Timeline for Pfaffian minors of skew symmetric matrix under perturbation
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Dec 17, 2016 at 21:07 | history | bounty ended | CommunityBot | ||
| S Dec 17, 2016 at 21:07 | history | notice removed | CommunityBot | ||
| Dec 14, 2016 at 12:50 | comment | added | Lev Borisov | You can probably make $D$ with positive minors by having $D_{ij}$ grow extremely fast in the lexicographic order so that one of the terms in each Pfaffian dominates the rest of the terms. | |
| Dec 13, 2016 at 20:50 | comment | added | Lev Borisov | If you have $A' = A + t D$, then Pfaffians of $A'$ will have as their leading terms as $t\to \infty$ the Pfaffians of $D$ times the appropriate power of $t$. So positivity for $A'$ follows from that of $D$. | |
| Dec 13, 2016 at 19:49 | comment | added | SiOn | @LevBorisov, could you please explain "the problem reduces to finding a skew-symmetric matrix with positive integer entries above the diagonal such that all Pfaffians are positive" in a bit detail. | |
| Dec 10, 2016 at 2:29 | comment | added | Lev Borisov | I suppose that existence could be proved by induction on $n$; there might also be some explicit construction(s). | |
| Dec 10, 2016 at 2:27 | comment | added | Lev Borisov | So by scaling $D= A'-A$ by a large integer, the problem reduces to finding a skew-symmetric matrix with positive integer entries above the diagonal such that all Pfaffians are positive. In fact, one can ignore the integrality and allow for arbitrary nonnegative entries (which can then be approximated by rationals and zeroes replaced by small rationals). | |
| Dec 9, 2016 at 23:46 | comment | added | SiOn | I really meant the upper diagonal entries! | |
| Dec 9, 2016 at 23:04 | comment | added | darij grinberg | "skew-symmetric matrix whose entries are positive integers" wut? How can two entries be positive integers if one of them is the negative of the other? | |
| S Dec 9, 2016 at 19:19 | history | bounty started | SiOn | ||
| S Dec 9, 2016 at 19:19 | history | notice added | SiOn | Draw attention | |
| Dec 6, 2016 at 20:31 | comment | added | David E Speyer | No, I don't know what the question is. But its only been two hours and maybe I'm being dense and everyone else understands this. | |
| Dec 6, 2016 at 20:26 | comment | added | SiOn | do you know the answer already? | |
| Dec 6, 2016 at 20:18 | comment | added | David E Speyer | Okay, I'll drop out and leave someone else to answer this. I am pretty sure the question is of the form: Given $A$ a skew symmetric matrix obeying CONDITIONS, do there exist skew symetric matrices $D$ and $A'$ obeying $A+D=A'$ and CONDITIONS? But I can't tell what those conditions are. | |
| Dec 6, 2016 at 20:12 | comment | added | SiOn | not really. This is not standard definition of perturbation. I am just calling this a perturbation. | |
| Dec 6, 2016 at 20:02 | comment | added | David E Speyer | You write "A perturbation of A, A′, is obtained by adding another skew-symmetric matrix whose entries are positive integers". Is this the definition of perturbation? Should the definition not include a condition that $A'-A$ is small? | |
| Dec 6, 2016 at 18:51 | comment | added | SiOn | Sorry I didn't get your question. Perturbation is done by skew symmetric matrices with positive integers entires while $A$ could be any skew symmetric matrix with positive real entries. | |
| Dec 6, 2016 at 18:44 | comment | added | David E Speyer | Okay. But your definition of perturbation is that $A'$ is a perturbation of $A$ if $A'-A$ is nonnegative, with no hypothesis that $A'-A$ be small. So, take any positive $B$ and then, for any $A$, if we take $t$ sufficiently large, $tB$ will be a perturbation of $A$. Is that what you wanted to ask for? | |
| Dec 6, 2016 at 18:38 | comment | added | SiOn | I really meant for a fixed definition of pfaffian, the pfaffian (as a number) is positive for all those minors. | |
| Dec 6, 2016 at 18:28 | comment | added | David E Speyer | By positive you mean positive above the diagonal right? But doesn't this question just come down to "is there a matrix $B$ all of whose Pfaffians are nonnegative"? Because, if there is, we can choose $t$ large enough that $tB-A$ is positive, and then $tB$ is a perturbation of $A$ by your definition. I assume you left something out of the conditions. | |
| Dec 6, 2016 at 17:47 | history | asked | SiOn | CC BY-SA 3.0 |