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Apr 7, 2011 at 9:30 vote accept Federico Poloni
Apr 6, 2011 at 8:11 answer added Suvrit timeline score: 3
Apr 6, 2011 at 2:16 comment added Dylan Thurston I don't know anything about separation of matrices, but the eigenvalues of $kU$ can be much closer to the eigenvalues of $S$ than the eigenvalues of $U$ with your assumptions. Eg, for some small $\epsilon$, you could have $S$ with an eigenvalue at $i - \epsilon$, $U$ with an eigenvalue at $(i + \epsilon)/2$, and $k=2$. This doesn't seem to fit your intuition. Am I misunderstanding what a "stable matrix" is?
Apr 5, 2011 at 18:21 history asked Federico Poloni CC BY-SA 2.5