Intuitive explanation of a flat DFT Given a vector
$${\bf x}= [a, 0, \dots, 0], a \neq  0, {\bf x} \in \mathbb{R}^n$$
If I compute its Discrete Fourier Transform (DFT), I get 
$${\bf DFT}( {\bf x}) = [a,a, \dots,  a]$$ I.e., a vector of all $a$s in $\mathbb{R}^n$ too --without complex component.
I know this is true by definition, but I cannot come up with an intuitive explanation. Could you please explain why this happens? I.e. why does a single spike at the beginning of my signal makes it flat in the frequency domain?
 A: The only signal that has all frequency amplitudes with same weight (flat spectra) and linear phase is an impulse.
Also using duality property of DFT, if DFT of a given function is a signal with only zero component, we could realize that the time-domain signal would have been a pure DC signal.
A: The spectrum is constant for two reasons. First, the magnitude of the DFT of an impulse is constant. Second, the phase of the DFT of an impulse is a linear function of the location of the impulse. Thus, the DFT of an impulse is of the form $Ae^{jt_0 \omega}$, where $A\in \mathbb{R}$ is the constant magnitude, $t_0$ is the location of the impulse in the time/image domain, $\omega$ is the frequency variable, and $j=\sqrt{-1}$. When the impulse is located at sample $t_0 = 0$, we have $e^{jt_0 \omega}=e^{j0\omega}=1$, and the spectrum has the constant value $A$.
A: Not sure if this is what you had in mind, but think about the continuous analog, a single spike as a delta function $a \delta(x)$, whose transform is
$$\int_{-\infty}^{\infty} dx \, a \delta(x) \, e^{i k x} = a$$
That is, the transform is a constant independent of $k$, just as in the discrete case.
