Editing Degenerate bilinear form (section)

==Terminology==
If ''f'' vanishes identically on all vectors it is said to be ''' totally degenerate'''. Given any bilinear form ''f'' on ''V'' the set of vectors

:<math>\{x\in V \mid f(x,y) = 0 \mbox{ for all } y \in V\}</math>

forms a totally degenerate [[linear subspace|subspace]] of ''V''. The map ''f'' is nondegenerate if and only if this subspace is trivial.

Geometrically, an [[isotropic line]] of the quadratic form corresponds to a point of the associated [[quadric surface|quadric hypersurface]] in [[projective space]]. Such a line is additionally isotropic for the bilinear form if and only if the corresponding point is a [[singular variety|singularity]]. Hence, over an [[algebraically closed field]], [[Hilbert's Nullstellensatz]] guarantees that the quadratic form always has isotropic lines, while the bilinear form has them if and only if the surface is singular.