Timeline for Can a vector-function $v:\mathbb{R}^n\to \mathbb{R}^n$ be an eigenvector of its own Jacobian matrix?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 5, 2018 at 20:48 | answer | added | Willie Wong | timeline score: 4 | |
| Dec 5, 2018 at 17:26 | comment | added | Gil Sanders | @WillieWong Wow, I didn't even thought about that, but I agree that if we interpret the vector as a row or as a column things are different! Now I would be very glad to have some references about the other case as well, which seems in fact somewhat different. Thank you! | |
| Dec 5, 2018 at 15:49 | comment | added | Willie Wong | Can you specify whether you meant a left or a right eigenvector? (In particular, is $\nabla v \cdot v$ meant as $(v \cdot \nabla)v$ or $\frac12(\nabla |v|^2)$?) The Jacobian matrix is in general not symmetric so that the left and right eigenvectors are not the same. I agree that the "natural" interpretation is $v\cdot \nabla v$ which is completely treated by Denis Serre, but just want to make sure. | |
| Dec 5, 2018 at 12:55 | vote | accept | Gil Sanders | ||
| Dec 5, 2018 at 11:49 | answer | added | Denis Serre | timeline score: 6 | |
| Dec 5, 2018 at 9:32 | history | edited | Peter Mueller | CC BY-SA 4.0 |
edited title
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| Dec 5, 2018 at 9:23 | comment | added | Jean Marie Becker | In your title, "eigenvalue" should be "eigenvector". | |
| Dec 4, 2018 at 19:55 | review | First posts | |||
| Dec 4, 2018 at 21:02 | |||||
| Dec 4, 2018 at 19:51 | history | asked | Gil Sanders | CC BY-SA 4.0 |