Does Matrix M Prove T is an Isomorphism Between Vector Spaces?

[atlantic]

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

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

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



My question is how I can show that T is an isomorphism. I know that this means that T 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 M is invertibel, and how?

 

[I like Serena]

Hi atlantic!

Doesn't Fourier guarantee a unique discrete transform between 8 points and 8 amplitudes on the cosine basis limited to the interval [0,pi]?

 

[atlantic]

But how do I prove this using the given information?

 

[I like Serena]

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(t_i) in ℝ⁸ and aⁿ in ℝ⁸ which are the coefficients of the cosines:
a_n = \sum_{i=1}^8 h(t_i) \cos n t_i
I think it identifies your matrix M.

 

[atlantic]

Bijection is not covered in my course Is there not any other way to make the proof?

 

[I like Serena]

Bijection is not covered in my course 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).