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<DIV>Why these with period 4 is simple enough: square any one of them, and the result is -1 times the identity matrix.</DIV>
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<DIV>Why only these is less obvious. I would suggest looking at eigenvalues.</DIV>
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<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: wouter meeussen <wouter.meeussen@pandora.be><BR>To: helms@uni-kassel.de<BR>Cc: Seqfan (E-mail) <seqfan@ext.jussieu.fr><BR>Sent: Thu, 8 Dec 2005 00:54:22 +0100<BR>Subject: Re: Periodic Antisymmetric Matrices counted by Hermitte Numbers<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_45f5308e-90e4-448e-aa89-331e8eba045a><PRE><TT>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.
</TT></PRE></DIV></DIV></DIV>
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