TITLE: On the sum of all elements of inverted correlation matrix
QUESTION [13 upvotes]: Assume I have a correlation matrix,$A$
 \begin{equation}
A_{i,j} =  
\begin{cases}
1,& \text{if}\ i=j\\
\rho_{i,j},& \text{otherwise}  \\
\end{cases}
\end{equation} 
Where $ 0\leq \rho_{i,j} \leq 1 $  and $A$ is positive definite. 
Now define a correlation matrix $\bar{A}$ as follows:
\begin{equation}
\overline{A} =  
\begin{cases}
1,& \text{if}\ i=j\\
\overline{\rho},& \text{otherwise} \\
\end{cases}
\end{equation}
Where $\overline{\rho}=\frac{\sum_{i \neq j} \rho_{i,j}}{n^{2}-n}$
I want to prove that the sum of elements in the inverse matrix $\overline{A}^{-1}$ is less than the the sum of elements in the inverse matrix ${A}^{-1}$. Any ideas?
I wrote a program that makes random matrices of this sort and it was true for all matrices in many different dimensions so it can't be just by chance. I just don't know how to prove it.
Thanks

REPLY [5 votes]: This is true because with respect to the partial ordering of positive semidefinite matrices, matrix inverse is convex on the set of all positive definite matrices.
More specifically, if $X,Y\succ0$ and $t\in[0,1]$, then $(1-t)X^{-1}+tY^{-1}\succeq\left[(1-t)X+tY\right]^{-1}$, because positive definite matrices are simultaneously diagonalisable by congruence and the inverse function on positive reals is convex.
Now observe that $\bar{A}=\frac1{n!}\sum_PP^TAP$, where the summation runs through all permutation matrices $P$. By assumption, $A$ is positive definite. Therefore $\bar{A}$ is positive definite too and its inverse exists. So, by convexity of matrix inverse of positive definite matrix,
\begin{aligned}
\frac1{n!}\sum_PP^TA^{-1}P
=\frac1{n!}\sum_P(P^TAP)^{-1}
\succeq\left(\frac1{n!}\sum_PP^TAP\right)^{-1}
=\bar{A}^{-1}.
\end{aligned}
Consequently,
$$
e^TA^{-1}e
=\frac1{n!}\sum_Pe^TP^TA^{-1}Pe
\ge e^T\bar{A}^{-1}e.
$$