Eigenvalues of an operator in $\ell_2$ I tried finding the eigenvalues of the following operator to no avail. I tried doing it by constructing the eigenvalues from the previous one.  Can someone help me out? $T : (\ell_2, \|\cdot\|_2) \rightarrow (\ell_2, \|\cdot\|_2)$ defined through
$$T(x_1, x_2, x_3, \ldots) := (x_2, x_3, x_4, \ldots).$$
 A: Suppose $\lambda$ is an eigenvalue. If $(x_1,x_2,\ldots)$ is a corresponding eigenvector, then 
$$(x_2,x_3,\ldots) = T(x_1,x_2,\ldots) = \lambda(x_1,x_2,\ldots,),$$
hence
$$\begin{align*}
x_2 &= \lambda x_1\\
x_3 &=\lambda x_2 = \lambda^2x_1\\
&\vdots\\
x_{n+1} &= \lambda x_n = \lambda^{n}x_1\\
&\vdots
\end{align*}$$
so your original vector is
$$(x_1,\lambda x_1,\lambda^2x_1,\ldots).$$
In particular, $x_1\neq 0$. 
But for this to be in $\ell_2$, it must satisfy that $\sum x_i^2\lt \infty$. Since
$$\sum_{i=0}^{\infty}(\lambda^ix_1)^2 = x_1^2\sum_{i=0}^{\infty}\lambda^{2i}$$
is a geometric series, one can determine exactly when it converges, and thus determine exactly what values of $\lambda$ will yield an eigenvector.
A: *

*With the notation of Arturo Magidin we have that $\sum_{i=0}^{\infty}\lambda^{2i}$ is convergent, hence $|\lambda|<1.$

*Now let $|\lambda|<1$ and define $x:=(1, \lambda, \lambda^2,...).$ Then $x \in \ell_2$ and $Tx= \lambda x.$
Consequence: $ \lambda$ is an eigenvalue of $T \iff |\lambda| <1.$
