Q:

If we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period/frequency

An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:

With the very surprising property of linearity in mind, let us return to the transmission of signals over electrical circuits. We have noted that the output signal corresponding to most input signals has a different shape or variation with time from the input signal. Figures II-1 and II-2 illustrate this. However, it can be shown mathematically (but not here) that, if we use a sinusoidal signal, such as that of Figure II-4, as an input signal to a linear transmission path, we always get out a sine wave of the same period, or frequency. The amplitude of the output sine wave may be less than that of the input sine wave; we call this attenuation of the sinusoidal signal. The output sine wave, may rise to a peak later than the input sine wave; we call this phase shift, or delay of the sinusoidal signal.

I'm trying to find the aforementioned proof that, if we use a sinusoidal signal as an input signal to a linear transmission path, then we always get out a sine wave of the same period, or frequency.
During my research, the closest thing to this that I have come across is slide 30 of this presentation:

I would greatly appreciate it if people could please take the time to either prove this or redirect me somewhere that has the proof.

A:

$y=G(x)$ is translation invariant, $G\Bigl(T_sx\Bigr)(t)=(T_sy)(t)=y(t+s)$ Together with the linearity this has the consequence that also differential operators are preserved,
$$\dot y(t)=\lim_{s\to 0}\frac{(T_sy)(t)-y(t)}s=\lim_{s\to 0}\frac{G\bigl(T_sx\bigr)(t)-G\bigl(x\bigr)(t)}{s}=G\left(\lim_{s\to 0}\frac{T_sx-x}s\right)(t)=G\bigl(\dot x\bigr)(t).$$
Now you can also apply this to the oscillator equation, $G(\ddot x+\omega^2x)=\ddot y+ω^2y$ and if $x$ is sinusoid with frequency $ω$, then so is $y$.
With $$G(\cos(ω\,\cdot\,))(t)=a\cos(ωt)+b\sin(ωt)$$ you also get the shifted
$$
G(\sin(ω\,\cdot\,))=G(\cos(ω\,\cdot\,-\tfrac\pi2))(t)=a\sin(ωt)-b\cos(ωt)
$$
so that indeed there are only two free parameters per frequency. To get the attenuation and phase, you only need to compute the polar coordinates $(A,\varphi)$ of the point $(a,-b)$.

