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Actually my question is about calculating the spectrum of any bounded operator. I know that if we have Banach spaces $\mathbb{E}$ and $\mathbb{F}$ over the field $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$, and a bounded operator $T: \mathbb{E} \rightarrow \mathbb{F}$, then the spectrum of $T$ is the set $$\sigma (T) = \{\lambda \in \mathbb{K}: \lambda I - T \text{ is not invertible}\}.$$ Also, I know the theorem stating that $\sigma (T) \subseteq B[0;||T||]$. I've seen similar questions here referencing things like point or residual spectrums, but I'm just thinking of the ordinary spectrum defined above.

It's clear that all eigenvalues of $T$ are in its spectrum, because if $\lambda$ is an eigenvalue of $T$, $\lambda I - T$ is going to have a kernel that is not $\{0\}$. But otherwise I've got no idea how to calculate the rest of the spectrum. My particular question is probably quite easy: finding the spectrum of the operator $R: \ell^2 (\mathbb{C}) \rightarrow \ell^2 (\mathbb{C})$ given by $$R(z_1, z_2, z_3, \dots) = (0, z_1, z_2, \dots).$$

$R$ has no eigenvalues and $-R = 0T-R$ is invertible so $0 \notin \sigma(R)$. And $||R||=1$, so if $\lambda \in \sigma(R)$ we must have $|\lambda| \leq 1$, but otherwise I don't think I can say more.

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  • $\begingroup$ Be aware, that R:l^2->l^2 is not invertible, since you can't find a corresponding sequence (z_1,z_2,....) so that R(z_1,z_2,....)=(b,z_1,z_2,....) in l^2 with b not zero. The map R is not surjective. 0 is actually in the spectrum of R. But I have to admit, that calculating the whole spectrum for operators is not that fun ;) $\endgroup$ Commented May 9, 2018 at 10:15
  • $\begingroup$ @MrMatzetoni, That's a good point about 0. But it's just occurred to me that the adjoint of R is L, the left-shift operator, and it's quite easy to show that the spectrum of L is the unit disc. Shouldn't that mean, since the spectrum of T is the conjugate set of the spectrum of its adjoint T* (also not hard to show, I think), that the spectrum of R is actually the conjugate set of the unit disc -- and therefore the whole of the unit disc, too? $\endgroup$
    – user477203
    Commented May 9, 2018 at 10:24
  • $\begingroup$ Basically, as you know that R has no eigenvalues, thus (lambda I - R) is injective for all lambda, you would have to check the surjectivity of all operators (lambda I - R) with |lambda|<=1. Your thought about the adjoint seems legit to me, this would mean sigma_p(R)={} and sigma(R)={lambda | |lambda|<=1 } $\endgroup$ Commented May 9, 2018 at 10:32
  • $\begingroup$ You might check this out: math.ubc.ca/~feldman/m511/spectralExamples.pdf $\endgroup$ Commented May 9, 2018 at 10:36
  • $\begingroup$ OK, that seems to be saying that the 'point' spectrum of R is empty, but the 'residual' spectrum is the unit disc, although the open one. But then since the spectrum is closed, the 'full' spectrum must be the disc, surely. But I'm a little uncertain as to the role of these other kinds of spectrums. Thanks for the tips -- and maybe an answer will make things clearer too. $\endgroup$
    – user477203
    Commented May 9, 2018 at 10:49

1 Answer 1

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Consider $S:=R^*$, then $S(z_1, z_2, z_3, \dots) = (z_2, z_3, z_4, \dots).$

If $ | \mu| <1$ and $z=(1, \mu, \mu^2,...)$, then $Sz= \mu z$, hence

$\{\mu: |\mu|<1\} \subseteq \sigma(S)$. Since $||S||=1$ and $\sigma(S)$ is closed, we derive

$\{\mu: |\mu| \le 1\} =\sigma(S)$.

Consequence: $\{\mu: |\mu| \le 1\} =\sigma(R)$.

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  • $\begingroup$ That makes sense, thanks. $\endgroup$
    – user477203
    Commented May 9, 2018 at 11:17

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