Prove a complex matrix $A$ is unitary

Question

Let $A$ be a $n\times n$ complex matrix and $A^*$ the adjoint of $A$, i.e. $(A^*)_{ij}= \bar{A}_{ji}$.

Let $w_n=e^{2\pi i/n}$ be a $n$th root of 1. Suppose that the $ij$ th entry is defined by $A_{ij}=w^{ij}_n/ \sqrt{n}$. Prove $A$ is unitary.

We want to show that $A^*A=I$.

I tried writing out the definitions.

We have $A^*A = \sum_{j=1}^n (\sum_{i=1}^n \bar{A}_{ji}A_{ij}) =\sum_{j=1}^n (\sum_{i=1}^n \frac{\bar{w_n}^{ji}}{\sqrt{n}} \frac{w_n^{ij}}{\sqrt{n}}) $.

But from here I didn't get anywhere. I think I'm missing some obvious identities.

Thanks in advance!

Answer 1

Here $$(A^*A)_{jk} = \sum_{\ell=1}^n \frac{\overline w_n^{\ell j}}{\sqrt n} \frac{w_n^{\ell k}}{\sqrt n} = \frac 1 n \sum^n_{\ell=1} \overline w_n^{\ell j}w_n^{\ell k}.$$ If $j = k$, then $$(A^*A)_{jj} = \frac 1 n \sum^n_{\ell=1} (\overline w_n w_n)^{2\ell j} = \frac 1 n \sum^n_{\ell=1} 1^{2\ell j} = \frac 1 n \sum^n_{\ell = 1} 1 = 1.$$

If $j \neq k$, then $\overline w_n^{\ell j} w_n^{\ell k} = e^{2 \pi i \ell (k-j)/n}$ so $$(A^*A)_{jk} = \frac 1 n \sum^n_{\ell = 1} \left(e^{2\pi i(k-j)/n}\right)^\ell = \frac {e^{2\pi i(k-\ell)/n}} n \sum^{n-1}_{\ell=0}\left(e^{2\pi i(k-j)/n}\right)^\ell = \frac {e^{2\pi i(k-\ell)/n}} n \frac{e^{2\pi i (k-\ell)} - 1}{e^{2\pi i(k-\ell)/n} - 1}.$$ The numerator is zero since $2\pi i (k-\ell)$ is an integer multiple of $2 \pi i$.

Answer 2

$\left( A^* A\right)_{k,j}=\sum_{h=1}^n A^*_{kh}A_{hj}= \sum_{h=1}^n \bar A_{hk}A_{hj}=\frac{1}{\sqrt n\sqrt n}\sum_{h=1}^n \overline{e^{\frac{2\pi i}{n}hk}}e^{\frac{2\pi i}{n}hj}=\frac{1}{n}\sum_{h=1}^n {e^{-\frac{2\pi i}{n}hk}}e^{\frac{2\pi i}{n}hj}=\frac{1}{n}\sum_{h=1}^n {e^{(j-k)\frac{2\pi i}{n}h}}=\delta_{kj}\stackrel{\text{ if }j\neq k}{=}\bigg\{\frac{1}{n}\sum_{h=0}^{n-1} {(e^{(j-k)\frac{2\pi i}{n}})^h}=\frac{1}{n}\left( \frac{1-e^{(j-k)\frac{2\pi in}{n}}}{1-e^{(j-k)\frac{2\pi i}{n}}}\right)=0\bigg\} .$