Timeline for Diagonalization of symmetric matrices of functions
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 18, 2023 at 11:27 | comment | added | Igor Khavkine | This old answer by Peter Michor links to a survey of results that could help if you relax the premise of your question. Namely, you could potentially find a continuous mapping $V\to U$, which "blows up" the points where $G(p)$ is singular such that the diagonalization of $G$ is continuous over $V$. | |
| May 17, 2023 at 5:21 | vote | accept | user1234567890 | ||
| May 16, 2023 at 17:38 | comment | added | user1234567890 | @ChristianRemling no it's not. That's the case for matrices over a field, but not over a general ring, and that's what I'm asking. | |
| May 16, 2023 at 16:50 | comment | added | Christian Remling | Your matrix is symmetric, so is diagonalizable everywhere. (A different question that is more interesting was answered by @RobertBryant below.) | |
| May 16, 2023 at 16:15 | history | became hot network question | |||
| May 16, 2023 at 12:31 | answer | added | Robert Bryant | timeline score: 8 | |
| May 16, 2023 at 7:00 | history | asked | user1234567890 | CC BY-SA 4.0 |