Timeline for How do I get the permutation matrix of an element in the symmetric group?
Current License: CC BY-SA 4.0
4 events
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| Apr 13, 2020 at 13:57 | vote | accept | geoffrey | ||
| Apr 13, 2020 at 12:25 | comment | added | Henrik Schumacher |
I know. Btw., calling SparseArray and PermutationList without a second argument is asking for trouble. And Map and friends are slow. ;)
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| Apr 13, 2020 at 12:22 | comment | added | Roman |
You don't even need to define n if you do SparseArray@MapIndexed[{#2[[1]], #1} -> 1 &, PermutationList[Cycles[{{2, 4}}]]].
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| Apr 13, 2020 at 12:15 | history | answered | Henrik Schumacher | CC BY-SA 4.0 |