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Bilinear map

( Multilinear Algebra )

Definition

Vector spaces
Modules

Properties
Examples
Continuity and separate ..

Sufficient conditions fo..
Composition map

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

A bilinear map can also be defined for modules. For that, see the article pairing.

Definition

Vector spaces

Let

V, W

and

X

be three over the same base field

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

X,

and for all

v \in V

, the map

B_v

w \mapsto B(v, w)

is a linear map from

W

X.

In other words, when we hold the second entry of the bilinear map fixed while letting the first entry vary, yielding

B_w

, the result is a linear operator, and similarly for when we hold the first entry fixed.

Such a map $B$ satisfies the following properties.

For any $\lambda \in F$ , $B(\lambda v,w) = B(v, \lambda w) = \lambda B(v, w).$
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 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 (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 map.

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 map in each argument.

Properties

An immediate consequence of the definition is that whenever or . This may be seen by writing the zero vector 0_V as (and similarly for 0_W) 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.

If V, W, X are finite-dimensional, then so is . For $X = F,$ that is, 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 $\R$ carries an inner product, then the inner product is a bilinear map $V \times V \to \R.$
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 canonical evaluation map, 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 $\R^3$ is a bilinear map $\R^3 \times \R^3 \to \R^3.$
Let $B : V \times W \to X$ be a bilinear map, and $L : U \to W$ 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

b : X \times Y \to Z

be a bilinear map. Then b is said to be if the following two conditions hold:

for all $x \in X,$ the map $Y \to Z$ given by $y \mapsto b(x, y)$ is continuous;
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 map 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 $X, Y, \text{ and } Z$ are the 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, \text{ and } Z

be locally convex and let

C : L(X; Y) \times L(Y; Z) \to L(X; Z)

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:

give all three the topology of bounded convergence;
give all three the topology of compact convergence;
give all three the topology of pointwise convergence.

If $E$ is an equicontinuous subset of $L(Y; Z)$ then the restriction $C\big\vert_{L(X; Y) \times E} : L(X; Y) \times E \to L(X; Z)$ 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 $L(X; Y)$ and every sequence $\left(v_i\right)_{i=1}^{\infty}$ converging to $v$ in $L(Y; Z),$ the sequence $\left(v_i \circ u_i\right)_{i=1}^{\infty}$ converges to $v \circ u$ in $L(Y; Z).$

Bilinear map

( Multilinear Algebra )

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Bilinear map ( Multilinear Algebra )

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Bilinear map

( Multilinear Algebra )