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My problem is to show that $$2\lambda_{min}(PA) \leq \lambda_{min}((D-M)A) ,$$ where $A$ is arbitrary $nxn$ symmetric positive definite matrix and $P$ is diagonal matrix, which has $\frac{1}{nA_{ii}}$ on the $i$-th element on the diagonal and $D$ is diagonal matrix with $i$-th diagonal element equals to $2\sum_{j = 1}^n\frac{p_{ij}A_{jj}}{Det_{ij}}$ and $M$ is symmetric matrix with $M_{ij} = 2\frac{p_{ij}A_{ij}}{Det_{ij}}$, where $p_{ij}$ are probabilities with $p_{ii} = 0$ for all $i$, $p_{ij} = \frac{1}{n(n-1)}$ for $i\neq j$, and $Det_{ij} = A_{ii}A_{jj} - A_{ij}^2$ for $i \neq j$, and $Det_{ii} = 1$ for all $i$. $A_{ij}$ corresponds to the element in the $i$-th row and $j$-th column. I tried many experiments and this statement always holds, even with some special cases $2$ can be substituted by arbitrary large constant, but I cannot get any theoretical proof.

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(Edit. The OP has changed the question. This answer is no longer valid.)

The inequality doesn't hold in general. Consider the case where $P=A=I_3,\ p_{12}=p_{21}=\frac12$ and all other $p_{ij}$s are zero. Then $PA=I_3$ but $(D-M)A=$ is singular.

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  • $\begingroup$ I edited question just for special case of uniform sampling on which I ran all the experiments. For general case, I only assume proper sampling, which I forgot to mention. $\endgroup$ – Ethan Dec 22 '17 at 9:00

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