We will do many examples today to get intuition for what's Fourier transform is doing.

Let $N$ be a positive integer.

Let 'x-space' be $V_x = \Z/ N \Z = \{0,1, \cdots, N-1\}$, and then (rescaled) 'p-space' is also $V_p = \Z / N\Z$, then we can build the Fourier transformation kernel $K(x,p) = e^{2\pi i (x p / N)} : V_x \times V_p \to U(1)$ where $U(1)$ is the unit circle in complex number.

The kernel satisfies the orthonormal condition in both x and p variable. $\frac{1}{N} \sum_{x \in V_x} K(x,p_1) \overline{K(x,p_2)} = \delta(p_1, p_2)$ $\frac{1}{N} \sum_{p \in V_p} K(x_1,p) \overline{K(x_2,p)} = \delta(x_1, x_2)$

Given a function $f(x): V_x \to \C$, we can expand it as $f(x) = \sum_{p \in V_p} K(x,p) F(p)$ for some function $F(p): V_p \to \C$. Using the orthonormal condition, we get, for any $q \in V_p$, we have $(1/N) \sum_{x \in V_x} f(x) \overline{K(x,q)} = (1/N) \sum_{x \in V_x} \sum_{p \in V_p} K(x,p) F(p) \overline{K(x,q)} = F(q).$ that is $F(q) = (1/N) \sum_{x \in V_x} f(x) \overline{K(x,q)}.$

Let $Fun(V_x, \C)$ be the space of functions from $V_x$ to $\C$. Similarly define $Fun(V_p, \C)$. Concretely $Fun(V_x, \C) = \C^N$ and $Fun(V_p, \C) = \C^N$.

Fourier transformation $FT$, sends an element $f(x) \in Fun(V_x,\C)$ to an element $F(p) \in Fun(V_p, \C)$. $FT$ is a linear map.

If we define hermitian inner product on $Fun(V_x,\C)$ as $\langle f, g\rangle_x = (1/N) \sum_{x \in V_x} f(x) \overline{g(x)},$ and we define hermitian inner product on $Fun(V_p,\C)$ as $\langle F, G \rangle_p = \sum_{p \in V_p} F(p) \overline{G(p)}.$ then we find that Fourier transformation is compatible with the two inner products, namely $\langle f, g \rangle_x = \langle F, G \rangle_p, \quad F = FT(f), G = FT(g).$

A function $f(x)$ is determined by its values $f(0), f(1)$. Similarly for $F(p)$. We have relations $F(0) = (1/2) (f(0) + f(1)), \quad F(1) = (1/2) (f(0) - f(1)).$ So, we can reconstruct $f(x)$ from $F(p)$, by $f(0) = F(0) + F(1), \quad f(1) = F(0) - F(1).$

$1 + (-1) = 0.$ and less obviously $1 + e^{2\pi i / 3} + e^{2\pi i (2/3)} = 0$ more generally $\sum_{j=0}^{N-1} e^{2\pi i (j/N)} = 0$

How to see this? You can say, this is the sum of all the $N$-th roots of unity, and we have $z^N - 1 = \prod_{j=0}^{N-1} ( z - e^{2\pi i (j/N)}).$ hence by looking at the coefficient of $z^{N-1}$, we see the sum of all the roots is 0.

Or, draw these roots as vectors on the complex plane, they show up as evenly distributed on the unit circle, since the summands are invariant under rotation by $2\pi/N$, hence the result is invariant under such a rotation. And the only possible number is 0.

try it yourself.