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?
--Dan
On Mar 21, 2014, at 3:23 AM, Fred Lunnon <fred.lunnon@gmail.com<javascript:;>> wrote:
Given a finite set of complex matrices generating an infinite group, I need to establish that every element of the group has all eigenvalues within the unit circle.
Such problems must have been well investigated --- perhaps (shudder) by math physicists --- but I have no idea where to start looking.
[It's actually a semigroup of integer matrices with constraints on allowed products, and circle radius exponential in the number of product factors --- but those details are almost certainly unimportant.]
WFL
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