Bilinear form

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In mathematics, a bilinear form on a vector space V is a bilinear mapping V × V → F, where F is the field of scalars. That is, a bilinear form is a function B: V × V → F which is linear in each argument separately:

$\begin{array}{l} \text{1. }B(u + u',v) = B(u,v) + B(u',v)\text{,} \\[4pt] \text{2. }B(u,v + v') = B(u,v) + B(u,v')\text{,} \\[4pt] \text{3. }B(\lambda u,v) = B(u, \lambda v) = \lambda\,B(u,v)\text{.} \\[4pt] \end{array}$

Any bilinear form on Fⁿ can be expressed as

$B(\textbf{x},\textbf{y}) = \textbf{x}^{\mathrm{T}}A\textbf{y} = \sum_{i,j=1}^n a_{ij} x_i y_j$

where A is an n × n matrix.

The definition of a bilinear form can easily be extended to include modules over a commutative ring, with linear maps replaced by module homomorphisms. When F is the field of complex numbers C, one is often more interested in sesquilinear forms, which are similar to bilinear forms but are conjugate linear in one argument.

1 Coordinate representation
2 Maps to the dual space
3 Reflexivity and orthogonality
4 Different spaces
5 Relation to tensor products
6 On normed vector spaces
7 See also
8 References
9 External links

[edit] Coordinate representation

Let $C=\{e_{1},\ldots,e_{n}\}$ be a basis for a finite-dimensional space V. Define the $n\times n$ - matrix A by $(A i j) = B (e i, e j)$ . Then if the $n\times 1$ matrix x represents a vector v with respect to this basis, and analogously, y represents w, then:

$B(v,w) = x^{T} A y\,$

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 :

$A' = S T A S$

[edit] 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 nondegenerate 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.

[edit] Reflexivity and orthogonality

A bilinear form

B : V × V → F

is reflexive if

$B(v,w)=0\Longleftrightarrow B(w,v)=0$ .

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

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 : $W^{\perp}=\{v| B(v,w)=0\ \forall w\in W\}$

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 either:

symmetric : $B (v, w) = B (w, v)$ for all $v,w\in V$ ; or
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

$B^{\pm} = \frac{1}{2} (B \pm B^*)$

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

[edit] 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.

In finite dimensions, this is equivalent to the pairing being nondegenerate (the spaces necessarily having the same dimensions). For modules (instead of vector spaces), nondegenerate is a weaker notion: a pairing can be nondegenerate without being a perfect pairing, for instance $\mathbf{Z} \times \mathbf{Z} \to \mathbf{Z}$ via $(x,y) \mapsto 2xy$ is non-degenerate, but induces multiplication by 2 on the map $\mathbf{Z} \to \mathbf{Z}^*$

[edit] 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*).

[edit] On normed vector spaces

A bilinear form on a normed vector space is bounded, if there is a constant $C$ such that for all $u, v\in V$

$B(u,v) \le C \|u\| \|v\|.$

A bilinear form on a normed vector space is elliptic, or coercive, if there is a non-zero constant $c$ such that for all $u\in V$

$B(u,u) \ge c \|u\|^2.$

[edit] See also

[edit] References

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