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 $x \in \ell^2$.


1 Answer 1


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.

  • 1
    $\begingroup$ What is $P$ here? The original operator? If so, I don't see why $P^2=P$ is true. If $e_n$ are the standard basis vectors then $P(e_4)=e_5$ and $P^2(e_4)=P(e_5)=e_6$. $\endgroup$
    – Mark
    Sep 1, 2020 at 19:45
  • $\begingroup$ You're right, I misread the question. I'll edit the answer. $\endgroup$ Sep 1, 2020 at 20:15

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