In the paper Kalman filtering with intermittent observations by Sinopoli et al., I found the following linear matrix inequality (LMI)
$$
\begin{bmatrix}X – (1-\lambda)AXA^T & \sqrt{\lambda}F \\ \sqrt{\lambda}F^T & X^{-1}\end{bmatrix} > 0 \label{eq1} \tag{1}
$$
where $X – (1-\lambda)AXA^T > 0,F < 0,$ and $\lambda \in [0,1]$. The author of the paper then writes
Using one more time the Schur complement decomposition on the first element of the matrix we obtain
$$
\begin{bmatrix}X & \sqrt{\lambda}F & \sqrt{1-\lambda}A \\ \sqrt{\lambda}F^T & X^{-1} & 0 \\ \sqrt{1-\lambda}A^T & 0 & X^{-1}\end{bmatrix} > 0 \label{eq2} \tag{2}
$$
However, I'm not sure how the author went from the LMI in \eqref{eq1} to the LMI in \eqref{eq2}. For reference, the "Schur complement decomposition" referred to by the author is the following theorem:
For any symmetric matrix $M$ of the form
$$
M = \begin{bmatrix}A & B \\ B^T & C\end{bmatrix}
$$
if $C$ is invertible, then $M > 0$ if and only if $C > 0$ and $A – BC^{-1}B^T > 0$.
Update
Based on @Park's answer, since both $X$ and $X – (1-\lambda)AXA^T$ are positive definite, then $$\begin{pmatrix} X-(1-\lambda)AXA^{t} &\sqrt[]{\lambda}F & 0 \\ \sqrt[]{\lambda}F^{t} & X^{-1} & 0 \\ 0 & 0 & X^{-1} \\ \end{pmatrix}$$ is also positive definite, as this is a block-diagonal matrix, where the upper-left block and the $X^{-1}$ in the bottom-right block both have positive eigenvalues. However, it seems that @Park's answer relies on the following congruence transformation: $$Q\begin{pmatrix} X-(1-\lambda)AXA^{t} &\sqrt[]{\lambda}F & 0 \\ \sqrt[]{\lambda}F^{t} & X^{-1} & 0 \\ 0 & 0 & X^{-1} \\ \end{pmatrix}Q^T = \begin{pmatrix}X & \sqrt{\lambda}F & \sqrt{1-\lambda}A \\ \sqrt{\lambda}F^T & X^{-1} & 0 \\ \sqrt{1-\lambda}A^T & 0 & X^{-1}\end{pmatrix}$$ to show that $$\begin{pmatrix}X & \sqrt{\lambda}F & \sqrt{1-\lambda}A \\ \sqrt{\lambda}F^T & X^{-1} & 0 \\ \sqrt{1-\lambda}A^T & 0 & X^{-1}\end{pmatrix} > 0$$ I'm not sure what the block-matrix $Q$ should be, so would appreciate some direction.
Best Answer
(1) is written as $$\begin{bmatrix}X & \sqrt{\lambda}F \\ \sqrt{\lambda}F^T & X^{-1}\end{bmatrix} - \begin{bmatrix} (1-\lambda)AXA^T & 0 \\ 0 & 0\end{bmatrix} > 0$$ or $$\begin{bmatrix}X & \sqrt{\lambda}F \\ \sqrt{\lambda}F^T & X^{-1}\end{bmatrix} - \begin{bmatrix} \sqrt{1-\lambda}A \\ 0 \end{bmatrix}X[\sqrt{1-\lambda}\,A^T \quad 0] > 0\tag{3}$$ which, using Schur complement, is equivalent to $$\begin{bmatrix}X & \sqrt{\lambda}F & \sqrt{1-\lambda}A \\ \sqrt{\lambda}F^T & X^{-1} & 0 \\ \sqrt{1-\lambda}A^T & 0 & X^{-1}\end{bmatrix} > 0.$$ (Note: In (3), let $A = \mbox{first term of LHS}$, $B^T = [\sqrt{1-\lambda}\,A^T \quad 0]$ and $C = X^{-1}$ and apply Schur complement.)