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Oct 14, 2019 at 5:47 comment added user6671 @user142929 thanks for your comment
Oct 12, 2019 at 20:35 comment added user142929 I think (from the last paragraph of the article due to Lagarias) that maybe it is interesting consider $l(x)=e^x\log x$. For $L(x)$ or $l(x)$, maybe can be interesting to state Möbius inversion, and related asymptotic formulas, section Generalizations of the Wikipedia Möbius inversion formula as starting point. I refer also (section 3) of Manuel Benito and Luis M. Navas and Juan Luis Varona, Möbius inversion from the point of view of arithmetical semigroup flows, Biblioteca de la Revista Matemática Iberoamericana, Proceedings of the Segundas Jornadas de Teoría de Números (Madrid, 2007).
May 24, 2019 at 18:07 history edited user6671 CC BY-SA 4.0
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May 19, 2019 at 4:23 history edited user6671 CC BY-SA 4.0
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May 18, 2019 at 20:21 history edited user6671 CC BY-SA 4.0
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May 18, 2019 at 20:19 comment added user6671 @NathanielJohnston: Ok, I will update the question, with further properties.
May 18, 2019 at 20:13 comment added Nathaniel Johnston @orgesleka -- It definitely can't be true if you allow arbitrary complex eigenvalues. Try the diagonal matrix $A$ with diagonal entries $1$ and $i$. Then $L(|A|) = 1$ but $|L(A)| \approx 2.27$.
May 18, 2019 at 16:43 comment added user6671 @NathanielJohnston: Some of the matrices I am considering do not have real eigenvalues, but still $L(|A|) = |L(A)|$.
May 18, 2019 at 16:17 comment added user6671 @NathanielJonston: Ok, thanks for your comment. Would you mind writing your comment in a bit more detail as an answer? (If it is not asked too much than maybe with reference?)
May 18, 2019 at 16:14 comment added Nathaniel Johnston Oh I see. You're right, that property holds then simply by usual functional calculus. $|A|$ is the largest eigenvalue of $A$, whereas $|L(A)|$ is the largest eigenvalue of $L(A)$. But the eigenvalues of $L(A)$ are just $L$ applied to the eigenvalues of $A$. Since $L$ is monotone, the largest eigenvalue of $L(A)$ is just $L(|A|)$. This works as long as $A$ has positive real eigenvalues (but if that's not the case then even defining $L(A)$ in the first place is going to be somewhat icky, so I hope it's OK).
May 18, 2019 at 15:52 comment added user6671 @NathanialJohnston: Spectral norm. Can you give your examples? The matrices I am looking at seem to have this property.
May 18, 2019 at 15:42 comment added Nathaniel Johnston Can you clarify your third edit? What is the Euclidean norm of a matrix? Frobenius/Hilbert-Schmidt norm? Or operator/spectral norm? Or something else? I've tried a few guesses, but numerics quickly disprove $L(|A|) = |L(A)|$ for every guess that I've made for what the norm should be.
May 18, 2019 at 14:38 history edited user6671 CC BY-SA 4.0
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May 10, 2019 at 18:29 history edited Carlo Beenakker CC BY-SA 4.0
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May 8, 2019 at 7:49 history edited user6671 CC BY-SA 4.0
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Apr 30, 2019 at 13:38 comment added user6671 @MattF. You are right. The uniqueness of your characterization is given.
Apr 30, 2019 at 12:42 comment added user44143 WolframAlpha confirms the uniqueness of the characterization I proposed: wolframalpha.com/input/…. The alternative suggestion has $L(1)=e$.
Apr 30, 2019 at 11:28 comment added user6671 @MattF. I think you forgot the condition $L'(1) = e+1$. Thanks for your help.
Apr 30, 2019 at 11:14 comment added user44143 One way to characterize it is $x L'' - (2 x - 1) L' + (x - 1) L =x^2 - 3 x + 1$, $L(1)=1$, $L''(1)=e$
Apr 30, 2019 at 10:48 history asked user6671 CC BY-SA 4.0