# Spectrum of an operator on $\ell^2$

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.

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 $$x \in \ell^2$$.

• With $\ell^2$ I mean the real square summable sequences Aug 31, 2020 at 0:52
• This operator is compact, implying that all of its spectrum, other than $0$, consists of eigenvalues. This means that you should be solving $(T-\lambda)x=0$, or rather finding the values of $\lambda$ for which there is a non-zero solution. Aug 31, 2020 at 0:53
• @Conifold Oh yes... I also noticed that (in fact $1/2^k \to 0$) but I didn't think about this important property! Thank you! Aug 31, 2020 at 0:58
• The operator is not self adjoint and it looks like the only eigenvalue is 0. Think of the n by n matrix, all 0's, except a set of 1's above the diagonal. Aug 31, 2020 at 1:20

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 $$\sigma(T)=\{0\}$$.