linear algebra - Non-degenerate multilinear forms

Basic idea: a bilinear form $B$ is degenerate if there are two nonzero vectors $v$ and $w$ so that not only is
$$B(v,w)=0,$$
but also either vector is enough to kill $B$ without help from the other:
$$B(v,-) = 0 mbox{ and } B(-,w)=0.$$

Another way to say it: even if we perturb $v$ and $w$, $B(v otimes w)=0$ to first order. From this point of view, we see that a bilinear form is degenerate iff it is an element of the variety dual to the Segre embedding of $mathbb{P}V times mathbb{P}W$ in $mathbb{P}(V otimes W)$.

This motivation generalizes gracefully to the following definition from Gelfand, Kapranov, and Zelevinsky's book Discriminants, resultants, and multidimensional determinants:

A $p$-linear form $T$ is said to be degenerate if either of the following equivalent conditions holds:

• there exist nonzero vectors $beta_i$ so that, for any $1 leq j leq p$,
  $$ T left( beta_1, ldots , beta_{j-1} , x_j ,beta_{j+1}, ldots , beta_{p} right) = 0 mbox{ for all $x_{j}$;} $$




• there exist nonzero vectors $beta_{i}$ so that $T$ vanishes at $otimes beta_{i}$ along with every partial derivative with respect to an entry of some $beta_{i}$:
  $$ T mbox{ and } frac{partial T}{partial beta^{(j)}_{i}} mbox{ vanish at $otimes beta_{i}$.}$$

In certain favorable cases (when the dimensions of the vector spaces $V_i$ are not too different) the dual to the Segre is a hypersurface; in other words, there is a single polynomial---the hyperdeterminant---which vanishes exactly at the degenerate multilinear forms. This polynomial possesses many magical properties and is much subtler than determinants of bilinear forms.

I can attest that this definition is at least useful, if not standard, since it came up in a substantial way in an elementary question about coin flipping: http://arxiv.org/abs/1009.4188 .