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Nov 24, 2011 at 20:26 vote accept Alexander
Nov 22, 2011 at 18:46 history closed Bill Johnson
Yemon Choi
Andrés E. Caicedo
user6976
Ryan Budney
too localized
Nov 21, 2011 at 21:01 vote accept Alexander
Nov 24, 2011 at 20:26
S Nov 21, 2011 at 21:00 vote accept Alexander
Nov 21, 2011 at 21:01
Nov 21, 2011 at 21:00 vote accept Alexander
S Nov 21, 2011 at 21:00
Nov 21, 2011 at 10:45 comment added Christopher A. Wong What do you mean by canonical basis? There is certainly more than one choice of eigenbasis if you just have a two-dimensional eigenspace, for example.
Nov 21, 2011 at 10:04 history edited Alexander CC BY-SA 3.0
added 80 characters in body
Nov 20, 2011 at 23:36 comment added Yemon Choi I'm also not quite sure what you're looking for with your questions; perhaps I have misunderstood something. We know that for the non-zero eigenvalues, the eigenspaces are finite-dimensional; so all the theory for the finite-dimensional case applies on each such eigenspace. On the other hand, one can have quasinilpotent compact operators on Banach spaces and there all bets are off unless one knows more about the invariant subspaces of the operator (like the Volterra case mentioned in on eof the answers)
Nov 20, 2011 at 23:32 comment added Yemon Choi Ringrose's little book "Compact non-self-adjoint operators" has a good exposition of how things work in the Hilbert space case. The basics of the theory for compact operators on general Banach spaces is in Rudin's Functional Analysis, if I recall correctly.
Nov 20, 2011 at 20:53 answer added Anatoly Kochubei timeline score: 0
Nov 20, 2011 at 20:00 answer added Florian timeline score: 2
Nov 20, 2011 at 16:17 history asked Alexander CC BY-SA 3.0