Bilinear Map
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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 ${\displaystyle \mathbb {F} }$. 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 ${\displaystyle \lambda \in \mathbb {F} }$, ${\displaystyle B(\lambda v,w)=B(v,\lambda w)=\lambda B(v,w)}$.
• 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 and we have for all 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 with T an -bimodule, and for which any n in N, is an R-module homomorphism, and for any m in M, is an S-module homomorphism. This satisfies

B(r ? m, n) = r ? B(m, n)
B(m, n ? s) = 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

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

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

A matrix M determines a bilinear map into the reals by means of a real bilinear form , then associates of this are taken to the other three possibilities using duality and the musical isomorphism

If V, W, X are finite-dimensional, then so is . For , i.e. bilinear forms, the dimension of this space is (while the space of linear forms is of dimension ). To see this, choose a basis for V and W; then each bilinear map can be uniquely represented by the matrix , and vice versa. Now, if X is a space of higher dimension, we obviously have .

## Examples

• Matrix multiplication is a bilinear map .
• If a vector space V over the real numbers R carries an inner product, then the inner product is a bilinear map .
• In general, for a vector space V over a field F, a bilinear form on V is the same as a bilinear map .
• If V is a vector space with dual space V*, then the application operator, is a bilinear map from 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 defines a bilinear map .
• The cross product in R3 is a bilinear map .
• Let be a bilinear map, and be a linear map, then is a bilinear map on .

## Continuity and separate continuity

Suppose X, Y, 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 X, Y, 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 X, Y, and Z be locally convex Hausdorff spaces and let ${\displaystyle C:L\left(X;Y\right)\times L\left(Y;Z\right)\to L\left(X;Z\right)}$ be the composition map defined by ${\displaystyle C(u,v):=v\circ u}$. In general, the bilinear map 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 E is an equicontinuous subset of ${\displaystyle L\left(Y;Z\right)}$ then the restriction ${\displaystyle C{\big \vert }_{L\left(X;Y\right)\times E}:L\left(X;Y\right)\times E\to L\left(X;Z\right)}$ is continuous for all three topologies.[1]
• If Y is a barreled space then for every sequence ${\displaystyle \left(u_{i}\right)_{i=1}^{\infty }}$ converging to u in ${\displaystyle L\left(X;Y\right)}$ and every sequence ${\displaystyle \left(v_{i}\right)_{i=1}^{\infty }}$ converging to v in ${\displaystyle L\left(Y;Z\right)}$, the sequence ${\displaystyle \left(v_{i}\circ u_{i}\right)_{i=1}^{\infty }}$ converges to ${\displaystyle v\circ u}$ in ${\displaystyle L\left(Y;Z\right)}$. [1]