Ordering of positive semidefinite matrices Suppose $A \in \mathbb R^{n \times r}$ with $r < n$ and $M, X \in \mathbb R^{n \times n}$ are symmetric and positive definite. I am wondering whether the following relation is correct:
\begin{align*}
A(M + A^{\top}XA)^{-1} A^{\top} < X^{-1},
\end{align*}
where $<$ means the Loewner ordering.
 A: Your relation is indeed correct.
Note that 
\begin{align*}
A(M + A^{T}XA)^{-1} A^{T} < X^{-1} & \iff\\
X^{1/2}A(M + A^{T}XA)^{-1} A^{T}X^{1/2} < X^{1/2}X^{-1}X^{1/2} & \iff\\
(X^{1/2}A)(M + (X^{1/2}A)^T(X^{1/2}A))^{-1} (X^{1/2}A)^T < I.
\end{align*}
Now, if we define $N = X^{1/2}A$, this inequality becomes
$$
N(M + N^TN)^{-1}N^T < I
$$
Note that
\begin{align*}
N(M + N^TN)^{-1}N^T < I & \iff\\
N(M^{1/2}[I + M^{-1/2}N^T(M^{-1/2}N^T)^T]M^{1/2})^{-1}N^T < I & \iff\\
(M^{-1/2}N^T)^T(I + M^{-1/2}N^T(M^{-1/2}N^T)^T)^{-1}M^{-1/2}N^T < I.\\
\end{align*}
Now, define $P = M^{-1/2}N^T$.  The inequality is then
$$
P^T(I + PP^T)^{-1}P < I.
$$
It therefore suffices to prove the above (which is to say that it suffices to consider the $X = M = I$ case).
By this post, we can rewrite
$$
P^T(I + PP^T)^{-1}P = (P^TP)(I + P^TP)^{-1}.
$$
Thus, for any eigenvalue $x$ of $P^TP$, $\frac{x}{1+x}$ is an eigenvalue of $P^T(I + PP^T)^{-1}P$.  Since $P^TP$ is positive semidefinite, we have $x \geq 0$. It follows that the eigenvalues of $P^T(I + PP^T)^{-1}P$ are smaller than $1$, so that we indeed have $P^T(I + PP^T)^{-1}P < I$ as desired.
