Decomposable elements of $\Lambda^k(V)$

Question

I have a conjecture. I have a problem proving or disproving it.

Let $w \in \Lambda^k(V)$ be a $k$-vector. Then $W_w=\{v\in V: v\wedge w = 0 \}$ is a $k$-dimensional vector space if and only if $w$ is decomposable.

For example, for $u=e_1\wedge e_2 + e_3 \wedge e_4$ we have $W_u = 0$.

Answer 1

Let us assume $V$ has dimension $n$. Suppose $w$ is a $k$-form which is indecomposable, hence of pure type $v_{i_{1}}\wedge v_{i_{2}}...\wedge v_{i_{k}}$. Then $W_{w}$ is comprised by linear combinations of $v_{i_{j}}$, $j=1,2,...k$. So $W_{w}$ is precisely a $k$ dimensional vector subspace of $V$.