Consider the following matrix:
$${\bf M} = \begin{bmatrix}a&b\\0&c\end{bmatrix}$$
if $a,b,c \in \mathbb R$, we can find:
$${\bf M}^{-1} = \begin{bmatrix}1/a&-b/(ac)\\0&1/c\end{bmatrix}$$
I have by hand verified that if we have block-matrix, ${\bf A,B,C,0} \in \mathbb R ^{n\times n}$
$${\bf M} = \begin{bmatrix}\bf A&\bf B\\\bf 0&\bf C\end{bmatrix}$$ Then instead we get:
$${\bf M}^{-1} = \begin{bmatrix}{\bf A}^{-1}&-{\bf A}^{-1}{\bf B}{\bf C}^{-1}\\\bf 0&{\bf C}^{-1}\end{bmatrix}$$
Here the multiplication from left and right is very important.
Now to the question, how can this possibly be combined with Krylov subspace methods in general and Conjugate-Gradient in particular?
Let us assume we want to solve $$\min_{\bf v} \{\|{\bf Mv-d}\|_2^2\}$$ and that $\bf d$ is similarly partitioned ${\bf d} = [{\bf d_1}^T, {\bf d_2}^T]^T$
Own progress: I have managed to find the following factorization:
$${\bf M}^{-1} = \begin{bmatrix}{\bf A}^{-1}&-{\bf A}^{-1}{\bf B}{\bf C}^{-1}\\\bf 0&{\bf C}^{-1}\end{bmatrix} = \begin{bmatrix}{\bf A}^{-1}&\bf 0\\\bf 0&\bf I\end{bmatrix}\begin{bmatrix}\bf I&\bf -B\\\bf 0&\bf I\end{bmatrix}\begin{bmatrix}\bf I&\bf 0\\\bf 0&{\bf C}^{-1}\end{bmatrix}$$
Now how can this be used together with the workings of C-G..?