Editing Bilinear form (section)

==Properties==
===Non-degenerate bilinear forms===
{{further|Degenerate bilinear form}}
Every bilinear form {{math|''B''}} on {{mvar|V}} defines a pair of linear maps from {{mvar|V}} to its [[dual space]] {{math|''V''<sup>∗</sup>}}. Define {{math|''B''<sub>1</sub>, ''B''<sub>2</sub>: ''V'' → ''V''<sup>∗</sup>}} by
{{block indent|left=1.6|text={{math|1=''B''<sub>1</sub>('''v''')('''w''') = ''B''('''v''', '''w''')}}}}
{{block indent|left=1.6|text={{math|1=''B''<sub>2</sub>('''v''')('''w''') = ''B''('''w''', '''v''')}}}}
This is often denoted as
{{block indent|left=1.6|text={{math|1=''B''<sub>1</sub>('''v''') = ''B''('''v''', ⋅)}}}}
{{block indent|left=1.6|text={{math|1=''B''<sub>2</sub>('''v''') = ''B''(⋅, '''v''')}}}}
where the dot ( ⋅ ) indicates the slot into which the argument for the resulting [[linear functional]] is to be placed (see [[Currying]]).

For a finite-dimensional vector space {{mvar|V}}, if either of {{math|''B''<sub>1</sub>}} or {{math|''B''<sub>2</sub>}} is an isomorphism, then both are, and the bilinear form {{math|''B''}} is said to be [[Degenerate form|nondegenerate]].  More concretely, for a finite-dimensional vector space, non-degenerate means that every non-zero element pairs non-trivially with some other element:
:<math>B(x,y)=0 </math> for all <math>y \in V</math> implies that {{math|1=''x'' = 0}} and
:<math>B(x,y)=0 </math> for all <math>x \in V</math> implies that {{math|1=''y'' = 0}}.

The corresponding notion for a module over a commutative ring is that a bilinear form is '''{{visible anchor|unimodular}}''' if {{math|''V'' → ''V''<sup>∗</sup>}} is an isomorphism. Given a finitely generated module over a commutative ring, the pairing may be injective (hence "nondegenerate" in the above sense) but not unimodular. For example, over the integers, the pairing {{math|1=''B''(''x'', ''y'') = 2''xy''}} is nondegenerate but not unimodular, as the induced map from {{math|1=''V'' = '''Z'''}} to {{math|1=''V''<sup>∗</sup> = '''Z'''}} is multiplication by 2.

If {{mvar|V}} is finite-dimensional then one can identify {{mvar|V}} with its double dual {{math|''V''<sup>∗∗</sup>}}. One can then show that {{math|''B''<sub>2</sub>}} is the [[Transpose of a linear map|transpose]] of the linear map {{math|''B''<sub>1</sub>}} (if {{mvar|V}} is infinite-dimensional then {{math|''B''<sub>2</sub>}} is the transpose of {{math|''B''<sub>1</sub>}} restricted to the image of {{mvar|V}} in {{math|1=''V''<sup>∗∗</sup>}}). Given {{math|''B''}} one can define the ''transpose'' of {{math|''B''}} to be the bilinear form given by
{{block indent|left=1.6|text=<sup>t</sup>''B''('''v''', '''w''') = ''B''('''w''', '''v''').}}

The '''left radical''' and '''right radical''' of the form {{math|''B''}} are the [[kernel (algebra)|kernel]]s of {{math|''B''<sub>1</sub>}} and {{math|''B''<sub>2</sub>}} respectively;{{sfn|Jacobson|2009|page=346}} they are the vectors orthogonal to the whole space on the left and on the right.{{sfn|Zhelobenko|2006|page=11}}

If {{mvar|V}} is finite-dimensional then the [[rank (linear algebra)|rank]] of {{math|''B''<sub>1</sub>}} is equal to the rank of {{math|''B''<sub>2</sub>}}. If this number is equal to {{math|dim(''V'')}} then {{math|''B''<sub>1</sub>}} and {{math|''B''<sub>2</sub>}} are linear isomorphisms from {{mvar|V}} to {{math|''V''<sup>∗</sup>}}.  In this case {{math|''B''}} is nondegenerate. By the [[rank–nullity theorem]], this is equivalent to the condition that the left and equivalently right radicals be trivial. For finite-dimensional spaces, this is often taken as the ''definition'' of nondegeneracy:
{{block indent|left=1.6|text= '''Definition:''' ''B'' is '''nondegenerate''' if {{math|1=''B''('''v''', '''w''') = 0}} for all '''w''' implies {{math|1='''v''' = '''0'''}}.}}

Given any linear map {{math|1=''A'' : ''V'' → ''V''<sup>∗</sup>}} one can obtain a bilinear form ''B'' on ''V'' via
{{block indent|left=1.6|text=''B''('''v''', '''w''') = ''A''('''v''')('''w''').}}

This form will be nondegenerate if and only if {{math|''A''}} is an isomorphism.

If {{mvar|V}} is [[finite-dimensional]] then, relative to some [[basis (linear algebra)|basis]] for {{mvar|V}}, a bilinear form is degenerate if and only if the [[determinant]] of the associated matrix is zero. Likewise, a nondegenerate form is one for which the determinant of the associated matrix is non-zero (the matrix is [[non-singular matrix|non-singular]]). These statements are independent of the chosen basis. For a module over a commutative ring, a unimodular form is one for which the determinant of the associate matrix is a [[Unit (ring theory)|unit]] (for example 1), hence the term; note that a form whose matrix determinant is non-zero but not a unit will be nondegenerate but not unimodular, for example {{math|1=''B''(''x'', ''y'') = 2''xy''}} over the integers.

===Symmetric, skew-symmetric, and alternating forms===
We define a bilinear form to be
* '''[[Symmetric bilinear form|symmetric]]''' if {{math|1=''B''('''v''', '''w''') = ''B''('''w''', '''v''')}} for all {{math|'''v'''}}, {{math|'''w'''}} in {{mvar|V}};
* '''[[Alternating form|alternating]]''' if {{math|1= ''B''('''v''', '''v''') = 0}} for all {{math|'''v'''}} in {{mvar|V}};
* '''{{visible anchor|skew-symmetric bilinear form|text=skew-symmetric}}''' or '''{{visible anchor|antisymmetric bilinear form|text=antisymmetric}}''' if {{math|1=''B''('''v''', '''w''') = −''B''('''w''', '''v''')}} for all {{math|'''v'''}}, {{math|'''w'''}} in {{mvar|V}};
*; Proposition: Every alternating form is skew-symmetric.
*; Proof: This can be seen by expanding {{math|''B''('''v''' + '''w''', '''v''' + '''w''')}}.

If the [[Characteristic (algebra)|characteristic]] of {{mvar|K}} is not 2 then the converse is also true: every skew-symmetric form is alternating. However, if {{math|1=char(''K'') = 2}} then a skew-symmetric form is the same as a symmetric form and there exist symmetric/skew-symmetric forms that are not alternating.

A bilinear form is symmetric (respectively skew-symmetric) [[if and only if]] its coordinate matrix (relative to any basis) is [[Symmetric matrix|symmetric]] (respectively [[Skew-symmetric matrix|skew-symmetric]]). A bilinear form is alternating if and only if its coordinate matrix is skew-symmetric and the diagonal entries are all zero (which follows from skew-symmetry when {{math|char(''K'') ≠ 2}}).

A bilinear form is symmetric if and only if the maps {{math|''B''<sub>1</sub>, ''B''<sub>2</sub>: ''V'' → ''V''<sup>∗</sup>}} are equal, and skew-symmetric if and only if they are negatives of one another. If {{math|char(''K'') ≠ 2}} then one can decompose a bilinear form into a symmetric and a skew-symmetric part as follows
<math display="block">B^{+} = \tfrac{1}{2} (B + {}^{\text{t}}B)  \qquad  B^{-} = \tfrac{1}{2} (B - {}^{\text{t}}B) ,</math>
where {{math|<sup>t</sup>''B''}} is the transpose of {{math|''B''}} (defined above).

===Reflexive bilinear forms and orthogonal vectors===
{{block indent|left=1| '''Definition:''' A bilinear form {{math|''B'' : ''V'' × ''V'' → ''K''}} is called '''reflexive''' if {{math|1=''B''('''v''', '''w''') = 0}} implies {{math|1=''B''('''w''', '''v''') = 0}} for all '''v''', '''w''' in ''V''.}}
{{block indent|left=1| '''Definition:''' Let {{math|''B'' : ''V'' × ''V'' → ''K''}} be a reflexive bilinear form. '''v''', '''w''' in ''V'' are '''orthogonal with respect to ''B''''' if {{math|1=''B''('''v''', '''w''') = 0}}.}}

A bilinear form {{math|''B''}} is reflexive if and only if it is either symmetric or alternating.{{sfn|Grove|1997}}  In the absence of reflexivity we have to distinguish left and right orthogonality. In a reflexive space the left and right radicals agree and are termed the ''kernel'' or the ''radical'' of the bilinear form: the subspace of all vectors orthogonal with every other vector.  A vector {{math|'''v'''}}, with matrix representation {{math|''x''}}, is in the radical of a bilinear form with matrix representation {{math|''A''}}, if and only if {{math|1=''Ax'' = 0 ⇔ ''x''<sup>T</sup>''A'' = 0}}. The radical is always a subspace of {{math|''V''}}.  It is trivial if and only if the matrix {{math|''A''}} is nonsingular, and thus if and only if the bilinear form is nondegenerate.

Suppose {{mvar|W}} is a subspace.  Define the ''[[orthogonal complement]]''{{sfn|Adkins|Weintraub|1992|page=359}}
<math display="block"> W^{\perp} = \left\{\mathbf{v} \mid B(\mathbf{v}, \mathbf{w}) = 0 \text{ for all } \mathbf{w} \in W\right\} .</math>

For a non-degenerate form on a finite-dimensional space, the map {{math|''V/W'' → ''W''<sup>⊥</sup>}} is [[bijective]], and the dimension of {{math|''W''<sup>⊥</sup>}} is {{math|dim(''V'') − dim(''W'')}}.

===Bounded and elliptic bilinear forms===
'''Definition:''' A bilinear form on a [[normed vector space]]  {{math|(''V'', ‖⋅‖)}} is '''bounded''', if there is a constant {{math|''C''}} such that for all {{math|'''u''', '''v''' ∈ ''V''}},
<math display="block"> B ( \mathbf{u} , \mathbf{v}) \le C \left\| \mathbf{u} \right\| \left\|\mathbf{v} \right\| .</math>

'''Definition:''' A bilinear form on a normed vector space {{math|(''V'', ‖⋅‖)}} is '''elliptic''', or [[Coercive function#Coercive operators and forms|coercive]], if there is a constant {{math|''c'' > 0}} such that for all {{math|'''u''' ∈ ''V''}},
<math display="block"> B ( \mathbf{u} , \mathbf{u}) \ge c \left\| \mathbf{u} \right\| ^2 .</math>