Timeline for spectral radius monotonicity
Current License: CC BY-SA 3.0
9 events
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Feb 28, 2018 at 20:49 | comment | added | Hans | Just revisiting, the norm $|v|$ of a vector $v$ in your answer has to be the Euclidean norm $\sqrt{v^Tv}$ for the derivation to go through, right? | |
May 6, 2013 at 20:38 | comment | added | Hans | @Sergei Ivanov: I posed another related question regarding spectral radius of non-negative matrices mathoverflow.net/questions/129890/a-spectral-radius-inequality. Please take a look. | |
May 6, 2013 at 15:05 | comment | added | Hans | @Sergei Ivanov: Your confusion comes from my inexact statement of the problem "there are two real ..., such that...". Sorry about that. I have now changed it to "... and two arbitrary ... such that...". See the edited version. Can you add explaination, but it'd be better not to change the original statement much, to your answer to reflect your revised opinion of the proof, for posterity? Thank you, Sergei. | |
May 6, 2013 at 14:24 | history | edited | Sergei Ivanov | CC BY-SA 3.0 |
corrected notice
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May 6, 2013 at 14:19 | comment | added | Sergei Ivanov | @Hans: yes, sure. I misread the question as "there exist $a$ and $b$ such that...". I am not sure about monotonicity. And actually I now see a flaw in the argument: $S'T$ is not symmetric, so its spectral radius is not equal to the norm. Sorry about this confusion. | |
May 6, 2013 at 13:45 | comment | added | Hans | @Sergei Ivanov: I see about positive definite. Regarding $b\rightarrow\infty$, I understand it's the limit. I point is that the limit case does not prove the monotonicity of the spectral radius on finite parameter $a$. For monotonicity, one need to do more. Do you agree? | |
May 6, 2013 at 10:38 | comment | added | Sergei Ivanov | @Hans: sorry, I meant "positive definite", not "positive elementwise". The inequality follows by diagonalization. The transition from $b\to\infty$ to a large $b$ is basically the definition of limit. | |
May 6, 2013 at 6:33 | comment | added | Hans | @Sergei Ivanov: Nice (partial?) solution. Thank you. But I have two questions. 1. You proved the inequality for positive symmetric matrices with $b\rightarrow\infty$. Is the transition to finite $b>a$ obvious from your present result? 2. Could you please explicate $|S'v|>|v|,\,\forall v\in R^n\\0$? Is this where your symmetry condition on the matrices comes in? | |
May 5, 2013 at 22:18 | history | answered | Sergei Ivanov | CC BY-SA 3.0 |