For Linear Time-invariant systems mapping any complex sequence $x[n]$ to an output $y[n]$, the question is if there the kernel must be trivial, given that the LTI system does not map everything to the zero sequence.
So , the ideal low-pass filter (which is LTI) has a kernel of any frequencies above the cutoff.
Ok, so then I wanted to find conditions that would make the kernel trivial. The best I could come up with are:
- input signal is causal ($x[n] = 0$ for n<0) WLOG assume $x \neq 0$
- LTI system is causal ($y[n]$ depends only on values of $x[n-k]$ for positive $k$)
- LTI system is relaxed ($y[n] =0$ until $x[n]$ starts)
If these conditions are true, then I argue that the kernel is trivial.
proof: The LTI system can be written as a convolution, $x[n] * h[n]$, where $h[n] $ is the impulse response. Then y[n] at n=0 depends only on x. If $h \neq 0$, we are done. If not, there will be a output caused by x sometime later, and it cannot be canceled out by the impulse response to the signal at time n, because that would be delayed further.
A.These conditions appear sufficient but not necessary. Is there a sufficient and necessary condition? Can I remove some of the 3 conditions and still be sufficient to prove the kernel is trivial?
B. Is my proof correct/ suggestions to make it better.