Conceptual question about Discrete Fourier Transform

Question

On the wikipedia page for the discrete Fourier transform, the first sentence says:

In mathematics, the discrete Fourier transform (DFT) converts a finite list of equally spaced samples of a function into the list of coefficients of a finite combination of complex sinusoids, ordered by their frequencies, that has those same sample values.

But as I learned it, the Fourier matrix $\mathcal{F}$ takes as its inputs vectors of coefficients $(c_0, c_1, \dotsc, c_{n-1})$ and produces a vector $(f(0), f(2\pi/n), f(2\cdot 2\pi/n), \dotsc, f((n-1)\cdot 2\pi/n))$, where $$f(x) = c_0 + c_1 e^{ix} + c_2 e^{2ix} + \dotsb + c_{n-1}e^{(n-1)ix}.$$

So the above description seems to be describing what the inverse Fourier matrix does. Why is this?

Answer 1

It's little bit more than just convention and comes down to how you identify the characters of the underlying group. Of course once you make a choice, everything works out similarly.

In the case of DFT, the group is the finite cyclic group $\mathbb{Z}_n$. When talking about characters, i.e. group homomorphims from $\mathbb{Z}_n$ to the unit circle, you can go around the circle either counter-clockwise or clockwise. If the former, $+i$ appears in your formula. If the latter, $-i$.

Same thing applies to the Fourier transform on $\mathbb{R}$. Do you wind the line around the circle counter-clockwise or clockwise, i.e. $i$ or $-i$? Do you wind it with period $1$ or $2 \pi$?...etc.

Answer 2

The definition of Fourier matrix is inconsistent in the literature. Wolfram says $\mathcal F_{j,k}=\exp(2\pi i jk/n)$, consistent with your notion of $\mathcal F$. Wikipedia puts $-i$ there, consistent with its description of DFT.

At some level, this is a moot point. I have a concept of $i$ in my head, and so do you. But there is no way to know if my concept of $i$ corresponds to your $i$ or to your $-i$. Between two roots of $-1$, one is chosen as $i$, but the choice is completely arbitrary.