There is an easy theorem in linear algebra that "successive Schur complementations compose nicely": for instance, let's say I have a matrix $$ M_0= \begin{bmatrix} A & B & C \\ D & E & F\\ G & H & I \end{bmatrix}, $$ with all blocks $n\times n$ in this and all the following formulas. I take the Schur complement of the leading diagonal submatrix $A$, to get $$ M_1= \begin{bmatrix} \hat{E} & \hat{F}\\ \hat{H} & \hat{I} \end{bmatrix}, $$ and then again the Schur complement of $\hat{I}$ in $M_1$, to get a matrix $\tilde{E}$. Then $\tilde{E}$ is the same as the Schur complement of $\begin{bmatrix}A & C\\ G & I\end{bmatrix}$ in $M_0$.

More in general, if I remove two sets of (disjoint) entries consecutively, then the result is the same as removing their union from the original matrix.

This result is difficult to write down properly in a more general version ("let $S$ and $T$ be two disjoint subsets of $\{1,2,\dots,n\}$..."), because after the first complement the numbering of the remaining rows and columns changes. I do need the general version where $S$ and $T$ are arbitrary subsets, so I can't get away with a statement like "without loss of generality, $S$ contains the first/last $k$ elements of $\{1,2,\dots,n\}$".

One of the ideas that come to mind is introducing an external labelling: if the block rows and columns are labelled $a,b,c$, then after removing $\{a\}$ the remaining ones are still labelled $(b,c)$. However this looks artificial and is definitely not standard in linear algebra (it is more usual in Markov chains, where Schur complements also appear).

How would you state this lemma formally in a paper? Is there a way out of this that I am overlooking?

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