Dual basis

Jump to: navigation, search

to a basis $\{ e _ {1} \dots e _ {n} \}$ of a module $E$ with respect to a form $f$

A basis $\{ c _ {1} \dots c _ {n} \}$ of $E$ such that

$$f ( e _ {i} , c _ {i} ) = 1 ,\ f ( e _ {i} , c _ {j} ) = 0 ,$$

$$i \neq j ,\ 1 \leq i , j \leq n ,$$

where $E$ is a free $K$- module over a commutative ring $K$ with a unit element, and $f$ is a non-degenerate (non-singular) bilinear form on $E$.

Let $E ^ {*}$ be the dual module of $E$, and let $\{ e _ {1} ^ {*} \dots e _ {n} ^ {*} \}$ be the basis of $E ^ {*}$ dual to the initial basis of $E$: $e _ {i} ^ {*} ( e _ {i} ) = 1$, $e _ {i} ^ {*} ( e _ {j} )= 0$, $i \neq j$. To each bilinear form $f$ on $E$ there correspond mappings $\phi _ {f} , \psi _ {f} : E \rightarrow E ^ {*}$, defined by the equations

$$\phi _ {f} ( x) ( y) = f ( x, y) ,\ \ \psi _ {f} ( x) ( y) = f ( y, x) .$$

If the form $f$ is non-singular, $\phi _ {f} , \psi _ {f}$ are isomorphisms, and vice versa. Here the basis $\{ c _ {1} \dots c _ {n} \}$ dual to $\{ e _ {1} \dots e _ {n} \}$ is distinguished by the following property:

$$\psi _ {f} ( c _ {i} ) = e _ {i} ^ {*} \ \ ( i = 1 \dots n) .$$

Comments

A bilinear form $f$ on $E$ is non-degenerate (also called non-singular) if for all $x \in E$, $f ( x , y ) = 0$ for all $y$ implies $x = 0$ and for all $y \in E$, $f ( x , y ) = 0$ for all $x$ implies $y = 0$. Occasionally the terminology conjugate module (conjugate space) is used instead of dual module (dual space).

References

 [a1] P.M. Cohn, "Algebra" , 1 , Wiley (1982)
How to Cite This Entry:
Dual basis. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Dual_basis&oldid=46777
This article was adapted from an original article by E.N. Kuz'min (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article