The theorem states:
Let $T \in \mathbb{S}^m$ be positive definite and $X \in \mathbb{S}^n$ and $U \in \mathbb{R}^{n \times m}$. Then $M$ is positive semidefinite if and only if $X - UT^{-1}U^T$ is, where
$$ M =\begin{bmatrix} T & U^T \\ U & X \end{bmatrix} $$
The proof starts off by observing that
$$ M = \begin{bmatrix} I & 0 \\ UT^{-1} & I\end{bmatrix}\begin{bmatrix}T & 0 \\ 0 & X - UT^{-1}U^T\end{bmatrix} \begin{bmatrix} I & T^{-1}U^T \\ 0 & I\end{bmatrix}$$
and then arguing from there, but this is the part that is unclear to me.
What is it about the theorem statement that makes it occur to you to factorise the left hand side in terms of the block diagonal on the right in the first place? Does this factorisation incidentally show up somewhere else, or is there some way one can "see" that $M$ can be factorised in such a way.