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# Bilinear map

In mathematics, a bilinear map is a function combining elements of two vector spaces to yield an element of a third vector space, and is linear in each of its arguments. Matrix multiplication is an example.

## Definition

### Vector spaces

Let ${\displaystyle V,W}$ and ${\displaystyle X}$ be three vector spaces over the same base field ${\displaystyle F}$. A bilinear map is a function

${\displaystyle B:V\times W\to X}$
such that for all ${\displaystyle w\in W}$,the map ${\displaystyle B_{w}}$
${\displaystyle v\mapsto B(v,w)}$
is a linear map from ${\displaystyle V}$ to ${\displaystyle X,}$ and for all ${\displaystyle v\in V}$, the map ${\displaystyle B_{v}}$
${\displaystyle w\mapsto B(v,w)}$
is a linear map from ${\displaystyle W}$ to ${\displaystyle X.}$ In other words, when we hold the first entry of the bilinear map fixed while letting the second entry vary, the result is a linear operator, and similarly for when we hold the second entry fixed.

Such a map ${\displaystyle B}$ satisfies the following properties.

• For any ${\displaystyle \lambda \in F}$, ${\displaystyle B(\lambda v,w)=B(v,\lambda w)=\lambda B(v,w).}$
• The map ${\displaystyle B}$ is additive in both components: if ${\displaystyle v_{1},v_{2}\in V}$ and ${\displaystyle w_{1},w_{2}\in W,}$ then ${\displaystyle B(v_{1}+v_{2},w)=B(v_{1},w)+B(v_{2},w)}$ and ${\displaystyle B(v,w_{1}+w_{2})=B(v,w_{1})+B(v,w_{2}).}$

If ${\displaystyle V=W}$ and we have B(v, w) = B(w, v) for all ${\displaystyle v,w\in V,}$ then we say that B is symmetric. If X is the base field F, then the map is called a bilinear form, which are well-studied (see for example Scalar product, Inner product and Quadratic form).

### Modules

The definition works without any changes if instead of vector spaces over a field F, we use modules over a commutative ring R. It generalizes to n-ary functions, where the proper term is multilinear.

For non-commutative rings R and S, a left R-module M and a right S-module N, a bilinear map is a map B : M × NT with T an (R, S)-bimodule, and for which any n in N, mB(m, n) is an R-module homomorphism, and for any m in M, nB(m, n) is an S-module homomorphism. This satisfies

B(rm, n) = rB(m, n)
B(m, ns) = B(m, n) ⋅ s

for all m in M, n in N, r in R and s in S, as well as B being additive in each argument.

## Properties

An immediate consequence of the definition is that B(v, w) = 0X whenever v = 0V or w = 0W. This may be seen by writing the zero vector 0V as 0 ⋅ 0V (and similarly for 0W) and moving the scalar 0 "outside", in front of B, by linearity.

The set L(V, W; X) of all bilinear maps is a linear subspace of the space (viz. vector space, module) of all maps from V × W into X.

If V, W, X are finite-dimensional, then so is L(V, W; X). For ${\displaystyle X=F,}$ that is, bilinear forms, the dimension of this space is dim V × dim W (while the space L(V × W; F) of linear forms is of dimension dim V + dim W). To see this, choose a basis for V and W; then each bilinear map can be uniquely represented by the matrix B(ei, fj), and vice versa. Now, if X is a space of higher dimension, we obviously have dim L(V, W; X) = dim V × dim W × dim X.

## Examples

• Matrix multiplication is a bilinear map M(m, n) × M(n, p) → M(m, p).
• If a vector space V over the real numbers ${\displaystyle \mathbb {R} }$ carries an inner product, then the inner product is a bilinear map ${\displaystyle V\times V\to \mathbb {R} .}$
• In general, for a vector space V over a field F, a bilinear form on V is the same as a bilinear map V × VF.
• If V is a vector space with dual space V, then the application operator, b(f, v) = f(v) is a bilinear map from V × V to the base field.
• Let V and W be vector spaces over the same base field F. If f is a member of V and g a member of W, then b(v, w) = f(v)g(w) defines a bilinear map V × WF.
• The cross product in ${\displaystyle \mathbb {R} ^{3}}$ is a bilinear map ${\displaystyle \mathbb {R} ^{3}\times \mathbb {R} ^{3}\to \mathbb {R} ^{3}.}$
• Let ${\displaystyle B:V\times W\to X}$ be a bilinear map, and ${\displaystyle L:U\to W}$ be a linear map, then (v, u) ↦ B(v, Lu) is a bilinear map on V × U.

## Continuity and separate continuity

Suppose ${\displaystyle X,Y,{\text{ and }}Z}$ are topological vector spaces and let ${\displaystyle b:X\times Y\to Z}$ be a bilinear map. Then b is said to be separately continuous if the following two conditions hold:

1. for all ${\displaystyle x\in X,}$ the map ${\displaystyle Y\to Z}$ given by ${\displaystyle y\mapsto b(x,y)}$ is continuous;
2. for all ${\displaystyle y\in Y,}$ the map ${\displaystyle X\to Z}$ given by ${\displaystyle x\mapsto b(x,y)}$ is continuous.

Many separately continuous bilinear that are not continuous satisfy an additional property: hypocontinuity.[1] All continuous bilinear maps are hypocontinuous.

### Sufficient conditions for continuity

Many bilinear maps that occur in practice are separately continuous but not all are continuous. We list here sufficient conditions for a separately continuous bilinear to be continuous.

• If X is a Baire space and Y is metrizable then every separately continuous bilinear map ${\displaystyle b:X\times Y\to Z}$ is continuous.[1]
• If ${\displaystyle X,Y,{\text{ and }}Z}$ are the strong duals of Fréchet spaces then every separately continuous bilinear map ${\displaystyle b:X\times Y\to Z}$ is continuous.[1]
• If a bilinear map is continuous at (0, 0) then it is continuous everywhere.[2]

### Composition map

Let ${\displaystyle X,Y,{\text{ and }}Z}$ be locally convex Hausdorff spaces and let ${\displaystyle C:L(X;Y)\times L(Y;Z)\to L(X;Z)}$ be the composition map defined by ${\displaystyle C(u,v):=v\circ u.}$ In general, the bilinear map ${\displaystyle C}$ is not continuous (no matter what topologies the spaces of linear maps are given). We do, however, have the following results:

Give all three spaces of linear maps one of the following topologies:

1. give all three the topology of bounded convergence;
2. give all three the topology of compact convergence;
3. give all three the topology of pointwise convergence.
• If ${\displaystyle E}$ is an equicontinuous subset of ${\displaystyle L(Y;Z)}$ then the restriction ${\displaystyle C{\big \vert }_{L(X;Y)\times E}:L(X;Y)\times E\to L(X;Z)}$ is continuous for all three topologies.[1]
• If ${\displaystyle Y}$ is a barreled space then for every sequence ${\displaystyle \left(u_{i}\right)_{i=1}^{\infty }}$ converging to ${\displaystyle u}$ in ${\displaystyle L(X;Y)}$ and every sequence ${\displaystyle \left(v_{i}\right)_{i=1}^{\infty }}$ converging to ${\displaystyle v}$ in ${\displaystyle L(Y;Z),}$ the sequence ${\displaystyle \left(v_{i}\circ u_{i}\right)_{i=1}^{\infty }}$ converges to ${\displaystyle v\circ u}$ in ${\displaystyle L(Y;Z).}$ [1]