Re: [math-fun] matrix group with bounded eigenvalues
Yes, Asimov's Law only applies if the matrices are orthogonally diagonalisable (<==> real symmetric). For other matrices, the `maximum length scale factor' can exceed the magnitude of the largest eigenvalue (e.g. in the case of a simple shear, where the eigenvalues are both unity). Sincerely, Adam P. Goucher
----- Original Message ----- From: Dan Asimov Sent: 03/21/14 05:20 PM To: math-fun Subject: Re: [math-fun] matrix group with bounded eigenvalues
I think I’ve verified what Jim says 32 ways from Sunday. I get
(81/200)*(3+sqrt(5))
or about 2.12 for the largest eigenvalue. This is certainly counterintuitive from my naïve viewpoint.
—Dan
On Mar 21, 2014, at 9:36 AM, James Propp <jamespropp@gmail.com> wrote:
I don't think this Dan'a guess is true. Consider the non-commuting matrices .9 .9 .0 .9 and .9 .0 .9 .9 Each of them has eigenvalues strictly inside the unit disk, but their product doesn't.
(Unless I made a mistake, in which case, please let me know!)
Jim
On Friday, March 21, 2014, Dan Asimov <dasimov@earthlink.net> wrote:
If the eigenvalues are *strictly* inside the unit circle, then I would expect the matrices to each be a contraction mapping on their vector space (with some contraction constant L < 1 equal to the maximum |eigenvalue|).
So then wouldn't a composite of finitely many of these matrices also be a contraction mapping, with contraction constant equal to the maximum of the finitely many contraction constants pertaining to each matrix in the composite?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Sorry everybody --- I fouled up by oversimplifying the problem. So let's make it a semigroup generated by a given finite set of integer matrices (it seems experimentally that my constraints can be jettisoned), and I'm trying to show that every product of s generators has all eigenvalues bounded in modulus by c^s (where c is also given). A toy example, which I can actually decide by elementary means, is generated by 10 2x2 matrices shown below, where c = (1 + rt5)/2 is the golden section. Notice that the product [1 0] [1 1] = [1 1] [1 1] [0 1] = [1 2] has eigenvalue c^2 , despite both generators having unit eigenvalues. WFL ___________________________
matOOOO; [1 1] [1 0] matOOOI; [1 0] [1 0] matOOII; [1 0] [1 1] matOIOO; [1 1] [0 0] matOIOI; [1 0] [0 1] matOIIO; [0 1] [1 0] matOIII; [0 0] [1 1] matIIOO; [1 1] [0 1] matIIOI; [0 1] [0 1] matIIII; [0 1] [1 1]
The rate of growth of a matrix semigroup is often quite delicate. See http://www.ams.org/journals/mcom/2000-69-231/S0025-5718-99-01145-X/S0025-571... for a fun example. Victor Sent from my iPhone
On Mar 21, 2014, at 14:35, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Sorry everybody --- I fouled up by oversimplifying the problem.
So let's make it a semigroup generated by a given finite set of integer matrices (it seems experimentally that my constraints can be jettisoned), and I'm trying to show that every product of s generators has all eigenvalues bounded in modulus by c^s (where c is also given).
A toy example, which I can actually decide by elementary means, is generated by 10 2x2 matrices shown below, where c = (1 + rt5)/2 is the golden section.
Notice that the product
[1 0] [1 1] = [1 1] [1 1] [0 1] = [1 2]
has eigenvalue c^2 , despite both generators having unit eigenvalues.
WFL
___________________________
matOOOO; [1 1] [1 0] matOOOI; [1 0] [1 0] matOOII; [1 0] [1 1] matOIOO; [1 1] [0 0] matOIOI; [1 0] [0 1] matOIIO; [0 1] [1 0] matOIII; [0 0] [1 1] matIIOO; [1 1] [0 1] matIIOI; [0 1] [0 1] matIIII; [0 1] [1 1]
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
WDS wrote: << Now wants, given any finite set of matrices, to prove any product of n of the matrices has all eigenvalues e obeying |e|<c^n for some positive constant c. (Is that right? That is now your goal?) >> Not quite --- c was to be chosen in advance (and sharp). << The spectral norm and the Frobenius norm of the matrices are submultiplicative (so both should work), and the max |eigenvalue| is upper bounded by sqrt(FrobeniusNorm). >> Aha --- that's a good point about the spectral norm. In the example I gave, c is in fact the spectral norm of each factor. If it turns out to be the spectral norm of all the generators, then I'm home and dry! On 3/21/14, Victor S. Miller <victorsmiller@gmail.com> wrote:
The rate of growth of a matrix semigroup is often quite delicate. See http://www.ams.org/journals/mcom/2000-69-231/S0025-5718-99-01145-X/S0025-571... for a fun example.
Victor
Nice paper, but with woeful implications if Warren's idea doesn't do the trick. Random matrix theory is one of Terry Tao's hobbies --- 'nuff said? WFL
On Mar 21, 2014, at 14:35, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Sorry everybody --- I fouled up by oversimplifying the problem.
So let's make it a semigroup generated by a given finite set of integer matrices (it seems experimentally that my constraints can be jettisoned), and I'm trying to show that every product of s generators has all eigenvalues bounded in modulus by c^s (where c is also given).
A toy example, which I can actually decide by elementary means, is generated by 10 2x2 matrices shown below, where c = (1 + rt5)/2 is the golden section.
Notice that the product
[1 0] [1 1] = [1 1] [1 1] [0 1] = [1 2]
has eigenvalue c^2 , despite both generators having unit eigenvalues.
WFL
___________________________
matOOOO; [1 1] [1 0] matOOOI; [1 0] [1 0] matOOII; [1 0] [1 1] matOIOO; [1 1] [0 0] matOIOI; [1 0] [0 1] matOIIO; [0 1] [1 0] matOIII; [0 0] [1 1] matIIOO; [1 1] [0 1] matIIOI; [0 1] [0 1] matIIII; [0 1] [1 1]
participants (3)
-
Adam P. Goucher -
Fred Lunnon -
Victor S. Miller