Degenerate form

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In mathematics, a degenerate bilinear form f(x,y) on a vector space V is one such that the map from V to V* (the dual space of V) given by v ↦ f(-,v) is not a bijection. An equivalent definition when V is finite-dimensional that there exist some non-zero x in V

<math>f(x,y)=0</math> for all y ∈ V.

A nondegenerate form is one that is not degenerate. That is, f is nondegenerate iff v ↦ f(-,v) is a bijection, or equivalently in finite dimensions, iff

<math>f(x,y)=0</math> for all y ∈ V

implies that x = 0.

Note that in an infinite dimensional space, we can have a bilinear form f for which v ↦ f(-,v) is injective but not surjective (confer Hilbert's paradox). For example, on the space of continuous functions on a closed bounded interval, the form f(φ,ψ)=∫ψ(x)φ(x) dx is not surjective, because e.g. the Dirac delta functional is in the dual space but not of the required form. On the other hand, this bilinear form satisfies f(φ,ψ)=0 for all φ implies that ψ=0.

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 subspace of V. f is nondegenerate iff this subspace is trivial.

If V is finite-dimensional then, relative to some basis for V, a bilinear form is degenerate iff the determinant of the associated matrix is zero. Likewise, a nondegenerate form is one for which the associated matrix is non-singular. These statements are independent of the chosen basis.

Sometimes the words anisotropic, isotropic and totally isotropic are used for nondegenerate, degenerate and totally degenerate respectively. Although definitions of these latter words can vary slightly between authors.