TITLE: A Question on Exterior Forms
QUESTION [6 upvotes]: For the last few days, I have been trying to answer the following algebraic question in exterior algebra. The following question appears as an algebraic step in the context of existence of solutions of a certain system of PDE. I have asked a special case of the problem in 
Link: Inequalities Involving Wedge Product (Reference Request) 
with the hope that this particular case will do what I have in mind, but this did not turn out to be the case. Any help in this direction is welcome.
QUESTION: 
Let $N$ be a linear subspace of $\Lambda^2(\mathbb{R}^n)$ satisfying
$$
\omega\wedge\omega\neq 0,\text{ for all }\omega\in N,\omega\neq 0.
$$
Is it true that there exists a $a\in\Lambda^4(\mathbb{R}^n)$, $a\neq 0$ such that
$$
\langle a;\omega\wedge\omega\rangle>0,\text{ for all }\omega\in N,\omega\neq 0?
$$
Comment:
My guess is that there will be such an $a$. But I could not prove it.

REPLY [6 votes]: For what is worth, here is a simple coordinate example in dimension 6. Take
$$
 \omega_0 = dx_1\wedge dx_2 + dx_3\wedge dx_4 + dx_5\wedge dx_6 ,
$$
$$
 \omega_1 = dx_1\wedge dx_3 + dx_2\wedge dx_4 ,
$$ 
$$
\omega_2 = dx_3\wedge dx_5 + dx_4\wedge dx_6 ,
$$
$$
\omega_3 = dx_5\wedge dx_1 + dx_6\wedge dx_2 .
$$
The sum of squares of these forms is zero, but the square of any nontrivial linear combination is not.