Separated DARE pencils have quantitatively stable deflating graphs

By [F1], the regular base DARE pencil has an $n+n$ stable/outside split, and the ordered representation [F2] puts the stable roots in the leading blocks and the outside roots in the trailing blocks. Write trailing diagonal pairs as $(s_i,t_i)$ and leading ones as $(p_j,q_j)$. Disjoint homogeneous roots give $p_j{t}_i-s_i{q}_j\mathrm{\ne }0$ for every $i,j$, including $t_i=0$. In the system $S_{22}Y-Z{S}_{11}=C$, $T_{22}Y-Z{T}_{11}=D$, process rows $i=n,n-1,\dots ,1$ and within each row columns $j=1,\dots ,n$; upper triangularity makes all other terms involve already determined $Y_{kj}$ with $k>i$ or $Z_{i\mathrm{\ell }}$ with $\mathrm{\ell }<j$. The two unknowns $(Y_{ij},Z_{ij})$ have coefficient matrix $\left[\begin{array}{cc}{s}_i& -p_j\\ t_i& -q_j\end{array}\right]$ with determinant $p_j{t}_i-s_i{q}_j\mathrm{\ne }0$, so the lower-oriented generalized Sylvester map is bijective. Its minimum singular value [F3] is therefore $\gamma >0$, and its inverse norm is $\kappa ={\gamma }^{-1}$.

Define $C_Y=S_{11}+E_{11}+(S_{12}+E_{12})Y$ and $D_Y=T_{11}+F_{11}+(T_{12}+F_{12})Y$. Block multiplication gives $(S+E)[I;Y]=[I;Z]C_Y$ and $(T+F)[I;Y]=[I;Z]D_Y$ exactly when $\mathcal{L}(Y,Z)=(S_{22}Y-Z{S}_{11},T_{22}Y-Z{T}_{11})$ equals $(-E_{21}-E_{22}Y+Z{E}_{11}+Z(S_{12}+E_{12})Y,-F_{21}-F_{22}Y+Z{F}_{11}+Z(T_{12}+F_{12})Y)$. On the product radius-$r$ ball, the right side has norm at most $\delta +2\delta r+(h+\delta )r^2$: the constant pair is bounded by $\delta$, the two linear pairs by $\delta \mathrm{\parallel }Y{\mathrm{\parallel }}_F$ and $\delta \mathrm{\parallel }Z{\mathrm{\parallel }}_F$, and the quadratic pair by $(h+\delta )\mathrm{\parallel }Z{\mathrm{\parallel }}_F\mathrm{\parallel }Y{\mathrm{\parallel }}_F$. Splitting the quadratic difference into its $Y$ and $Z$ changes bounds the Lipschitz constant by $2\delta +2(h+\delta )r$. Applying ${\mathcal{L}}^{-1}$ gives a self-map and contraction under the stated inequalities; iteration from zero has geometrically summable increments, so completeness [F8] gives a limit in the closed ball, continuity gives a fixed point, and contraction gives uniqueness. For $r=2\kappa \delta$ and $\theta =\kappa \delta$, substitution gives self-map ratio $\kappa [\delta +2\delta r+(h+\delta )r^2]\mathrm{/}r=1\mathrm{/}2+2\theta +2\kappa h\theta +2{\theta }^2$ and contraction constant $2\theta +4\kappa h\theta +4{\theta }^2$; the displayed strict condition implies both. For $\delta =0$, the exact pair $Y=Z=0$ solves the graph equations.

Set $R_Y=\left[\begin{array}{cc}I& 0\\ Y& I\end{array}\right]$ and $L_Z=\left[\begin{array}{cc}I& 0\\ -Z& I\end{array}\right]$. They are invertible shears of determinant one, and the graph equations from step 2.1 give $L_Z(S+E)R_Y=\left[\begin{array}{cc}{C}_Y& S_{12}+E_{12}\\ 0& A^{\mathrm{\prime }}\end{array}\right]$ and $L_Z(T+F)R_Y=\left[\begin{array}{cc}{D}_Y& T_{12}+F_{12}\\ 0& B^{\mathrm{\prime }}\end{array}\right]$, where $A^{\mathrm{\prime }}=S_{22}+E_{22}-Z(S_{12}+E_{12})$ and $B^{\mathrm{\prime }}=T_{22}+F_{22}-Z(T_{12}+F_{12})$. Unitary invariance, $\mathrm{\parallel }M{\mathrm{\parallel }}_2\le \mathrm{\parallel }M{\mathrm{\parallel }}_F$ [F5], and submultiplicativity [F6] give $\mathrm{\parallel }{C}_Y-S_{11}{\mathrm{\parallel }}_2,\mathrm{\parallel }{A}^{\mathrm{\prime }}-S_{22}{\mathrm{\parallel }}_2\le a$ and $\mathrm{\parallel }{D}_Y-T_{11}{\mathrm{\parallel }}_2,\mathrm{\parallel }{B}^{\mathrm{\prime }}-T_{22}{\mathrm{\parallel }}_2\le b$. Every leading stable root is finite, so $T_{11}$ is invertible and $K_s=T_{11}^{-1}{S}_{11}$ is upper triangular with diagonal moduli below one. Every trailing outside root has nonzero first homogeneous coordinate, so $S_{22}$ is invertible and $K_u=S_{22}^{-1}{T}_{22}$ is upper triangular with diagonal moduli below one, including zero for infinity; thus both are Schur stable. By [F7], $W_j={\sum }_{k\ge 0}(K_j^{\ast }{)}^k{K}_j^k$ converges for $j=s,u$, is positive definite, and satisfies $W_j-K_j^{\ast }{W}_j{K}_j=I$. The two Neumann margins and [F12] make $D_Y,A^{\mathrm{\prime }}$ invertible with inverse norms at most $\mathrm{\parallel }{T}_{11}^{-1}{\mathrm{\parallel }}_2\mathrm{/}(1-\mathrm{\parallel }{T}_{11}^{-1}{\mathrm{\parallel }}_{2}b)$ and $\mathrm{\parallel }{S}_{22}^{-1}{\mathrm{\parallel }}_2\mathrm{/}(1-\mathrm{\parallel }{S}_{22}^{-1}{\mathrm{\parallel }}_{2}a)$. The identities $D_Y^{-1}{C}_Y-K_s=D_Y^{-1}[(C_Y-S_{11})-(D_Y-T_{11})K_s]$ and $A^{\mathrm{\prime }-1}{B}^{\mathrm{\prime }}-K_u=A^{\mathrm{\prime }-1}[(B^{\mathrm{\prime }}-T_{22})-(A^{\mathrm{\prime }}-S_{22})K_u]$ give exactly the stated ${ϵ}_s,{ϵ}_u$ bounds.

Put $P_Y=(I+Y^{\ast }Y{)}^{-1\mathrm{/}2}$. Since $V=[U{V}_2]$ is unitary, $(U+V_{2}Y{)}^{\ast }(U+V_{2}Y)=I+Y^{\ast }Y$, so ${\widehat{U}}^{\ast }\widehat{U}=I$ and $U^{\ast }\widehat{U}=P_Y$. Therefore $(\widehat{U}-U{)}^{\ast }(\widehat{U}-U)=2(I-P_Y)$. The largest eigenvalue of $Y^{\ast }Y$ is $y^2$, so [F3], [F5], and [F9] give the exact norm $\mathrm{\parallel }\widehat{U}-U{\mathrm{\parallel }}_2=\beta (y)$. The scalar function $g(t)=t-2(1-(1+t{)}^{-1\mathrm{/}2})$ has $g(0)=0$ and $g^{\mathrm{\prime }}(t)=1-(1+t{)}^{-3\mathrm{/}2}\ge 0$ for $t\ge 0$, proving $\beta (y)\le y\le r$. Also $(I-U{U}^{\ast })\widehat{U}=V_{2}Y{P}_Y$, whose largest singular value is $y\mathrm{/}\sqrt{1+y^2}$ by [F3] and [F5].

If $K$ is either quotient in step 3.1 and $\mathrm{\parallel }H{\mathrm{\parallel }}_2\le ϵ$, then $W-(K+H{)}^{\ast }W(K+H)=I-(K^{\ast }WH+H^{\ast }WK+H^{\ast }WH)$. The subtracted Hermitian term has norm at most $\mathrm{\parallel }W{\mathrm{\parallel }}_2(2\mathrm{\parallel }K{\mathrm{\parallel }}_2ϵ+{ϵ}^2)<1$, so the left side is positive definite; an eigenvector of $K+H$ with eigenvalue $\lambda$ gives $(1-\mathrm{\mid }\lambda {\mathrm{\mid }}^2)v^{\ast }Wv>0$, hence $\mathrm{\mid }\lambda \mathrm{\mid }<1$. Applied to $D_Y^{-1}{C}_Y$ and $A^{\mathrm{\prime }-1}{B}^{\mathrm{\prime }}$, this makes the leading perturbed roots finite and stable, and the trailing roots outside, with zero eigenvalues of $A^{\mathrm{\prime }-1}{B}^{\mathrm{\prime }}$ representing infinity. The shears have determinant one, so [F13] preserves the homogeneous determinant under their left/right transformations; [F10] factors it across the diagonal blocks. Since $D_Y$ and $A^{\mathrm{\prime }}$ are invertible, both diagonal pencils are regular and their product is a regular pencil with the claimed $n+n$ root split. The regular base and perturbed determinant polynomials have finite root sets by [F14]; choose ${\lambda }_0\in \mathbb{C}$ outside their union. Put $\mathcal{R}=(N^{\mathrm{\prime }}-{\lambda }_0{M}^{\mathrm{\prime }}{)}^{-1}{M}^{\mathrm{\prime }}$ for $N^{\mathrm{\prime }}=L_Z(S+E)R_Y$ and $M^{\mathrm{\prime }}=L_Z(T+F)R_Y$. Then $N^{\mathrm{\prime }}-\lambda M^{\mathrm{\prime }}=(N^{\mathrm{\prime }}-{\lambda }_0{M}^{\mathrm{\prime }})[I-(\lambda -{\lambda }_0)\mathcal{R}]$, so [F13] expresses the pencil determinant as a nonzero constant times $\det (I-(\lambda -{\lambda }_0)\mathcal{R})$. Jordan form [F11] factors the latter into ${\prod }_j(1-(\lambda -{\lambda }_0){\mu }_j)$ over the eigenvalues ${\mu }_j$ of $\mathcal{R}$, with algebraic multiplicity. Thus each finite root maps injectively to $\mu =1\mathrm{/}(\lambda -{\lambda }_0)$, and the polynomial degree is the number of nonzero ${\mu }_j$, counted with algebraic multiplicity. Its homogeneous form is $\det (\beta N^{\mathrm{\prime }}-\alpha M^{\mathrm{\prime }})={\beta }^{2n}\det (N^{\mathrm{\prime }}-(\alpha \mathrm{/}\beta )M^{\mathrm{\prime }})$; by [F15], its multiplicity at $[1:0]$ is $2n$ minus that degree, exactly the multiplicity of the zero eigenvalue of $\mathcal{R}$. The matrix $\mathcal{R}$ is block upper triangular with diagonal blocks $(C_Y-{\lambda }_0{D}_Y{)}^{-1}{D}_Y$ and $(A^{\mathrm{\prime }}-{\lambda }_0{B}^{\mathrm{\prime }}{)}^{-1}{B}^{\mathrm{\prime }}$, whose spectra are disjoint because their pencil-root groups are disjoint and the projective map is injective. For a generalized eigenvector $[x;y]$ belonging to a leading-block eigenvalue $\mu$, its trailing equation is $({\mathcal{R}}_{22}-\mu I{)}^{k}y=0$; disjoint spectra force $y=0$. Conversely, Jordan form [F11] supplies a basis of generalized eigenvectors of the leading $n\times n$ block, which embed as $[x;0]$ and span exactly the full leading-root generalized eigenspace. Undoing the graph shears and unitary factors identifies this right subspace with $V[I;Y]$ and its matching left deflating graph with $Q[I;Z]$.

Write $\mathrm{\Delta }U=\widehat{U}-U$. Under the strict chart margin, ${\widehat{U}}_1=U_1(I+U_1^{-1}\mathrm{\Delta }{U}_1)$ and $\mathrm{\parallel }{U}_1^{-1}\mathrm{\Delta }{U}_1{\mathrm{\parallel }}_2\le \mathrm{\parallel }{U}_1^{-1}{\mathrm{\parallel }}_2\beta (y)<1$. By [F12], ${\widehat{U}}_1$ is invertible and $\mathrm{\parallel }{\widehat{U}}_1^{-1}{\mathrm{\parallel }}_2\le (\mathrm{\parallel }{U}_1^{-1}{\mathrm{\parallel }}_2^{-1}-\beta (y){)}^{-1}$. Using $U_2=X{U}_1$ gives ${\widehat{U}}_2{\widehat{U}}_1^{-1}-X=(\mathrm{\Delta }{U}_2-X\mathrm{\Delta }{U}_1){\widehat{U}}_1^{-1}$; [F6] yields the stated chart estimate. Since $\beta$ is increasing and $\beta (y)\le \beta (r)$, the stronger margin $\mathrm{\parallel }{U}_1^{-1}{\mathrm{\parallel }}_2\beta (r)<1$ gives the computable weaker bound. An independently rotated orthonormal basis of the same space is $\widehat{U}{R}_0$ for unitary $R_0$ and has the same chart because $({\widehat{U}}_1{R}_0{)}^{-1}=R_0^{-1}{\widehat{U}}_1^{-1}$. The proof assumes no structure on $E,F$, so it gives no Riccati residual, symmetry, positivity, or post-symmetrization stability certificate.

As $\delta \to 0$ with $r=2\kappa \delta$, $\theta \to 0$, $a,b\to 0$, and ${ϵ}_s,{ϵ}_u\to 0$; hence the contraction, Neumann, and both Stein margins in the Statement hold for all sufficiently small $\delta$. If $U_1$ is invertible, also $\mathrm{\parallel }{U}_1^{-1}{\mathrm{\parallel }}_2\beta (r)<1$ for all sufficiently small $\delta$, so the computable chart margin holds.