Let $F$ be an $n \times n$ matrix with $F_{j,k} = \frac{1}{\sqrt n} \bar \omega^{(j-1)(k-1)}$. Show that the matrix $F$ is unitary. Let $F$ be an $n \times n$ matrix with $F_{j,k} = \frac{1}{\sqrt n} \bar \omega^{(j-1)(k-1)}$, where $j,k=1,2,3,...n$ and $\omega=e^{\frac{2\pi i}{n}}$. Show that the matrix $F$ is  unitary.
So, we need to show $FF^{*}=I$? And use $F^{*}=\bar F^{T}$?
 A: To find $FF^*_{i,j}$, let's multiply the $i^{\text{th}}$ row by the conjugate of the $j^{\text{th}}$ column.
$$FF^*_{i,j}=\sum_{k=1}^n F_{ik}\bar{F_{kj}}=\sum_{k=1}^n \frac{1}{\sqrt{n}}\bar\omega^{(i-1)(k-1)} \cdot \frac{1}{\sqrt{n}}\omega^{(k-1)(j-1)}=\frac 1 n \sum_{k=1}^n\bar\omega^{(i-1)(k-1)}\omega^{(k-1)(j-1)}$$
Since $\omega$ is a root of unity, its conjugate is $\omega^{-1}$, so we can change $\bar\omega^{(i-1)(k-1)}$ to $\omega^{(i-1)(1-k)}$:
$$FF^*_{ij}=\frac 1 n \sum_{k=1}^n\omega^{(i-1)(1-k)}\omega^{(k-1)(j-1)}=\frac 1 n \sum_{k=1}^n \omega^{(i-1)(1-k)+(k-1)(j-1)}=\frac 1 n \sum_{k=1}^n \omega^{jk-ik-j+i}=\frac 1 n \sum_{k=1}^n \omega^{i-j}\omega^{jk-ik} \\ =\frac {\omega^{i-j}} n \sum_{k=1}^n \omega^{jk-ik}=\frac {\omega^{i-j}} n \sum_{k=1}^n (\omega^{j-i})^k$$
Now, if $j \neq i \implies j-i \neq 0$, then $\omega^{j-i}$ is an $n^{\text{th}}$ root of unity, meaning the sum of its powers from $1$ to $n$ is $0$, so $FF^*_{i,j}$ for $j \neq i$ is $0$.
Otherwise, when $j=i \implies j-i=0$, then $\omega^{j-i}=1$, meaning the sum of its powers from $1$ to $n$ is $n$, giving us $\frac {\omega^{i-i}} n \sum_{k=1}^n (\omega^{i-i})^k=\frac 1 n\cdot n=1$, so $FF^*_{i,j}$ for $j = i$ is $1$.
The above two sentences proves $FF^*=I$, so $F$ is unitary.
