Timeline for Invariant subspaces of permutation matrix [closed]
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7 events
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| Oct 3, 2013 at 22:50 | comment | added | Jason Starr | "Is this also true for fields that are not algebraically closed?" No, my comment was for algebraically closed fields. Over a non-algebraically closed field, you can try to use Galois descent: the eigenvalues of $\sigma$ are $n^{\text{th}}$ roots of unity. The Galois group acts on these roots. For each Galois orbit, the sum over that orbit of the projectors onto the corresponding eigenspaces will be Galois invariant. | |
| Oct 3, 2013 at 19:13 | comment | added | Joker | "the invariant subspaces are direct sums of one-dimensional invariant subspaces" . Is this also true for fields that are not algebraically closed? | |
| Oct 3, 2013 at 18:08 | history | closed |
Alain Valette Andrey Rekalo Benjamin Steinberg Karl Schwede Anthony Quas |
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| Oct 3, 2013 at 17:13 | review | Close votes | |||
| Oct 3, 2013 at 18:13 | |||||
| Oct 3, 2013 at 17:03 | review | First posts | |||
| Oct 3, 2013 at 17:33 | |||||
| Oct 3, 2013 at 16:58 | comment | added | Jason Starr | By "invariant", do you mean "mapped back to itself"? If so, since every permutation matrix is semisimple (at least in characteristic 0), then the invariant subspaces are direct sums of one-dimensional invariant subspaces. So if you know all the one-dimensional invariant subspaces, then you know all the invariant subspaces. | |
| Oct 3, 2013 at 16:48 | history | asked | Joker | CC BY-SA 3.0 |