TITLE: Linear algebra lemma
QUESTION [5 upvotes]: 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 $K$.

REPLY [3 votes]: First note that if a 2-form is degenerate, it is 0 on some 4-subspace (take a lagrangian subspace of the quotient by the kernel).
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.