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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.
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.
