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A Isomorphism between a linear space and its dual

  1. Mar 27, 2016 #1

    Geofleur

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    I have been trying to prove the following theorem, for a finite dimensional vector space ## X ## and its dual ## X^* ##:

    Let ## f:X\rightarrow X^* ## be given by ## f(x) = (x|\cdot) ##, where ## (x|\cdot) ## is linear in the first argument and conjugate linear in the second (so I am using the mathematicians' convention here). Then ## f ## is bijective if and only if [## (y|x)=0 ## for all ## x \in X ## implies that ## y = 0 ##].

    I have been able to show the "only if" part - that ## f ## being bijective implies the statement about the inner product. For the "if" part of the proof (where we assume that [## (y|x)=0 ## for every ## x \in X ## implies ## y = 0 ##] and prove that ## f ## is bijective), I am able to prove that ## f ## is injective, but I am running into a problem in proving that ## f ## is surjective. Here is my attempt:

    Let ##x^*## be an arbitrary linear functional in ## X^*##. Because the space ## X ## is finite dimensional, we may choose an orthonormal basis ## \{e_i\}_{i=1}^n ## and write an arbitrary vector in ## X ## as ## x = \sum_{i=1}^{n}x_i e_i ##. Then we have ## x^*(x) = x^*(\sum_{i=1}^{n} x_i e_i) = \sum_{i=1}^{n}x_i x^*(e_i) ##. Now we need to find a ## y \in X ## such that ## f(y) = x^*##, i.e., such that ## (y|x) = x^*(x) ## for all ## x \in X ##. For any ## y \in X ## we can write ## y = \sum_{j=1}^{n} y_j e_j ##, so maybe choosing the ## y_j ## appropriately will do the job. We have ## (y|x) = (\sum_{j=1}^{n} y_j e_j | \sum_{i=1}^{n} x_i e_i) = \sum_{j} \sum_{i} y_j \overline{x}_i (e_j|e_i) = \sum_{i} y_i \overline{x}_i ##. The problem is that I want this to equal ## x^*(x) = \sum_{i=1}^{n}x_i x^*(e_i) ##, but the former expression involves ## \overline{x}_i ## while the latter involves ## x_i ##. If I had used the physicists' convention and made the inner product conjugate linear in the first argument instead, this problem would not have occurred. But it seems strange that a theorem like this would be dependent upon a convention like that. Any insights?
     
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  3. Mar 27, 2016 #2

    Samy_A

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    Isn't the problem that in ##(y|x) = x^*(x)##, the LHS is conjugate linear in x, while the RHS is linear in x?

    In other words, shouldn't f be defined as ##f(x) = (\cdot|x)##?
     
  4. Mar 27, 2016 #3

    micromass

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    Another is by saying ##X## and ##X^*## have the same dimension.
     
  5. Mar 27, 2016 #4

    Geofleur

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    @Samy_A: That makes sense - it means I would need to make a correction in the book I am reading, though (Analysis, Manifolds, and Physics, Choquet-Bruhat).

    @micromass: Do you mean I was wrongfully assuming that ## X ## and ## X^* ## have the same dimension? I thought that all I had assumed is that the action of ## x^* ## is completely determined by its effect on the basis vectors of ## X ##. Did I sneak something else in without realizing it?
     
  6. Mar 27, 2016 #5

    micromass

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    No, it's correct. In finite dimensions, ##X## and ##X^*## have the same dimension. But once you know that, the proof is done. See the alternative theorem. This says that a linear map between two spaces of the same finite dimension is an isomorphism iff it is surjective iff it is injective.
     
  7. Mar 27, 2016 #6

    Geofleur

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    Ah - I understand! Thanks :-D
     
  8. Mar 27, 2016 #7

    micromass

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    If you're interested in quantum mechanics or functional analysis, you probably will want to think about how much this result generalizes to infinite dimensions...
     
  9. Mar 28, 2016 #8

    WWGD

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    Are you assuming a default inner product defined?
     
  10. Mar 28, 2016 #9

    Geofleur

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    I think so? I am defining a sequilinear mapping ## X \times X \rightarrow \mathbb{C} ## such that ## (x,y) \mapsto (x|y) ##, where

    ## (x|y) = \overline{(y|x)} ##, and
    ## (\alpha x + \beta y|z) = \alpha(x|z) + \beta(y|z) ##.

    I ended up just writing a note in the margin pointing out that ## (x|\cdot) \in X^* ## if one uses the physics convention; otherwise, we have ## (\cdot|x) \in X^* ## as Samy_A suggested.
     
    Last edited: Mar 28, 2016
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