Let's define the matrix $M=C(sI-A)^{-1}B$ and $G=(I-LM)(I-KM)^{-1}$, where $A$, $B$, $C$, $L$, $K$ are matrices of suitable dimensions, $s$ is a complex scalar variable and $I$ is the identity matrix. I would like to prove that $\bar{G}=(K-L)C(sI-A-BKC)^{-1}B+I$ is equal to $G$.
I tried to apply the matrix inversion lemma to $\bar{G}$ but without success. Any idea on how I could prove it?
Let us set $D:=sI-A$ and rewrite the defining relationships:
$$\tag{1}M=CD^{-1}B$$
$$\tag{2a}G=(I-LM)(I-KM)^{-1} \iff G(I-KM)=I-LM \iff $$
$$\tag{2b}G-I=(GK-L)M \iff G-I=(GK-L)CD^{-1}B$$
(the last expression comes from $(1)$). We have to prove that
$$G\stackrel{?}{=}(K-L)C(D-BKC)^{-1}B+I \iff $$
$$\tag{3}G-I\stackrel{?}{=}(K-L)C(D-BKC)^{-1}B$$
Otherwise said, taking into account the other expression of $G-I$ in $(2b)$, we have to establish that:
$$\tag{4}(GK-L)CD^{-1}\stackrel{?}{=}(K-L)C(D-BKC)^{-1}$$
Let us develop aside:
$$\tag{5}(D-BKC)^{-1}=(D(I - D^{-1}BKC))^{-1}=(I - D^{-1}BKC)^{-1}D^{-1}$$
Taking $(5)$ into account, $(4)$ amounts to establish that:
$$\tag{6}(GK-L)C\stackrel{?}{=}(K-L)C(I - D^{-1}BKC)^{-1} \iff$$
$$\tag{7}(GK-L)C(I - D^{-1}BKC)\stackrel{?}{=}(K-L)C \iff$$
$$\tag{8}(GK-L)(C - MKC)\stackrel{?}{=}(K-L)C \iff$$
$$\tag{9}(GK-L)(I - MK)C\stackrel{?}{=}(K-L)C\iff$$
$$\tag{10}(GK-L)(I - MK)\stackrel{?}{=}(K-L)\iff$$
$$\tag{11}GK-GKMK-L+LMK\stackrel{?}{=}K-L$$
But, due to $(2)$, $(G-GKM)K=(I-LM)K$
which establishes the result.