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Homework Help: Linear algebra, isomorphism

  1. Oct 27, 2011 #1
    Information:
    The vector-space [itex]\mathcal{F}([0,\pi],\mathbb{R})[/itex] consists of all real functions on [itex][0,\pi][/itex]. We let [itex]W[/itex] be its subspace with the basis [itex]\mathcal{B}[/itex] = {[itex]1,cost,cos(2t),cos(3t),...,cos(7t)[/itex]}.

    [itex]T: W \rightarrow \mathbb{R} ^8[/itex] is the transformation where: [itex]T(h) = (h(t_1), h(t_2),...,h(t_8))[/itex], [itex]h \in W[/itex] and [itex]t_i = (\pi(2i-1))/16[/itex] , [itex]i\in[/itex]{[itex]1,2,...8[/itex]}

    The call the standard basis for [itex]\mathbb{R} ^8[/itex] for [itex]\mathcal{C}[/itex]. The change-of-coordinates matrix of [itex]T[/itex] from [itex]\mathcal{B}[/itex] to [itex]\mathcal{C}[/itex] is an invertibel matrix we call [itex]M[/itex].



    My question is how I can show that [itex]T[/itex] is an isomorphism. I know that this means that [itex]T[/itex] must be a invertibel linear transformation, but how do I show this? Does it have anything to do with the fact that the change-of-coordinates matrix [itex]M[/itex] is invertibel, and how?
     
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  3. Oct 27, 2011 #2

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    Hi atlantic! :smile:

    Doesn't Fourier guarantee a unique discrete transform between 8 points and 8 amplitudes on the cosine basis limited to the interval [0,pi]?
     
  4. Oct 28, 2011 #3
    But how do I prove this using the given information?
     
  5. Oct 28, 2011 #4

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    You need to show that T is a bijection that conserves multiplication.

    I haven't worked it out yet, but the Fourier series gives a bijection between h(ti) in ℝ8 and an in ℝ8 which are the coefficients of the cosines:
    [tex]a_n = \sum_{i=1}^8 h(t_i) \cos n t_i[/tex]
    I think it identifies your matrix M.
     
  6. Oct 28, 2011 #5
    Bijection is not covered in my course:frown: Is there not any other way to make the proof?
     
  7. Oct 28, 2011 #6

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    I believe invertible linear transformation covers it.
    If you can find an invertible transformation matrix M that does the job between B and C, it is trivial that it is a bijection.

    And I just checked and it turns out that for vector spaces to be isomorphic you do not need conservation of multiplication (actually that is implicit since they are linear).
     
    Last edited: Oct 28, 2011
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