$F^HF$ and inverse of $F=[e^{i\frac{2\pi kl}{n}}]_{k,l=0}^{n-1}$ Let n=1,2,3,... and $i^2=-1$ and:
$$F=[e^{i\frac{2\pi kl}{n}}]_{k,l=0}^{n-1}\in\Bbb{C}^{n,n}$$
Find $F^HF$ and $F^{-1}$.
In this quite challenging (at least for me) problem I started from finding the matrix $F^HF$. In order to do that you need find $F^H$ first. I think the equation of this matrix is $F^H=[e^{-i\frac{2\pi kl}{n}}]_{k,l=0}^{n-1}$. So now let's think about matrix $F^HF$. To do that we need to know what is the k'th row of $F^H$. It looks like this:
$[e^{-i\frac{2\pi k0}{n}} e^{-i\frac{2\pi k1}{n}} e^{-i\frac{2\pi k2}{n}}... e^{-i\frac{2\pi k(n-1)}{n}}]$.
Now what about j'th column of $F$? According to definition it should look like that:
$[e^{i\frac{2\pi 0j}{n}} e^{i\frac{2\pi 1j}{n}} e^{i\frac{2\pi 2j}{n}}... e^{i\frac{2\pi (n-1)j}{n}}]^T$
So let's think about indices k,j of $F^HF$. It should be sum of multiplication of corresponding elements of k'th row of $F^H$ and j'th column of $F$. So the value at indices k,j of $F^HF$ should look like that:
$e^{-i\frac{2\pi k0}{n}}*e^{i\frac{2\pi 0j}{n}} + e^{-i\frac{2\pi k1}{n}}*e^{i\frac{2\pi 1j}{n}}+...+e^{-i\frac{2\pi k(n-1)}{n}}*e^{i\frac{2\pi (n-1)j}{n}}$
So we can write $F^HF$ down as:
$[\sum_{m=0}^{n-1}e^{\frac{i2\pi m}{n}(j-k)}]_{k,j=0}^{n-1}$
Have I done everything right? Or maybe I completely screwed up this part of the problem?
Also, how to proceed with finding $F^{-1}$?
 A: In case someone struggles with something similair in the future, I think I've managed to solve this problem so I'm posting my answer:
It turns out that my result that $F^HF=[\sum_{m=0}^{n-1}e^{\frac{i2\pi m}{n}(j-k)}]_{k,j=0}^{n-1}$ is probably true and it can be written in a simpler form. If you look at this equation long enough you see that it is geometric series inside it, and we know a simple equation that enables us to solve it. For $m=0$ we always have $e^0=1$ so $a_1=1$, $q=e^{\frac{i2\pi(j-k)}{n}}$, and we have $n$ terms in the series. So plugging that in into equation $S_n=\frac{a_1(1-q^n)}{1-q}$ (notice: we can't use that equation for $q=1$ and that is the case when $j=k$, so we'll deal with this later, now we are only looking at values not on diagonal of the matrix) we have $S_n=\frac{(1-q^n)}{1-q}$ when plugging $a_1$ and now we see something interesting. Let's look at $1-q^n$. If $q=e^{\frac{i2\pi(j-k)}{n}}$, then $q^n=e^{i2\pi(j-k)}$. As $j-k$ always give us an integer, it is basically equal to $e^{i2\pi}=1$. So $1-q^n=1-1=0$. That means that values that are not on the diagonal of the matrix are always equal to $0$ so the $F^HF$ is a diagonal matrix. What is on its diagonal? When $j=k$ we have:
$\sum_{m=0}^{n-1}e^0=n*1=n$, that means that $F^HF=diag(n)$.
Finding the $F^{-1}$ is now trivial, it's just $\frac{1}{n}F^H$, because $(\frac{1}{n}F^H)F=\frac{1}{n}(F^HF)=\frac{1}{n}*diag(n)=diag(1)=I_n$
Notice: I'm not sure about my answer, it would still be helpful if someone verified it.
