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Difference between revisions of "Bilinear mapping"

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''bilinear function''
 
''bilinear function''
  
A mapping $f$ from the product $V\times A$ of a left unitary $A$-module $V$ and of right unitary $B$-module $W$ into an $(A,B)$-[[Bimodule|bimodule]] $H$, satisfying the conditions
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A mapping $f$ from the product $V\times W$ of a left unitary $A$-module $V$ and of right unitary $B$-module $W$ into an $(A,B)$-[[Bimodule|bimodule]] $H$, satisfying the conditions
  
 
$$f(v+v',w)=f(v,w)+f(v',w);$$
 
$$f(v+v',w)=f(v,w)+f(v',w);$$

Latest revision as of 17:52, 8 December 2014

2020 Mathematics Subject Classification: Primary: 15-XX [MSN][ZBL]

bilinear function

A mapping $f$ from the product $V\times W$ of a left unitary $A$-module $V$ and of right unitary $B$-module $W$ into an $(A,B)$-bimodule $H$, satisfying the conditions

$$f(v+v',w)=f(v,w)+f(v',w);$$

$$f(v,w+w')=f(v,w)+f(v,w');$$

$$f(av,w)=af(v,w);$$

$$f(v,wb)=f(v,w)b;$$ where $v,v'\in V,\ w,w'\in W,\ a\in A,\ b\in B$ are arbitrarily chosen elements, and $A$ and $B$ are rings with a unit element. The tensor product $V\otimes W$ over $\Z$ has the natural structure of an $(A,B)$-bimodule. Let $\def\phi{\varphi}\phi:V\times W\to V\otimes W$ be a canonical mapping; any bilinear mapping $f$ will then induce a homomorphism of $(A,B)$-bimodules $\tilde f:V\otimes W\to H$ for which $f=\tilde f\circ\phi$. If $A=B$ and $A$ is commutative, then the set $L_2(V,W,H)$ of all bilinear mappings $V\times W\to H$ is an $A$-module with respect to the pointwise defined operations of addition and multiplication with elements in $A$, while the correspondence $f\mapsto \tilde f$ establishes a canonical isomorphism between the $A$-module $L_2(V,W,H)$ and the $A$-module $L(V\otimes W,H)$ of all linear mappings from $V\otimes W$ into $H$.

Let $V$ and $W$ be free modules with bases $v_i,\ i\in I$, and $w_j$, $j\in J$, respectively. A bilinear mapping $f$ is fully determined by specifying $f(v_i,w_j)$ for all $i\in I$, $j\in J$, since for any finite subsets $I'\subset I$, $J'\subset J$, the following formula is valid:

$$f\Big(\sum_{i\in I'}a_iv_i,\sum_{j\in J'}w_jb_j\Big) = \sum_{i\in I',j\in J'} a_if(v_i,w_j)b_j.\label{1}$$ Conversely, after the elements $h_{ij},\;i\in I,\;j\in J$, have been chosen arbitrarily, formula (1), where $f(v_i,w_j) = h_{ij}$, defines a bilinear mapping from $V\times W$ into $H$. If $I$ and $J$ are finite, the matrix $(f(v_i,w_j))$ is said to be the matrix of $f$ with respect to the given bases.

Let a bilinear mapping $f:V\times W\to H$ be given. Two elements $v\in V$, $w\in W$ are said to be orthogonal with respect to $f$ if $f(v,w) = 0$. Two subsets $X\subset V$ and $Y\subset W$ are said to be orthogonal with respect to $f$ if any $x\in X$ is orthogonal to any $y\in Y$. If $X$ is a submodule in $V$, then

$$X^\perp = \{w\in W:f(x,w) = 0\textrm{ for all } x\in X \},$$ which is a submodule of $W$, is called the orthogonal submodule or the orthogonal complement to $X$. The orthogonal complement $Y^\perp$ of the submodule $Y$ in $W$ is defined in a similar way. The mapping $f$ is said to be right-degenerate (left-degenerate) if $V^\perp \ne \{0\}$ ($W^\perp \ne \{0\}$). The submodules $V^\perp $ and $W^\perp $ are called, respectively, the left and right kernels of the bilinear mapping $f$. If $V^\perp = \{0\}$ and $W^\perp = \{0\}$, then $f$ is said to be non-degenerate; otherwise it is said to be degenerate. The mapping $f$ is said to be a zero mapping if $V^\perp = W$ and $W^\perp = V$.

Let $V_i,\ i\in I$, be a set of left $A$-modules, let $W_i,\ i\in I$, be a set of right $B$-modules, let $f_i$ be a bilinear mapping from $V_i\times W_i$ into $H$, let $V$ be the direct sum of the $A$-modules $V_i$, and let $W$ be the direct sum of the $B$-modules $W_i$. The mapping $f:V\times W\to H$, defined by the rule

$$f\Big(\sum_{i\in I}v_i,\sum_{i\in I}w_i\Big) = \sum_{i\in I} f_i(v_i,w_i),$$ is a bilinear mapping and is said to be the direct sum of the mappings $f_i$. This is an orthogonal sum, i.e. the submodule $V_i$ is orthogonal to the submodule $W_j$ with respect to $f$ if $i\ne j$.

The bilinear mapping $f$ is non-degenerate if and only if $f_i$ is non-degenerate for all $i\in I$. Moreover, if $f$ is non-degenerate then one has

$$V_i^\perp = \sum{j\ne i}W_j, \quad W_i^\perp = \sum{j\ne i}V_j,$$ If $A=B=H$, a bilinear mapping is called a bilinear form.

References

[Bo] N. Bourbaki, "Elements of mathematics. Algebra: Algebraic structures. Linear algebra", 1, Addison-Wesley (1974) pp. Chapt.1;2 MR0354207
[La] S. Lang, "Algebra",

Addison-Wesley (1974) MR0277543 Zbl 0984.00001

How to Cite This Entry:
Bilinear mapping. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Bilinear_mapping&oldid=35497
This article was adapted from an original article by V.L. Popov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article