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I am trying to understand what it means exactly for the spectrum of an operator to be closed. I am considering infinite dimensional operators.

Say we know, for example, that the point spectrum of some operator is made up of 4 values, $\{1,2,3,4\}$ does that mean that all numbers in $[1,4]$ are in the spectrum (in the residual or continuous spectrum) since by definition the spectrum is closed? Alternatively, is the $\{1,2,3,4\}$ set considered closed but just disjoint so that we don't necessarily have all of $[1,4]$ in the spectrum.

Can the spectrum be disjoint EDIT: not disjoint, disconnected?

I am afraid that I might be misunderstanding this idea.

I had a problem where the eigenvalues in the point spectrum were dense in $[0,1]$ and we concluded that the spectrum is made up of the closure of $[0,1]$. So I am wondering if the density was the important aspect of that, or if it works that way in general.

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    $\begingroup$ $\{1,2,3,4\}$ is already closed, so its closure is $\{1,2,3,4\}$ and not $[1, 4]$ $\endgroup$ – mercio Jul 30 '18 at 6:41
  • $\begingroup$ Thank you very much. I appreciate the clarification on that. $\endgroup$ – MathIsHard Jul 30 '18 at 6:42
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Let $T(e_1)=e_1,T(e_2)=2e_2,T(e_3)=3e_3,T(e_4)=4e_4,Te_j=e_j$ for all $j>4$ where $\{e_j\}$ is the standard orthonormal basis for $l^{2}$. Then $1,2,3,4$ are eigen values of $T$. If $\lambda \notin \{1,2,3,4\}$ you can solve the equation $Tx-\lambda x=y$. You will get a unique solution for any $y \in l^{2}$. Hence $T-\lambda I$ is a bijection. Automatically the inverse is continuous (by OMT) so $\lambda $ is n at in the spectrum. Thus $\sigma (T)=\{1,2,3,4\}$. [The solution of $Tx-\lambda x=y$ is given by $x_j=\frac {y_j} {1-\lambda} $ if $j >4$ and $x_j=\frac {y_j} {j-\lambda} $ if $j \leq 4$]. Incidentally, you are using the word disjoint for disconnected. The statement $A$ is disjoint has no meaning.

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  • $\begingroup$ Thank you very much for the help. If I am understanding correctly then, the set $\{1,2,3,4\}$ is closed and it is okay that the spectrum is the union closed sets? $\endgroup$ – MathIsHard Jul 30 '18 at 6:07
  • $\begingroup$ You are right. I meant disconnected... $\endgroup$ – MathIsHard Jul 30 '18 at 6:09
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    $\begingroup$ Yes, spectrum need not be a connected set. $\endgroup$ – Kavi Rama Murthy Jul 30 '18 at 6:09
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The answer of Kavi Rama is not O.K.

Example: Let $T: l^2 \times \mathbb C^4 \to l^2 \times \mathbb C^4$ be defined by

$T((\xi_1, \xi_2, \xi_3,...),(z_1,z_2,z_3,z_4))=(0, \xi_1, \xi_2, \xi_3,...),(z_1,2z_2,3z_3,4z_4))$.

Then the point spectrum of $T$ is $\{1,2,3,4\}$, but the spectrum of $T$ is given by

$$\{ \lambda \in \mathbb C: |\lambda| \le 1\} \cup \{1,2,3,4\}.$$

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  • $\begingroup$ Thank you for your answer. What part of Kavi Rama's answer and your's is different? $\endgroup$ – MathIsHard Jul 30 '18 at 6:47
  • $\begingroup$ Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$? $\endgroup$ – MathIsHard Jul 30 '18 at 6:48

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