Answer
See explanation
meaning that the matrix $W(s)$ is the Schur complement of the matrix $
$A-s I_{n} in the system matrix $\operatorname{in} \operatorname{eq}(8)$
Work Step by Step
Top equation:
$\left(A-s I_{n}\right) x+B u=0$
Bottom equation:
$C x+u=y$
From top eq we get:
$\left(A-s I_{n}\right) x=-B u$
As $A-s I_{n}$ is invertible, we obtain:
$x=\left(A-s I_{n}\right)^{-1}(-B u)=-\left(A-s I_{n}\right)^{-1} B u$
Substitute $x$ in bottom equation with what we get:
$C\left(-\left(A-s I_{n}\right)^{-1} B u\right)+u=y$
$u-C\left(A-s I_{n}\right)^{-1} B u=y$
$\mathrm{So}, y=\left(I_{m}-C\left(A-s I_{n}\right)^{-1} B\right) u$
If $W(s)=I_{m}-C\left(A-s I_{n}\right)^{-1} B,$ then $y=W(s) u,$ meaning that the matrix $W(s)$ is the Schur complement of the matrix $A-s I_{n}$ in the system matrix $\operatorname{in} \operatorname{eq}(8)$
