Proposition 41.3.2 (Schur Complement Condition for Positivity, Invertible Case).label Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$. Suppose that $P$ is invertible, then
- (P1)
If $P$ is also positive, then
\[\begin{bmatrix}P&A \\ A^{*}&Q\end{bmatrix} = \begin{bmatrix}P^{1/2}&0 \\ A^{*}P^{-1/2}&I\end{bmatrix}\begin{bmatrix}I&0 \\ 0&Q - A^{*}P^{-1}A\end{bmatrix}\begin{bmatrix}P^{1/2}&P^{-1/2}A \\ 0&I\end{bmatrix}\] - (P2)
$\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^{*}P^{-1}A \ge 0$.
The operator $Q - A^{*}P^{-1}A$ is the Schur complement of $P$. Similarly, if $Q$ is invertible, then
- (Q1)
If $Q$ is also positive, then
\[\begin{bmatrix}P&A \\ A^{*}&Q\end{bmatrix} = \begin{bmatrix}I&AQ^{-1/2}\\ 0&Q^{1/2}\end{bmatrix}\begin{bmatrix}P - AQ^{-1}A^{*}&0 \\ 0&I\end{bmatrix}\begin{bmatrix}I&0 \\ Q^{-1/2}A^{*}&Q^{1/2}\end{bmatrix}\] - (Q2)
$\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$ if and only if $Q \ge 0$ and $P - AQ^{-1}A^{*} \ge 0$.
Proof, [Exercise 3.2, Po02]. (P1): Since $P$ is positive and invertible, so is $P^{1/2}$. Thus
and
(P2): If $P$ is also positive and $Q - A^{*}P^{-1}A \ge 0$, then by (P1), $\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right]$ is of the form $MDM^{*}$, where $D$ is positive. Thus the given matrix is also positive.
On the other hand, if $\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$, then as a diagonal entry, $P$ is also positive. In addition,
is also of the form $MDM^{*}$, where $D$ is positive, so the diagonal matrix, and hence $Q - A^{*}P^{-1}A$ are also positive.$\square$
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