# Finding a bilinear map that differs on two given points.

Let $V,W$ be finite dimensional real vector spaces, and let $(v,w) \not= (x,y)$ be two points in $V \times W$. Is it possible to construct a bilinear map $\alpha: V \times W \to \mathbb{R}$ such that $\alpha (v,w) \not= \alpha (x,y)$? If so, how do I define such a map?

The reason I want to know is that this is the last step in my verification that $\operatorname{Bilin}(V,W;\mathbb{R})^*$ is the tensor product of $V$ and $W$. Any help would be appreciated, thank you.

-

This is not possibly if you have two points of the form $(ax,y)$ and $(x,ay)$ for a scalar $a$, due to bilinearity. If you don't have this property then there exist indices $i,j$ such that $v_iw_j\neq x_iy_j$ (I assume that all vectors are coordinate vectors in some $\mathbb R^n$). A bilinear map is "nothing else" than a $n\times m$ matrix, where $n=dim V$ and $m=dim W$. If you choose the matrix with a one at the entry $(i,j)$ and zeros everywhere else, it does the job.
The reason I asked this question is because I was trying to show that the map $\pi: V\times W \to B^*$, where $B$ is the vector space of all bilinear maps $V \times W \to \mathbb{R}$, is injective, where $\pi (v,w)(\alpha) = \alpha(v,w)$. So obviously by what you you say, this cannot be injective. Since that map corresponds to the tensor product map $V\times W \to V \otimes W$, am I right in saying that $(v,w) \mapsto v \otimes w$ is not injective? –  Paul Slevin May 1 '12 at 16:31