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  • $\begingroup$ This is definitely correct (as a result), since the vectors stay in place, but how does it work exactly? And, even more relevant, does it work for the critical second example? $\endgroup$ Commented Nov 16, 2022 at 12:00
  • $\begingroup$ If the second example is {{0, 1, 2, 4}, {1, 0, 3, 5}, {2, 3, 0, 1}, {4, 5, 1, 0}} then it sorts to {{0, 1, 1, 2}, {1, 0, 3, 4}, {1, 3, 0, 5}, {2, 4, 5, 0}}. I will edit my answer to make the algorithm clearer $\endgroup$ Commented Nov 16, 2022 at 14:01
  • $\begingroup$ What do you mean by “the vectors stay in place” $\endgroup$ Commented Nov 16, 2022 at 23:12
  • $\begingroup$ Bad wording...The numbers of the vector stay "together". Obviously, this can't work with just sorting all numbers of the matrix - it's just a lucky accident that (0123),(0145),(0246) and (0356) stay together after sorting. OK, I reword it: if a matrix contains the vector (a1,a2,...) before, it must also contain this vector after sorting (in any order of the vector elements). $\endgroup$ Commented Nov 17, 2022 at 13:02
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    $\begingroup$ @IntroductionToProbability Your solution does not have the same eigenvalues as the starting matrix. The "staying together" means the rows can change position and the numbers in a row change position within the row. Numbers should not leave their row, only be rearranged within it. $\endgroup$ Commented Mar 27, 2023 at 12:59