Say we have a positive definite maxtrix, $M$, of size $m$. Then is there any relationship between $A=eMe'$ and $B=eM^{-1}e'$ for large $m$? I would like to claim something like $AB \sim m^2$ for large $m$. Is such a relationship possible?
Many thanks for your hints.
This is not a complete answer, but gives some of the ideas.
Since $M$ is positive definite, we can write $M=QDQ^T$ where $Q$ is orthonormal and $D$ is diagonal. Then $M^{-1}=QD^{-1}Q^T$. Since $D$ is diagonal, $D^{-1}$ is just the inverse of the diagonal elements.
Suppose that $e=(1,\cdots,1)$. Then $eMe^T=(eQ)D(eQ)^T$. On the other hand, $eM^{-1}e^T=(eQ)D^{-1}(eQ)^T$. If we write $eQ=\sum a_ie_i$, then $eMe^T=\sum a_i^2d_i$ and $eM^{-1}e^T=\sum \frac{a_i^2}{d_i}$. Observe that since $Q$ is orthonormal, $eQe^T=(e,e)=n$. Therefore, $\sum a_i=n$.
Observe that $AB=\left(\sum a_i^2d_i\right)\left(\sum \frac{a_i^2}{d_i}\right)\geq \left(\sum (a_i\sqrt{d_i})\left(\frac{a_i}{\sqrt{d_i}}\right)\right)^2=\left(\sum a_i^2\right)^2$ by the Cauchy-Schwarz inequality. Since $\sum a_i=n$, the smallest that $\sum a_i^2$ coud be is when each $a_i=1$, so the sum would be $n$. In other words, $AB\geq n^2$. $AB$ could, however, be arbitrarily large if, perhaps one $d_i$ is especially small and another is reasonably large.