For a system $H$ operating on infinite-length discrete-time signals, an eigenvector is a signal which, when passing through the system, is modified only by a scalar factor. So suppose $s[n]$ is an eigenvector to system $H$, then $H\{s[n]\}=\lambda_s s[n]$, where $\lambda_s$ is a complex number.It is a remarkable property that all LTI systems have a set of eigenvectors, and furthermore that these eigenvectors are of the form $e^{j\omega n}$. This means that, whatever $\omega$ may be, if $e^{j\omega n}$ is input into any LTI system, the output will simply be a scaled version of the input (the amount of scaling depending on the nature of the system, of course):
Complex harmonic sinusoids are eigenvectors of discrete-time infinite length LTI systems. When given as inputs to any LTI system, the output is simply a scaled version of the input.
Proving this special property of LTI systems is straightforward; we simply express the output in terms of the input ($s_\omega[n]=e^{j\omega}n$) and system impulse response ($h[n]$) through the convolution sum, and simplify:
$\begin{align*}s_\omega[n] \ast h[n]&=\sum_{m=-\infty}^{\infty} s_\omega [n-m]\,h[m]\\&=\sum_{m=-\infty}^{\infty} e^{j \omega (n-m)} \,h[m] \\&=\sum_{m=-\infty}^{\infty} e^{j \omega n} \, e^{-j \omega m} \, h[m]\\&=\left( \sum_{m=-\infty}^{\infty} h[m] \, e^{-j \omega m} \right) e^{j \omega n} \\&= \lambda_\omega \, s_\omega[n] ~~\checkmark\end{align*}$
The frequency response of an lti system
Take a close look at the value $\lambda_\omega$ above. It is intimately related to the system, as we would expect, by the system's impulse response:
$\lambda_\omega=\sum_{m=-\infty}^{\infty} h[m]\, e^{-j \omega m}$
This is simply the DTFT of the impulse response! It is so important, we give it a special notation $H(\omega)$, and a special name as well. $H(\omega)$ is called the
frequency response , for it explains how the system responds (i.e., in what way it scales) to particular input frequencies $\omega$. This is also apparent by seeing the DTFT as an inner product of $h[n]$ with the sinusoidal $e^{-j \omega m}$. The value of the inner product (which is the value $H(\omega)$ grows or shrink in ways corresponding to the similarity of $h[n]$ and $e^{-j \omega m}$.
Although we cannot display the system matrix involved (as it is infinite in length), the DTFT diagonalizes the system $H$ for infinite length signals, just as the DFT does with finite length ones (LINK). Expressing this diagonalization in terms of the DTFTs of the system, we have that if $y[n]=h[n]\ast x[n]$, then $Y(\omega)=H(\omega)X(\omega)$.
