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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?

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    $\begingroup$ 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 $$\det(\Sigma-D)\det(\Sigma+D)$$ $\endgroup$ – 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
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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.

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