Timeline for Spectrum of a product of a symmetric positive definite matrix and a positive definite operator
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jul 26, 2023 at 15:13 | vote | accept | SAKLY | ||
| Jul 24, 2023 at 19:52 | comment | added | Jochen Glueck | @ChristianRemling: Apparently one can indeed construct such an example by taking an infinite diagonal operator that consists of matrices which almost commute; see my answer for details. | |
| Jul 24, 2023 at 19:50 | answer | added | Jochen Glueck | timeline score: 1 | |
| Jul 24, 2023 at 19:05 | comment | added | Christian Remling | @JochenGlueck: Yes, I had assumed $A: H\to H$. | |
| Jul 24, 2023 at 18:43 | comment | added | Christian Remling | @JochenGlueck: Another attempted rescue operation would be to interpret "discrete" = pure point spectrum, which is actually quite common (though unfortunate) terminology in physics. | |
| Jul 24, 2023 at 17:14 | comment | added | SAKLY | @JochenGlueck Yes, if possible, we can restrict ourselves to this particular case for the operator $A$. | |
| Jul 24, 2023 at 16:39 | comment | added | Jochen Glueck | @SAKLY: Ok. Please note, though, that this definition of discrete spectrum is rather uncommon (as pointed out by Christian Remling) and implies that $A$ is a linear combination of finitely many orthogonal projections. Could you please confirm again that you really have this quite restrictive class of operators in mind for $A$? | |
| Jul 24, 2023 at 12:33 | comment | added | SAKLY | @JochenGlueck Yes, that's what I mean. | |
| Jul 24, 2023 at 3:03 | comment | added | Jochen Glueck | @ChristianRemling: Regarding your discussion with the OP: I believe they wan't $M$ to be an element of $\mathbb{R}^{2 \times 2}$ and then interpret it as an operator on $H \times H$. Of course, the eigenvalues of this operator don't have finite multiplicity - but the spectrum is discrete as a topological set and maybe this is what they mean. | |
| Jul 24, 2023 at 2:58 | comment | added | Jochen Glueck | @ChristianRemling: Yes, sure - but as you noted, a bounded operator in infinite dimensions cannot have this property for every spectral value. So my intention was to ask the OP for clarification. | |
| Jul 23, 2023 at 22:36 | comment | added | Christian Remling | @SAKLY: You wrote your Hilbert space is infinite-dimensional, so there are certainly (positive, if desired) operators without any discrete spectrum. | |
| Jul 23, 2023 at 22:26 | comment | added | Christian Remling | @JochenGlueck: $\sigma_d$ = isolated eigenvalues of finite multiplicity, I have the advantage of never having seen any other definition. | |
| Jul 23, 2023 at 22:10 | comment | added | Jochen Glueck | What definition of "discrete spectrum" do you use? | |
| Jul 23, 2023 at 19:32 | review | Close votes | |||
| Aug 6, 2023 at 20:43 | |||||
| Jul 23, 2023 at 19:29 | comment | added | SAKLY | Excuse me. $M$ is assumed to be a $2\times 2$ symmetric positive definite matrix so it has discrete spectrum which consists of its eigenvalues. | |
| Jul 23, 2023 at 19:10 | comment | added | Christian Remling | $A=1$, and then take any $M$ with non-discrete spectrum. (Here $\sigma(A)=\{ 1\}$ is discrete as a set, though actually the spectrum is not discrete in the more natural sense of empty essential spectrum. But then a bounded self-adjoint operator always has non-empty essential spectrum.) | |
| Jul 23, 2023 at 18:18 | history | asked | SAKLY | CC BY-SA 4.0 |