Periodic Antisymmetric Matrices counted by Hermitte Numbers

[franktaw at netscape.net]

Why these with period 4 is simple enough: square any one of them, and the result is -1 times the identity matrix.
 
Why only these is less obvious.  I would suggest looking at eigenvalues.
 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645
 
 
-----Original Message-----
From: wouter meeussen <wouter.meeussen at pandora.be>
To: helms at uni-kassel.de
Cc: Seqfan (E-mail) <seqfan at ext.jussieu.fr>
Sent: Thu, 8 Dec 2005 00:54:22 +0100
Subject: Re: Periodic Antisymmetric Matrices counted by Hermitte Numbers


Hi Gottfried,

sorry for the long interuption. Busy at work.
I was just pointing out that these matrices form a combinatorial construct that 
is counted by the
(unsigned) Hermite Numbers. Somewhat rare I think.
formula: 2^(k/2) Mod[k+1,2] k!/k!!
They seem to be equivalent to all permutations of 2n with the restriction that 
all cycles have
length two (pairings): example:
{1,5},{2,7},{3,4},{6,8} stands for the matrix

0  0  0  0  x  0  0  0
0  0  0  0  0  0  x  0
0  0  0  x  0  0  0  0
0  0 -x  0  0  0  0  0
-x 0  0  0  0  0  0  0
0  0  0  0  0  0  0  x
0 -x  0  0  0  0  0  0
0  0  0  0  0 -x  0  0

and now each +x can be choosen as +1 or -1. Hence the 2^(k/2).

The crux is to grok why all these and only these (antisymmetric) matrices are 
periodic, and why with
period 4.

W.
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