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Mar 3, 2019 at 16:32 history edited Jonas Adler CC BY-SA 4.0
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Mar 3, 2019 at 12:19 vote accept Jonas Adler
Mar 3, 2019 at 11:31 answer added Federico Poloni timeline score: 5
Mar 3, 2019 at 11:18 comment added Jochen Glueck Thank you for the update. But what about nilpotent nonzero matrices? Those have spectral radius zero.
Mar 3, 2019 at 11:17 comment added Jonas Adler I updated the proof of positive definiteness, the new one also gives an explicit bound.
Mar 3, 2019 at 11:17 history edited Jonas Adler CC BY-SA 4.0
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Mar 3, 2019 at 11:14 comment added Jochen Glueck By the way, sub-additivity is probably also going to be a problem since infima do not respect subadditivity, in general.
Mar 3, 2019 at 11:11 comment added Jochen Glueck Well, I think the situation is actually a bit more involved. Multiplying the norm by a constant $a > 0$ is not really a problem since the constant $a$ cancels in the definition of the induced matrix norm.
Mar 3, 2019 at 10:47 comment added Jonas Adler You are right, perhaps that is not at all true. I do think we could find such $c_1$, $c_2$ though, I'll have to look around for a proof. We would also have to "normalize" away a constant since the proof obviously fails by considering the family of norms $\|x\| = a \|x\|_2$ as $a \to \infty$.
Mar 3, 2019 at 10:24 comment added Jochen Glueck I'm not sure I can follow your argument for positive definiteness. Why should $c_1$ and $c_2$ be independent of $X$?
Mar 3, 2019 at 10:15 history edited Jonas Adler CC BY-SA 4.0
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Mar 3, 2019 at 10:10 history asked Jonas Adler CC BY-SA 4.0