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Consider two positive-definite Toeplitz matrices $M_1$ and $M_2$ both with dimension $2^j \times 2^j$. Their matrix elements are: $$M_1[x,y] = \frac{\text{sin}(\pi(x-y)/2^j)}{\pi(x-y)} \qquad M_2[x,y] = \frac{1}{2^n}\frac{\text{sin}(\pi(x-y)/2^j)}{\text{sin}(\pi(x-y)/2^n)}$$ where $1<j<n$. One observes that two matrices have the same trace and as $n \rightarrow \infty$ $M_2$ approaches $M_1$. Now it appears $M_1$ belongs to a famous class of matrices, whose eigenvectors are discrete prolate spheroidal sequences and whose eigenvalues decay very quickly. By numerical simulations, I observed $M_2$'s eigenvalues decay even faster, particularly I suspect $\lambda_{1}(M_2) > \lambda_{1}(M_1)$ and $\lambda_k(M_2) < \lambda_k(M_1)$ for $k>1$, ie. $M_2$'s largest eigenvalue is larger but all others are smaller.

My question is if this statement is true; if so, could anyone give some hints on how to prove it? I was looking at Terry Tao's blog on https://terrytao.wordpress.com/2008/10/28/when-are-eigenvalues-stable/ and some work on eigenvalue perturbation (eg. perturbation theory in quantum mechanics) but I'm not sure if this is the right direction. Thanks in advance!

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  • $\begingroup$ An additional comment is the sum of eigenvalues of both matrices is 1, and $|M_1-I/2^j| < |M_2 - I/2^j|$. Would this be sufficient to argue that $M_1$ has a flatter eigenvalue distribution? $\endgroup$ Commented May 9, 2022 at 10:14

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