Is anyone familiar with the following, or anything close to it?

Lemma. Suppose $A$, $B$ are nonzero finite-dimensional vector spaces over an infinite field $k$, and $V$ a subspace of $A\otimes_k B$ such that

(1) For every nonzero $a\in A$ there exists nonzero $b\in B$ such that $a\otimes b\in V$,

and likewise,

(2) For every nonzero $b\in B$ there exists nonzero $a\in A$ such that $a\otimes b\in V$.

Then

(3) $\dim_k(V) \geq \dim_k(A) + \dim_k(B) - 1$.

Remarks: The idea of (1) and (2) is that the spaces $A$ and $B$ are minimal for "supporting" $V$; that is, if we replace $A$ or $B$ by any proper homomorphic image, and we map $A\otimes B$ in the obvious way into the new tensor product, then that map will not be one-one on $V$. The result is equivalent to saying that if one is given a finite-dimensional subspace $V$ of a tensor product $A\otimes B$ of arbitrary vector spaces, then one can replace $A$, $B$ by images whose dimensions sum to $\leq \dim(V) + 1$ without hurting $V$.

In the lemma as stated, if we take for $A$ a dual space $C^*$, and interpret $A\otimes B$ as $\mathrm{Hom}(C,B)$, then the hypothesis again means that $C$ and $B$ are minimal as spaces "supporting" $V$, now as a subspace of $\mathrm{Hom}(C,B)$; namely, that restricting to any proper subspace of $C$, or mapping onto any proper homomorphic image of $B$, will reduce the dimension of $V$.

In the statement of the lemma, where I assumed $k$ infinite, I really only need its cardinality to be at least the larger of $\dim_k A$ and $\dim_k B$.

The proof is harder than I would have thought; my write-up is 3.3K. I will be happy to show it if the result is new.

On the maximal dimension of a completely entangled subspace..."by K. Parathasarathy; ias.ac.in/mathsci/vol114/nov2004/Pm2342.pdf --- in particular, your subspaces have the "separable" state property, while the cited paper considers "full entangled" subspaces. $\endgroup$ – Suvrit Sep 6 '13 at 15:04