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Oct 1, 2017 at 4:43 vote accept Alexander Burstein
Aug 11, 2017 at 17:50 answer added esg timeline score: 5
Aug 10, 2017 at 19:44 comment added Alexander Burstein @FedorPetrov My coauthor and I also looked at those eigenvectors for small $n$, but couldn't find a general formula for a given $k$ (except that $k=n$ is easy). Probably, some inclusion-exclusion going on there, but the exact formula is hard to guess.
Aug 10, 2017 at 19:30 comment added Fedor Petrov The eigenvectors for $n$ at most 5 form a pretty nice orthogonal basis, but I do not see a general formula from them.
Aug 10, 2017 at 19:23 comment added Suvrit @AlexanderBurstein btw the positive definiteness of the matrix $A_m$ in Sec. 4 of your paper can follow "more directly" by recalling that the Pascal matrix is positive definite (proof: $\binom{p+q}{p} = c\int_0^{2\pi} (1+e^{i\theta})^p(1+e^{-i\theta})^qd\theta$) and invoking Schur's theorem on psdness of the elementwise product of psd matrices.
Aug 10, 2017 at 18:47 history edited Alexander Burstein CC BY-SA 3.0
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Aug 10, 2017 at 18:34 comment added Alexander Burstein To see where this result might have been useful (and also, speaking of traces of matrices), take a look at Section 4 of this paper. @FedorPetrov
Aug 10, 2017 at 18:32 comment added Alexander Burstein There is a paper by Amdeberhan that proves the determinant of this matrix is the product of those eigenvalues, but that does not seem to help in finding the individual eigenvalues. @FedorPetrov
Aug 10, 2017 at 10:13 comment added Fedor Petrov By the way, computing the trace of this matrix and the sum of eigenvalues we get a famous identity $\sum \binom{2i}{i}\binom{2(n-i)}{n-i}=4^n$.
Aug 10, 2017 at 8:04 history edited Carlo Beenakker CC BY-SA 3.0
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Aug 10, 2017 at 7:38 history edited Alexander Burstein
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Aug 10, 2017 at 7:36 history edited Emil Jeřábek
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Aug 10, 2017 at 7:34 history edited Alexander Burstein
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Aug 10, 2017 at 7:30 review First posts
Aug 10, 2017 at 7:37
Aug 10, 2017 at 7:26 history asked Alexander Burstein CC BY-SA 3.0