Confused as to how to prove the basis of dft is orthonormal I have been stuck for hours trying to prove that the basis of discrete fourier transform is orthonormal can anyone point me in the direction of how to do so
 A: Let $\omega = e^{-2\pi i/N}$. The Fourier matrix looks like
$$
\boldsymbol{F}=\frac{1}{\sqrt{N}}
\begin{bmatrix}
     1 &               1 &        1 & \ldots & 1 \\
     1 &          \omega & \omega^2 & \ldots & \omega^{2(N-1)} \\
     1 &        \omega^2 & \\
\vdots &          \vdots &          & \ddots \\
     1 & \omega^{2(N-1)} &          &        & \omega^{(N-1)^2}
\end{bmatrix},
$$
and you want to show that $\boldsymbol{F}^*\boldsymbol{F}=\boldsymbol{I}$. Note that
$$
\boldsymbol{F}^*=\frac{1}{\sqrt{N}}
\begin{bmatrix}
     1 &                1 &           1 & \ldots & 1 \\
     1 &      \omega^{-1} & \omega^{-2} & \ldots & \omega^{-2(N-1)} \\
     1 &      \omega^{-2} & \\
\vdots &           \vdots &          & \ddots \\
     1 & \omega^{-2(N-1)} &          &        & \omega^{-(N-1)^2}
\end{bmatrix}.
$$
Simply perform the matrix multiplication, and be sure to use the formula for the geometric sum,
$$
1+r+r^2+\ldots+r^{N-1}=
\begin{cases}
\frac{r^N-1}{r-1} & \text{if } r\neq 1\\
N & \text{if } r=1
\end{cases}.
$$
If you still need more details, just say so in a comment, and I will elaborate when I have time.
