Actually my question is about calculating the spectrum of any bounded operator. I know that if we have Banach spaces $\mathbb{E}$ and $\mathbb{F}$ over the field $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$, and a bounded operator $T: \mathbb{E} \rightarrow \mathbb{F}$, then the spectrum of $T$ is the set $$\sigma (T) = \{\lambda \in \mathbb{K}: \lambda I - T \text{ is not invertible}\}.$$ Also, I know the theorem stating that $\sigma (T) \subseteq B[0;||T||]$. I've seen similar questions here referencing things like point or residual spectrums, but I'm just thinking of the ordinary spectrum defined above.
It's clear that all eigenvalues of $T$ are in its spectrum, because if $\lambda$ is an eigenvalue of $T$, $\lambda I - T$ is going to have a kernel that is not $\{0\}$. But otherwise I've got no idea how to calculate the rest of the spectrum. My particular question is probably quite easy: finding the spectrum of the operator $R: \ell^2 (\mathbb{C}) \rightarrow \ell^2 (\mathbb{C})$ given by $$R(z_1, z_2, z_3, \dots) = (0, z_1, z_2, \dots).$$
$R$ has no eigenvalues and $-R = 0T-R$ is invertible so $0 \notin \sigma(R)$. And $||R||=1$, so if $\lambda \in \sigma(R)$ we must have $|\lambda| \leq 1$, but otherwise I don't think I can say more.