BILINEAR FORM

In mathematics, a 'bilinear form' on a vector space ''V'' over a field ''F'' is a mapping ''V'' × ''V'' → ''F'' which is linear in both arguments. That is, ''B'' : ''V'' × ''V'' → ''F'' is bilinear if the maps
:$v\; mapsto\; B(v,\; w)$
:$v\; mapsto\; B(w,\; v)$
are linear for each ''w'' in ''V''. This definition applies equally well to modules over a commutative ring with linear maps being module homomorphisms.
Note that a bilinear form is a special case of a bilinear operator.
When ''F'' is the field of complex numbers 'C' one is often more interested in sesquilinear forms. These are similar to bilinear forms but are conjugate linear in one argument instead of linear.

Contents

Coordinate representation

Maps to the dual space

Reflexivity and orthogonality

Different spaces

Relation to tensor products

On normed vector spaces

See also

External links

Coordinate representation

Let $C=\{e\_\{1\},ldots,e\_\{n\}\}$ be a basis for a finite-dimensional space ''V''. Define the $n\; imes\; n$ - matrix ''A'' by $(A\_\{ij\})=B(e\_\{i\},e\_\{j\})$. Then if the $n\; imes\; 1$ matrix x represents a vector ''v'' with respect to this basis, and analogously, ''y'' represents ''w'', then:
:$B(v,w)\; =\; x^\{T\}\; A\; y,$
Suppose '' C' '' is another basis for ''V'', with :

with ''S'' an invertible $n\; imes\; n$ - matrix.
Now the new matrix representation for the symmetric bilinear form is given by :
$A\text{'}\; =S^\{T\}\; A\; S$

Maps to the dual space

Every bilinear form ''B'' on ''V'' defines a pair of linear maps from ''V'' to its dual space ''V''
★ . Define $B\_1,B\_2colon\; V\; o\; V^$
★ by
:$B\_1(v)(w)\; =\; B(v,w)$
:$B\_2(v)(w)\; =\; B(w,v)$
This is often denoted as
:$B\_1(v)\; =\; B(v,\{cdot\})$
:$B\_2(v)\; =\; B(\{cdot\},v)$
where the ($cdot$) indicates the slot into which the argument is to be placed.
If either of ''B''₁ or ''B''₂ is an isomorphism, then both are, and the bilinear form ''B'' is said to be 'nondegenerate'.
If ''V'' is finite-dimensional then one can identify ''V'' with its double dual ''V''
★
★ . One can then show that ''B''₂ is the transpose of the linear map ''B''₁ (if ''V'' is infinite-dimensional then ''B''₂ is the transpose of ''B''¹ restricted to the image of ''V'' in ''V''
★
★ ). Given ''B'' one can define the ''transpose'' of ''B'' to be the bilinear form given by
:$B^$
★ (v,w) = B(w,v).
If ''V'' is finite-dimensional then the rank of ''B''₁ is equal to the rank of ''B''₂. If this number is equal to the dimension of ''V'' then ''B''₁ and ''B''₂ are linear isomorphisms from ''V'' to ''V''
★ . In this case ''B'' is nondegenerate. By the rank-nullity theorem, this is equivalent to the condition that the kernel of ''B''₁ be trivial. In fact, for finite dimensional spaces, this is often taken as the ''definition'' of nondegeneracy. Thus ''B'' is nongenerate if and only if
:$B(v,w)=0mbox\{\; for\; all\; w\}Rightarrow\; v=0.$
Given any linear map ''A'' : ''V'' → ''V''
★ one can obtain a bilinear form ''B'' on ''V'' via
:$B(v,w)\; =\; A(v)(w)$
This form will be nondegenerate if and only if ''A'' is an isomorphism.

Reflexivity and orthogonality

A bilinear form
:''B'' : ''V'' × ''V'' → ''F''
is ''reflexive'' if
:$B(v,w)=0Longleftrightarrow\; B(w,v)=0$.
Reflexivity allows us to define orthogonality: two vectors ''v'' and ''w'' are ''orthogonal'' with respect to the reflexive bilinear form if and only if :
:$B(v,w)=0$ or $B(w,v)=0$
The radical of a bilinear form is the subset of all vectors orthogonal with every other vector. A vector ''v'', with matrix representation ''x'', is in the radical if and only if :
$A\; x=\; 0\; Longleftrightarrow\; x^\{T\}\; A=0$
The radical is always a subspace of ''V''. It is trivial if and only if the matrix ''A'' is nonsingular, and thus if and only if the bilinear form is nondegenerate.
Suppose ''W'' is a subspace. Define :

When the bilinear form is nondegenerate, the map $Wleftarrow\; W^\{perp\}$ is bijective, and the dimension of $W^\{perp\}$ is dim(''V'')-dim(''W'').
One can prove that ''B'' is reflexive if and only if it is :

★ 'symmetric' : $B(v,w)=B(w,v)$ for all $v,win\; V$; OR

★ 'alternating' if $B(v,v)=0$ for all $vin\; V$
Every alternating form is skew-symmetric ( $B(v,w)=-B(w,v)$). This may be seen by expanding ''B''(''v''+''w'',''v''+''w'').
If the characteristic of ''F'' is not 2 then the converse is also true (every skew-symmetric form is alternating). If, however, char(''F'') = 2 then a skew-symmetric form is the same thing as a symmetric form and not all of these are alternating.
A bilinear form is symmetric (resp. skew-symmetric) if and only if its coordinate matrix (relative to any basis) is symmetric (resp. 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 char(''F'') ≠ 2).
A bilinear form is symmetric if and only if the maps $B\_1,B\_2colon\; V\; o\; V^$
★ are equal, and skew-symmetric if and only if they are negatives of one another. If char(''F'') ≠ 2 then one can always decompose a bilinear form into a symmetric and a skew-symmetric part as follows
:$B^\{pm\}\; =\; \{1over\; 2\}(B\; pm\; B^$
★ )
where ''B''
★ is the transpose of ''B'' (defined above).

Different spaces

Much of the theory is available for a bilinear mapping
:''B'': ''V'' × ''W'' → ''F''.
In this situation we still have linear mappings of ''V'' to the dual space of ''W'', and of ''W'' to the dual space of ''V''. It may happen that both of those mappings are isomorphisms; assuming finite dimensions, if one is an isomorphism, the other must be. When this occurs, ''B'' is said to be a 'perfect pairing'.

Relation to tensor products

By the universal property of the tensor product, bilinear forms on ''V'' are in 1-to-1 correspondence with linear maps ''V'' ⊗ ''V'' → ''F''. If ''B'' is a bilinear form on ''V'' the corresponding linear map is given by
:$votimes\; wmapsto\; B(v,w).$
The set of all linear maps ''V'' ⊗ ''V'' → ''F'' is the dual space of ''V'' ⊗ ''V'', so bilinear forms may be thought of as elements of
:$(Votimes\; V)^\{$
★ } cong V^{
★ }otimes V^{
★ }.
Likewise, symmetric bilinear forms may be thought of as elements of ''S''²''V''
★ (the second symmetric power of ''V''
★ ), and alternating bilinear forms as elements of Λ²''V''
★ (the second exterior power of ''V''
★ ).

On normed vector spaces

A bilinear form on a normed vector space is 'bounded', if there is a constant $C$ such that for all $u,\; vin\; V$
:$B(u,v)\; le\; C\; |u|\; |v|.$
A bilinear form on a normed vector space is 'elliptic', or 'coercive', if there is a non-zero constant $c$ such that for all $uin\; V$
:$B(u,u)\; ge\; c\; |u|^2.$

See also

★ bilinear operator

★ multilinear form

★ quadratic form

★ sesquilinear form

★ inner product space

External links

★