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Nov 11, 2016 at 14:33 comment added Michael Barr Yes that's the right matrix. Thanks. BTW, they add up, as the must to 5, which is a good check.
Nov 11, 2016 at 11:31 answer added Pietro Majer timeline score: 5
Nov 11, 2016 at 7:48 comment added Anthony Quas For $n=20$, there are three eigenvalues outside the unit disk; for $n=50$, there are six and for $n=100$, there are eight.
Nov 11, 2016 at 5:19 history reopened Yemon Choi
Carlo Beenakker
Stefan Kohl
Willie Wong
Andrés E. Caicedo
Nov 10, 2016 at 19:52 review Reopen votes
Nov 10, 2016 at 23:39
Nov 10, 2016 at 19:39 comment added Carlo Beenakker the conjecture seems false: for $n=5$ I find the eigenvalues, 3.06664, 1.00351, 0.501031, 0.272797, 0.15602, so there are two eigenvalues greater than 1. Just to be sure we are talking about the same matrix: $$\left( \begin{array}{ccccc} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{2} & 1 & \frac{2}{3} & \frac{2}{4} & \frac{2}{5} \\ \frac{1}{3} & \frac{2}{3} & 1 & \frac{3}{4} & \frac{3}{5} \\ \frac{1}{4} & \frac{2}{4} & \frac{3}{4} & 1 & \frac{4}{5} \\ \frac{1}{5} & \frac{2}{5} & \frac{3}{5} & \frac{4}{5} & 1 \\ \end{array} \right)$$
Nov 10, 2016 at 19:30 history closed Will Jagy
Franz Lemmermeyer
Michael Albanese
Alex Degtyarev
Dietrich Burde
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Nov 10, 2016 at 18:39 comment added Yemon Choi @FedericoPoloni I think this is needlessly harsh. Michael Barr is not some apathetic undergraduate
Nov 10, 2016 at 18:12 review Close votes
Nov 10, 2016 at 19:35
Nov 10, 2016 at 17:51 comment added Federico Poloni Downvoted for clear lack of effort. You didn't typeset the matrix in Latex, and you don't want to recompute the result for a 3x3 matrix?
Nov 10, 2016 at 17:42 history asked Michael Barr CC BY-SA 3.0