Relationship between a vector and its discrete fourier transform Suppose we have a real vector $a=(a_0,\ldots,a_{n-1})$ and let $b=(b_0,\ldots,b_{n-1})$ be its discrete fourier trasnform. So $$\mathscr{F}(a_k)=b_k=\sum_{i=0}^{n-1}{a_i}e^{\frac{-j2\pi i k}{n}}$$Let $j=\sqrt{-1}$ here .
I'm trying to show that
$$n\sum^{n-1}_{i=0}a_i^2=\sum^{n-1}_{i=0}|b_i|^2$$
But I'm not exactly sure how to approach this problem, we can try substituting the Fourier transform (in the form by euler's formula)
$$n\sum^{n-1}_{i=0}a_i^2=\sum^{n-1}_{i=0}\left|\sum^{n-1}_{q=0}a_q\left[\cos\left(\frac{2\pi iq}{n}\right)-j\sin\left(\frac{2\pi iq}{n}\right)\right]\right|^2$$
but this doesn't really help. Is there some geometric intuition or fact I'm missing about this question? For example the Sum of nth roots of unity is 0, but unfortunately we have to deal with the coefficients $a_k$ for each $b_k$. Any ideas is appreciated, thanks!
 A: Instead of considering the sum, let's consider the matrix form.
Let the discrete Fourier transform matrix to be $F$, which can be expressed using $\omega$ $n$th unit root. $\omega^n=1$
$$
F_{ik} = \exp(-j\frac{2\pi}{n} ki) =\omega^{ki},\;\omega=\exp(-j\frac{2\pi}{n})
$$
In matrix form it looks like this
$$
F=\begin{bmatrix}
1 & 1 & 1 & ... & 1 \\
1 & \omega & \omega^2 & ... & \omega^{n-1}\\
1 & \omega^2 & \omega^4 & ... & \omega^{2(n-1)}\\
 & & & ...\\
1 & \omega^{n-1} & \omega^{2(n-1)} & ... & \omega^{(n-1)^2}\\
\end{bmatrix}
$$
Then DFT transform in your definition is
$$
b=Fa
$$
What is to be proved is the following ($\bar b$ means complex conjugate)
$$
\sum_i |b_i|^2=\bar b^T b = n a^Ta=n\sum_i a_i^2
$$
Then we just need to prove
$$
\forall a, \bar b^T b = \bar a^T\bar F^T F a = a^T\bar F F a=na^Ta\\
\bar F^T F = n I_n
$$
The first line used the fact $a$ is real and $F$ is symmetric.
So we just need to show the $F$ in your definition satisfy $\bar FF = nI$. (in standard definition $F$ is normalized and thus unitary, in your case it's not)
Note that $\bar F$ just substitute $\omega\to \omega^{-1}$
Thus for a pair of index $i,k$ if $i==k$
$$
(\bar FF)_{ii}=\sum_{l=0}^{n-1} \omega^{-li}\omega^{li} = \sum_{l=0}^{n-1}1=n
$$
If $i\neq k$, the last equation used the rotational symmetry of the sum of $n$ different $n$th unit root.
$$
(\bar FF)_{ik}=\sum_{l=0}^{n-1} \omega^{-li}\omega^{lk} = \sum_{l=0}^{n-1}\omega^{l(k-i)}=0
$$
So we get
$$
\bar FF = nI\\
\bar b^T b = na^Ta
$$

For the sum, with nonzero integer $\Delta$,
$$
\begin{align}
\sum_{l=0}^{n-1}\omega^{l\Delta} = &\omega^{-\Delta}\sum_{l=0}^{n-1}\omega^{(l+1)\Delta}\\
=&\omega^{-\Delta}\sum_{l=1}^{n}\omega^{l\Delta}\\
=&\omega^{-\Delta}(\sum_{l=1}^{n-1}\omega^{l\Delta} + 1)\; (\mbox{note 
 } 
 \omega^n=1)\\
=&\omega^{-\Delta}\sum_{l=0}^{n-1}\omega^{l\Delta}
\end{align}
$$
