Timeline for A game on equiangular polygons

10 events

when toggle format	what		by	license	comment
Jul 28, 2013 at 9:09	vote	accept	CommunityBot
Jul 28, 2013 at 8:07	history	edited	Pietro Majer	CC BY-SA 3.0	added 30 characters in body
Jul 28, 2013 at 7:54	comment	added	Pietro Majer		@Sally: The first statement just means that $\mathbb{Z}^n$ is not contained in any proper linear subspace of $\mathbb{Z}^n$ (in particular, it is not contained in the eigenspace of all eigenvalues less than 1 in absolute value). Second statement: $L$ is singular iff $\lambda_k=0$ for some $1\le k \le n$, that is $\cos(2\pi k/n)=1/2$, so $n=6k$ or $5n=6k$.
Jul 28, 2013 at 6:03	comment	added	user30300		@Pietro: Please explain two parts: whay "$\Bbb{Z}^n$ spans linearly $\Bbb{R}^n$" ? and in the last paragraph, when $L$ is singular how do you conclude that $6 \| n$ ?
Jul 27, 2013 at 18:55	comment	added	Pietro Majer		Indeed, I was editing to elaborate on this point. So everything reduces to checking when 0, 1, or -1 is an eigenvalue of L.
Jul 27, 2013 at 16:29	history	edited	Pietro Majer	CC BY-SA 3.0	added 306 characters in body
Jul 27, 2013 at 16:21	history	edited	Pietro Majer	CC BY-SA 3.0	added 306 characters in body
Jul 27, 2013 at 15:52	comment	added	Barry Cipra		If $A$ stays bounded, you'll wind up repeating things, which requires eigenvalues that are roots of unity. Doesn't this pretty much finish things?
Jul 27, 2013 at 14:56	history	edited	Pietro Majer	CC BY-SA 3.0	added 82 characters in body
Jul 27, 2013 at 14:42	history	answered	Pietro Majer	CC BY-SA 3.0