Spectrum of an $\ell^2$-operator I need to find the spectrum $\sigma(T)$ of the following operator on $\ell^2$ (real sequences):
$$T(x)=(x_1,x_2,0,0,x_4,x_5,\dots) \;\; \forall x=(x_1,x_2,\dots) \in \ell^2$$
but I'm having some problems (yesterday I posted a similar question and it's been pointed out that the operator was compact but this is not)

Take $y \in \ell^2$ and suppose $(T-\lambda)x=y$, we have the following relations:
$$(1-\lambda)x_1=y_1 \\
(1-\lambda)x_2=y_2\\
-\lambda x_3=y_3\\
-\lambda x_4=y_4\\
x_4-\lambda x_5=y_5\\
x_5-\lambda x_6=y_6 \\
\vdots$$
from these we obtain
$$x_1=(1-\lambda)^{-1}y_1\\
x_2=(1-\lambda)^{-1} y_2\\
x_3=-\lambda^{-1} y_3\\
x_4=-\lambda^{-1} y_4\\
x_5=-\left(\lambda^{-1}y_5+\lambda^{-2} y_4 \right)\\
x_6=-\left(\lambda^{-1} y_6+\lambda^{-2}y_5+\lambda^{-3} y_4 \right)\\
\vdots$$
now I don't know how to determine whether or not $x \in \ell^2$.
 A: EDIT: This is not applicable to the question. See below.
Denote the operator as $P$. It should be clear that $P$ is a projection; $P^2=P^*=P$. For any projection, the spectrum is a subset of $\{0,1\}$, with at most one of these excepted. We can define an explicit inverse in all other cases: $(P-\lambda I)^{-1}=\frac{I}{-\lambda}+\frac{P}{\lambda-\lambda^2}$. This can be verified via the following computation: \begin{align*}
(P-\lambda I)\frac{1}{-\lambda}(I+\frac{P}{\lambda-1})&=\frac{1}{-\lambda}[P-\lambda I+\frac{P^2}{\lambda -1}-\frac{\lambda P}{\lambda -1}]\\
&=\frac{1}{-\lambda}[P-\lambda I-P]\\
&=I.
\end{align*}
The fact that this is a two-sided inverse follows from starring the entire equation, which does nothing but reverse the order of multiplication.
EDIT HERE: The operator $T$ above takes $e_1\mapsto e_1,e_2\mapsto e_2,e_3\mapsto e_3$, and then shifts forward on the next indices. This means that $T$ has a decomposition as $I_3\oplus R$, where $I_3$ is the $3\times 3$ identity operator and $R$ is the right shift operator. The spectrum of the direct sum of finitely many operators is the union of their spectra. The spectrum of the identity operator is $1$, and the spectrum of the right shift operator is the closed unit disk $\{z\in\mathbb C\,|\,|z|\le 1\}$. So the spectrum of $T$ is the closed unit disk.
