# Number of vectors such that the projection is decomposable

Let $V$ be a vector space of dimension $n\geq 6$ over the finite field $\mathbb{F}_q$. Let $\omega\in\bigwedge^{n-3}V$ be a nonzero element. Define the annihilator subspace of $\omega$ by $V_\omega=\{z\in V: \omega\wedge z= 0\}$. It is well known that $0\leq \dim V_\omega \leq n-3$. Suppose for this given $\omega$ that we have $\dim V_\omega= n-k$ where $k\geq 5$. I want to prove that $$\left|\{x\in V: \omega\wedge x\in \bigwedge^{n-2}V\text{ is decomposable}\}\right|\leq q^{n-k}(q^{k-3}-1)(q^2+1).$$

By a decomposable element of $\bigwedge^{d}V$, we mean a nonzero element that can be written as $v_1\wedge\ldots\wedge v_d$ for some linearly independent elements $\{v_1,\ldots,v_d\}$ of $V$. I can prove this when $k=6$ but I am not able to prove this in general.

• You probably mean "annihilator" and "$= 0$" should replace "$\neq 0$" in your definition. – Luc Guyot Jan 16 '18 at 17:24
• yes. Thanks, It was a typo I just changed it. – Singh Jan 17 '18 at 7:50