Timeline for Derivative of singular value decomposition of $I + \alpha X$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2019 at 13:35 | review | Close votes | |||
| Sep 27, 2019 at 15:00 | |||||
| Sep 27, 2019 at 13:14 | comment | added | Carlo Beenakker | you just want to do degenerate second order perturbation theory; any quantum mechanics textbook will explain how, you basically first want to diagonalise $W$ in the subspace of the degenerate eigenvalues of $D$; with a specific information on how $X$ looks like there is not much more that one can say here. | |
| Sep 27, 2019 at 11:23 | comment | added | lcv | Eigenvalues and singular values are not the same thing. Although for hermitian matrices the latter are the absolute values of the former. Anyway it seems that your problem can be solved with standard (degenerate) perturbation theory. | |
| Sep 27, 2019 at 11:14 | comment | added | Dirk B. | Thanks for pointing that out. Yes, $X$ is hermitian. First order pertubation theory gives zero as all the entries on the diagonal of $X$ are zero. I updated the question to include that. | |
| Sep 27, 2019 at 11:11 | history | edited | Dirk B. | CC BY-SA 4.0 |
added that X is hermitian and that it's diagonal is zero
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| Sep 27, 2019 at 11:00 | comment | added | Carlo Beenakker | is $X$ Hermitian? (you say "symmetric complex", do you mean Hermitian?); and why does first order perturbation theory give zero? | |
| Sep 27, 2019 at 9:55 | review | First posts | |||
| Sep 27, 2019 at 10:01 | |||||
| Sep 27, 2019 at 9:52 | history | asked | Dirk B. | CC BY-SA 4.0 |