# Linear algebra lemma

The following Lemma is in Beauville-Donagi, and I always took it for granted. Now I've tried to find a proof, but got stuck. They say it is a really simple lemma, so I may just be overlooking something easy.

Let $V$ be a vector space of dimension $6$, and let $W \subset \bigwedge^2 V^*$ be a subspace of dimension $2$. Assume that every form in $W$ is degenerate. Then there is a subspace $K \subset V$ of dimension $4$ such that each form in $W$ restricts to $0$ on $K$.

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Now, assume not. Pick two elements that span $W$. If either of them has 4-d kernel, it is 0 on any 4-d subspace, and we can use whichever on the other vanishes on.
Thus, every element of $W$ has 2-d kernel. If two elements had different kernels, then one of their linear combinations would have no kernel. Thus, they all kill the same 2-d subspace. Thus, we've reduced to the statement that any two 2-forms on a 4-d space $Z$ have a common Lagrangian subspace. Pick any line $L$; this is isotropic for both, since all lines are. Consider the intersections of the symplectic orthogonals of $L$ for the two 2-forms. These are 3-d, so their intersection is a 2-space. Now you win.