Consider the discrete fourier transform over a finite field $GF(q)$. Let also $\omega$$\in$$GF(q)$ be an element of order $n$ and which is an $n$-th root of unity.
Definition 1. Let $v$ = ($v_0$, $v_1$, ... ,$v_{n-1}$) be a vector over $GF(q)$ and $\omega$ be an element of $GF(q)$ of order $n$. The fourier transform of the vector $v$ is the vector $V$ = ($V_0$, $V_1$, ... ,$V_{n-1}$) with components given by
$V_j$ = $\sum_{i=0}^{n-1}w^{ij}v_i$, $j = 0, ... , n-1$
Definition 2. The vector v is related to the vector V by
$v_i$ = $\frac1n\sum_{j=0}^{n-1}w^{-ij}V_j$, where n is interpreted as an integer of the field.
*The double parentheses on subscripts indicate modulo n.
Question: Could someone explain the proof of following theorem? It is simply written that immediate substitutions prove the theorem.
Theorem:(Modulation and translation properties). If {$v_i$}<->{$V_j$} is a Fourier transform pair, then the following are Fourier transform pairs: {$w^iv_i$}<->{$V_{((j+1))}$} and {$v_{((i-1))}$} <-> {$w^jV_{j}$}
Now suppose that sequence $x$ is such that $x_i = w^i v_i$ for all $i$.
Replace the $x_i$ in Equation (1) by $w^i v_i$. Combine the $w^{ij}$ and $w^i$ terms to get $w^{i(j+1)}$. Call this Equation (2).
Look at Definition 1 as given to you. It tells you the value of $V_{j}$. Replace $j$ by $j+1$ on both sides of the equation to get a definition of $V_{j+1}$. Call this Equation (3). Note that $j+1$ might exceed $n-1$, and if so, you need to use the property that Definition 1 can be applied even when $j > n$ as long as we remember that $w^n = 1$ and so $V_{n+j}=V_j$, that is, $V_j = V_{((j))}$ for all integers $j \in \mathbb Z$.
Compare Equations (2) and (3) to see if you can reach the conclusion that you need to reach.
The other theorem can be proved similarly starting from Definition 2 instead of Definition 1.