The stabilizing CARE solution is the Hamiltonian stable invariant graph

Block multiplication gives $H\left[\begin{array}{c}I\\ X\end{array}\right]=\left[\begin{array}{c}A-GX\\ -Q-A^{\ast }X\end{array}\right]\mathrm{.}$ By [F1], $-Q-A^{\ast }X=XA-XGX=X(A-GX)$, proving the graph identity.

Set $T_X=\left[\begin{array}{cc}I& 0\\ X& I\end{array}\right]$, whose inverse is $\left[\begin{array}{cc}I& 0\\ -X& I\end{array}\right]$. Using [F1], $T_X^{-1}H{T}_X=\left[\begin{array}{cc}A-GX& -G\\ 0& -(A-GX{)}^{\ast }\end{array}\right]\mathrm{.}$ The eigenvalues of the first diagonal block lie in the open left half-plane by [F1]. If $\lambda$ is a root of its characteristic polynomial, then $-\bar{\lambda }$ is a root of the characteristic polynomial of $-(A-GX{)}^{\ast }$, since $\det (-\bar{\lambda }I+(A-GX{)}^{\ast })=\overline{\det (-\lambda I+(A-GX))}=0$; these reflected roots lie in the open right half-plane. By [F5] each diagonal characteristic polynomial has degree $n$ and splits into $n$ roots counted with algebraic multiplicity. The characteristic polynomial factorization [F3] and root/eigenvalue connection [F4] therefore give exactly $n$ eigenvalues of $H$ in each open half-plane and none on the imaginary axis. The first coordinate subspace is invariant and its restriction has the stable spectrum of $A-GX$. Any generalized eigenvector of the displayed triangular matrix for a stable eigenvalue has trailing component in the kernel of a power of $(-(A-GX{)}^{\ast }-\lambda I)$; that matrix is invertible because the trailing spectrum is in the open right half-plane, so the trailing component is zero. By [F6] these generalized eigenspaces have total dimension equal to the stable algebraic multiplicity $n$ from [F5], so they fill the first coordinate subspace, which is therefore the stable invariant subspace. Applying $T_X$ maps it to $\mathrm{Ran}\left[\begin{array}{c}I\\ X\end{array}\right]$.