Can anyone show me that both $\cos t$ and $\sin t$ are eigen signals. Here is a little bit background of eigen-function.
The output of a continuous-time, linear time-invariant system is denoted by $T\{z(t)\}$ where $x(t)$ is the input signal. A signal $z(t)$ is called eigen-signal of the system $T$ , when $T\{z(t)\} = \gamma z(t)$, where $\gamma$ is a complex number, in general, and is called an eigenvalue of $T$. EDIT: Suppose the impulse response of the system $T$ is real and even.
On
As a sort of hint:
I suppose you have heard of convolution \begin{equation} y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau \end{equation} Then what happens if our input is of the form $x(t)=e^{st}$? In that case $y(t)=H(s)e^{st}$ with $H(s)=\int_{-\infty}^{\infty}e^{-s\tau}d\tau$, so $e^{st}$ is an eigenfunction of the LTI system with eigenvalue $H(s)$. Now for your question, can we for example write $\cos$ in a more useful way? Yes, by using Euler's formula \begin{equation} e^{j\theta}=\cos(\theta)+j\sin(\theta) \implies cos(\theta)=\frac{1}{2}e^{j\theta}+\frac{1}{2}e^{-j\theta} \end{equation} Then you could already see everything you need just from there, but it might be easier to first generalize our previous result to Fourier series. There the idea is that you represent a signal as a sum of harmonically related eigenfunctions. Why? Because as we saw, with such an input the output is fairly easy to compute! Have a look at the equations below: \begin{equation} x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0 t} \to y(t)=\sum_{k=-\infty}^{\infty}a_kH(jk\omega_0)e^{jk\omega_0 t}=\sum_{k=-\infty}^{\infty}b_ke^{jk\omega_0 t} \end{equation}
I hope this helps, but I would suggest to have a look at the book and/or lectures by Alan Oppenheim, which is what I used.
The output of the LTI system will be the convolution of its impulse response with the input (see). Since the impulse response of a Lyapunov stable LTI system is a finite sum of complex exponentials (see example 1 here), and the sin and cos functions may also be represented as a sum of complex exponentials (see), it is clear that the integrand in the convolution operation is also an exponential. The integral of an exponential is a scaling of that same exponential. Thus, an input with a sinusoidal (complex exponential) form yields an output also with a sinusoidal (complex exponential) form.
I have to argue with your T{z(t)} notation though, because in many cases T{sin(t)} = a*sin(t+b). The "phase shift" b cannot be ignored. Of course, this is just a notation issue, because generally speaking, your problem is supposing z(t) = a*e^(s*t) and T{z} will result in only a scaling of the complex amplitude of the phasor z(t). For example, T{z} could yield a*e^(b)*e^(s*t) = a*e^(s*t+b), i.e. a phase shift, while there is still a clear eigenvalue of e^b. Note that a, b, and s are all in general complex.
The fact that the LTI system is Lyapunov stable is also important. Without Lyapunov stability, it can have an impulse response of the form t*e^-t which would break this argument. I assume that whatever "even" means in regards to the impulse response is meant to handle that case.