I am trying to figure out if the system in the diagram (part a) is time knvariant or not. When the input is shifted the output is shifted as well, so I am thinking the system is time invariant, but I wanted to ask for help to get more concrete proof.
For continuous time:
A system is time-invariant, if the coefficients of the differential equation describing the system are constants. That is, they don't depend on time. $$\ddot{y}(t)+2\dot{y}(t)+8y(t)=0 \: \: \: \: \: \text{time-invariant} \tag1$$
$$\ddot{y}(t)+2t\dot{y}(t)+8ty(t)=0 \: \: \: \: \text{time-variant} \tag2$$
For discrete time:
A system is time-invariant, if the coefficients of the difference equation describing the system are constants. $$y[k+2] = -2y[k+1]-8y[k] \: \: \: \: \text{time-invariant} \tag3$$ $$y[k+2] = -2ky[k+1] -8ky[k] \: \: \: \: \text{time-variant} \tag4$$
In your exercise, the system is described through the system's impulse response \$h[n]\$ and is clearly time-invariant.
