Given a symmetric positive-definite matrix $\Sigma$, consider the space $\mathcal{D}$ of diagonal matrices such that $\forall D\in\mathcal{D}$, the matrix $\Sigma-D\Sigma^{-1}D$ is positive definite. What is the connectedness of $\mathcal{D}$? Is it simply connected?



Characterizing space that preserves positive-definiteness property
Let $\Gamma$ be a convex subset of the set of symmetric real matrices. Then $\mathcal{D}=\{D\in\Gamma;\Sigma-D\Sigma^{-1}D>0\}$ is a convex set. 


Proof: we use the following known result.


(\*) Let $S>0$ and $T$ be real symmetric matrices s.t. $T$ has $k$ positive, $l$ negative and $n-k-l$ zero eigenvalues. Then $ST$ has $k$ positive, $l$ negative and $n-k-l$ zero eigenvalues.


Let $D\in\Gamma$. From (\*) we deduce that: $S=\Sigma-D\Sigma^{-1}D>0$ iff $spectrum(\Sigma^{-1}S)\subset {\mathbb{R}^+}^\*$ iff $spectrum(I-(\Sigma^{-1}D)^2))\subset {\mathbb{R}^+}^\*$. Since $spectrum(\Sigma^{-1}D)\subset \mathbb{R}$, we obtain:


$S>0$ iff $spectrum(\Sigma^{-1}D)\subset (-1,1)$ iff $spectrum(\Sigma^{-1/2}D\Sigma^{-1/2})\subset (-1,1)$ and $\mathcal{D}=\{D\in\Gamma;spectrum(\Sigma^{-1/2}D\Sigma^{-1/2})\subset (-1,1)\}$ is, of course, a convex set. 


