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?
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