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 spectrum of Carleman matrix Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 02/22/2009, 09:29 PM (This post was last modified: 06/28/2017, 11:29 AM by Gottfried. Edit Reason: log(0) instead of zeta(1) for logical reasons ) bo198214 Wrote:Hey Gottfried, did you ever thought about the spectrum of the Carleman matrix of $\exp_b$? ... oh so many times... Quote:For finite matrices the spectrum is just the set of eigenvalues. However for an infinite matrix or more generally for a linear operator $A$ on a Banach space the spectrum is defined as all values $\lambda$ such that $A-\lambda I$ is not invertible. We saw that the eigenvalues of the truncated Carleman matrices of $\exp$ somehow diverge. So it would be very interesting to know the spectrum of the infinite matrix. As this also has consequences on taking non-integer powers of those matrices. Or is the spectrum just complete $\mathbb{C}$ because $A$ is not invertible itself? Well, I've had at most 100 explanative pages about infinite matrices in my hands, so I cannot cite an authoritative statement about the cardinality, as well as about the characteristics of the non-invertibility. But what I decided to use as hypotheses: 1) I didn't discuss about a single eigenvalue or some arbitrarily formed set yet. I used always the restriction, that a) an infinite set of eigenvalues exists, which has also a unique structure: it is defined by the consecutive powers of one base-parameter. and that b) there are *infinitely many* such sets of eigenvalues, according to the infinitude of fixpoints: each fixpoint defines one of such sets. While one single eigenvalue defines "$A-\lambda I$ is non-invertible", the definition of such a set is more informative, since with this we can say $A * W = W * dV(\lambda)$ where dV(x) denotes, as usual, a diagonalmatrix of consecutive powers of x - and each of these powers is an eigenvalue so that $A-\lambda^0 I$, $A-\lambda I$, $A-\lambda^2 I$, $A-\lambda^3 I$,... are all non-invertible. The complete set of eigenvalues I "know of" is thus the set of all fixpoints and all of their (nonnegative) integer powers, which is then N x N. And W is an infinite set of invariant column-vectors according to the selection of lambda and its power. There may be other eigenvalues, but I didn't think about this yet. -------------------------------------------- 2) The idea, that the spectrum is complete $\mathbb{C}$ because of non-invertibility seems to be a non sequitur. I see the noninvertibility of A for one single reason. First: A is decomposbale into two well known triangular matrices Binomial P and Stirling-kind2 S2 (factorially scaled). Both factors P and S2 are invertible, so $A = S2 * P\sim$ and formally hte inverse is possible $A^{-1} = P^{-1}\sim * S2^{-1} = P^{-1}\sim * S1$ where S1 is the matrix of Stirlingnumbers 1st kind, also factorially scaled. The reason why A is not invertible, is that in $P^{-1}\sim * S1$ the dotproduct first row by second column is infinite and exactly gives log(0). If by some measure this multiplication can be avoided in a more complex matrix-operation, then the whole formula behaves like if A^-1 would exist (we discussed this in the context of fixpoint-shift) Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

 Messages In This Thread spectrum of Carleman matrix - by bo198214 - 02/22/2009, 01:18 PM RE: spectrum of Carleman matrix - by Gottfried - 02/22/2009, 09:29 PM RE: spectrum of Carleman matrix - by bo198214 - 02/22/2009, 10:53 PM RE: spectrum of Carleman matrix - by Gottfried - 02/23/2009, 03:52 AM

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