Timeline for Reference request: continuity of Cholesky factor
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| May 26, 2020 at 8:34 | vote | accept | ABIM | ||
| May 25, 2020 at 16:47 | history | became hot network question | |||
| May 25, 2020 at 12:22 | history | edited | ABIM |
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| May 25, 2020 at 11:42 | vote | accept | ABIM | ||
| May 25, 2020 at 11:42 | |||||
| May 25, 2020 at 11:15 | comment | added | ABIM | Thanks. This will be very interesting! | |
| May 25, 2020 at 11:04 | comment | added | Federico Poloni | Higham's book Functions of matrices. It has a definition on how to extend an arbitrary scalar function to matrices (like you probably already studied with the exponential of a nonsymmetric matrix) in Chapter 1, including some remarks on branches and the principal square root, and then a chapter devoted to the properties of matrix square roots. | |
| May 25, 2020 at 10:36 | comment | added | ABIM | No worries, I worked out what I needed from the answer as you pointed out. However, now I'm interested (purely out of curiousity) do you have a reference to this principle matrix squre-root s, purely our of scientific interest. | |
| May 25, 2020 at 10:33 | vote | accept | ABIM | ||
| May 25, 2020 at 11:42 | |||||
| May 25, 2020 at 10:32 | answer | added | Federico Poloni | timeline score: 2 | |
| May 25, 2020 at 10:26 | comment | added | Federico Poloni | There is a thing called "the (principal) matrix square root", which is defined for all matrices (possibly nonsymmetric) with no real negative eigenvalues and no nontrivial Jordan blocks in zero. It is continuous, but it is nontrivial to prove it. However, it is not what you are asking about here, so I have changed the title. Calling a Cholesky factor "square root" is slightly improper, although I have already heard it in various contexts. | |
| May 25, 2020 at 9:57 | history | edited | ABIM | CC BY-SA 4.0 |
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| May 25, 2020 at 9:56 | comment | added | ABIM | @FedericoPoloni What do you mean? In the case of semi-definite, ut symmetric matrices, would it be easier? | |
| May 25, 2020 at 9:55 | comment | added | Federico Poloni | I have edited the title. The matrix square root is another thing, and proving that it is continuous is nontrivial (for non-symmetric matrices) in my view. | |
| May 25, 2020 at 9:54 | history | edited | Federico Poloni | CC BY-SA 4.0 |
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| May 25, 2020 at 9:45 | answer | added | Carlo Beenakker | timeline score: 3 | |
| May 25, 2020 at 9:34 | history | edited | YCor | CC BY-SA 4.0 |
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| May 25, 2020 at 8:55 | history | edited | ABIM | CC BY-SA 4.0 |
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| May 25, 2020 at 8:45 | history | asked | ABIM | CC BY-SA 4.0 |