[math-fun] Re: powers of PseudoAntisymmetric (-1,0,1)- Matrices
"wouter meeussen" <wouter.meeussen@pandora.be> writes:
One counter-example to spoil things a bit:
What is this a counterexample of?
for n=7, a small sample of (-1,0,1)-matrices whose powers are 'same', shows that not all are pseudo-Antisymmetric (= AntiSymmetric + Diagonal)
What about [ 0 1 ] [ 1 0 ] ? Isn't that a non-PA (-1,0,1)-matrix whose powers are PA (-1,0,1)-matrices? But the (powerlength = divisor of 12) remains valid. That's not true of the permutation matrix [ 0 1 0 0 0 ] [ 0 0 1 0 0 ] [ 0 0 0 1 0 ] [ 0 0 0 0 1 ] [ 1 0 0 0 0 ]. Earlier, you wrote
Consider
the (-1,0,1)-matrices T with properties : Det[T] not zero (invertible), all powers T^k are also invertible (-1,0,1) matrices.
Properties: powerlength of T divides 12, Det[t] is 1 or -1, T is pseudoAntisymmetric,
What are you asserting or conjecturing about those matrices and those properties? By the way, a more standard term for "powerlength" would be "order", or "multiplicative order" if you use the word "order" for some other property. Certainly any matrix of finite multiplicative order must have unit determinant, but I can't see any other likely conjecture. Dan
hi Dan, yes, I should stick to the PA-matrix variety. reading http://nyjm.albany.edu:8000/~mark/derham.html , I got spurred to look for samples of 6x6 and 7x7 matrices (type (-1,0,1) and PA) to test if De Rham's conjecture could be demonstrated to fail. To generate bigger samples, I tested products instead of powers of the few I got, and weeded out the non-(-1,0,1) chaff, to remain with valid extra samples plus a few (to me surprising) non-PA ones. Hence my oops... sorry for that. Can you count how many nxn (-1,0,1) matrices wether PA or not, are periodic (aka of finite multiplicative order)? W. ----- Original Message ----- From: "Dan Hoey" <Hoey@aic.nrl.navy.mil> To: "wouter meeussen" <wouter.meeussen@pandora.be>; "math-fun" <math-fun@mailman.xmission.com> Sent: Monday, November 17, 2003 8:38 PM Subject: [math-fun] Re: powers of PseudoAntisymmetric (-1,0,1)- Matrices
"wouter meeussen" <wouter.meeussen@pandora.be> writes:
One counter-example to spoil things a bit:
What is this a counterexample of?
for n=7, a small sample of (-1,0,1)-matrices whose powers are 'same', shows that not all are pseudo-Antisymmetric (= AntiSymmetric + Diagonal)
What about [ 0 1 ] [ 1 0 ] ? Isn't that a non-PA (-1,0,1)-matrix whose powers are PA (-1,0,1)-matrices?
But the (powerlength = divisor of 12) remains valid.
That's not true of the permutation matrix
[ 0 1 0 0 0 ] [ 0 0 1 0 0 ] [ 0 0 0 1 0 ] [ 0 0 0 0 1 ] [ 1 0 0 0 0 ].
Earlier, you wrote
Consider
the (-1,0,1)-matrices T with properties : Det[T] not zero (invertible), all powers T^k are also invertible (-1,0,1) matrices.
Properties: powerlength of T divides 12, Det[t] is 1 or -1, T is pseudoAntisymmetric,
What are you asserting or conjecturing about those matrices and those properties?
By the way, a more standard term for "powerlength" would be "order", or "multiplicative order" if you use the word "order" for some other property.
Certainly any matrix of finite multiplicative order must have unit determinant, but I can't see any other likely conjecture.
Dan
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=Dan Hoey Certainly any matrix of finite multiplicative order must have unit determinant, but I can't see any other likely conjecture.
To be precise, by "unit" I assume you must mean root-of-unity, since, eg, a det=-1 might be even order?
participants (3)
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Dan Hoey -
Marc LeBrun -
wouter meeussen