If the matrices form a group, then their eigenvalues must lie on the unit circle, as otherwise the inverse would have an eigenvalue outside. A sufficient condition is that each of the generating matrices is unitary (with respect to some choice of basis), since then the group they generate is a subgroup of the unitary group. But this is not a necessary condition, as the matrix [[1,1],[0,1]] satisfies the requirement, but is not unitary in any basis. -- Gene
________________________________ From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, March 21, 2014 8:58 AM Subject: Re: [math-fun] matrix group with bounded eigenvalues
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> 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