Timeline for Eigenvalues invariant under 90° rotation
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2022 at 18:09 | vote | accept | Sascha | ||
| Jul 31, 2022 at 16:18 | answer | added | Antoine Labelle | timeline score: 5 | |
| Jul 31, 2022 at 15:00 | comment | added | Michael Engelhardt | @FredHucht - if it weren't for that factor $i$ in front of $A+A^T $, which completely changes things ... but if we leave out the factor $i$, the matrix is Hermitean, and the eigenvalues real, so the claimed property again can't arise ... | |
| Jul 31, 2022 at 14:54 | comment | added | Michael Engelhardt | The new factor 2 in front of $B$ makes things rather more pathological: For $N=2$ and $N=4$, the matrices now are deficient, consisting of $2\times 2$ Jordan blocks, all with eigenvalue 0; so yes, in a trivial sense, you could say that, if 0 is an eigenvalue, also $e^{i\pi /2} 0$ is an eigenvalue. For $N=3$, the eigenvalues are nonzero, and the claimed property doesn't (and can't) hold. | |
| Jul 31, 2022 at 14:33 | comment | added | Fred Hucht | Related? mathoverflow.net/questions/426279 | |
| Jul 31, 2022 at 13:46 | comment | added | Sascha | @MichaelEngelhardt sorry there was a 2 missing in front of $B$, now it should be fine. | |
| Jul 31, 2022 at 13:45 | history | edited | Sascha | CC BY-SA 4.0 |
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| Jul 31, 2022 at 2:36 | comment | added | Michael Engelhardt | This property of the eigenvalues does not seem to hold for $N=2$, $N=3$ or $N=4$ (I haven't tried any higher). Of course, for $N=2$ or $N=3$, this can only be true if all eigenvalues are zero (but they aren't). For $N=4$, I had some hope, but alas, no: The eigenvalues are $i\sqrt{3} $, $-i\sqrt{3} $, $0$, $0$. | |
| Jul 30, 2022 at 12:02 | history | edited | Sascha | CC BY-SA 4.0 |
added 99 characters in body; edited title
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| Jul 30, 2022 at 10:25 | history | asked | Sascha | CC BY-SA 4.0 |