I'm confused about two problems from Oppenheim's signals and systems course with regards to proving time invariance. This involves, for any given system, proving that shifting output in time is equivalent to shifting input in time.
The first problem concerns proving the following system in time-invariant
$$y[n] = \sum_{k = -\infty}^{\infty} x[k] = T[x[n]] \tag{1}$$
I get that passing in a shifted input leads to the following $$T[x[n-n_0]] = \sum_{k = -\infty}^{\infty} x[k-n_0] \tag{2}$$
however I can't figure out why (2) is equal to $y[n-n_0]$, because if I substitute $n-n_0$ into equation in (1), I would think it is still equal to (1)
The second problem concern's this system,
$$y[n] = \sum_{k = -\infty}^{n} x[k] = T[x[n]] \tag{3}$$
Here once again I understand how passing a shifted signal as input leads to
$$T[x[n-n_0]] = \sum_{k = -\infty}^{n} x[k-n_0] \tag{4}$$
According to the answers, shifting the output leads to
$$y[n-n_0] = \sum_{l = -\infty}^{n-n0} x[l] \tag{5}$$
which I think is just substituting $n-n_0$ for $n$ in the second expression in (3). However, I need to prove that this is somehow equal to to (4). So I'm unsure about the equivalence between (4) and (5) as well as the need for switching variables. Thank you for any help or hints.
Best Answer
For the first problem, note that the output is a constant number, i.e., $y[n]=y[n-n_0]$, for all $n,n_0\in\mathbb{Z}$.
For the second problem, you can show the equivalence of (4) and (5) by setting $k-n_0=l$ in (4) and perform the summation over index $l$ noting that it ranges from $-\infty -n_0=-\infty$ to $n-n_0$