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Relating to Duality in Vector Spaces

  1. Jul 24, 2009 #1
    This subject came up in some notes on linear algebra I'm reading and I don't get it. Please help me understand.

    --

    First, the relevant background and notation relating to my question:

    Let S be a nonempty set and F be a field. Denote by l(S) the family of all F-valued functions on S and l_c(S) the family of all functions mapping S to F with finite support; that is, those functions on S which are nonzero at only finitely many elements of S.

    Now let V be a vector space with (over F) with basis B. Define M : l_{c}(B) \rightarrow V by M(v) = sum_{e \in B}v(e)e.

    Note that l(B) (under pointwise operations) is a vector space, l_c(B) is a vector subspace of l(B), and M is an isomorphism.

    If we write v for M(v) we see that v = sum_{e \in B}v(e)e.

    We will go further and use M to identify V with l_c(B) and write v = sum_{e \in B}v(e)e. That is, in a vector space with basis we will treat a vector as a scalar valued function on its basis.

    --

    And now here is the question: according to the above, what is the value of f(e) when e and f are elements of the basis B? (And most importantly, why?)
     
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  3. Jul 24, 2009 #2

    Hurkyl

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    Writing [tex]\hat{f}[/tex] for [tex]M^{-1}(f)[/tex]....

    The relevant equation here is that
    [tex]f = \sum_{e \in B} \hat{f}(e) e[/tex]

    Knowing that B is a basis, what does that tell you?
     
  4. Jul 26, 2009 #3
    If [tex] f = \sum_{e \in B} \hat{f}(e)e [/tex] then since B is a basis we must have [tex] \hat{f}(e) = 0 [/tex] for all e in B.

    Yes? No?
     
    Last edited: Jul 26, 2009
  5. Jul 26, 2009 #4

    Hurkyl

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    Well, you can check yourself! What happens if you plug that in for [tex]\hat{f}[/tex]?
     
  6. Jul 27, 2009 #5
    Well if [tex] \hat{f}(e) = 0 [/tex] for all [tex] e \in B [/tex], I'm lead to unsatisfactory conclusion that [tex] f = 0 [/tex]. (Unsatisfactory since f is, by assumption, a member of the basis B; 0 is never a member of a linearly independent set.)

    Hrmm... If we write the following:

    [tex] f = (\sum_{e \in B - \{f\}}\hat{f}(e)e) + \hat{f}(e)f [/tex]

    it seems clear that [tex] \hat{f}(e) = 0 [/tex] whenever [tex] e \in B - \{f\} [/tex] (that is [tex]e \neq f [/tex]) and [tex] \hat{f}(e) = 1 [/tex] whenever [tex] e = f [/tex]; more concisely: [tex] \hat{f}(e) = \delta_{ef} [/tex] (Kronecker delta).
     
    Last edited: Jul 27, 2009
  7. Jul 27, 2009 #6

    Hurkyl

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    That looks reasonable. That formula for [tex]\hat{f}(e)[/tex] is a well-defined function of both f and e, and [tex]M(\hat{f}) = f[/tex] as desired.

    So, your question is settled, then?
     
  8. Jul 27, 2009 #7

    Hurkyl

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    Incidentally, can you see how an element of lc(B) behaves as if it was a "linear combination in the elements of B"? And how M-1 acts as the function that "writes a vector as a linear combination of basis elements"?
     
  9. Jul 27, 2009 #8
    Not quite sure what you mean.

    And actually I have another 'exercise' with which I need help; it appears to be related to the above question.

    Let v be a nonzero vector in a vector space V and E be a basis for V which contains the vector v. Then there exists a linear functional [tex] \phi \in V^* [/tex] (the dual of V) such that [tex] \phi(v) = 1 [/tex] and [tex] \phi(e) = 0 [/tex] for every [tex] e \in E - \{v\} [/tex].

    Now this function [tex] \phi [/tex] seems somewhat similar to the [tex] f, \hat{f} [/tex] about which we spoke of earlier. But I'm having trouble seeing it all fits together; in particular how [tex] l_c(B) [/tex] relates to the dual of V, and also how to define [tex] \phi [/tex].

    By the way, in case you were wondering, the 'goal' of the section of the notes from which I'm drawing these questions is to prove what the author is calling the "Riesz-Fréchet theorem for vector spaces".
     
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