Timeline for Spectral abscissa of symmetric matrix with skew-symmetric perturbation
Current License: CC BY-SA 4.0
12 events
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| Feb 21, 2020 at 15:53 | comment | added | Jochen Glueck | Welcome to MathOverflow! I don't quite see what kind of result you are looking for. By choosing $A=0$ and by choosing a skew-symmetric matrix $S$ that has only $i\|S\|$ and $-i\|S\|$ as eigenvalues, you can see that the result that you mentioned (below the paragraph that is written italic) is optimal. | |
| Feb 21, 2020 at 15:07 | history | edited | user98563 | CC BY-SA 4.0 |
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| Feb 21, 2020 at 14:57 | history | edited | user98563 | CC BY-SA 4.0 |
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| Feb 21, 2020 at 14:44 | history | edited | user98563 | CC BY-SA 4.0 |
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| Feb 21, 2020 at 14:37 | history | edited | user98563 | CC BY-SA 4.0 |
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| Feb 21, 2020 at 14:27 | history | edited | user98563 | CC BY-SA 4.0 |
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| Feb 21, 2020 at 14:16 | comment | added | user98563 | You are right. Sorry about that, I corrected the formulation. | |
| Feb 21, 2020 at 14:14 | history | edited | user98563 | CC BY-SA 4.0 |
Fixed mistake
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| Feb 21, 2020 at 13:40 | comment | added | David Handelman | The result for normal matrices is misstated---take diagonal matrices. The best I can see is that for any eigenvalue $\lambda$ of $A$, there exists an eigenvalue $\mu$ of $B$ such that $|\lambda - \mu| \leq \epsilon$. If $A$ and $B$ are symmetric, then the eigenvalues are real and can be ordered, and a stronger result is available. | |
| Feb 21, 2020 at 12:51 | history | edited | user98563 | CC BY-SA 4.0 |
Improved formatting
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| Feb 21, 2020 at 11:00 | review | First posts | |||
| Feb 21, 2020 at 11:14 | |||||
| Feb 21, 2020 at 10:59 | history | asked | user98563 | CC BY-SA 4.0 |