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.

Coordinate representation

Let $C=\{e_{1},\ldots ,e_{n}\}$ be a basis for V, assuming the latter is of finite dimension. Define the $n\times n$ - matrix A by $(A_{ij})=B(e_{i},e_{j})$ . Then if the $n\times 1$ matrix x represents a vector v with respect to this basis, and analogously, y represents w, $B(v,w)$ is given by :

$x^{T}Ay$

Suppose C' is another basis for V, with : ${\begin{bmatrix}e'_{1}&\cdots &e'_{n}\end{bmatrix}}={\begin{bmatrix}e_{1}&\cdots &e_{n}\end{bmatrix}}S$ with S an invertible $n\times n$ - matrix. Now the new matrix representation for the symmetric bilinear form is given by :

$A'=S^{T}AS$

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 $B_{1},B_{2}\colon V\to V^{*}$ 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 B₁ or B₂ 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 B₂ is the transpose of the linear map B₁ (if V is infinite-dimensional then B₂ is the transpose of B¹ 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 B₁ is equal to the rank of B₂. If this number is equal to the dimension of V then B₁ and B₂ 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 B₁ 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)=0{\mbox{ 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 said to be reflexive if $\forall v,w\in V:B(v,w)=0\Longleftrightarrow B(w,v)=0$ .

Reflexivity allows us to define orthogonality : two vectors v and w are said to be 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 : $Ax=0\Longleftrightarrow x^{T}A=0$ 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 : $W^{\perp }=\{v|B(v,w)=0\ \forall w\in W\}$

When the bilinear form is nondegenerate, the map $W\leftarrow W^{\perp }$ 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 $v,w\in V$
alternating if $B(v,v)=0$ for all $v\in V$

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 $B_{1},B_{2}\colon V\to V^{*}$ 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 }={1 \over 2}(B\pm B^{*})

where B* is the transpose of B (defined above).

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

v\otimes w\mapsto 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

(V\otimes V)^{*}\cong V^{*}\otimes V^{*}.

Likewise, symmetric bilinear forms may be thought of as elements of S²V* (the second symmetric power of V*), and alternating bilinear forms as elements of Λ²V* (the second exterior power of V*).