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Apr 26, 2015 at 7:32 history edited Ricardo Andrade CC BY-SA 3.0
replaced tag 'spectral'
Jun 1, 2011 at 8:37 answer added Vedarun timeline score: 0
May 31, 2011 at 18:22 comment added Pietro Majer To clarify, can you confirm: $A'=A+\epsilon H$; $A$ and $H$ are symmetric and $H$ is rank-one, thus $H=u\otimes u$. The second eigenvector of $A'$ is prescribed. It is not necessarily the second eigenvector of $A$, though. Also, it seems implicitly assumed that the second eigenvalue (in decreasing order?) of $A'$ is simple.
May 31, 2011 at 15:44 answer added Igor Rivin timeline score: 2
May 31, 2011 at 15:03 comment added Michael Renardy It is not too hard to characterize rank one perturbations which make a given vector an eigenvector. I assume what makes the problem difficult is to guarantee that this eigenvector is the "second" one. You did not state the criterium for ordering, however.
May 31, 2011 at 7:21 history edited Denis Serre CC BY-SA 3.0
added 11 characters in body
May 31, 2011 at 6:37 history asked Vedarun CC BY-SA 3.0