power of a matrix limit A^n
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From: Arturo Magidin on 15 Jun 2010 02:15 On Jun 15, 12:51 am, adamk <ad...(a)adamk.net> 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". > What then is wrong with the fact that if k is > an eigenvalue and v is a non-zero eigenvector: > > A*v=k*v , then A^n*v=k^n*v , so that, for > > |k|<1 , as n->oo, k^n->0, so that (some hand-waving) > > A^n*v= k^n*v-> 0 , so that A^n*v=0 . ? Nothing. That *will* happen for the eigenvector corresponding to the eigenvalue (1/2) in my example. But since not every vector is a linear combination of eigenvectors (there is no basis of eigenvectors), you cannot argue from the fact that it works for eigenvectors that A^n*v -- > 0 *for all v* (which is what you would need to argue that "A^n goes to 0"). It will hold for any v which is in the span of the eigenspaces, but this need not be the entire space. I mean, I gave an *explicit example*. Did you try to think about what happens with that example and why your argument does not tell you that A^n "goes to 0" ? (My example does have one issue: for some values of n>0, it is *not* true that all eigenvalues of A^n have absolute value less then 1; I suspect you can fix that by using the companion matrix of a suitable irreducible nonlinear polynomial instead of the companion matrix for x^2+1 as I did). -- Arturo Magidin
From: Arturo Magidin on 15 Jun 2010 02:34 On Jun 15, 12:51 am, adamk <ad...(a)adamk.net> 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". > > What then is wrong with the fact that if k is > an eigenvalue and v is a non-zero eigenvector: > > A*v=k*v , then A^n*v=k^n*v , so that, for > > |k|<1 , as n->oo, k^n->0, so that (some hand-waving) > > A^n*v= k^n*v-> 0 , so that A^n*v=0 . ? First, the claim "so that A^n *v = 0" is *false*, with or without handwaving. If k is not zero, then A^n*v will *not* be zero for *any* value of n; the *limit* may be equal to 0, but A^n*v never will be. Second: you are showing that for any *eigenvector* you have A^n*v --> 0; in particular, you have that if w is a vector that lies in the linear span of the eigenspaces of A, then A^n*w --> 0 as n-->oo. But what about vectors that *do not* lie in the span of the eigenspaces? That's what happens in the (explicit) example I gave: the only vectors for which your argument works are those of the form (0,0,z); for any vector of the form (x,y,z) with x or y nonzero, A^n*v will *not* go to 0 as n-->oo. The error lies in thinking that because "all eigenvalues" satisfy the desired property that somehow implies that you have enough eigenvectors to cover the entire space. This need not happen. Because: how do you go from "it works for eigenvectors" to "it works for all vectors", in general? What if the matrix has *no* eigenvectors, so that the condition is vacuously true? Now, my example does have a wrinkle: there are values of n for which A^n does *not* satisfy the desired properties (some eigenvalues do not have absolute value less than 1); e.g., A^4 has eigenvalues 1 (twice) and 1/4. But if you pick the companion matrix of an irreducible quadratic whose *complex* roots r and r' (complex conjugate) are such that r^n is never a real number, then the complex eigenvalues of A^n will be r^n, r'^n and (1/2)^n, and so the only real eigenvalue will be (1/2)^n. -- Arturo Magidin
From: HardySpicer on 15 Jun 2010 02:46 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". > > -- > Arturo Magidin I meant to say that my matrix is always symmetric - is that ok now? Hardy
From: Arturo Magidin on 15 Jun 2010 02:48 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? -- Arturo Magidin
From: HardySpicer on 15 Jun 2010 05:19
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? > > -- > Arturo Magidin I wasn't sure what diagonalized meant. 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. Hardy |