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Dual Basis

by Diophantus
Tags: basis, dual
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Diophantus
#1
Mar11-07, 05:26 PM
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Whilst trying to refresh myself on what a dual space of a vector space is I have confused myself slightly regarding conventions. (I am only bothered about finite dimensional vector spaces.)

I know what a vector space, a dual space and a basis of a vector space are but dual bases:

I seem to recall something about the delta function being used. Does every basis have a dual basis or just standard the bases (i.e. (1,0,..,0), (0,1,0,...,0) , ... (0, ... ,0,1) )? Does the delta function rule have to apply to a dual basis or is it just a condition which ensures that a dual basis of an orthonormal basis is also orthonormal?
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matt grime
#2
Mar11-07, 05:35 PM
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Given a basis e_i of a vector space V, the dual basis of V* is defined by f_i that satsify the rule f_i(e_j)=delta_{ij}. As the definition indicates, every basis V goves rise to a dual basis of V*.

Orthonormality doesn't enter into the question (these are just vector spaces, not inner product spaces).
Diophantus
#3
Mar11-07, 06:04 PM
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Ah, so the delta function ensures a 1-1 correspondance beteween the two sets of bases. Thanks.

mathwonk
#4
Mar15-07, 09:28 AM
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Dual Basis

the infinite dimensional case is more interesting. I.e. the functions dual to a basis of V do not then give absis for V*. I.e. a linear combination of those dual functions must vanish on all but a finite number of the original basis vectors, but a general linear functiion can do anything on them.

so how do you find a natural basis for ALL linear functions? I do not know the answer. perhaps there is no nice basis.

e.g. as an analogy think of a linear function on the positive integers with values in the set {0,...,9} as an infinte decimal.

then how would you find a "basis" for all infinite decimals, such that every other infinite decimal is a finite Z linear combination of those?

this is not a perfect analogy but gives some idea. i.e. a big problem is the basis would have to be uncountable.
HallsofIvy
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Mar15-07, 01:33 PM
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The dual of an infinite dimensional vector space is not, in general, isomorphic to the original space (but the dual of the dual is!).
matt grime
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Mar15-07, 04:57 PM
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Quote Quote by HallsofIvy View Post
The dual of an infinite dimensional vector space is not, in general, isomorphic to the original space (but the dual of the dual is!).
Oh no it is not. In general the double dual doesn't even have the same cardinality as the original for infinite dimensional vector spaces. There is, in general, a canonical inclusion of V into V**, and if V is a pure injective module for a ring then it is a split injection.


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