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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 21, 2013 at 14:51 comment added Michal R. Przybylek @Piotr, could you write down the whole proof?
Jun 21, 2013 at 13:41 comment added Piotr Migdal @Michal Edited.
Jun 21, 2013 at 13:41 history edited Piotr Migdal CC BY-SA 3.0
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Jun 20, 2013 at 21:28 comment added Michal R. Przybylek @Piotr, you have shown one (the trivial) direction. What about the other? (it is in Mark's answer)
Jun 20, 2013 at 15:25 comment added Piotr Migdal @Thibaut $A$ is an operator, not a vector. $v$ is a vector.
Jun 20, 2013 at 12:56 comment added 5th decile But why should an eigenvector of $A^n$ be an eigenvector of $A$?
Jun 20, 2013 at 12:43 history answered Piotr Migdal CC BY-SA 3.0