Timeline for Is anything known about the eigenspectrum of the regular representation of the permutation group?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 25, 2015 at 14:57 | comment | added | Dima Pasechnik | a similar argument works for elements of order 3; as an eigenvalue $\lambda$ and its conjugate must occur with the same multiplicity, we see that the number of eigenvalues equal to 1 must be equal to twice the number of eigenvalues $\zeta$, for $\zeta$ a fixed primitive cubic root of unity. | |
| Apr 25, 2015 at 14:47 | comment | added | Dima Pasechnik | for the elements of order 2, it's trivial that the number of eigenvalues equal to 1 equals the number of eigenvalues equal to -1, as the trace of each non-identity element equals to 0. | |
| Apr 25, 2015 at 2:57 | comment | added | user6818 | But now see what the comment above by ARupinksi says! I am now confused! | |
| Apr 25, 2015 at 1:36 | comment | added | user6818 | Yes. But from knowing an irreducible representation of $S_n$ (say given as the partition of $n$ to which it corresponds to) how do I infer its eigenspectrum from it? | |
| Apr 25, 2015 at 0:07 | history | answered | Dima Pasechnik | CC BY-SA 3.0 |