# Degenerate form

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In mathematics, specifically linear algebra, a **degenerate** bilinear form ƒ(*x,y*) on a vector space *V* is one such that the map from to (the dual space of ) given by 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

## Non-degenerate forms

A **nondegenerate** or **nonsingular** form is one that is not degenerate, meaning that is an isomorphism, or equivalently in finite dimensions, if and only if

## 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 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 is injective but not surjective. For example, on the space of continuous functions on a closed bounded interval, the form

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

## 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

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 words can vary slightly between authors.{{ safesubst:#invoke:Unsubst||$N=Dubious |date=__DATE__ |$B=
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Beware that a vector such that is called isotropic for the quadratic form associated with the bilinear form and the existence of isotropic lines does not imply that the form is degenerate.