Timeline for Is the Perron-Frobenius dimension of a G-Set given by its cardinality?

10 events

when toggle format	what		by	license	comment
Apr 3, 2019 at 22:17	history	edited	Benen Harrington	CC BY-SA 4.0	deleted 1 character in body
Apr 3, 2019 at 20:15	comment	added	Frieder Ladisch		@PhilTosteson: fair enough! Sorry I overlooked this.
Apr 3, 2019 at 15:17	comment	added	Phil Tosteson		I agree this is a much more satisfying answer. I did identify/use that eigenvector though.
Apr 3, 2019 at 14:49	history	edited	Benen Harrington	CC BY-SA 4.0	added 12 characters in body
Apr 3, 2019 at 13:53	comment	added	Frieder Ladisch		The diagonal element corresponding to $K$ is exactly the number of cosets in $G/H$ that are fixed by $K$. I think this answer is better than the other one since it makes explicit the common Perron-Frobenius eigenvector for all the $e_H$'s, namely $e_{\{1\}}$. For nontrivial subgroups $K$, the corresponding eigenvector is not $e_K$, but can be computed by Möbius inversion, this has been done by David Gluck (1981, Illinois J. Math. 25, no.1, pp.63-67).
Apr 3, 2019 at 11:00	history	edited	Benen Harrington	CC BY-SA 4.0	added 99 characters in body
Apr 3, 2019 at 9:48	history	edited	Benen Harrington	CC BY-SA 4.0	deleted 8 characters in body
Apr 3, 2019 at 9:37	history	edited	Benen Harrington	CC BY-SA 4.0	added 105 characters in body
Apr 3, 2019 at 9:30	review	First posts
Apr 3, 2019 at 9:51
Apr 3, 2019 at 9:29	history	answered	Benen Harrington	CC BY-SA 4.0