Consider the next theorem:

Let be $E$ is an $n$-dimensional vector space over $\mathbb R$ and $\alpha$ a 2-vector. Then there is a basis $\sigma_1,\sigma_2,\ldots,\sigma_n$ such that $$\alpha=\sigma_1 \wedge\sigma_2 + \sigma_3 \wedge \sigma_4 + \ldots + \sigma_{2r-1}\wedge\sigma_{2r}$$

One can prove this theorem in the following way. $\alpha$ represents a skew-symmetric bilinear form and its matrix looks like a Jordan normal matrix in suitable basis $\sigma_1,\sigma_2,\ldots,\sigma_n$. I'd like to know, whether one can prove the theorem above without referring to Jordan theorem and deduce the Jordan theorem from this fact.

Thanks in advance.


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