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

2$\begingroup$ With a Schur complement formula argument you can switch to positive definiteness of $$\begin{bmatrix}\Sigma&D\\D&\Sigma\end{bmatrix}$$ and then from block matrix determinant argument you can investigate connectedness of $$\det(\SigmaD)\det(\Sigma+D)$$ $\endgroup$ – percusse Nov 13 '14 at 4:32

$\begingroup$ So I suppose you're suggesting to look at the leading principal minors? $\endgroup$ – Al Nejati Nov 13 '14 at 21:14
Let $\Gamma$ be a convex subset of the set of symmetric real matrices. Then $\mathcal{D}=\{D\in\Gamma;\SigmaD\Sigma^{1}D>0\}$ 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 $nkl$ zero eigenvalues. Then $ST$ has $k$ positive, $l$ negative and $nkl$ zero eigenvalues.
Let $D\in\Gamma$. From (*) we deduce that: $S=\SigmaD\Sigma^{1}D>0$ 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.