$K>0\in \mathbb{R}^{n\times n}$, $P>0 \in \mathbb{R}^{n\times n}$ are diagonal positive definite matrices. And $R\geq 0\in \mathbb{R}^{m\times m}$ is positive semi-definite matrix. Let $B\in \mathbb{R}^{n\times m}$, $n>m$ . Let $A\in\mathbb{R}^{n\times n}$ be a rank-deficient matrix.

I am trying to find the conditions on $K$ (existance) such that, the matrix \begin{eqnarray} \underbrace{\begin{bmatrix} K-P+B R B^\top & KA\\ A^\top K &A^\top K A \end{bmatrix}}_{M}\geq 0 \end{eqnarray}.

PS: I have tried Schur's compliment analysis on $M$, It didn't give me any useful result. So , I have tried the following \begin{eqnarray} \underbrace{\begin{bmatrix} K-P+B R B^\top & KA\\ A^\top K &A^\top K A \end{bmatrix}}_{M}= \begin{bmatrix} I&O\\O&A^\top \end{bmatrix} \underbrace{\begin{bmatrix} K-P+B R B^\top & K\\ K & K \end{bmatrix}}_{N} \begin{bmatrix} I&O\\O&A \end{bmatrix} \end{eqnarray}

This implies $M\geq 0$ if $N>0$. Note that this only sufficient but not necessary. Using Schur's compliment analysis on $N$ we get \begin{eqnarray} K&>&0\\ K-P+B R B^\top-K(K)^{-1}K&>&0 \end{eqnarray} The above second equation implies $BRB^\top-P>0$, which is a contradiction as $P>0$ and $BRB^\top\geq 0$. This does not imply that $M\geq 0$ as the contradiction, that is derived from $N$ is only sufficient but not Necessary.