Bilinear product of two non-zero vectors is non-zero?

Question

Suppose we have a finite dimensional vector space over a field $F$ with an arbitrary bilinear form $\cdot: F\times F\to F$ satisfying bilinearity axioms. Can we always claim that the product of two non-zero vectors is non-zero? If so how do I prove it?

Answer 1

We can find bilinear operators where the product of nonzero vectors is zero. A trivial example would be the function that maps everything to zero—I think that satisfies all of the bilinearity axioms.

A second example is the dot product: the dot product in $\mathbb{R}^2$ is a bilinear operator, and there exist pairs of perpendicular nonzero vectors. Their dot product is zero, but neither vector is the zero vector.

We can also build other examples. Suppose your vector space is $\mathbb{R}^2$. Let $e_1, e_2$ be an orthonormal basis, and define a bilinear operator by:

$$B(v,w) = (v\cdot e_1)(w\cdot e_2)$$

Essentially: take the dot product of the first vector with $e_1$, and the second vector with $e_2$, and multiply the resulting numbers together.

You can show that this is a bilinear operator because dot products are linear.

However, $e_1$ and $e_2$ are nonzero vectors where $B(e_2, e_1) = 0$.

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One useful related property that bilinear operators can have is nondegeneracy: a bilinear operator is nondegenerate if the statement $\forall y, B(x,y)=0$ holds only when $x=0$.

The dot product in $\mathbb{R}^2$ is a bilinear operator with this property: if $x$ is a vector where for every $y \in \mathbb{R}^2$, $x \cdot y = 0$, then $x$ is necessarily the zero vector.