Find all eigenvalue for $S$ Let $V$ be the vector space of all converging real sequences, that is
$$V = \lbrace (a_1,a_2,a_3,\ldots) |\forall i \in \mathbb {N}: a_i \in\ \mathbb{R}\;\mathrm{and}\; \exists\! \lim_{n\to\infty}a_n \rbrace$$
We are given this linear transformation:
$$ S:V \rightarrow  V,\; S(a_1,a_2,a_3,\ldots) = (a_2,a_3,a_4,\ldots)$$

Find all of the eigenvalues of $S$.


MY METHOD: In fact, $S$ "moves left". The linear transformation
deletes the first member in the sequence. I tried to find any
$\lambda$ $\in\ \mathbb{R}$, such that there is a sequence  with at
least one $a_i \neq 0$ that will satisfy: 
$S(a_1,a_2,a_3,a_4,\ldots) = (\lambda a_1, \lambda a_2, 
 \lambda a_3, \lambda a_4)$, 
but without success to this moment

 A: You need to solve $$S(a_1,a_2,a_3,\ldots) = \lambda(a_1,a_2,a_3,\ldots)$$ i.e. $$(a_2,a_3,a_4,\ldots) = (\lambda a_1,\lambda a_2,\lambda a_3,\ldots).$$
Now equate each element of those two sequences. We need $a_2 = \lambda a_1$, $a_3 = \lambda a_2$, $a_4 = \lambda a_3$, ....
Substitute the first equation into the second to get $a_3 = \lambda a_2 = \lambda(\lambda a_1) = \lambda^2 a_1$.
Substitute the second equation into the third to get $a_4 = \lambda a_3 = \lambda(\lambda^2 a_1) = \lambda^3 a_1$.
So the first four terms of the eigensequence are $a_1$, $a_2 = \lambda a_1$, $a_3 = \lambda^2 a_1$, and $a_4 = \lambda^3 a_1$.
Can you figure out the rest of the eigensequence? Don't forget that $\lambda$ needs to be such that the eigensequence converges.
A: The equation $Sa = \lambda a$, for $a = (a_i)_{i \in \mathbb N} \in V$, implies that
$$\lambda a_i = a_{i+1}, \quad \forall i \in \mathbb N.$$
Therefore, $(1, \lambda, \lambda^2, \ldots)$ is an eigenvector if and only if $-1 < \lambda \le 1$.
