Let $V$ be a finite dimensional vector space over $\mathbb{C}$. Then we can form the exterior algebra $\wedge V$. Let $L\subset \wedge^3 V$ be a linear subspace. I am working on the following claim.
Claim: (Assume $\dim V\geq 5$) Suppose that for any $T\in \wedge^2 V$, there exists some nonzero $u\in V$ such that $u\wedge T \in L$. Then there must be a nonzero $u_0 \in V$ such that $u_0\wedge (\wedge^2 V) \subset L$.
The other direction of the claim is obviously true since we can just pick $u_0$ for every $T \in \wedge^2 V$. My intuition is that $\wedge^3 V$ does not need many generators and when you can pick different $u$s for different $T$s, you end up generating all of $\wedge^3 V$. However, I am having trouble realizing this intuition. I can try to give a lower bound on the dimension of $L$ using the condition in the claim but that is insufficient as far as I can tell. It is such a concrete problem but my friends and I can barely make any progress on it.
If anyone can give a proof of this claim or find a counter example, that would be extremely helpful.