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Suppose $A$ is an $n\times n$ real symmetric positive definite matrix. Let $A^{(-1)}_{i,j}$ be the $n\times n$ matrix the entries $(i,i),\,(i,j),\,(j,i),\,(j,j)$ of which equal to the corresponding entries of the inverse of the row $i$ column $j$ principal minor of $A$, and all other entries zero. Let me put the verbal defintion in the last sentence into more sepcific terms. Let $a_{ij}$ be the $(i,j)$ entry of $A$. Let $C_{i,j}$ be the $2\times2$ matrix $C_{i,j} :=\begin{bmatrix}a_{ii} & a_{ij} \\ a_{ij} & a_{jj}\end{bmatrix}$, which is the row $i$ column $j$ principal minor of $A$. Let $\big[C_{i,j}^{-1}\big]_{kl},\,k,l\in\{1,2\}$ be the $(k,l)$ entry of the inverse of $C_{i,j}$. Then the $(p,q), \,p,q\in\{1,2,\cdots,n\}$ entry of $A^{(-1)}_{i,j}$ is then $$\big[A_{i,j}^{(-1)}\big]_{pq} :=[C_{i,j}^{-1}]_{11}\delta_{ip}\delta_{iq}+[C_{i,j}^{-1}]_{12}(\delta_{ip}\delta_{jq}+\delta_{iq}\delta_{jp})+[C_{i,j}^{-1}]_{22}\delta_{jp}\delta_{jq}.$$

Now we are clear about what exactly $A_{i,j}^{(-1)}$ is, define $$B:=\frac{1}{n-1}\sum_{i<j}A_{i,j}^{(-1)},\quad P:=\big(\text{diag}(A)\big)^{-1},$$ and $\lambda_{\min}(M)$ to be the minimal eigenvalue of matrix $M$ (with all real eigenvalues).

Random test suggests the following inequality. Is it true in general? $$\lambda_{\min}(PA)\le \lambda_{\min}(BA)$$


It is true for $n=2$.


This is an equivalent and simplified reformulation of a problem in math.exchange.com.

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  • $\begingroup$ For $n=2$ are you just saying that $1 \leq \min (\sigma(PA))?$ $\endgroup$
    – Igor Rivin
    Dec 29, 2017 at 4:32
  • $\begingroup$ @IgorRivin: Yes, but with your inequality direction reversed. Note I just changed symbol for the minimal eigenvalue in my question. $\endgroup$
    – Hans
    Dec 29, 2017 at 5:03
  • $\begingroup$ Why A_ij invertible? $\endgroup$
    – M. Lin
    Dec 29, 2017 at 5:59
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    $\begingroup$ @PietroMajer: It is not the entrywise inverse of $M_{lk}$ but the matrix inverse of $[M_{lk}]$. I have added a detailed and specific definition of $A_{i,j}^{(-1)}$ in the question Please review and.let me know if my definition is clear to you. $\endgroup$
    – Hans
    Jan 4, 2018 at 0:24
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    $\begingroup$ @PietroMajer: No worries. I appreciate your attention to the question. I would appreciate it even more if you would spend some time pondering the problem and intrigue your colleagues with this problem in order to find a solution. $\endgroup$
    – Hans
    Jan 4, 2018 at 16:31

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