Searching an analogues for Schur complement When I'm trying to solve a matrix inequality set of the following: 
\begin{equation}
\begin{array}{l}
A-BC^{-1}B^{T}>0\\
C>0
\end{array}
\end{equation}
Where $A$ is a given $p\times p$ positive-definite matrix, $B$ is an unknown $p\times q$ matrix and $C$ is an unknown $q\times q$ matrix.
At first look, I cannot say that this problem can be represented in a linear form (in the matrix variables $B,C$) and that the convex combination of a given two solutions for the inequalities is also a solution. But Schur complement provides a simple tool to change the inequalities into an equivalent form which is linear in the variables and thus also convex:
\begin{equation}
\begin{bmatrix}
A & B\\
B^{T} & C
\end{bmatrix}>0 
\end{equation}
My question is:
If there some analog transformation that produce a linear (in the variables) form of the following:
\begin{equation}
\begin{array}{l}
A-BC^{-1}DC^{-1}B^{T}>0\\
C>0
\end{array}
\end{equation}
Where $A$ is a given $p\times p$ positive-definite matrix, $B$ is an unknown $p\times q$ matrix, $C$ is an unknown $q\times q$ matrix and $D$ is a given $q\times q$ positive-definite matrix.
In some sense, if Schur complement help with 2'nd order multiplication, what can help with the 4'th order?
Thanks, Y. 
 A: I will assume that $B$ is given, otherwise $B = 0$ would be a trivial solution in which case $C$ can be any positive definite matrix.
Since $C$ has to be positive definite means that it and its inverse is full rank and symmetric, therefore $C^{-1}D\,C^{-1}$ has to be positive definite as well if $D$ is positive definite. Therefore you could use the substitution $P^{-1} = C^{-1}D\,C^{-1}$, so
$$
A - B\,P^{-1}B^\top \succ 0 \\
P \succ 0
$$
and solve this LMI for $P$. A corresponding positive definite solution for $C$ can then be found from $P$ by using the continuous time algebraic Riccati equation (CARE). Namely the relation between $P$ and $C$ can also be written as $P = C\,D^{-1}C$, which is a very simplified version of a CARE, which is normally defined as
$$
A^\top X + X\,A - X\,B\,R^{-1}B^\top X + Q = 0,
$$
but when using $A=0$, $X = C$, $B = I$, $R = D$ and $Q = P$ we again get $P = C\,D^{-1}C$. This should have a solution, since a CARE requires that $R$ is positive definite and $Q$ is semi-positive definite, which is the case since $P$ and $D$ are positive definite.
