# Kernel of an LTI system

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:

1. input signal is causal ($$x[n] = 0$$ for n<0) WLOG assume $$x[0] \neq 0$$
2. LTI system is causal ($$y[n]$$ depends only on values of $$x[n-k]$$ for positive $$k$$)
3. 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[0]. If $$h[0] \neq 0$$, we are done. If not, there will be a output caused by x[0] sometime later, and it cannot be canceled out by the impulse response to the signal at time n, because that would be delayed further.

My questions:

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.