Decomposable elements of $\Lambda^k(V)$ 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 $W_u = 0$.

Roughly another 2 years later let me try giving a different proof.
Suppose first $w$ is a decomposable $k$-vector. Then there exist $k$ linearly independent vectors $\{e_i\}$, $i=1,\ldots k$ such that $w=e_1\wedge\cdots\wedge e_k$. Complete $\{e_i\}$ to a basis of $V$, it is then clear that $e_i \wedge w=0$ for $i=1,\ldots k$, $e_i\wedge w\neq 0 $ for $i=k+1, \ldots n$, hence $\mathrm{dim}(W_w)= k$.
Suppose now $\mathrm{dim}(W_w)= k$, and let $\{e_i\}$, $i=1,\ldots k$ be a basis of $W_w$. By using the following fact:

Let $w$ be a $k$-vector, $u$ a non-vanishing vector. Then $w \wedge u =0 $ if and only if $w= z\wedge u $ for some $(k-1)$-vector $z$.

one establishes that $w$ is of the form $w=c e_1 \wedge\cdots\wedge e_k$ for some constant $c$.
