# Linear algebra, isomorphism

1. Oct 27, 2011

### 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?

2. Oct 27, 2011

### 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]?

3. Oct 28, 2011

### atlantic

But how do I prove this using the given information?

4. Oct 28, 2011

### 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(ti) 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.

5. Oct 28, 2011

### atlantic

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

6. Oct 28, 2011

### I like Serena

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