It is $F_n = (\exp(2 \pi i /n)^{j\cdot k}) \in \mathbb{C}^{n,n}$ with $j,k \in \{1,...,n\}$.
What is $F_n^H \cdot F_n$?
I already figured out that $F_n^H = \overline{F_n}$ and I think the product is n-times the unit matrix.
But how can i show / compute this?
Let $C$ be the multiplication result, so it is $c_{k,j} = \displaystyle\sum\limits_{j=1}^n \overline{f_{kj}} \cdot f_{jk}$. Which is $\displaystyle\sum\limits_{j=1}^n \exp(2 \pi i /n)^{- (k\cdot j)} \cdot \exp(2 \pi i /n)^{j\cdot k}$ (sorry if the indices are wrong).
Since $\exp(2 \pi i /n)^{- (k\cdot j)} \cdot \exp(2 \pi i /n)^{j\cdot k} = 1 $ every entry of the resulting matrix would be $\displaystyle\sum\limits_{j=1}^n 1 = n$ and not only those on the principal diagonal, right? Where is my fallacy?
\begin{align} c_{k,j}&= \sum_{\color{red}l=1}^n \bar{F}_{\color{red}lk}F_{\color{red}lj} \\ &= \sum_{l=1}^n \exp\left(\frac{-2\pi i lk}{n} \right) \exp\left(\frac{2\pi i lj}{n} \right)\\ &=\sum_{l=1}^n \exp \left(\frac{2\pi il(j-k)}{n} \right) \end{align}
Now, let's consider two cases, if $j=k$, then we end up up $\sum_{l=1}^n \exp(0)=\sum_{l=1}^n 1= n$.
If $j \ne k$, then
\begin{align} \sum_{l=1}^n \exp\left( \frac{2\pi i l (j-k)}{n}\right) &= \exp\left( \frac{2\pi i(j-k)}{n}\right)\cdot \frac{1-\exp(2\pi i(j-k))}{1-\exp\left( \frac{2\pi i(j-k)}{n}\right)} \\ &=\exp\left( \frac{2\pi i(j-k)}{n}\right)\cdot \frac{1-1}{1-\exp\left( \frac{2\pi i(j-k)}{n}\right)} \\ &=\exp\left( \frac{2\pi i(j-k)}{n}\right)\cdot \frac{0}{1-\exp\left( \frac{2\pi i(j-k)}{n}\right)} \\ &= 0 \end{align}