# Bilinear map

(Redirected from Bilinear operator)

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

$B:V\times W\to X$

such that for all $w\in W$ , the map $B_{w}$

$v\mapsto B(v,w)$

is a linear map from $V$  to $X$ , and for all $v\in V$ , the map $B_{v}$

$w\mapsto B(v,w)$

is a linear map from $W$  to $X$ . 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 $B$  satisfies the following properties.

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

If V = W and we have B(v, w) = B(w, v) 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 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

A first 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.

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

{\begin{matrix}{\begin{aligned}&V\times V{\overset {M}{\longrightarrow }}\mathbb {R} \\&(v,w)\mapsto v'Mw\\&M_{ij}=M(b_{i},b_{j})\\&M\in V^{*}\otimes V^{*}\\&M=M_{st}\beta ^{s}\otimes \beta ^{t}\\~\\~\end{aligned}}&{\begin{aligned}&V\times V^{*}{\overset {M}{\longrightarrow }}\mathbb {R} \\&(v,f)\mapsto v'Mf'\\&M_{i}^{j}=M(b_{i},\beta ^{j})\\&M\in V^{*}\otimes V\\&M=M_{s}^{t}\beta ^{s}\otimes b_{t}\\&M_{s}^{t}=M_{su}g^{ut}\\&MG^{-1}\end{aligned}}\\{\begin{aligned}&V^{*}\times V{\overset {M}{\longrightarrow }}\mathbb {R} \\&(f,w)\mapsto fMw\\&M_{j}^{i}=M(\beta ^{i},b_{j})\\&M\in V\otimes V^{*}\\&M=M_{t}^{s}b_{s}\otimes \beta ^{t}\\&M_{t}^{s}=g^{su}M_{ut}\\&G^{-1}M\end{aligned}}&{\begin{aligned}&V^{*}\times V^{*}{\overset {M}{\longrightarrow }}\mathbb {R} \\&(f,g)\mapsto fMg'\\&M^{ij}=M(\beta ^{i},\beta ^{j})\\&M\in V\otimes V\\&M=M^{st}b_{s}\otimes b_{t}\\&M^{st}=g^{su}M_{uv}g^{vt}\\&G^{-1}MG^{-1}\end{aligned}}\end{matrix}}

If V, W, X are finite-dimensional, then so is L(V, W; X). For X = F, i.e. 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 R carries an inner product, then the inner product is a bilinear map V × VR.
• 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 R3 is a bilinear map R3 × R3R3.
• Let B : V × WX be a bilinear map, and L : UW be a linear map, then (v, u) ↦ B(v, Lu) is a bilinear map on V × U.

## Continuity and separate continuity

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

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

Many separately continuous bilinear that are not continuous satisfy an additional property: hypocontinuity. 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 $b:X\times Y\to Z$  is continuous.
• If X, Y, and Z are the strong duals of Fréchet spaces then every separately continuous bilinear map $b:X\times Y\to Z$  is continuous.
• If a bilinear map is continuous at (0, 0) then it is continuous everywhere.

### Composition map

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