Characterizing zero sets of bilinear maps Let $V,W$ be vector spaces and $X\subset V\times W.$ If $X$ is the zero set of a collection of bilinear maps then it satisfies the following properties:

*

*$(0,w),(v,0)\in X$ for all $v,w.$

*If $(v,w)\in X$ then $(\alpha v,w),(v, \alpha w)\in X$ for any scalar $\alpha.$

*If $(v_1,w),(v_2,w)\in X$ then $(v_1+v_2,w)\in X$ and the same for the second coordinate.

Now suppose a set $X\subsetneq V\times W$ satisfies conditions 1-3. Then is there always a non-zero bilinear map vanishing on $X?$ If so, is $X$ the zero set of a collection of bilinear maps?
These questions can be phrased in terms of the subspace spanned by $X$ when viewed in $V\otimes W,$ but that's as far as I've been able to get.
 A: This will not always work. Take $V=W=\mathbb R^2$, and then something like $v_1=(1,0)$, $v_2=(0,1)$, $v_3=(1,1)$, $v_4=(1,2)$, $v_5=(2,1)$, and let $X$ be the smallest set that contains $(v_j, v_j)$ ($j=1,2,3$), $(v_4,v_5)$ and satisfies your conditions. In other words, the vectors here can be multiplied by arbitrary numbers, but that's it, since condition (3) never applies.
Then for example $(v_4,v_4)\notin X$, but any bilinear form that vanishes on $X$ is identically equal to zero (check that the four vectors above span the tensor product $V\otimes V$ when thought of as elements of this space, or, easier still perhaps, work with the matrix representation of the bilinear form with respect to the standard basis $v_1,v_2$).
A more highbrow answer is also possible: If $K=\mathbb C$, we can take
$$
X = \{ (a\overline{v}, bv): a,b\in\mathbb C, v\in V \} .
$$
Here $\overline{v}$ means the complex conjugate of $v$, taken componentwise, with respect to a fixed basis. If $B$ denotes the bilinear form, then this choice of $X$ makes the sesquilinear form $S(v,w)=B(\overline{v},w)$ vanish on all $(v,v)$, so $S$ is identically zero by polarization.
