Mi problema es demostrar que $$2\lambda{min}(PA) \leq \lambda{min}((D-M)A) ,$$ 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} = R {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 $2$ puede ser sustituida por grande constante arbitraria, pero no consigo ninguna prueba teórica.
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Chris Ballance
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