Timeline for Eigenvalues of permutations of a real matrix: can they all be real?
Current License: CC BY-SA 3.0
6 events
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| Jul 5, 2013 at 9:24 | comment | added | Wolfgang | For $c(M)$, that is the same as diag(1,1,1,1,0). I find $c(M)=22$ here: 6 pairs $\pm i$ and 16 pairs of 3rd roots of unity. BTW, I really doubt that this can lead to a "similar non-singular matrix": if all EVs of a matrix are 0, a small perturbation will produce "lots" of complex roots. :( Believe me, it seems hopeless to start with a singular matrix! | |
| Jul 5, 2013 at 8:49 | comment | added | Felix Goldberg | However, it might be possible to play a bit with it and obtain a similar non-singular matrix. | |
| Jul 5, 2013 at 8:48 | comment | added | Felix Goldberg | @Wolfgang I mean a different kind of matrix. I've added now an example that shows what I mean. The downside is that the matrix is singular. | |
| Jul 5, 2013 at 8:48 | history | edited | Felix Goldberg | CC BY-SA 3.0 |
added 128 characters in body
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| Jul 5, 2013 at 7:36 | comment | added | Wolfgang | Looks interesting. But if I am right that e.g. $A_5=\pmatrix{0&1&0&0&0\cr-1&0&1&0&0\cr0&-1&0&1&0\cr0&0&-1&0&1\cr0&0&0&-1&0}$, then for odd $n$, $A_n$ is not invertible. And I find $c(A_5)=110$ (the EV $\pm i$ alone occur 14 times) and $c(A_6)=1274.$ Are you sure your program runs through all the $n!$ permutations? I use the numtoperm function in Pari. | |
| Jul 4, 2013 at 22:38 | history | answered | Felix Goldberg | CC BY-SA 3.0 |