it would help if we think of $\omega$ as the map
$$\omega : \mathbb R^n \times \mathbb R^n \to\mathbb R,
\\
\omega (v, v') \mapsto v^T M v',$$
where $M$ is an $n$-by-$n$ real skew-symmetric matrix.
Then we could use this fact about the spectral theory of skew-symmetric matrices to deduce that there exists an orthogonal matrix
$$ Q = [ u_1 | \dots | u_n ]$$ (where the orthonormal vectors $u_1, \dots, u_n$ are the columns of $Q$), and a block-diagonal matrix of the form
$$ \Sigma = \begin{bmatrix}
\begin{matrix}0 & \lambda_1\\ -\lambda_1 & 0\end{matrix} &  0 & \cdots & 0 \\
0 & \begin{matrix}0 & \lambda_2\\ -\lambda_2 & 0\end{matrix} &  & 0 \\
\vdots &  & \ddots & \vdots \\
0 & 0 & \cdots & \begin{matrix}0 & \lambda_r\\ -\lambda_r & 0\end{matrix} \\
& & & & \begin{matrix}0 \\ & \ddots \\ & & 0 \end{matrix}
\end{bmatrix},$$
with $\lambda_1, \dots , \lambda_r > 0$ such that
$$ M = Q M Q^T.$$
So if we define the dual basis of $V^\star$ like this:
$$ e^1(v) = \sqrt{\lambda_1} u_1^T v, \ \ \ \ \ e^2(v) = \sqrt{\lambda_1} u_2^T v,\\
e^3(v) = \sqrt{\lambda_2} u_3^T v, \ \ \ \ \ e^4(v) = \sqrt{\lambda_2} u_4^T v,\\
\vdots\\
e^{2r-1}(v) = \sqrt{\lambda_r} u_{2r-1}^T v, \ \ \ \ \ e^{2r}(v) = \sqrt{\lambda_r} u_{2r}^T v, \\
e^{2r+1}(v) = u_{2r + 1}^T v, \\ \ \ \ \ \vdots \ \ \ \ \\ e^{n}(v) = u_n^T v,
$$
then we would have
$$\omega(v, v') = \left( e^1(v) e^2(v') - e^2(v) e^1(v') \right) + \dots  + \left( e^{2r - 1}(v) e^{2r}(v') - e^{2r}(v) e^{2r - 1}(v') \right) ,$$
which is to say that 
$$ \omega = e^1 \wedge e^2 + \dots e^{2r - 1} \wedge e^{2r}.$$
