Rational complex exponent and proof that a matrix is unitary

Question

I need to prove that a square matrix $A$ with elements $a_{jk}$ has orthogonal rows and columns: $$a_{jk} = exp\Big(\frac{2\pi i}{n}jk\Big)_{j,k=0}^{n-1}$$

It's possible to prove it by showing that $A$ multiplied by its conjugate transpose $B=A^*$ is an identity matrix $I$ times some number $\alpha$: $$AA^*=AB=\alpha I = C$$

Elements of the matrix $C$ can be computed as: $$C_{jk}=(AB)_{jk} = \sum_{m=0}^{n-1}A_{jm}B_{mk} = \sum_{m=0}^{n-1}exp\Big(\frac{2\pi i}{n}jm\Big)exp\Big(-\frac{2\pi i}{n}mk\Big)$$

If $j=k$ (diagonal elements) then $C_{jk}=\sum_{m=0}^{n-1}exp(0)=n$, however I can't figure out how to calculate $C_{jk}$ when $j \neq k$ because rules of exponentiation are somewhat different for complex exponents than for real ones.

I think the solution is related to the identity that $n$-th roots of unity, for $n > 1$, add up to $0$: ${\displaystyle \sum _{k=0}^{n-1}e^{2\pi i{\frac {k}{n}}}=0.}$

QUESTION:

How to prove $C_{jk}=0$ for $j \neq k$ and what exponentiation rules are still applicable for rational complex exponents?

Answer 1

Observe \begin{align} \sum^{n-1}_{m=0} \exp\left(\frac{2\pi i}{n} jm \right)\exp\left(-\frac{2\pi i}{n} mk \right) = \sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{n} (j-k)m \right). \end{align} Assume $j \neq k$ and WLOG assume $j>k$ which means \begin{align} (j-k) m = l m = nq+r_m \end{align} where $0\leq r_m < n$. Thus, it follows \begin{align} \sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{n} (j-k)m \right) = \sum^{n-1}_{m=0} \exp\left(2\pi i q+ \frac{2\pi i}{n}r_m \right)= \sum^{n-1}_{m=0} \exp\left( \frac{2\pi i}{n}r_m\right). \end{align}

Case 1 $l$ is coprime to $n$

Then it's not hard to see that $r_m$ will take on very value $0, 1, \ldots , n-1$ which mean we could rewrite the sum as \begin{align} \sum^{n-1}_{k=0}\exp\left( \frac{2\pi i}{n}k\right)=0. \end{align}

Case 2 $l$ is not coprime to $n$.

In this case, we have that $l = dk$ and $n=dp$ where $d =\gcd (n, l)$ and $\gcd(k, p) =1$. Hence \begin{align} \sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{n} lm \right) = \sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{p}km \right)= d \sum^{p-1}_{m=0}\exp\left(\frac{2\pi i}{p}km \right)=0. \end{align}

Answer 2

If $0\le j,m\lt n$ and $j\ne m$, then $e^{2\pi i(j-m)}=1$ and $e^{2\pi i(j-m)\frac 1n}\ne1$. Therefore, using the formula for the sum of a geometric series, $$ \begin{align} \left(A\overline{A}^T\right)_{j,m} &=\sum_{k=0}^{n-1}e^{2\pi i\frac{jk}n}e^{-2\pi i\frac{km}n}\\ &=\sum_{k=0}^{n-1}e^{2\pi i(j-m)\frac kn}\\ &=\frac{e^{2\pi i(j-m)}-1}{e^{2\pi i(j-m)\frac 1n}-1}\\[9pt] &=0\tag{1} \end{align} $$ Whereas, if $j=m$, then $e^{2\pi i(j-m)\frac kn}=1$. Therefore, $$ \begin{align} \left(A\overline{A}^T\right)_{j,m} &=\sum_{k=0}^{n-1}e^{2\pi i\frac{jk}n}e^{-2\pi i\frac{km}n}\\ &=\sum_{k=0}^{n-1}e^{2\pi i(j-m)\frac kn}\\[9pt] &=n\tag{2} \end{align} $$ Therefore, $$ A\overline{A}^T=nI\tag{3} $$