Timeline for Writing a matrix as a sum of two invertible matrices
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 28, 2019 at 1:29 | comment | added | LSpice | @NoamD.Elkies, your link goes to Solmsen - The world of the dead in Book 6 of the Æneid, which, while undoubtedly fascinating, probably isn't as relevant as Lord - Matrices as sums of invertible matrices (MSN). :-) | |
| Sep 7, 2013 at 5:48 | vote | accept | CommunityBot | ||
| Sep 6, 2013 at 18:43 | answer | added | user38122 | timeline score: 19 | |
| Sep 6, 2013 at 16:23 | comment | added | The Masked Avenger | The argument I remember also hinged on the 2 by 2 case, and then inducted on the order n of the matrix, but requiring the ring to have every element either be a unit or the sum of two units. I'm not sure that it can be done otherwise. | |
| Sep 6, 2013 at 9:53 | answer | added | user30230 | timeline score: 6 | |
| Sep 6, 2013 at 5:23 | comment | added | Noam D. Elkies | Sorry, good catch: $A,B,C,D$ in the last line should be $a,b,c,d$. So what I should ask is: $({a\phantom.b\atop c\phantom.d})$ is the sum of two invertible matrices over $A$, one of which is $({\alpha\phantom.\beta\atop \gamma\phantom.\delta})$, iff $\alpha\delta - \beta\gamma = (\alpha+a)(\delta+d) - (b+\beta)(c+\gamma) = 1$. Can these two simultaneous equation be solved in $A$? | |
| Sep 6, 2013 at 4:50 | comment | added | The Masked Avenger | Your double use of A is unfortunate. Could you rewrite your last equation, or remind us how you overload A? | |
| Sep 6, 2013 at 4:29 | comment | added | Noam D. Elkies | There's a paper "Matrices as Sums of Invertible Matrices" (N.J.Lord, Math Magazine 60 #1 (1987), 33-35 = jstor.org/stable/269013 ) that does it for a field. But commutative rings seem harder. Try $n=2$: Let $k$ be the two-element field, and $A = k[a,b,c,d]$. Are there $\alpha,\beta,\gamma,\delta \in A$ such that $\alpha\delta - \beta\gamma = (\alpha+A)(\delta+D) - (B+\beta)(C+\gamma) = 1$ ? | |
| Sep 6, 2013 at 1:06 | comment | added | The Masked Avenger | It is true for many rings. The terminology is somethig like "clean".There is also some USENET news posts on Sums of Invertible Matrices. A web search may help further. | |
| Sep 5, 2013 at 22:28 | comment | added | Alexandre Eremenko | I agree that my proposal does not work for all rings. | |
| Sep 5, 2013 at 22:22 | comment | added | Ricardo Andrade | @Alexandre Eremenko: What if the ring is not a division ring? Additionally, over a field with two elements, the identity matrix can only be written as a sum of two invertible matrices as follows: $\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix} + \begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix}$ | |
| Sep 5, 2013 at 22:05 | comment | added | Alexandre Eremenko | Use upper triangular matrices with non zero diagonals | |
| Sep 5, 2013 at 21:42 | review | Close votes | |||
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| Sep 5, 2013 at 21:32 | review | First posts | |||
| Sep 5, 2013 at 21:38 | |||||
| Sep 5, 2013 at 21:28 | history | edited | user39321 | CC BY-SA 3.0 |
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| Sep 5, 2013 at 21:23 | comment | added | Ryan Budney | You might want to consider the answer in the $n=1$ case as an example of how you could handle the general case. I've voted to close. | |
| Sep 5, 2013 at 21:21 | comment | added | Ryan Budney | What's a giver ring? giver $\to$ given ? | |
| Sep 5, 2013 at 21:15 | history | asked | user39321 | CC BY-SA 3.0 |