Timeline for Similarity of a matrix with its transpose
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 29, 2022 at 7:55 | comment | added | Kapil | That is the "5-lemma" which says that if $Q$ and $R$ are invertible, then so is $P$. | |
| Apr 29, 2022 at 6:36 | comment | added | Fedor Petrov | thank you, and another probably silly question: why is this $P$ invertible? | |
| Apr 29, 2022 at 6:27 | history | edited | Kapil | CC BY-SA 4.0 |
Explain the calculation of P
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| Apr 29, 2022 at 6:24 | comment | added | Fedor Petrov | Sorry, I missed this point, thinking that $P$ also has coefficients in $F[\lambda]$. Could you please elaborate on how to get $P$ knowing $Q$ and $R$? | |
| Apr 29, 2022 at 6:16 | comment | added | Kapil | @FedorPetrov However, $P$ is a map of finite dimensional vector spaces over $F$ and thus is a given matrix over $F$ (which need not be infinite). | |
| Apr 29, 2022 at 6:02 | comment | added | Fedor Petrov | of course, but $F[\lambda]$ itself is infinite | |
| Apr 29, 2022 at 5:35 | comment | added | Kapil | @FedorPetrov I think the algorithm for SNF is for any PID where Bezout's identity can be explicitly computed. So it can be done over any $F[\lambda]$, as far as I can see. Am I missing something? | |
| Apr 29, 2022 at 5:28 | comment | added | Fedor Petrov | But this method works only over infinite field, right? | |
| Apr 29, 2022 at 4:15 | history | answered | Kapil | CC BY-SA 4.0 |