Timeline for Is the Perron-Frobenius dimension of a G-Set given by its cardinality?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 3, 2019 at 14:10 | comment | added | alpoge | Sorry if I’ve misunderstood, but let z_i be all the eigenvalues, with z_1\in \R the real one of maximal absolute value. Write \zeta_i := z_i/|z_i|, so that \zeta_i\in S^1 for all i and \zeta_1 = 1. Now choose n_k\to \infty for which (\zeta_1^{n_k}, \zeta_2^{n_k}, ...) are all within \eps of (1, ..., 1) [via the pigeonhole principle]. Then tr(A^{n_k}) = z_1^{n_k} (1 + \sum_{i > 1} \zeta_i^{n_k} (|z_i|/z_1)^{n_k}) > z_1^{n_k}/2 once \eps is sufficiently small and n_k is sufficiently large. Now Phil’s argument works to show that z_1 = #|X|. Let me know if I’ve overlooked something! | |
| Apr 2, 2019 at 3:18 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
added 6 characters in body
|
| Apr 2, 2019 at 0:04 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
deleted 107 characters in body
|
| Apr 1, 2019 at 23:47 | comment | added | darij grinberg | Hmm. Afraid I cannot comment much on this; I don't know these results. | |
| Apr 1, 2019 at 23:36 | comment | added | Phil Tosteson | @darijgrinberg I added a potential fix | |
| Apr 1, 2019 at 23:35 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
added 726 characters in body
|
| Apr 1, 2019 at 22:27 | comment | added | Phil Tosteson | Ah, thanks. I misunderstood what the OP said Perron--Frobenius implied. It does seem reasonable that you can fix this, but I don't see how at the moment. | |
| Apr 1, 2019 at 22:25 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
added 326 characters in body
|
| Apr 1, 2019 at 21:53 | comment | added | darij grinberg | Hmm. What if several eigenvalues with equal absolute values cancel each other? I suspect they won't be able to do so consistently, but I don't see a good source for that. | |
| Apr 1, 2019 at 21:42 | comment | added | Phil Tosteson | What does irreducible mean? I think you can write any matrix $A$ in Jordan form and compute traces of powers. If the largest generalized eigenvalue of $a$ (in absolute value) is real, then you get a dominant contribution from it: $c \lambda^n$ where $c$ is the dimension of the generalized eigenspace. | |
| Apr 1, 2019 at 20:47 | vote | accept | Chris H | ||
| Apr 1, 2019 at 20:09 | comment | added | darij grinberg | Is it guaranteed that the maximal positive eigenvalue dominates the sum even if the matrix is not irreducible? | |
| Apr 1, 2019 at 14:09 | history | edited | Phil Tosteson | CC BY-SA 4.0 |
added 67 characters in body
|
| Apr 1, 2019 at 13:48 | history | answered | Phil Tosteson | CC BY-SA 4.0 |