Does the Schur complement preserve the partial order? Let 
$$\begin{bmatrix}
A_{1} &B_1  \\ B_1'  &C_1
\end{bmatrix} \quad \text{and} \quad \begin{bmatrix}
A_2 &B_2 \\ B_2'  &C_2
\end{bmatrix}$$
be symmetric positive definite and conformably partitioned matrices. If 
$$\begin{bmatrix}
A_{1} &B_1  \\ B_1'  &C_1
\end{bmatrix}-\begin{bmatrix}
A_2 &B_2 \\ B_2'  &C_2
\end{bmatrix}$$
is positive semidefinite, is it true 
$$(A_1-B_1C^{-1}_1B_1')-(A_2-B_2C^{-1}_2B_2')$$ also positive semidefinite? Here, $X'$ means the transpose of $X$.

Yes, it does. The assumption $$\begin{bmatrix}
A_{1} &B_1  \\ B_1^T  &C_1
\end{bmatrix}-\begin{bmatrix}
A_2 &B_2 \\ B_2^T  &C_2
\end{bmatrix} \geq 0$$ implies that for any vector $\begin{pmatrix} x & y \end{pmatrix}$,
$$ \begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A_{1} &B_1  \\ B_1^T  &C_1
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \geq \begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A_2 &B_2 \\ B_2^T  &C_2
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix}~~~~~(*)$$ But for any partitioned matrix, 
$$\begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A &B  \\ B^T  &C
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = (x + A^{-1} B y)^T A (x + A^{-1} B y) + y^T(C-B^T A^{-1} B)y.$$ Moreover, if the partitioned matrix on the left-hand side is positive definite, then each of the two terms on the right=hand side is positive. Thus picking arbitrary $y$ and $x = -A_1^{-1}B_1y$ in (*) gives
$$y^T(C_1-B_1^T A_1^{-1} B_1)y \geq \mbox{ something positive} + y^T(C_2-B_2^T A_2^{-1} B_2)y,$$ which implies 
$$y^T(C_1-B_1^T A_1^{-1} B_1)y \geq  y^T(C_2-B_2^T A_2^{-1} B_2)y,$$
which implies the conclusion you want.
