Timeline for Measurability of eigenvalues-eigenvectors of a positive compact operator
Current License: CC BY-SA 4.0
14 events
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| Jun 19, 2023 at 20:49 | history | edited | user127022 | CC BY-SA 4.0 |
added 1425 characters in body
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| Mar 23, 2023 at 0:53 | comment | added | John | The answer here by user2048 seems to suggest a way to answer both questions. | |
| S Dec 21, 2022 at 18:02 | history | bounty ended | CommunityBot | ||
| S Dec 21, 2022 at 18:02 | history | notice removed | CommunityBot | ||
| S Dec 20, 2022 at 9:21 | history | suggested | Christophe Leuridan | CC BY-SA 4.0 |
Typo corrected
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| Dec 20, 2022 at 7:57 | review | Suggested edits | |||
| S Dec 20, 2022 at 9:21 | |||||
| S Dec 13, 2022 at 16:34 | history | bounty started | user127022 | ||
| S Dec 13, 2022 at 16:34 | history | notice added | user127022 | Authoritative reference needed | |
| Dec 10, 2022 at 17:25 | comment | added | Christian Remling | @JochenGlueck: The point is well taken, but I think the argument should work anyway since one can also remove degeneracies by arbitrarily small (in operator norm) perturbations. But min-max gives this more conveniently and it also shows somewhat more quantitatively that $\|\lambda_k(S)-\lambda_k(T)\|\le \|S-T\|$. | |
| Dec 10, 2022 at 10:31 | comment | added | Jochen Glueck | @Echo: Re your 2nd comment: I don't see how continuity with respect to the Hausdorff metric alone can give the continuity of the eigenvalue maps: one can't reconstruct the list of eigenvalues of $T$ from the set $\sigma(T)$ since $\sigma(T)$ does not contain information about eigenvalue multiplicity. | |
| Dec 10, 2022 at 10:20 | comment | added | Jochen Glueck | @Echo: I'm having difficulties to follow your first comment. The ranks of the spectral projections can jump, so they can't be continuos. To get continuity one would need to decompose the spectral projections associated to non-simple eigenvalues into rank-1 projections in a continuous way. I don't see why this is possible. | |
| Dec 10, 2022 at 6:16 | comment | added | user473423 | The map $T\mapsto \sigma(T)$ is continuous in the Hausdorff-metric, hence the eigenvalue maps are continuous. | |
| Dec 9, 2022 at 21:18 | comment | added | user473423 | I guess if you write each operator as a weighted sum of spectral projections, you will see that those summands depend continuously on the operator. | |
| Dec 9, 2022 at 18:07 | history | asked | user127022 | CC BY-SA 4.0 |