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. 



Is this lemma in elementary linear algebra new?
This is a nice lemma: I know a good deal of similar results but this one is unknown to me. 


I believe it is suitable, as an answer, to give a proof that works with no restriction on the cardinality of the underlying field $F$. I will frame the answer in terms of matrix spaces. 
Thus, we have a linear subspace $V \subset M\_{n,p}(F)$ such that, for every 
non-zero vector $X \in F^n$, the space $V$ contains a rank $1$ matrix with column space $F X$
and, for every non-zero vector $Y \in F^p$, the space $V$ contains a rank $1$ matrix with row space $F Y^t$. Note that those assumptions are unchanged in multiplying $V$ by
invertible matrices (be it on the left or on the right). 


The proof works by induction on $p$. The case where $p=1$ or $n=1$ is obvious.
Assume now that $p>1$ and $n>1$. The discussion is split into two cases, where
the standard basis of $F^p$ is denoted by $(e\_1,\dots,e\_p)$. 


Case 1: $V e\_p=F^n$.
Then, one writes every matrix $M$ of $V$ as 
$M=\begin{bmatrix}
A(M) & C(M)
\end{bmatrix}$ where $A(M) \in M\_{n,p-1}(F)$ and $C(M) \in F^n$. 
With our assumptions, we find rank $1$ matrices $M\_1,\dots,M\_{p-1}$
in $V$ with respective row spaces $F e\_1^t,\dots,F e\_{p-1}^t$. 
Then, $M\_1,\dots,M\_{p-1}$ are linearly independent and all belong to 
the kernel of $V \ni M \mapsto C(M)$. Using the rank theorem, one deduces that 
$\dim V \geq (p-1)+\dim C(V)=(p-1)+n$. 


Case 2 : $V e\_p \subsetneq F^n$. 
Multiplying $V$ on the left by a well-chosen invertible matrix, we lose no generality in assuming that $V e\_p \subset F^{n-1} \times \{0\}$. In other words, every matrix $M$
of $V$ may be written as 
$$M=\begin{bmatrix}
A(M) & C(M) \\
R(M) & 0
\end{bmatrix}$$
where $A(M)$ is an $(n-1) \times (p-1)$ matrix, $R(M)$ is a row matrix and $C(M)$ is a column matrix. Then, we note that $A(V)$ satisfies the same set of assumptions as $V$:
indeed, if we take a non-zero row $L \in M\_{1,p-1}(F)$, then we know that $V$
contains a rank $1$ matrix $M\_1$ whose row space is spanned by $\begin{bmatrix}
L & 1
\end{bmatrix}$. Obviously the last row of $M\_1$ is zero whence $A(M\_1)$
is non-zero and its row space is included in $F L$. 
One works likewise to obtain the remaining part of the condition. 
Thus, by induction one finds 
$$\dim A(V) \geq (n-1)+(p-1)-1.$$
Finally, we know that $V$ must contain a non-zero matrix $M\_2$ with $A(M\_2)=0$ and $C(M\_2)=0$, and that it must contain a non-zero matrix $M\_3$ with $A(M\_3)=0$ and $R(M\_3)=0$. Obviously, $M\_2$ and $M\_3$ are linearly independent vectors in the kernel
of $V \ni M \mapsto A(M)$. Using the rank theorem, one concludes that 
$$\dim V \geq 2+\dim A(V) \geq 2+(n-1)+(p-1)-1=n+p-1.$$ 


