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Let $M = (m_{ij})$ be $n \times n$ symmetric positive definite matrix. Then it can be proven that $$ M^{1/2}A M^{1/2} \succeq M^{1/2}D M^{1/2}\succ 0 $$ so \begin{equation} \lambda_{\min}(M^{1/2}A M^{1/2}) \geq \lambda_{\min}(M^{1/2}D M^{1/2}), \tag{!} \end{equation} where D is diagonal matrix with diagonal elements equal to $\frac{1}{m_{ii}}$ and A is matrix with diagonal elements equal to $$a_{ii} = \frac{1}{n-1}\sum_{j = 1,i\neq j}^n \frac{m_{jj}}{m_{ii}m_{jj} - m_{ij}m_{ji}} $$ and non-diagonal elements equal to $$a_{ij} = \frac{-1}{(n-1)} \frac{m_{ij}}{m_{ii}m_{jj} - m_{ij}m_{ji}}. $$

But in reality, ratio of these two minimal eigenvalues can be arbitrary large, for instance $$ M = \begin{bmatrix} 1& 0.99 \\[0.3em] 0.99 & 1 \end{bmatrix}$$ adding $9s$ to $0.99$, ratio $\frac{\lambda_{\min}(M^{1/2}A M^{1/2})}{\lambda_{\min}(M^{1/2}D M^{1/2})}$ is not bounded. For $n = 2$, answer is that $M$ should be close to singular. And their ratio is only equal to $1$ only for diagonal matrix. My intuition is that similar structure with $2 \times 2$ submatrices (close to singular) should apply for higher dimensions. Also my experiments suggest so. Does anybody have idea how to proof that for matrix with some specific structure, this ratio can be large (at least 2).

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Consider the family of matrices $M:=M_t:=N^2$, where $N:=P+tI$, $P$ is the $n\times n$ matrix with all entries equal $1$ and $I$ is the $n\times n$ identity matrix, so that $M^{1/2}=N$. Let $t\downarrow0$, so that $N\succ0$ (but barely). Note that $P^2=nP$, and the eigenvalues of $P$ are $n$ and $0$. Then $m_{ii}=d:=n+2t+t^2$ and $m_{ij}=e:=n+2t$ if $i\ne j$. So, $A=-cP+(b+c)I$ and $D=\frac1d\,I$, where $b:=\frac{d^2}{d^2-e^2}$ and $c:=-\frac b{n-1}$.

Next, $M^{1/2}D M^{1/2}=NDN=\frac1d\,M=\frac1d\,(eP+t^2I)$, so that the RHS of (!) is $\frac{t^2}d\sim\frac{t^2}n$.

On the other hand, $M^{1/2}A M^{1/2}=NAN=\frac{1+O(t)}{2(n-1)}\,(nI-P)$, so that the LHS of (!) is $\frac{1+O(t)}2\sim\frac12$, which much greater than the RHS of (!).

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  • $\begingroup$ Thank you, very good example. Do you think, that there can be find some feature of matrix $M$, which "drives" this difference? $\endgroup$
    – Ethan
    Jan 24, 2018 at 17:55
  • $\begingroup$ I am glad you liked the answer. This family of matrices $M$ seemed the easiest to try for this problem, which is what I did. I am not sure, though, about general conditions for this phenomenon. $\endgroup$ Jan 24, 2018 at 18:19

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