Timeline for Eigenvectors as continuous functions of matrix - diagonal perturbations
Current License: CC BY-SA 3.0
13 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 5, 2015 at 22:47 | vote | accept | Beni Bogosel | ||
| May 28, 2015 at 12:17 | answer | added | Surb | timeline score: 4 | |
| May 24, 2015 at 15:10 | comment | added | Christian Remling | The standard reference for this kind of thing is Kato, Perturbation theory for linear operators. | |
| May 24, 2015 at 15:10 | comment | added | Christian Remling | You don't need $D$ diagonal for this, symmetric is enough. The claim on the eigenvector follows for example by writing the spectral projection as a contour integral of the resolvent. | |
| May 24, 2015 at 14:28 | comment | added | Allen Knutson | Joe Silverman is correct; this is a job for the implicit function theorem (where the function being defined implicitly is $M \mapsto $its lowest eigenvector). Said theorem requires some derivative to be onto and this eventually turns into the simpleness of the eigenvalue. | |
| May 24, 2015 at 11:18 | comment | added | Joe Silverman | The key here, I think, is your assumption that the eigenvalue $\lambda$ is simple. ($A$ needn't be symmetric PD.) Then for $D$ in a small enough neighborhood, there is a continuous (even analytic) map $D\to\lambda(D)$ with $\lambda(0)=\lambda$ and $\lambda(D)$ an eigenvalue of $A+D$. The point is that if you take a simple root of a polynomial, then for sufficiently small perturbations of the coefficients, you get an analytic perturbation of the root. Presumably the same holds for the associated eigenvector, again since the eigenspace is one dimensional. | |
| May 24, 2015 at 10:53 | comment | added | Beni Bogosel | $A$ is constant, $D$ is diagonal and variable (with $N$ parameters, where $N \times N$ is the size of $A$). Ideally, the perturbation $D$ should only be continuous, but if higher regularity assumptions are needed, I don't mind. | |
| May 24, 2015 at 10:52 | history | edited | Beni Bogosel | CC BY-SA 3.0 |
added 262 characters in body
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| May 24, 2015 at 10:40 | comment | added | Federico Poloni | What is constant and what is varying there? Is $D$ a matrix-valued function? Of how many parameters? How regular? | |
| May 24, 2015 at 9:55 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
edited tags
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| May 24, 2015 at 8:24 | history | asked | Beni Bogosel | CC BY-SA 3.0 |