Degenerate bilinear form

In mathematics, specifically linear algebra, a degenerate bilinear form f(x, y) on a vector space V is a bilinear form such that the map from V to V^∗ (the dual space of V) given by v ↦ (x ↦ f(x, v)) is not an isomorphism. An equivalent definition when V is finite-dimensional is that it has a non-trivial kernel: there exist some non-zero x in V such that

$f(x,y)=0\,$ for all $y \in V.$

Non-degenerate forms

A nondegenerate or nonsingular form is one that is not degenerate, meaning that $v \mapsto (x \mapsto f(x,v))$ is an isomorphism, or equivalently in finite dimensions, if and only if

$f(x,y)=0\,$ for all $y \in V$ implies that x = 0.

Using the determinant

If V is finite-dimensional then, relative to some basis for V, a bilinear form is degenerate if and only if the determinant of the associated matrix is zero – if and only if the matrix is singular, and accordingly degenerate forms are also called singular forms. Likewise, a nondegenerate form is one for which the associated matrix is non-singular, and accordingly nondegenerate forms are also referred to as non-singular forms. These statements are independent of the chosen basis.

Related notions

There is the closely related notion of a unimodular form and a perfect pairing; these agree over fields but not over general rings.

Examples

The most important examples of nondegenerate forms are inner products and symplectic forms. Symmetric nondegenerate forms are important generalizations of inner products, in that often all that is required is that the map $V \to V^*$ be an isomorphism, not positivity. For example, a manifold with an inner product structure on its tangent spaces is a Riemannian manifold, while relaxing this to a symmetric nondegenerate form yields a pseudo-Riemannian manifold.

Infinite dimensions

Note that in an infinite dimensional space, we can have a bilinear form ƒ for which $v \mapsto (x \mapsto f(x,v))$ is injective but not surjective. For example, on the space of continuous functions on a closed bounded interval, the form

$f(\phi,\psi) = \int\psi(x)\phi(x) dx$

is not surjective: for instance, the Dirac delta functional is in the dual space but not of the required form. On the other hand, this bilinear form satisfies

$f(\phi,\psi)=0\,$ for all $\,\phi$ implies that $\psi=0.\,$

Terminology

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

$\{x\in V \mid f(x,y) = 0 \mbox{ for all } y \in V\}$

forms a totally degenerate subspace of V. The map ƒ is nondegenerate if and only if this subspace is trivial.

Sometimes the words anisotropic, isotropic and totally isotropic are used for nondegenerate, degenerate and totally degenerate respectively, although definitions of these latter can be a bit ambiguous: a vector $x\in V$ such that $f(x,x)=0$ is called isotropic for the quadratic form associated with the bilinear form $f$, but such vectors can arise even if the bilinear form has no nonzero isotropic vectors.

Geometrically, an isotropic line of the quadratic form corresponds to a point of the associated quadric hypersurface in projective space. Such a line is additionally isotropic for the bilinear form if and only if the corresponding point is a 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.