# Dual basis

*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