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Double Dual Isomorphism Proof

  1. Sep 14, 2006 #1
    I am trying to prove that T'' = T (where T'' is the double transpose of T) by showing that the the dual of the dual of a linear finite vector space is isomorphic to the original vector space.

    i.e., T: X --> U (A linear mapping)
    The transpose of T is defined as the following:
    T': U' --> X' (Here U' is the dual of U and X' is the dual of X)
    And the double transpose of T is defined as:
    T'': X'' --> U'' (Here X'' is the dual of the dual of X and U'' is the dual of the dual of U)
    And since X'' is isomorphic to X and U'' is isomorphic to U, (This is the part I still need to prove)
    T'' = T
     
    Last edited: Sep 14, 2006
  2. jcsd
  3. Sep 14, 2006 #2

    matt grime

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    Hold on there. I got confused because you're using * to mean transpose when * means dual, and then you talk about dual spaces.... confusing.

    Anyway, just use the definition of transpose (which is what, in your course?) and dual.

    (Ohj, and the double dual of a (linear) vector space is not isomorphic to the original space, in general, by the way.)
     
    Last edited: Sep 14, 2006
  4. Sep 14, 2006 #3
    I have edited my original post in hopes it will make things a little clearer.
     
  5. Sep 14, 2006 #4

    matt grime

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    The editing still omits the crucial information about what *your* definition of the 'transpose/dual' of a linear map is. In my world it is taulogically trivial that the double dual is what you started with. But as you haven't defined the 'transpose/dual' it is impossible to offer advice.
     
  6. Sep 14, 2006 #5
    The definition of the dual is the linear space formed by the set of linear functions on a linear space X.
     
  7. Sep 14, 2006 #6

    matt grime

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    Now, given a linear map X-->Y how do you define the map on the duals Y*--->X*? Do it twice and what do you have?
     
  8. Sep 14, 2006 #7
    Alright, thanks. I understand now.
     
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