Timeline for A problem of matrix polynomial expansion
Current License: CC BY-SA 4.0
11 events
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| Sep 21, 2020 at 14:27 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Sep 21, 2020 at 13:06 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Sep 21, 2020 at 13:05 | comment | added | fs l | Yes, I see it, thanks for your reply! | |
| Sep 21, 2020 at 13:04 | comment | added | Carlo Beenakker | I added some details, including the special case $b=|a-c|/2$ when $A$ is not diagonalizable. | |
| Sep 21, 2020 at 13:04 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Sep 21, 2020 at 11:50 | comment | added | fs l | Ok, I see. But this case is for diagonalizable matrix right? What if $A$ is not diagonalizable? | |
| Sep 21, 2020 at 11:14 | comment | added | Carlo Beenakker | you diagonalize the matrix $A=U\Lambda U^{-1}$, with diagonal $\Lambda$, and then substitute $A^p=U\Lambda^p U^{-1}$, hence ${\rm tr}\,\sigma A^r DA^{n-1-r}={\rm tr}\,\tilde\sigma\Lambda^r \tilde{D}\Lambda^{n-1-r}$, with $\tilde\sigma=U^{-1}\sigma U$ and $\tilde{D}=U^{-1}DU$. | |
| Sep 21, 2020 at 11:06 | comment | added | fs l | Thanks for providing this method by using trace trick! But I am a bit confused how to derive the trace equality $(tr(\sigma A^r D A^{n-1-r}))$? Could you give me some hints? Thanks! | |
| Sep 21, 2020 at 11:02 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Sep 21, 2020 at 10:57 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Sep 21, 2020 at 10:48 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |