Degenerate form
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(Redirected from Nondegenerate form)
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For other uses, see Degeneracy.
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- for all
- for all implies that x = 0.
There is the closely related notion of a unimodular form and a perfect pairing; these agree over fields but not over general rings.
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.
[edit] 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- for all implies that
[edit] 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 vectorsSometimes 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.
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.
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