Let $T:\ell^2\rightarrow\ell^2$ be the operator
$$T(x_j)_{j\in\mathbb{N}}=(y_j)_{j\in\mathbb{N}} \ ,$$
where $y_1=x_2,y_2=x_3,y_3=x_1, y_j=x_j$ for all $j\geq 4$.
I want to describe the spectrum of $T$.
Of course $\|T\|=1$, therefore $\sigma(T)\subseteq \{\lambda\in\mathbb{C}\ |\ |\lambda|\leq 1\}$.
Take $\lambda\neq 0$ and $\ell_2\ni x\neq 0$ such that $x\in\ker(T-\lambda I)$. Working out the calculations I came up with the relation $x_2=\lambda x_1$ and $x_3=\frac{x_1}{\lambda}, x_j=0\forall j\geq 4$, supposing here that $x_1\neq 0$ w.l.o.g., which will let me obtain all eigenvalues of $T$ (and therefore the whole spectrum as I believe $T$ is compact because $T(B(0,1))=B(0,1)\subseteq\ell_2$). However, using the fact that $0=(T-\lambda I)x\in\ell_2$ gives me a polynomial of the form $\lambda^8-2\lambda^3+1$ which could be solved with some brute force, but I still believe there is an easy way here.
Best Answer
The spectrum is actually the set of three cube roots of $1$. Since $T(u,u^{2},u^{3},0,0,...)=u(u,u^{2},u^{3},0,0,...)$ it follows that $u \in \sigma(T)$ for any cube root $u$ of $1$ . If $\lambda $ is not a cube root of $1$ it is easy to solve the equation $Tx-\lambda x=y$ for $x$ in terms of $y$. (You will get $x_1=\frac {\lambda^{2}y_1+\lambda y_2+y_3} {1-\lambda^{3}}$, $x_n=\frac {y_n} {1-\lambda}$ for $n \geq 4$ and I will let you write down $x_2,x_3$). You can now check that $\|x\| \leq C\|y\|$ for some $C \in (0,\infty)$, independent of $y$) which show that $T-\lambda I$ has a bounded inverse.
Note also that $T^{3}=I$ implies that the spectrum is a subset of the set of all cube roots of unity by the Spectral Mapping Theorem.