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Linear algebra, isomorphism

  • Thread starter atlantic
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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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  • #2
I like Serena
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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]?
 
  • #3
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But how do I prove this using the given information?
 
  • #4
I like Serena
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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.
 
  • #5
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Bijection is not covered in my course:frown: Is there not any other way to make the proof?
 
  • #6
I like Serena
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Bijection is not covered in my course:frown: Is there not any other way to make the proof?
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).
 
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