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