A surprisingly difficult fact about bilinear maps

Question

Here's a problem you can give to a linear algebra student to haze them, as I was given by my friend: let $V$, $W$, $X$ be finite-dimensional $\mathbb C$-vector spaces, and $b:V\times W\to X$ bilinear and strongly nondegenerate in the sense that if $b(v,w) = 0,$ then $v = 0$ or $w = 0$. Show that $\dim V+\dim W \leq \dim X+1$.

Perhaps I'm naïve, but this looks like it should fall to direct linear-algebraic methods – but all the proofs my friend knows use high-tech tools such as the Segre embedding, dimension theory, or cohomology. I don't know any algebraic geometry, and while I do understand the cohomology argument it's not very enlightening to me – is there a more elementary argument, or is the cohomology argument in some sense "necessary"?

Supposedly this holds over any algebraically closed field $k$, if that helps. You do need it to be algebraically closed, though; for example, taking $V = W = X = \mathbb C$ as $\mathbb R$-vector spaces, complex multiplication is a strongly nondegenerate $\mathbb R$-bilinear map $V\times W\to X$ and this dimension inequality fails to hold.

Answer 1

Below is a summary of useful observations/notes made in the comments (now moved to chat).

Feel free to update this as useful comments are added to the chat or anywhere on this post.

• This notion of "strongly non-degenerate" is distinct from the usual definition of degeneracy for bilinear forms. Typically, a bilinear form $b$ is considered "non-degenerate" when the only $v \in V$ such that $b(v,w) = 0$ for all $w \in W$ is $0 \in V$ (and likewise reversing the role of $v$ and $w$) (BG).
• The usual notion of "non-degenerate" is insufficient for the inequality: for instance, one can consider the bilinear form $b(v, w) = v_1v_2 - w_1w_2$ with $V = W = \Bbb C^2$ and $X = \Bbb C$. (BG)
• The map $v \mapsto b(v, \cdot)$ must be injective and that the maps of the form $b(v, \cdot)$ from $W$ to $X$ are also injective. So, the image of $v \mapsto b(v, \cdot)$ is a subspace of $\mathcal L(W,X)$ whose elements are all injective. (BG; See existing answer by Levent)
• "Strongly non-degenerate" also has another standard meaning (when considering bilinear functionals on an infinite-dimensional Banach space, namely that the ‘flat map’ from the Banach space into the dual is an isomorphism, rather than just being injective). (peek-a-boo)
• A counterexample machine that works over any non-algebraically-closed field. Let $K$ be a field and $L/K$ a nontrivial finite extension. Then multiplication in $L$ defines a strongly nondegenerate bilinear map $L\times L\to L$ of $K$-vector spaces which violates the dimension inequality. At least when $V\cong W\cong X$, this idea relates the theorem to the existence of nontrivial division algebras over $K$. (Snacc)
• Let $(V,W,X)=(\mathbb C^m,\mathbb C^n,\mathbb C^d)$. The given condition is eqv. to the existence of some rectangular matrices $B_1,\ldots,B_d\in M_{m,n}(\mathbb C)$ such that for each nonzero $n$-vector $w$, the augmented matrix $\pmatrix{B_1w&\cdots&B_dw}\in M_{m,d}(\mathbb C)$ has full row rank $m$. In turn, it is eqv. to the existence of some $C_1,\ldots,C_n\in M_{m,d}(\mathbb C)$ such that every nonzero linear combination of these $n$ matrices has full row rank. So, the linear span of those $C_j$s is a subspace consisting of matrices of the same fixed rank (except the zero matrix) (user1551)
• In linear algebra literature, the classical paper about dimensions of such subspaces is “Spaces of matrices of fixed rank” by Westwick (1987), which does involve algebraic geometry. (user1551)

Answer 2

I think the following turns Ben Grossmann's comment into a rigorous answer.

Ben notes that the map $V\rightarrow L(W,X),\; v\mapsto b(v,\cdot)$ is an injective map and each $b(v,\cdot)\in L(W,X)$ itself is an injective map from $W$ to $X$. Analogously, $w\mapsto b(\cdot,w)$ is also injective and each $b(\cdot,w)$ is injective.

We first observe that this implies $\dim V, \dim W \leq \dim X$, since otherwise there is no injective map $W\rightarrow X$ (resp. $V\rightarrow X$).

Hence, the set of non-injective maps and the set of rank-deficient maps coincides in $L(W,X)$. However, the set $$ Z = \{ f\in L(W,X) \mid \mathrm{rk} f \leq \dim W - 1 \} $$ is a homogeneous algebraic variety of dimension $(\dim W - 1) (\dim X + 1)$. See this answer or the Wikipedia page. In other words, the projective variety $\mathbb{P}(Z)\subset \mathbb{P}(L(W,X))$ is of dimension $(\dim W -1)(\dim X + 1) - 1$. Hence, it intersects with any linear subspace of dimension \begin{align} \operatorname{codim} \mathbb{P}(Z) &= \dim X \dim W - 1 - (\dim W -1)(\dim X + 1)+1 \\ & =\dim X - \dim W+1. \end{align} However, the projectivization of the image of $V$ under $v\mapsto b(v,\cdot)$ is a $\dim V-1$ dimensional subspace of injective maps. This shows that $$ \dim V - 1 < \dim X - \dim W +1 $$ and this finishes the proof.

Answer 3

[This is just a restatement of the problem.]

Let ${ V, W, X }$ be finite dimensional ${ \mathbb{C} - }$vector spaces. Let ${ b : V \times W \to X }$ be a bilinear map. Let ${ b }$ be strongly nondegenerate, that is

$${ b(v, w) = 0 \implies v = 0 \text{ or } w = 0 . }$$

We are to show

$${ \text{To show:} \quad \dim(V) + \dim(W) \leq \dim(X) + 1 . }$$

Fix bases ${ \mathscr{V} = (v _1, \ldots, v _m), }$ ${ \mathscr{W} = (w _1, \ldots, w _n) , }$ and ${ \mathscr{X} = (\mathbf{x} _1, \ldots, \mathbf{x} _p) , }$ of ${ V, }$ ${ W , }$ and ${ X }$ respectively.

Note that the bilinear map ${ b }$ can be written as

$${ {\begin{aligned} &\, b \left(\sum _i x _i v _i, \sum _j y _j w _j \right) \\ = &\, \sum _{i, j} x _i y _j b(v _i, w _j) \\ = &\, \sum _{i, j} x _i y _j \left( \mathscr{X} B _{i, j} \right) \\ = &\, \mathscr{X} \sum _{i, j} x _i y _j B _{i, j} \\ = &\, \mathscr{X} \sum _{i, j} x _i y _j \left( \sum _k e _k b _{i, j, k} \right) \\ = &\, \mathscr{X} \sum _{i, j, k} x _i y _j b _{i, j, k} e _k . \end{aligned}} }$$

Note that in the last sum the indices are over ${ i \in [m], }$ ${ j \in [n] , }$ and ${ k \in [p] . }$

Hence the bilinear map can be viewed as the map

$${ {\begin{aligned} &\, \mathbb{C} ^m \times \mathbb{C} ^n \to \mathbb{C} ^p , \\ &\, (X, Y) \mapsto \begin{pmatrix} \sum _{i, j} x _i y _j b _{i, j, 1} \\ \vdots \\ \sum _{i, j} x _i y _j b _{i, j, p} \end{pmatrix} . \end{aligned}} }$$

Let ${ B _k := (b _{i, j, k}) _{i, j} .}$ Hence the above bilinear map is

$${ {\begin{aligned} &\, \mathbb{C} ^m \times \mathbb{C} ^n \to \mathbb{C} ^p, \\ &\, (X, Y) \mapsto \begin{pmatrix} X ^T B _1 Y \\ \vdots \\ X ^T B _p Y \end{pmatrix} \end{aligned}} . }$$

Note that the strongly nondegenerate condition can be written as

$${ X ^T B _1 Y = 0, \ldots, X ^T B _p Y = 0 \implies X = 0 \text{ or } Y = 0 . }$$

We are to show

$${ \text{To show:} \quad m + n \leq p + 1 . }$$

Hence the problem can be written as

To show: Let ${ B _1, \ldots, B _p \in \mathbb{C} ^{m \times n} }$ be such that

$${ X ^T B _1 Y = 0, \ldots, X ^T B _p Y = 0 \implies X = 0 \text{ or } Y = 0 . }$$

Then ${ m + n - 1 \leq p . }$

Note that

$${ \text{vec}(AXB) = (B ^T \otimes A) \text{vec}(X) . }$$

Hence the condition can be written as

$${ {\begin{aligned} &\, (Y ^T \otimes X ^T) \text{vec}(B _1) = 0, \ldots, (Y ^T \otimes X ^T) \text{vec}(B _p) = 0 \\ \implies &\, X = 0 \text{ or } Y = 0 \end{aligned}} }$$

that is

$${ {\begin{aligned} &\, \text{vec}(B _1) ^T (Y \otimes X) = 0, \ldots, \text{vec}(B _p) ^T (Y \otimes X) = 0 \\ \implies &\, Y \otimes X = 0 \end{aligned}} }$$

that is

$${ \begin{pmatrix} \text{vec}(B _1) ^T \\ \vdots \\ \text{vec}(B _p) ^T \end{pmatrix} (Y \otimes X) = 0 \implies Y \otimes X = 0 . }$$

Hence the problem can be written as

To show: Let ${ M \in \mathbb{C} ^{p \times mn} . }$ Suppose for ${ Y \in \mathbb{C} ^n , }$ ${ X \in \mathbb{C} ^m , }$

$${ M(Y \otimes X) = 0 \implies Y \otimes X = 0 . }$$

Then ${ m + n - 1 \leq p . }$