# Is this lemma in elementary linear algebra new?

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.

• Welcome to Mathoverflow! – Mark Sapir Sep 5 '13 at 21:09
• I don't have time to think about this right now, but it seems strikingly familiar to the following theorem of Hopf. If $f:A \otimes B\to C$ is a linear map which is injective on each factor separately, then $\dim f(A\otimes B) \geq \dim A + \dim B - 1.$ However, this theorem is true over $\mathbb{C}$ but false over $\mathbb{R}$ (the proof is given by algebraic topology), so maybe it is only a superficial observation. – Jack Huizenga Sep 6 '13 at 0:50
• This feels like a statement from projective geometry. $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ somehow "intersects enough" $\mathbb{P}(A) \times \mathbb{P}(B)$ so that $\dim(\mathbb{P}(V)) \geq \dim(\mathbb{P}(A) \times \mathbb{P}(B))$. – Martin Brandenburg Sep 6 '13 at 1:23
• Probably also related: Flanders' theorem (§8.3 in Prasolov's Linear Algebra book www2.math.su.se/~mleites/Prasolov/prasLinAlg/pr-linAlg-main.dvi ). – darij grinberg Sep 6 '13 at 2:30
• Somehow it feels that the following "dual" result is very closely related, and could for some fields also yield your inequality (by splitting the tensor product into "separable" and "joint" spaces, whose dimensions add up to $d_Ad_B$): 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. – Suvrit Sep 6 '13 at 15:04

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 $$\dim V \geq 2+\dim A(V) \geq 2+(n-1)+(p-1)-1=n+p-1.$$
• This is a nice proof. Although it seems to be algebraic, I think it is geometric in disguise. Perhaps someone can formulate this proof coordinate-free, and/or in terms of the the intersection of $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ with $\mathbb{P}(A) \times \mathbb{P}(B) \hookrightarrow \mathbb{P}(A \otimes B)$? – Martin Brandenburg Sep 6 '13 at 19:20