Timeline for Solving $\text{trace}\left[\left(I + pY\right)^{-1} \left(I - p^{2}Y\right)\right] = 0$ for scalar $p$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2017 at 10:22 | comment | added | Rodrigo de Azevedo | @Suvrit I edited my answer and used the Schur decomposition instead, as you recommended. | |
| Jan 19, 2017 at 10:21 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 |
Schur > Jordan
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| Jan 18, 2017 at 22:42 | comment | added | Suvrit | Yes, what I meant by not computable is that JNF is notorious for being sensitive to numerical concerns, and hence I would not embed it as a subroutine anywhere; as noted in my comment, using your observation, we might as well use Schur decomposition, which is easy to compute and also works, so not only do we not care about diagonalizable, we also do not care about JNF here :-) | |
| Jan 18, 2017 at 20:21 | comment | added | Rodrigo de Azevedo | @Suvrit Why "not computable"? What does that mean? I have computed the Jordan decomposition using paper & pencil. When I learned it, I was told that it was extremely sensitive to perturbations. Examples convinced me of that claim. | |
| Jan 18, 2017 at 19:25 | comment | added | Suvrit | +1: since the JNF is not computable, it suffices to use "Schur decomposition" which is much easier to compute --- the rest of the argument applies unchanged. (link: en.wikipedia.org/wiki/Schur_decomposition ) | |
| Jan 18, 2017 at 19:21 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 |
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| Jan 18, 2017 at 19:13 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 |
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| Jan 18, 2017 at 18:53 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 |
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| Jan 18, 2017 at 18:23 | history | answered | Rodrigo de Azevedo | CC BY-SA 3.0 |