Does the kernel of every alternating form contain a decomposable element?

Question

Let $V$ be a real $n$-dimensional vector space, and let $1 < k < n$.

Let $\alpha \in \bigwedge^k (V^*) \cong (\bigwedge^k V)^*$. Thinking of $\alpha$ as a linear functional $\bigwedge^k V \to \mathbb{R}$, does $\ker \alpha$ always contain a (non-zero) decomposable element?

Answer 1

Let $(v_1, \ldots, v_{k+1})$ be a linearly independent family of vectors of $V$.

Let $W=\mathbb{R}v_k+\mathbb{R}v_{k+1}$, and consider the map $\beta: w \in W \longmapsto \alpha(v_1 \wedge \ldots \wedge v_{k-1} \wedge w$.

$\beta$ is a linear form on $W$ (which has dimension $2$), hence has a non-trivial kernel element $w$, and $\alpha(v_1 \wedge \ldots \wedge v_{k-1} \wedge w)=0$, and $v_1 \wedge \ldots \wedge v_{k-1} \wedge w \neq 0$.