Skip to main content
9 events
when toggle format what by license comment
Sep 12, 2017 at 11:38 vote accept PThomasCS
Dec 22, 2015 at 1:26 comment added Anthony Quas I think the idea is that if the spectrum is simple, then the eigenvectors and eigenvalues vary continuously. Hence if the limiting diagonal matrix has almost surely got distinct eigenvalues, then the eigenspaces of the limiting matrix are one-dimensional and are spanned by elements of the standard basis. By continuity, this is also true for nearby matrices.
Dec 21, 2015 at 17:29 answer added Liviu Nicolaescu timeline score: 1
Dec 21, 2015 at 16:42 comment added PThomasCS @LiviuNicolaescu "because in this case, with probability one, the spectrum of $M_n$ is simple." Can you elaborate on why this implies the result?
Dec 21, 2015 at 15:44 comment added Liviu Nicolaescu You need to make some assumptions on the nature of randomness of the matrices $M_n$ that I gather are finite dimensional. If you assume that the entries of $M_n$ are continuous random variables then the result is true because in this case, with probability one, the spectrum of $M_n$ is simple.
Dec 21, 2015 at 14:58 history edited PThomasCS CC BY-SA 3.0
added 28 characters in body
Dec 21, 2015 at 14:51 history edited PThomasCS CC BY-SA 3.0
added 95 characters in body
Dec 21, 2015 at 14:25 review First posts
Dec 21, 2015 at 14:51
Dec 21, 2015 at 14:21 history asked PThomasCS CC BY-SA 3.0