From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Geometric sum of matrixes
Date: 8 Oct 1999 19:49:32 GMT
Newsgroups: sci.math
To: Hankel O'Fung
Keywords: Spectral Radius Formula
In article <37FD8346.9F9DC9B3@polyu.edu.hk>,
Hankel O'Fung writes:
> Robert Israel wrote:
> > It does converge if all eigenvalues of A are less than 1 in absolute value:
> > by the Spectral Radius Formula lim_{n -> infinity} || A^n ||^(1/n) < 1.
> The inequality is true, but I don't see how is it related to the convergence
> of the series. Perhaps some elaboration is needed.
> > Moreover, in that case the solution of I+ATA'=T is unique: note that
> > if ASA'=S, then A^n S (A')^n = S, and since || A^n || = || (A')^n || -> 0
> > we get S = 0.
> Hmmm, we don't have ||B||=||B'|| or ||A^n||=||(A')^n|| in general
> ( consider the maximum column sum norm and A=[1/2 0; 1 0] ),
> although || A^n || -> 0 does imply that || (A')^n || -> 0 and S=0.
As a functional analyst by training, I naturally use the operator norm. In
this case it is true that ||B|| = ||B'||. Moreover, the Spectral Radius
Formula says (for the operator norm) lim_{n -> infinity} ||A^n ||^(1/n)
= sup { |r|: r in spectrum of A } (call this R(A)). In our case R(A) < 1.
If R(A) < r < 1, then for n sufficiently large ||A^n||^(1/n) < r so
||(A')^n|| = ||A^n|| < r^n, and thus ||A^n (A')^n|| < r^(2n), which
implies the convergence of the series.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z