Lemma Let be a Fourier Matrix then we have Proof Lets examine For these terms we have Misplaced && = \sum_{i=0}^{n-1} \omega^{i(j-k)}\\ & = \sum_{i=0}^{n-1} \left ( \omega^{j-k} \right )^n \end{align*}$$ Which splits into cases if $j = k$ then $\omega^{j-k} = 1$ and we have this sum is $n$. Whereas if $j \not = k$ from the [[Sum of roots of unity|sum of roots of unity]] we have this sum is 0. Therefore $$ F_n(\omega^{-1}) F_n(\omega) = n I_n$$ giving us the desired result.