Is it true that $w\wedge v=0\implies w=v\wedge\phi$ for some $\phi\in\bigwedge^{k-1}V$?

Question

In particular, what if $v=0$ and $w\ne0$? Then $w\wedge v=0$, but $w$ cannot equal $v\wedge\phi$ for any $\phi$, because $v\wedge\phi=0$ for all $\phi$.

I'm considering $v\in V$, $w\in\bigwedge^kV$, and $\phi\in\bigwedge^{k-1}V$.

Answer 1

As you have shown, the statement in the title is not true.

However, it is true if $v \neq 0$ (at least if $V$ is finite-dimensional). To see this, first complete $\{v\}$ to a basis, i.e. choose a basis $\{v_1, \dots, v_n\}$ for $V$ such that $v_1 = v$. There is an induced basis for $\bigwedge^kV$ given by $\{v_{i_1}\wedge\dots\wedge v_{i_k} \mid 1 \leq i_1 < \dots < i_k \leq n\}$ so

$$w = \sum_{1 \leq i_1 < \dots < i_k \leq n}a_{i_1,\dots,i_k}v_{i_1}\wedge\dots\wedge v_{i_k}.$$

If $i_1 = 1$ then $v\wedge v_{i_1}\wedge\dots\wedge v_{i_k} = v\wedge v\wedge\dots\wedge v_{i_k} = 0$, so

$$v\wedge w = \sum_{1 < i_1 < \dots < i_k \leq n}a_{i_1,\dots,i_k}v\wedge v_{i_1}\wedge\dots\wedge v_{i_k} = \sum_{1 < i_1 < \dots < i_k \leq n}a_{i_1,\dots,i_k}v_1\wedge v_{i_1}\wedge\dots\wedge v_{i_k}.$$

As $\{v_{j_1}\wedge\dots\wedge v_{j_{k+1}} \mid 1 \leq j_1 < \dots < j_{k+1} \leq n\}$ is a basis for $\bigwedge^{k+1}V$ and $v\wedge w = 0$, we see that $a_{i_1,\dots,i_k} = 0$ for $i_1 > 1$. Therefore

\begin{align*} w &= \sum_{1 < i_2 < \dots < i_k \leq n}a_{1,i_2,\dots,i_k}v_1\wedge v_{i_2}\wedge\dots\wedge v_{i_k}\\ &= v_1\wedge\left(\sum_{1 < i_2 < \dots < i_k \leq n}a_{1,i_2,\dots,i_k}v_{i_2}\wedge\dots\wedge v_{i_k}\right)\\ &= v\wedge\phi \end{align*}

where

$$\phi = \sum_{1 < i_2 < \dots < i_k \leq n}a_{1,i_2,\dots,i_k}v_{i_2}\wedge\dots\wedge v_{i_k} \in \bigwedge\nolimits^{k-1}V.$$