Let $M$ be a $3\times 3$ real invertible diagonal matrix and $H$ a Hilbert space of infinite dimension (for example, we can take $H$ as the space of square integrable functions over a bounded lipschitz domain $\Omega\subset\mathbb{R}^3$).
I want to determine the essential spectrum of $M$ as a multiplication operator acting from $H^3$ to itself.
Does it consist of the eigenvalues of $M$? Or its numerical range ?
Thank you.
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2$\begingroup$ When you ask the same question on mathoverflow and math stack exchange (or any other to stack exhange sites), could you please add a link in each question to the other when you do this kind of thing, so people don't spend effort trying to answer a question on one site that's already been answered on the other? $\endgroup$– Will SawinCommented Aug 30, 2021 at 22:31
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$\begingroup$ @WillSawin I deleted the question on math stack exchange site. $\endgroup$– SAKLYCommented Aug 31, 2021 at 7:37
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$H^3$ is $\mathbb{C}^3 \otimes H$ and the action of $M$ on it is just the usual action on the first tensor factor. It is a completely straighforward exercise to prove that the $\lambda$-eigenspace of $M\otimes\operatorname{id}_H$ is exactly $\operatorname{Eig}_\lambda(M) \otimes H$ and similarly for generalized eigenspaces. In particular: The spectrum of $M\otimes\operatorname{id}_H$ is exactly the same as the spectrum of $M$, i.e. discrete consisting precisely of the eigenvalues.
answered Aug 31, 2021 at 7:44
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$\begingroup$ thank you. Does that result still the same for arbitrary $3\times 3$ matrix ? $\endgroup$– SAKLYCommented Sep 1, 2021 at 19:30