Timeline for Is simultaneous similarity of matrices independent from the base field?
Current License: CC BY-SA 4.0
22 events
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| Aug 18, 2020 at 11:01 | vote | accept | Anton Klyachko | ||
| Aug 15, 2020 at 1:00 | comment | added | Donu Arapura | @LSpice I don't agree. While it's true this was asked and answered before, the sites are different. Some of us don't usually look at the other one. | |
| Aug 14, 2020 at 18:26 | review | Close votes | |||
| Aug 16, 2020 at 14:11 | |||||
| Aug 14, 2020 at 18:07 | comment | added | LSpice | I’m voting to close this question because @DavidESpeyer points out it's a duplicate of math.stackexchange.com/questions/305696 . | |
| Aug 14, 2020 at 16:31 | comment | added | Anton Klyachko | @LSpice, I have changed the letter:) | |
| Aug 14, 2020 at 16:27 | history | edited | Anton Klyachko | CC BY-SA 4.0 |
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| Aug 14, 2020 at 16:26 | answer | added | Anton Klyachko | timeline score: 5 | |
| Aug 14, 2020 at 14:21 | comment | added | David E Speyer | As everyone is saying, this follows from Noether-Deuring. See mathoverflow.net/questions/28469/hilbert-90-for-algebras for a quick proof. I also asked this question a while back math.stackexchange.com/questions/305696 . | |
| Aug 14, 2020 at 14:17 | comment | added | darij grinberg | Yes, this follows from the Noether-Deuring theorem, as @JeremyRickard has said. You don't even need to consider infinite-dimensional algebras for that; it suffices to use the subalgebra of the matrix ring generated by the matrices $A_1, A_2, \ldots, A_m$. | |
| Aug 14, 2020 at 14:14 | comment | added | LSpice | I like this question, but I have never before seen someone so notationally bold as to use $G$ for a field. (My personal field alphabet has $E$ or $K$ as the next letter after $F$.) Now what do we call algebraic groups over $G$? // Also, is it any help to look at a matrix $\widetilde T$ over $F$ conjugating $A_1 \oplus \dotsb \oplus A_m$ to $B_1 \oplus \dotsb \oplus B_m$? | |
| Aug 14, 2020 at 14:07 | history | edited | YCor |
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| Aug 14, 2020 at 13:35 | comment | added | Fedor Petrov | @GeoffRobinson ah, $Im(S)$ is invariant for all $A_i$, thus for all matrices, thus it is the whole space $F^n$, and $S$ is invertible. This seems to avoid the uniqueness argument. | |
| Aug 14, 2020 at 12:58 | comment | added | Geoff Robinson | Irreducible in my proof above meant under left multiplication by all of $M_{n}(G)$. | |
| Aug 14, 2020 at 12:52 | comment | added | Geoff Robinson | @Jeremy Rickard: I think it probably does, but I always remember the irreducible case | |
| Aug 14, 2020 at 12:50 | comment | added | Jeremy Rickard | Doesn't the Noether-Deuring theorem work anyway? The hypotheses give us two finite-dimensional modules over the free algebra $F\langle x_1,\dots,x_m\rangle$ that become isomorphic when you extend scalars to $G$, so by Noether-Deuring they are already isomorphic over $F$. | |
| Aug 14, 2020 at 12:45 | comment | added | Geoff Robinson | @FedorPetrov :It's Schur's Lemma I think. The $A_{i}$ span $M_{n}(F)$ over $F$ and certainly span $M_{n}(G)$ over $G$. Hence the $B_{i}$ span $M_{n}(G)$ as well. Thus, over $G$ , if there is a non-zero matrix $S$ with $A_{i}S = SB_{i}$ for each $i$. Then $ImS$ is an invariant subspace of the vectors space $G^{n}$ of column vectors., so is the whole space $G^{n}$. Thus $S$ is invertible, and the we have $TS^{-1}A_{i}ST^{-1} = A_{i}$ for each $i$, so $ST^{-1}$ is scalar. | |
| Aug 14, 2020 at 12:21 | comment | added | Fedor Petrov | @Geoff why is it unique? | |
| Aug 14, 2020 at 11:52 | history | edited | Anton Klyachko | CC BY-SA 4.0 |
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| Aug 14, 2020 at 11:30 | comment | added | Geoff Robinson | I think that if the $A_{i}$ span ${\rm M}_{n}(F)$, then I the answer is yes, essentially as in the Noether-Doering Theorem, there is a solution to the linear equations in the smaller field, because there is one in the bigger field. This solution is then unique up to non-zero scalar multiples, so since the solution in the bigger field gives an invertible matrix, so does the one in the smaller field. | |
| Aug 14, 2020 at 11:12 | comment | added | Anton Klyachko | I have edited that. In these cases the answer is yes. | |
| Aug 14, 2020 at 11:10 | history | edited | Anton Klyachko | CC BY-SA 4.0 |
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| Aug 14, 2020 at 10:04 | history | asked | Anton Klyachko | CC BY-SA 4.0 |