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Dual Basis

  1. Mar 11, 2007 #1
    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?
  2. jcsd
  3. Mar 11, 2007 #2

    matt grime

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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).
  4. Mar 11, 2007 #3
    Ah, so the delta function ensures a 1-1 correspondance beteween the two sets of bases. Thanks.
  5. Mar 15, 2007 #4


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    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.
  6. Mar 15, 2007 #5


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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!).
  7. Mar 15, 2007 #6

    matt grime

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    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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