power of a matrix limit A^n
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From: Ronald Bruck on 15 Jun 2010 12:58 In article <42c5f96b-15a3-4157-97f9-02d8d04a8a63(a)h13g2000yqm.googlegroups.com>, TCL <tlim1(a)cox.net> wrote: > On Jun 15, 12:11 am, HardySpicer <gyansor...(a)gmail.com> wrote: > > If I have a matrix A whose eigenvalues have all magnitude less than 1, > > then does A^n approach zero for large integer n? > > > > Hardy > > In fact,A^n approaches zero iff all eigenvalues of A have magnitude > less than 1. In fact, isn't there a theorem that a self-mapping T of a separable metric space has a UNIQUE fixed-point f, and T^nx --> f for all x, iff there is an equivalent metric so it's a strict contraction? i.e. d(Tx,Ty) < d(x,y) when x \ne y? Something like that, anyway. Proved in the early 60's? I'd have to dig through my notes. -- Ron Bruck
From: Arturo Magidin on 15 Jun 2010 13:25 On Jun 15, 4:19 am, HardySpicer <gyansor...(a)gmail.com> wrote: > On Jun 15, 6:48 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > On Jun 15, 1:46 am, HardySpicer <gyansor...(a)gmail.com> wrote: > > > > On Jun 15, 5:13 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > On Jun 14, 11:11 pm, HardySpicer <gyansor...(a)gmail.com> wrote: > > > > > > If I have a matrix A whose eigenvalues have all magnitude less than 1, > > > > > then does A^n approach zero for large integer n? > > > > > Not necessarily; take > > > > > ( 0 -1 0 ) > > > > ( 1 0 0 ) > > > > ( 0 0 .5 ) > > > > > The only eigenvalue is 1/2, A^n does not approach 0. The 2 x 2 > > > > submatrix corresponding to the upper left corner alternates between > > > > > ( 0 -1 ) > > > > ( 1 0 ), > > > > > ( -1 0 ) > > > > ( 0 -1 ), > > > > > ( 0 1 ) > > > > (-1 0 ), > > > > > and > > > > > ( 1 0 ) > > > > ( 0 1 ), > > > > > so the matrix never "approaches 0". > > > > > If A is diagonalizable, then the answer is "yes" for suiitable notion > > > > of "approach 0". > > > > I meant to say that my matrix is always symmetric - is that ok now? > > > Since symmetric matrices are always diagonalizable, what does the last > > sentence in my previous message tell you? > I wasn't sure what diagonalized meant. A matrix A is "diagonalizable" if you can find a basis of eigenvectors for the space. Equivalently, if there exists an invertible matrix Q such that QAQ^{-1} is a diagonal matrix. > I thought you meant in some > form of Jordan format with a matrix of eigenvalues in the centre > between the matrix of eigen-vectors and its inverse. So... you know what the Jordan canonical form is, but you don't know what it means for a matrix to be "diagonalizable"? (And, suggestion: if you aren't sure what the response means, ASK!) -- Arturo Magidin
From: Arturo Magidin on 15 Jun 2010 13:26 On Jun 15, 6:21 am, r...(a)trash.whim.org (Rob Johnson) wrote: > In article <98a219af-8682-44b3-bb6c-c2712befb...(a)q12g2000yqj.googlegroups..com>, > > > > Arturo Magidin <magi...(a)member.ams.org> wrote: > >On Jun 14, 11:11=A0pm, HardySpicer <gyansor...(a)gmail.com> wrote: > >> If I have a matrix A whose eigenvalues have all magnitude less than 1, > >> then does A^n approach zero for large integer n? > > >Not necessarily; take > > >( 0 -1 0 ) > >( 1 0 0 ) > >( 0 0 .5 ) > > >The only eigenvalue is 1/2, A^n does not approach 0. The 2 x 2 > >submatrix corresponding to the upper left corner alternates between > > >( 0 -1 ) > >( 1 0 ), > > >( -1 0 ) > >( 0 -1 ), > > >( 0 1 ) > >(-1 0 ), > > >and > > >( 1 0 ) > >( 0 1 ), > > >so the matrix never "approaches 0". > > >If A is diagonalizable, then the answer is "yes" for suiitable notion > >of "approach 0". > > The eigenvalues of your matrix are { i, -i, 1/2 }; If you are working over the complex number. But I was working over the real numbers, where the only eigenvalue is 1/2. > two of the > eigenvalues have magnitude 1 or greater. I don't see how it > satisfies the OP's conditions. The OP did not specify the field he was working on; I was working over R. -- Arturo Magidin
From: Arturo Magidin on 15 Jun 2010 13:27 On Jun 15, 11:52 am, Ronald Bruck <br...(a)math.usc.edu> wrote: > In article > <98a219af-8682-44b3-bb6c-c2712befb...(a)q12g2000yqj.googlegroups.com>, > > > > Arturo Magidin <magi...(a)member.ams.org> wrote: > > On Jun 14, 11:11 pm, HardySpicer <gyansor...(a)gmail.com> wrote: > > > If I have a matrix A whose eigenvalues have all magnitude less than 1, > > > then does A^n approach zero for large integer n? > > > Not necessarily; take > > > ( 0 -1 0 ) > > ( 1 0 0 ) > > ( 0 0 .5 ) > > > The only eigenvalue is 1/2, A^n does not approach 0. The 2 x 2 > > submatrix corresponding to the upper left corner alternates between > > > ( 0 -1 ) > > ( 1 0 ), > > > ( -1 0 ) > > ( 0 -1 ), > > > ( 0 1 ) > > (-1 0 ), > > > and > > > ( 1 0 ) > > ( 0 1 ), > > > so the matrix never "approaches 0". > > But this matrix has two eigenvalues of absolute value 1, so does not > satisfy the conditions. As I pointed out elsewhere, the OP did not specify what field he was working on. The roots i and -i of the characteristic polynomial are only eigenvalues if you are working over the complex numbers, not if you are working over the real numbers. > The answer to the OP is "yes". The spectral radius r_\sigma(T) is > lim_n ||T^n||^(1/n), and is less than 1; therefore T^n --> 0. Like the original poster, you seem to be assuming that you are working over an algebraically closed field. That was not part of the premises. -- Arturo Magidin
From: Arturo Magidin on 15 Jun 2010 13:28
On Jun 15, 11:48 am, Ray Vickson <RGVick...(a)shaw.ca> wrote: > On Jun 14, 9:11 pm, HardySpicer <gyansor...(a)gmail.com> wrote: > > > If I have a matrix A whose eigenvalues have all magnitude less than 1, > > then does A^n approach zero for large integer n? > > Yes. This follows from the Jordan Canonical Form; Not every matrix over any field has a Jordan Canonical form; it does if the characteristic polynomial splits, but may not have it if it doesn't. -- Arturo Magidin |