Timeline for An (open?) problem about a sequence of nested principal sub-matrices and their determinants
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2015 at 15:58 | history | reopened |
Alex Degtyarev Stefan Kohl♦ Joonas Ilmavirta Johannes Hahn Joseph O'Rourke |
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| Apr 19, 2015 at 9:40 | review | Reopen votes | |||
| Apr 19, 2015 at 15:58 | |||||
| Apr 19, 2015 at 9:24 | history | edited | teide4 | CC BY-SA 3.0 |
I reformulated my question because it was banned (and I do not understand why).
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| Apr 18, 2015 at 22:13 | comment | added | Aaron Meyerowitz | Just to show "not always" $$A = \left[ \begin{array}{ccc} 1 & 2 & 3 \\ 2 & 4 & 1 \\ 1 & 12 & 3 \end{array} \right]$$ has determinant $50$. | |
| Apr 18, 2015 at 21:49 | history | closed |
Will Jagy coudy Dima Pasechnik Ricardo Andrade Joonas Ilmavirta |
Not suitable for this site | |
| Apr 18, 2015 at 20:12 | comment | added | Richard Stanley | For a positive definite matrix, every principal minor is positive. | |
| Apr 18, 2015 at 18:39 | comment | added | teide4 | Note that this is not just a sequence of minors. They are: (1) nested; (2) "symmetric": the indices of the columns and the rows coincide; (3) "central": the diagonal of the minor lies on the diagonal of the matrix. I think that I took care enough to avoid this confusion. | |
| Apr 18, 2015 at 18:22 | review | Close votes | |||
| Apr 18, 2015 at 21:49 | |||||
| Apr 18, 2015 at 18:17 | comment | added | The Masked Avenger | If you compute the adjoint of the matrix, and count the number of nonzero terms of its permanent, that number times n! is what I think is the number of sequences will be, however you may have to compute all the minors to prove this guess. | |
| Apr 18, 2015 at 18:16 | comment | added | Mirko | repost from MSE math.stackexchange.com/questions/1240497/… (posted originally only three hours earlier, and already having two answers posted) | |
| Apr 18, 2015 at 18:11 | history | edited | darij grinberg |
edited tags
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| Apr 18, 2015 at 18:11 | comment | added | Alex Degtyarev | Each time asking "Under which conditions" you should specify in what terms you want an answer. Your desired conclusion is a condition! | |
| Apr 18, 2015 at 18:05 | comment | added | The Masked Avenger | Actually, some permutation have all zero entries in the diagonal. In fact, J-I has such a regular sequence where all but the 1 by 1 matrix has a zero diagonal. For any regular matrix, determinant computation by Laplace expansion will give you a sequence. In fact, it will give you at least n! many of them, if you count by original index sets. | |
| Apr 18, 2015 at 17:57 | review | First posts | |||
| Apr 18, 2015 at 18:11 | |||||
| Apr 18, 2015 at 17:55 | history | asked | teide4 | CC BY-SA 3.0 |