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Consider a matrix $X$ to be \begin{equation} X=P-PA^\top\left(APA^\top + Q\right)^{-1}AP, \end{equation} where $P\in\Re^n$ is a positive definite matrix, $A\in\Re^n$ is a non-singular matrix, $Q\in\Re^n$, such that \begin{equation} Q=\begin{bmatrix}Q_1 & 0 \\ 0 & 0 \end{bmatrix} \end{equation} where $Q_1\in\Re^r$ is a positive definite matrix and $r<n$.

My question is: Is it possible to prove that matrix $X$ is a positive semi-definite matrix?

Many thanks

Steve

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Yes, $X$ is psd. Note that $APA^T+Q\succeq APA^T\succ 0$ because $Q$ is psd and so $$ (APT^T+Q)^{-1}\preceq (APA^T)^{-1}=(A^T)^{-1}P^{-1}A^{-1}. $$ Thus, for all $v$, \begin{align*} v^TPA^T(APT^T+Q)^{-1}APv&=(APv)^T(APT^T+Q)^{-1}(APv)\\ &\leq(APv)^T(A^T)^{-1}P^{-1}A^{-1}APv\\ &=v^TP(A^T)(A^T)^{-1}P^{-1}A^{-1}APv\\ &=v^TPv. \end{align*} The claim follows.

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This looks an awful lot like use-case for the Schur Complement.

In particular, $X$ is the Schur complement of the following block matrix:

$$Z=\begin{bmatrix} P & PA^T \\ AP & APA^T+Q \end{bmatrix}.$$

A basic property of Schur complements is that when $P>0$, then $X\geq 0$ iff $Z\geq 0$.

By decomposing $$Z = \begin{bmatrix} 1 & 0 \\ 0 & A \end{bmatrix} \begin{bmatrix} P & P \\ P & P \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & A^T \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & Q \end{bmatrix},$$ it is not hard to see that $Z\geq 0$. Hence, $X\geq 0$.

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