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Nov 7, 2019 at 8:22 comment added Gerry Myerson First of all, you shouldn't post to two (or more!) sites at all. You post to one site and, if a week goes by with no useful response, then you may post to another (if the post is on-topic on both sites). When/if you do, you include a link to the other post in each of the two posts.
Nov 7, 2019 at 7:02 comment added alex.jordan @GerhardPaseman You might be interested in the answer I posted on math.se (follow Gerry's link) that describes a way to generate all of these matrices if you can first identify certain triangles in the plane (or tetrahedrons for the $n=4$ case, etc.).
Nov 7, 2019 at 5:59 comment added Ri-Li @GerryMyerson Thanks a lot for informing me. By the way, I don't know about this much. So can you pls put some light here? e.g where and whom do I have to inform?
Nov 7, 2019 at 4:13 comment added Gerry Myerson Also posted to m.se math.stackexchange.com/questions/3424001/… without notice to either site – that's an abuse.
Nov 6, 2019 at 18:44 comment added LSpice "It has a strong influence with Cremona groups but I won't disclose that"—I understand the impulse to protect one's research directions, but I think that, in general, if you want to motivate other people to do some of your research work for you then you need at least to give them some idea of what it's working towards.
Nov 6, 2019 at 18:17 answer added Gerhard Paseman timeline score: 1
Nov 6, 2019 at 17:37 comment added Ri-Li Can you please give me solutions for $k=2$ and $k=3$ writing elaborately. I will disclose the motivation of thinking about this problem. It has a strong influence with Cremona groups but I won't disclose that once I go with that a bit more. I can chat and discuss those things once I get this part done and you will realise column part is not so important here. Although in my given examples the column sum is also preserved but this is not the case in general.
Nov 6, 2019 at 16:57 comment added Gerhard Paseman I retract a previous remark. Since you only have row and not column constraints there will be superexponentially many lower triangular matrices which are solutions. I would be interested in solutions with column constraints with matrices that are not triangular. Gerhard "Let's Make It Even Harder" Paseman, 2019.11.06.
Nov 6, 2019 at 16:52 comment added Gerhard Paseman For k=1 and arbitrary orders, the count is half the number of permutation matrices. This could become a good question for MathOverflow if you provide more motivation and originating context. Since you are asking for a relatively small determinant, my guess is the restriction is tight enough that given k there will be few order 3 matrices satisfying the relation. Gerhard "Up To Permutation, Of Course" Paseman, 2019.11.06.
Nov 6, 2019 at 16:33 history asked Ri-Li CC BY-SA 4.0