Spectrum of the operator $Tx=(0,0,x_2/2^2,x_3/2^3,...)$ on $\ell^2$

Question

Consider the operator $T:\ell^2 \to \ell^2$ defined by

$$T(x)=\left(0,0,\frac{x_2}{2^2},\frac{x_3}{2^3},\dots,\frac{x_n}{2^n}, \dots \right),\\ \forall x=(x_1,x_2,x_3,x_4, \dots, x_n, \dots) \in \ell^2$$

I want to determine its spectrum $\sigma(T)$ but it's getting messy.

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My attempt: take $x,y \in \ell^2$ and $(T-\lambda)x=y$, we have the following relations:

$$-\lambda x_1=y_1 \\ -\lambda x_2=y_2\\ \frac{x_2}{4}-\lambda x_3=y_3\\ \frac{x_3}{8}-\lambda x_4=y_4\\ \vdots\\ \frac{x_n}{2^n}-\lambda x_{n+1}=y_{n+1} $$

from these we have

$$x_1=-\frac{y_1}{\lambda}\\ x_2=-\frac{y_2}{\lambda}\\ x_3=-\frac{y_3}{\lambda}-\frac{y_2}{4\lambda^2}\\ x_4=-\frac{y_4}{\lambda}-\frac{y_3}{8\lambda^2}-\frac{y_2}{32\lambda^2}\\ \vdots$$

now I don't know how to determine whether or not $x \in \ell^2$.

Answer 1

Your operator is compact, since it is a limit of finite-rank. So all nonzero elements of the spectrum are eigenvalues. If $Tx=\lambda x$ with $\lambda\ne0$, then \begin{align} 0&=\lambda x_1,\\ 0&=\lambda x_2,\\ \frac{x_n^2}{2^n}&=\lambda x_{n+1},\qquad n>1. \end{align} From $x_2=0$ this implies that $x_n=0$ for all $n$, so $\lambda$ cannot be an eigenvalue. Thus $\sigma(T)=\{0\}$.