The stabilizing DARE solution is the stable deflating graph of a regular symplectic pencil

The matrix $I+GX$ is invertible. Indeed, if $(I+GX)v=0$, multiplication by $v^{\ast }X$ gives $v^{\ast }Xv+(Xv{)}^{\ast }G(Xv)=0.$ Both summands are nonnegative. By [F3], $v^{\ast }Xv=0$ implies $Xv=0$, and the original equation then gives $v=0$. Also $S=R+B^{\ast }XB\succ 0$ by [F1]. The identity $(I+GX{)}^{-1}=I-B{S}^{-1}{B}^{\ast }X$ is verified by multiplying the right side by $I+GX$: the extra term is $B[R^{-1}-S^{-1}-S^{-1}(B^{\ast }XB)R^{-1}]B^{\ast }X=0$, since $S=R+B^{\ast }XB$. Therefore $A_c=(I+GX{)}^{-1}A$.

The DARE and the identity in step 1.1 imply $X-Q=A^{\ast }X(I+GX{)}^{-1}A=A^{\ast }X{A}_c$. Consequently, $N\left[\begin{array}{c}I\\ X\end{array}\right]=\left[\begin{array}{c}A\\ X-Q\end{array}\right]=\left[\begin{array}{c}I+GX\\ A^{\ast }X\end{array}\right]A_c=M\left[\begin{array}{c}I\\ X\end{array}\right]A_c\mathrm{.}$ This proves the deflating graph identity.

Put $T_X=\left[\begin{array}{cc}I& 0\\ X& I\end{array}\right]$ and $L_X=\left[\begin{array}{cc}I& 0\\ -A^{\ast }X(I+GX{)}^{-1}& I\end{array}\right]$. By [F6] both are block triangular with identity diagonal and determinant one. For arbitrary homogeneous coordinates $[\alpha :\beta ]$, $L_X(\beta N-\alpha M)T_X=\left[\begin{array}{cc}(I+GX)(\beta A_c-\alpha I)& -\alpha G\\ 0& \beta I-\alpha A_c^{\ast }\end{array}\right]\mathrm{.}$ The lower-left block vanishes by $X-Q=A^{\ast }X{A}_c$; the lower-right block uses $(I+GX{)}^{-1}G=G(I+XG{)}^{-1}$ and $I-X(I+GX{)}^{-1}G=(I+XG{)}^{-1}$. Taking determinants and using [F5] and [F7] gives $\det (\beta N-\alpha M)=\det (I+GX)\det (\beta A_c-\alpha I)\det (\beta I-\alpha A_c^{\ast })\mathrm{.}$ The first homogeneous factor has $n$ roots, counted with multiplicity; all are finite and equal to eigenvalues of $A_c$, hence strictly inside the disk by [F1] and [F8]. The roots of $A_c^{\ast }$ are the conjugates of those of $A_c$, since $\det (\mu I-A_c^{\ast })=\overline{\det (\stackrel{‾}{\mu }I-A_c)}$; hence each nonzero eigenvalue of $A_c^{\ast }$ has modulus less than one. The second factor has $n$ roots, counted with multiplicity: each finite root is the reciprocal of a nonzero eigenvalue of $A_c^{\ast }$ and therefore has modulus greater than one, while every zero eigenvalue of $A_c^{\ast }$ contributes a root at $\beta =0$, which is infinite. The root counts follow from [F9], with finite eigenvalues identified by [F8]. Neither factor vanishes on $\mathrm{\mid }\alpha \mathrm{\mid }=\mathrm{\mid }\beta \mathrm{\mid }$, so the pencil is regular with the stated root split. The homogeneous determinant is invariant under these transforms because [F5] gives multiplicativity and both shear determinants equal one by [F6]. Finally, $(I+GX)$ is invertible by step 1.1, so $M[I;X]$ has full column rank. A Jordan basis of $A_c$ from [F10], together with $N[I;X]=M[I;X]A_c$ from step 2.1, shows that the graph carries all $n$ stable finite root chains; the determinant factorization shows there are exactly $n$ stable roots in total, so this is the stable deflating graph. This uses no inverse of $A$.