power of a matrix limit A^n
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From: Stephen Montgomery-Smith on 15 Jun 2010 13:56 Arturo Magidin wrote: > 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 I think that the OP must have meant over the complex numbers. First of all, he is talking about convergence, so he must be working in a field that has topology. Secondly, he is talking about things with magnitude less than one, so the field has to have a sense of magnitude. While there are fields (like p-adics) that have a sense of "norm" I hardly think this is in general use for linear algebra. So it is likely that he is talking about the reals, which is an ordered field. But then the usage of the word "magnitude" suggests complex rather than real, because otherwise he would have been more likely to use the word "absolute value." For the OP, the assertion he wants to have as true is nicely stated in the web page: http://en.wikipedia.org/wiki/Spectral_radius
From: Arturo Magidin on 15 Jun 2010 14:24 On Jun 15, 12:56 pm, Stephen Montgomery-Smith <step...(a)math.missouri.edu> wrote: > Arturo Magidin wrote: > > 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. > I think that the OP must have meant over the complex numbers. Well, apparently the OP also meant symmetric matrices only (see his later message); but he did not say so. > First of > all, he is talking about convergence, so he must be working in a field > that has topology. Secondly, he is talking about things with magnitude > less than one, so the field has to have a sense of magnitude. While > there are fields (like p-adics) that have a sense of "norm" I hardly > think this is in general use for linear algebra. What's wrong with the real numbers? They have a topology, they have a sense of magnitude, and it is a fairly common field for doing linear algebra. > So it is likely that > he is talking about the reals, which is an ordered field. But then the > usage of the word "magnitude" suggests complex rather than real, because > otherwise he would have been more likely to use the word "absolute value." > > For the OP, the assertion he wants to have as true is nicely stated in > the web page: > > http://en.wikipedia.org/wiki/Spectral_radius Yes, the assertion is true if the matrix is diagonalizable, or if the characteristic polynomial splits, or if the matrix has a Jordan form (which of course includes the diagonalizable case); just not in general for even the real numbers. Of course, adding hypothesis may make it true (like adding the missing "matrix is symmetric"), but *as stated* it was false. -- Arturo Magidin
From: HardySpicer on 15 Jun 2010 15:58 On Jun 16, 5:25 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > 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 Just names..there are so many in matrix theory. Ok so it is Jordan form. That earlier example I agree has eigen values of unity so does not apply however. Hardy
From: HardySpicer on 15 Jun 2010 16:00 On Jun 16, 5:56 am, Stephen Montgomery-Smith <step...(a)math.missouri.edu> wrote: > Arturo Magidin wrote: > > 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 > > I think that the OP must have meant over the complex numbers. First of > all, he is talking about convergence, so he must be working in a field > that has topology. Secondly, he is talking about things with magnitude > less than one, so the field has to have a sense of magnitude. While > there are fields (like p-adics) that have a sense of "norm" I hardly > think this is in general use for linear algebra. So it is likely that > he is talking about the reals, which is an ordered field. But then the > usage of the word "magnitude" suggests complex rather than real, because > otherwise he would have been more likely to use the word "absolute value." > > For the OP, the assertion he wants to have as true is nicely stated in > the web page: > > http://en.wikipedia.org/wiki/Spectral_radius That's true, they can be complex or real for my example. Hardy
From: Arturo Magidin on 15 Jun 2010 16:08
On Jun 15, 2:58 pm, HardySpicer <gyansor...(a)gmail.com> wrote: > On Jun 16, 5:25 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > 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!) > > Just names.. .... which have meanings. > there are so many in matrix theory. They tend to be important details, you know. > Ok so it is Jordan > form. I find it amazing that you would know what the "Jordan form" of a matrix is, but you do not know what it means for a matrix to be diagonalizable. > That earlier example I agree has eigen values of unity so does > not apply however. The example only has "eigenvalues of unity" if you are working over the complex numbers, something that you never specified. If you work over the real numbers, then the *ONLY* eigenvalue of that matrix is 0.5, which satisfies the conditions you explicitly placed (it does not, of course, satisfy the apparently large number of conditions you were *thinking of* but didn't bother to mention out loud). As has been pointed out, if your matrix has a Jordan canonical form (which is certainly the case if you are working over an algebraically closed field, e.g., the complex numbers), then the conclusion will follow. But not every matrix over the real numbers has a Jordan canonical form (over the real numbers). -- Arturo Magidin |