Timeline for If $A,B$ are upper triangular matrices such that $AX=XA\implies BX=XB$ for upper triangular $X$, is $B$ a polynomial in $A$?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2018 at 2:14 | vote | accept | Spot | ||
| Jul 10, 2018 at 1:50 | answer | added | David E Speyer | timeline score: 30 | |
| Jul 9, 2018 at 20:22 | history | edited | Spot | CC BY-SA 4.0 |
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| Jul 9, 2018 at 18:42 | comment | added | MTyson | @LSpice Every diagonal $B$ is polynomial in any regular diagonal $A$. You just need a polynomial sending each $A_{ii}$ to $B_{ii}$, which is furnished by Lagrange interpolation. | |
| Jul 9, 2018 at 17:49 | comment | added | LSpice | Although, speaking of @MeisamSoleimaniMalekan's example, suppose that $A$ is regular (= eigenvalues pairwise distinct) diagonal. Then the implication holds whenever $B$ is diagonal, but surely it's not the case that every diagonal $B$ is polynomial in any regular diagonal $A$, right? | |
| Jul 9, 2018 at 17:48 | comment | added | LSpice | @MeisamSoleimaniMalekan, doesn't one usually speak of positivity and square roots only for symmetric matrices, which are upper triangular only if diagonal? | |
| Jul 9, 2018 at 17:42 | answer | added | Alex Gavrilov | timeline score: 8 | |
| Jul 9, 2018 at 14:19 | answer | added | Folkmar Bornemann | timeline score: 8 | |
| Jul 9, 2018 at 9:59 | comment | added | MSMalekan | What happens if $A$ is positive, and $B=A^\frac12$? | |
| Jul 9, 2018 at 5:07 | comment | added | zibadawa timmy | @GerhardPaseman I think that doesn't work, at least not in a particularly obvious/trivial way. $\begin{pmatrix} 0& 1\\ 0 & 0\end{pmatrix}$ commutes with any matrix of the form $\begin{pmatrix}a & b\\ 0 & a\end{pmatrix}$, but $\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$ commutes with such a matrix if and only if $b=0$. | |
| Jul 9, 2018 at 3:18 | comment | added | Spot | @FrancoisZiegler Ah! That's actually quite helpful. I kept searching for "double commutant" and getting stuff on Von Neumann algebras. Thanks! | |
| Jul 9, 2018 at 2:54 | comment | added | Francois Ziegler | You are asking if $\mathbb F[A]$ is its own double centralizer in $T_n(\mathbb F)$. (It’s only a reference request if you know it’s true. I don’t.) | |
| Jul 8, 2018 at 23:23 | answer | added | Igor Rivin | timeline score: 4 | |
| Jul 8, 2018 at 21:15 | comment | added | Gerhard Paseman | Can't something like transpose be used to expand the class of matrices commuting with A, and then close under other algebraic operations? Gerhard "Taking New Slant On It?" Paseman, 2018.07.08. | |
| Jul 8, 2018 at 20:05 | comment | added | LSpice | Just an observation: by induction, you can assume that $B$ has non-$0$ entries only in its last column. (I don't know how to prove the general result, or even whether it's true.) | |
| Jul 8, 2018 at 19:10 | review | First posts | |||
| Jul 8, 2018 at 19:12 | |||||
| Jul 8, 2018 at 19:06 | history | asked | Spot | CC BY-SA 4.0 |