Degenerate bilinear form

In mathematics, specifically linear algebra, a degenerate bilinear form

f

(

x

,

y

)

{\displaystyle f(x,y)}

on a vector space

V

{\displaystyle V}

is a bilinear form such that the map from

V

{\displaystyle V}

to

V

∗

{\displaystyle V^{*}}

(the dual space of

V

{\displaystyle V}

) given by

v

↦

(

x

↦

f

(

x

,

v

)

)

{\displaystyle v\mapsto (x\mapsto f(x,v))}

has a non-trivial kernel, i.e. there exist some non-zero

x

{\displaystyle x}

in

V

{\displaystyle V}

such that

f

(

x

,

y

)

=

0

{\displaystyle f(x,y)=0}

for all

y

∈

V

{\displaystyle y\in V}

.

An equivalent definition when

V

{\displaystyle V}

is finite-dimensional is that the previous map is not an isomorphism.

Nondegenerate forms

A nondegenerate or nonsingular form is a bilinear form that is not degenerate, meaning that

v

↦

(

x

↦

f

(

x

,

v

)

)

{\displaystyle v\mapsto (x\mapsto f(x,v))}

is an isomorphism, or equivalently in finite dimensions, if and only if

f

(

x

,

y

)

=

0

{\displaystyle f(x,y)=0}

for all

y

∈

V

{\displaystyle y\in V}

implies that

x

=

0

{\displaystyle 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

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 fields but not over general rings.