Can I apply the Schur complement on both sides of an inequality I have the following matrix inequality that I need to express as an LMI
$ A^T Q^{-1} A - Q + \sum_i A_i^T Q^{-1} A_i < 0$
($Q > 0$, $A = XQ+YB$, $A_i = X_i Q+Y_i B$)
which I rewrote to
$ A^T Q^{-1} A - Q  < - \sum_i A_i^T Q^{-1} A_i$
Now I want to use Schur's complement and express this inequality as
$
\begin{bmatrix}
    -Q & A^T \\
    A & -Q 
\end{bmatrix}
< \sum_i
\begin{bmatrix}
    0 & A_i^T \\
    A_i & Q 
\end{bmatrix}
$
and therefore
$
\begin{bmatrix}
    -Q & A^T \\
    A & -Q 
\end{bmatrix}
- \sum_i
\begin{bmatrix}
    0 & A_i^T \\
    A_i & Q 
\end{bmatrix}
 < 0 $ 
which is a form my solver, solving for Q and B, should be able to handle.
Can I use the Schur complement in this way and are these two inequalities equivalent?
 A: You can't use Schur's complement in that way. Namely when you have the matrix inequalities
$$
A - B^\top C^{-1} B \succ 0, \quad C \succ 0,
$$
then you can combine this into the following one matrix inequality
$$
M =
\begin{bmatrix}
A & B^\top \\ B & C
\end{bmatrix} \succ 0.
$$
This is because $M$ can be decomposed as follows
$$
M = 
\underbrace{
\begin{bmatrix}
I & B^\top C^{-1} \\ 0 & I
\end{bmatrix}
}_{T^\top}
\begin{bmatrix}
A - B^\top C^{-1} B & 0 \\ 0 & C
\end{bmatrix}
\underbrace{
\begin{bmatrix}
I & 0 \\ C^{-1} B & I
\end{bmatrix}
}_{T}.
$$
Namely $M \succ 0$ means that $x^\top M\,x > 0\ \forall\, x\neq0$. So when using $x=T\,y$ one gets $x^\top M\,x = y^\top T^\top M\,T\,y$. It holds that $x\neq0$ iff $y\neq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^\top M\,T \succ 0$.
But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that
$$
T^\top \left(
\begin{bmatrix}
Q & -A^\top \\
-A & Q 
\end{bmatrix}
+ \sum_i
\begin{bmatrix}
0 & A_i^\top \\
A_i & Q 
\end{bmatrix}
\right) T
$$
gives back the original matrix inequalities as blocks on the diagonal.
