Lemma 41.3.9 (Schur Complement Condition for Positivity).label Let $H$ be a complex Hilbert space, $x, y \in B(H)$, and $z \in G(B(H))$, then
\[\begin{bmatrix}x&y \\ y^{*}&z\end{bmatrix} \ge 0\]
if and only if $z \ge 0$ and $x - yz^{-1}y^{*} \ge 0$.
Proof. ($\Leftarrow$): Suppose that $z \ge 0$ and $x - yz^{-1}y^{*} \ge 0$, then since $z$ is invertible,
\begin{align*}\begin{bmatrix}x&y \\ y^{*}&z\end{bmatrix}&= \begin{bmatrix}I&yz^{-1}\\ 0&I\end{bmatrix}\begin{bmatrix}x - yz^{-1}y^{*}&0 \\ 0&z\end{bmatrix}\begin{bmatrix}I&0 \\ z^{-1}y^{*}&I\end{bmatrix} \\&= \begin{bmatrix}I&0 \\ z^{-1}y^{*}&I\end{bmatrix}^{*}\begin{bmatrix}x - yz^{-1}y^{*}&0 \\ 0&z\end{bmatrix}\begin{bmatrix}I&0 \\ z^{-1}y^{*}&I\end{bmatrix}\end{align*}
Since the middle factor is positive, the given matrix is positive as well.
($\Rightarrow$): Now suppose that the given matrix is positive, then for any $\xi \in H$,
\[\dpn{z\xi, \xi}{H}= \angles{\begin{bmatrix}x&y \\ y^{*}&z\end{bmatrix}\begin{bmatrix}0 \\ \xi\end{bmatrix}, \begin{bmatrix}0 \\ \xi\end{bmatrix}}_{H^2}\ge 0\]
so $z \ge 0$. As such,
\begin{align*}\begin{bmatrix}x - yz^{-1}y^{*}&0 \\ 0&z\end{bmatrix}&= \begin{bmatrix}I&-yz^{-1}\\ 0&I\end{bmatrix}\begin{bmatrix}x&y \\ y^{*}&z\end{bmatrix}\begin{bmatrix}I&0 \\ -z^{-1}y^{*}&I\end{bmatrix} \\&= \begin{bmatrix}I&0 \\ -z^{-1}y^{*}&I\end{bmatrix}^{*}\begin{bmatrix}x&y \\ y^{*}&z\end{bmatrix}\begin{bmatrix}I&0 \\ -z^{-1}y^{*}&I\end{bmatrix} \\\end{align*}
Since the middle factor is positive, the left hand side is also positive. Therefore $x - yz^{-1}y^{*} \ge 0$ as well.$\square$
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