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

$${\displaystyle B:V\times W\to X}$$

${\displaystyle w\in W}$

${\displaystyle B_{w))$

$${\displaystyle v\mapsto B(v,w)}$$

${\displaystyle V}$

${\displaystyle X,}$

${\displaystyle v\in V}$

${\displaystyle B_{v))$

$${\displaystyle w\mapsto B(v,w)}$$

${\displaystyle W}$

${\displaystyle X.}$

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 (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 × N → T with T an (R, S)-bimodule, and for which any n in N, m ↦ B(m, n) is an R-module homomorphism, and for any m in M, n ↦ B(m, n) 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

An immediate consequence of the definition is that B(v, w) = 0_X whenever v = 0_V or w = 0_W. This may be seen by writing the zero vector 0_V as 0 ⋅ 0_V (and similarly for 0_W) 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(e_i, f_j), 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 ${\displaystyle V\times V\to \mathbb {R} .}$ The product vector space has one dimension.
• In general, for a vector space V over a field F, a bilinear form on V is the same as a bilinear map V × V → F.
• 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 × W → F.
• 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 ${\displaystyle X,Y,}$ and ${\displaystyle Z}$ 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]