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S Jul 23, 2014 at 14:09 history suggested CommunityBot CC BY-SA 3.0
removed obviously incorrect statement from the problem definition
Jul 23, 2014 at 13:41 review Suggested edits
S Jul 23, 2014 at 14:09
S Jul 20, 2014 at 10:00 history suggested CommunityBot CC BY-SA 3.0
added some more information
Jul 20, 2014 at 9:54 review Suggested edits
S Jul 20, 2014 at 10:00
Nov 4, 2013 at 20:35 history edited john stark CC BY-SA 3.0
clarification
Nov 4, 2013 at 20:29 history edited john stark CC BY-SA 3.0
clarification
Nov 4, 2013 at 19:44 history edited john stark CC BY-SA 3.0
typos, layout
Nov 4, 2013 at 17:05 comment added john stark Cyclic = circulant indeed. Also, the matrix R is formed from some Fourier transform, so the eigenvalues converge to this transform as $N$ grows. See the OP for an update of the text.
Nov 4, 2013 at 17:04 history edited john stark CC BY-SA 3.0
corrected question according to comment
Nov 3, 2013 at 23:12 comment added Federico Poloni I have some difficulty in interpreting the question. Is $R$ a function of $N$? How does it vary with $N$? By "cyclic" you mean "circulant" as in the title?
Nov 3, 2013 at 22:58 history edited john stark CC BY-SA 3.0
cleaned it up and simplified
Nov 3, 2013 at 20:40 review Suggested edits
Nov 3, 2013 at 21:05
Nov 3, 2013 at 11:09 comment added john stark yes, for each element. From Matlab, it appears as if the solution is $(R+\lambda I)^{-1}$ for some positive value $\lambda$. If true in general, the problem reduces into finding $\lambda$. And I agree, the large limit is not very helpful, but laid my complete problem down anyway.
Nov 2, 2013 at 20:55 comment added Carlo Beenakker expectation of a matrix? you mean the average of every single matrix element? unlikely that the large-$N$ limit will be of much help for that....
Nov 2, 2013 at 19:47 history asked john stark CC BY-SA 3.0