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Aug 25, 2020 at 20:42 vote accept soupy
Aug 30, 2017 at 8:03 comment added Federico Poloni Note that, in a generalization of this argument to a non-diagonalizable matrix, if $v_1$ is an eigenvector and $v_2$ is the second vector of a Jordan basis, then their duals swap their roles: $f_2$ is the eigenvector and $f_1$ is the generalized eigenvector. I think that the generalized proof will have to account for this property in some way.
Jul 30, 2017 at 23:16 comment added Ted Shifrin Since you need the eigenvalue in there, I don't quite see it quickly, @Qiaochu.
Jul 30, 2017 at 23:02 comment added Qiaochu Yuan You can run basically the same argument with the expression $(A - I)^k$ and generalized eigenvectors.
Jul 30, 2017 at 22:54 comment added Ted Shifrin What if there is no eigenbasis?
Jul 30, 2017 at 22:18 history answered Qiaochu Yuan CC BY-SA 3.0