Positive semidefinite matrix using Schur Complement

Question

Let

$$ G =\begin{pmatrix} \Sigma & \Sigma - \text{diag} \lbrace s \rbrace \\ \Sigma - \text{diag} \lbrace s \rbrace & \Sigma \end{pmatrix} $$

where $s \in \mathbb{R}_{\geq 0}^p$ is some nonnegative vector, and $\Sigma$ is a positive definite covariance matrix.

I need to argue that $G$ is positive semidefinite using the Schur complement. That is, $G$ is positive semidefinite iff the Schur complement $$ 2\text{diag} \lbrace s \rbrace-\text{diag} \lbrace s \rbrace \Sigma^{-1} \text{diag} \lbrace s \rbrace $$ is positive semidefinite. However I am not sure how to prove this.

Answer 1

You can write down the nonstrict version of the Schur complement formula

$$ G\succeq 0 \iff \left\{\begin{array}{cc}\Sigma\succeq 0\\ \Sigma-(\Sigma-\textrm{diag}(s))\Sigma^{\dagger}(\Sigma-\textrm{diag}(s))\succeq0\\ (I-\Sigma\Sigma^\dagger)(\Sigma-\textrm{diag}(s))=0\end{array}\right. $$ The second condition expands to $$ \Sigma-(\Sigma-\textrm{diag}(s))\Sigma^{\dagger}(\Sigma-\textrm{diag}(s))= \textrm{diag}(s)\Sigma^{\dagger}\Sigma + \Sigma\Sigma^{\dagger}\textrm{diag}(s)-\textrm{diag}(s)\Sigma^{\dagger}\textrm{diag}(s) $$ if $\Sigma$ is also invertible then you are done.