Timeline for Are unitarily equivalent permutation matrices permutation similar?
Current License: CC BY-SA 4.0
5 events
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| Dec 10, 2019 at 21:24 | comment | added | Benjamin Steinberg | Alternatively the trace of $A^m$ counts the number of elements whose orbit has size dividing $m$ and so if they are linearly conjugate they have the same number of cycles of each size. | |
| Dec 10, 2019 at 19:06 | comment | added | YCor | @shuhalo Why? I assume that they're linearly conjugate, which is even weaker than unitarily conjugate, and obtain that they're conjugate as permutations, which is exactly what you're asking. | |
| Dec 10, 2019 at 17:36 | comment | added | shuhalo | I agree with the proof. But this only partially answer the question, does it? | |
| Dec 10, 2019 at 16:01 | history | edited | YCor | CC BY-SA 4.0 |
simplified
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| Dec 10, 2019 at 15:50 | history | answered | YCor | CC BY-SA 4.0 |