Timeline for Matrices with real spectrum
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2016 at 6:22 | vote | accept | Delio Mugnolo | ||
| May 9, 2016 at 6:21 | history | edited | Delio Mugnolo | CC BY-SA 3.0 |
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| Feb 11, 2015 at 9:54 | comment | added | user25199 | Following from @Noah's comment, there is a significant quantity of physics literature exploring this possibility. See eg arxiv.org/abs/1402.1082 | |
| Feb 10, 2015 at 22:57 | comment | added | Carlo Beenakker | thank you for the comments --- I was just referring to the random matrix situation, where with probability 1 any matrix that has all real eigenvalues is similar to a diagonal matrix. | |
| Feb 10, 2015 at 22:54 | comment | added | Geoff Robinson | I meant to say :In which case A can not be similar to a diagonal matrix in general, as I implicitly said above | |
| Feb 10, 2015 at 22:53 | answer | added | Denis Serre | timeline score: 2 | |
| Feb 10, 2015 at 22:50 | comment | added | Noah Stein | Are you familiar with pseudospectra? Some of the basic examples in the theory show that real or not, eigenvalues are misleading when your matrix is not normal. | |
| Feb 10, 2015 at 22:46 | answer | added | Peva Blanchard | timeline score: 4 | |
| Feb 10, 2015 at 22:45 | comment | added | Geoff Robinson | @Carlo Beenakker: In what sense are you using the expression similaar? I thought $A$ and $B$ being similar meant $B = TAT^{-1}$ for some invertible matrix $T$. | |
| Feb 10, 2015 at 22:44 | comment | added | Alex Degtyarev | @CarloBeenakker How can it be similar to a symmetric ($=$ diagonalizable) matrix if it has large Jordan blocks? Say, $[[1,1],[0,1]]$ is certainly not similar to $I$. | |
| Feb 10, 2015 at 22:05 | history | edited | Delio Mugnolo | CC BY-SA 3.0 |
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| Feb 10, 2015 at 22:01 | comment | added | Carlo Beenakker | it is certainly similar to a symmetric matrix, namely to the diagonal matrix containing the real eigenvalues on the diagonal. (ah, and omit the space after @ if you want the ping to work) | |
| Feb 10, 2015 at 21:50 | comment | added | Delio Mugnolo | @ Carlo Beenakker: Nice result, thanks, but actually I was thinking of what to do once I am already given a matrix with real spectrum. | |
| Feb 10, 2015 at 21:49 | comment | added | Delio Mugnolo | @ Geoff Robinson: Right, sorry. I have corrected my question. | |
| Feb 10, 2015 at 21:48 | history | edited | Delio Mugnolo | CC BY-SA 3.0 |
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| Feb 10, 2015 at 21:37 | comment | added | Carlo Beenakker | Can anything be said? Well, if you choose the $n^2$ matrix elements from independent normal distributions, the probability that all eigenvalues are real is $2^{-n(n-1)/4}$ (a result due to Ginibre). | |
| Feb 10, 2015 at 21:31 | comment | added | Geoff Robinson | Re the last question: Clearly not in general. Consider a Jordan block of size greater than $1$ with one real eigenvalue. | |
| Feb 10, 2015 at 21:11 | history | asked | Delio Mugnolo | CC BY-SA 3.0 |