Editing Degenerate bilinear form

{{Other uses|Degeneracy (disambiguation){{!}}Degeneracy}}
In [[mathematics]], specifically [[linear algebra]], a '''degenerate bilinear form''' {{nowrap|''f''{{hairsp}}(''x'', ''y''{{hairsp}})}} on a [[vector space]] ''V'' is a [[bilinear form]] such that the map from ''V'' to ''V''<sup>∗</sup> (the [[dual space]] of ''V''{{hairsp}}) given by {{nowrap|''v'' ↦ (''x'' ↦ ''f''{{hairsp}}(''x'',&thinsp;''v''{{hairsp}}))}} is not an [[isomorphism]].  An equivalent definition when ''V'' is [[dimension (vector space)|finite-dimensional]] is that it has a non-trivial kernel: there exist some non-zero ''x'' in ''V'' such that

:<math>f(x,y)=0\,</math> for all <math>\,y \in V.</math>

==Nondegenerate forms==
A '''nondegenerate''' or '''nonsingular''' form is a [[bilinear form]] that is not degenerate, meaning that <math>v \mapsto (x \mapsto f(x,v))</math>  is an [[isomorphism]], or equivalently in finite dimensions, [[if and only if]]<ref>{{cite web|url=https://www.dpmms.cam.ac.uk/study/IB/LinearAlgebra/2008-2009/bilinear-08.pdf |year=2008 |first=T. A. |last=Fisher |title=Linear Algebra: Non-degenerate Bilinear Forms |website=Department of Pure Mathematics and Mathematical Statistics |publisher=Cambridge University |access-date=26 May 2024}}</ref>
:<math>f(x,y)=0</math> for all <math>y \in V</math> implies that <math>x = 0</math>.

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

==Related notions==
If for a [[quadratic form]] ''Q'' there is a non-zero vector ''v'' ∈ ''V'' such that ''Q''(''v'') = 0, then ''Q'' is an [[isotropic quadratic form]]. If ''Q'' has the same sign for all non-zero vectors, it is a [[definite quadratic form]] or an '''anisotropic quadratic form'''.

There is the closely related notion of a [[unimodular form]] and a [[perfect pairing]]; these agree over [[field (mathematics)|fields]] but not over general [[ring (mathematics)|rings]].

==Examples==
The study of real, quadratic algebras shows the distinction between types of quadratic forms. The product ''zz''* is a quadratic form for each of the [[complex number]]s, [[split-complex number]]s, and [[dual number]]s. For ''z'' = ''x'' + &epsilon; ''y'', the dual number form is ''x''<sup>2</sup> which is a '''degenerate quadratic form'''. The split-complex case is an isotropic form, and the complex case is a definite form.

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

==Infinite dimensions==
{{Disputed section|Which dual space?|date=May 2025}}
Note that in an infinite-dimensional space, we can have a bilinear form ƒ for which <math>v \mapsto (x \mapsto f(x,v))</math> is [[injective]] but not [[surjective]].  For example, on the space of [[continuous function]]s on a closed bounded [[interval (mathematics)|interval]], the form 
:<math> f(\phi,\psi) = \int\psi(x)\phi(x) \,dx</math> 
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 
:<math>f(\phi,\psi)=0</math> for all <math>\phi</math> implies that <math>\psi=0.\,</math>
In such a case where ƒ satisfies injectivity (but not necessarily surjectivity), ƒ is said to be ''weakly nondegenerate''.{{cn|date=May 2025}}

==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.

== See also ==
* {{annotated link|Indefinite inner product space}}
* {{annotated link|Dual system}}
* {{annotated link|Linear form}}

== References ==
{{reflist}}

{{Duality and spaces of linear maps}}
{{Topological vector spaces}}

[[Category:Bilinear forms]]
[[Category:Functional analysis]]