A separated simple Floquet multiplier has a left/right-conditioned finite enclosure

Since $w^{\ast }A={\lambda }_0{w}^{\ast }$, $W$ is invariant; since $Av={\lambda }_{0}v$ and $w^{\ast }v=1$, every vector splits uniquely as $\alpha v+y$ with $y\in W$. The restriction has no eigenvalue ${\lambda }_0$ by algebraic simplicity. Also $Q$ is the projection onto $W$. For $n\ge 2$, its range is nonzero and $Q$ acts as the identity there, so $q\ge 1$.

Let $\lambda \in D$ be an eigenvalue of $A+E$ with eigenvector $x=\alpha v+y$, $y\in W$. If $\alpha =0$, projection onto $W$ gives $y=(\lambda I_W-A_W{)}^{-1}QEy$. Thus $\mathrm{\parallel }y{\mathrm{\parallel }}_2\le Rq\delta \mathrm{\parallel }y{\mathrm{\parallel }}_2$, while $Rq\delta \le R{q}^2\delta <1$, forcing $y=0$, contrary to $x\mathrm{\ne }0$. Hence $\alpha \mathrm{\ne }0$; rescale the eigenvector so $\alpha =1$.

Projecting $(A+E)(v+y)=\lambda (v+y)$ onto $W$ gives $y=(\lambda I_W-A_W{)}^{-1}QE(v+y)$. Therefore $\mathrm{\parallel }y{\mathrm{\parallel }}_2\le Rq\delta (1+\mathrm{\parallel }y{\mathrm{\parallel }}_2)$ and $\mathrm{\parallel }y{\mathrm{\parallel }}_2\le Rq\delta \mathrm{/}(1-Rq\delta )$.

Applying $w^{\ast }$ to the same eigen-equation gives $\lambda -{\lambda }_0=w^{\ast }Ev+w^{\ast }Ey$. The first term, which [F3] identifies as the derivative along any differentiable normalized simple-eigenpair branch through $({\lambda }_0,v,w)$, has modulus at most $\kappa \delta$, while the second is at most $\kappa \delta \mathrm{\parallel }y{\mathrm{\parallel }}_2$. Substituting step 3.1 gives the stronger denominator $1-Rq\delta$. Since $q\ge 1$, $1-R{q}^2\delta \le 1-Rq\delta$ and both are positive, so replacing by the smaller denominator yields the stated bound. If $A+E$ is monodromy, [F5] identifies $\lambda$ as one of its Floquet multipliers. No monodromy property of $A$ alone is needed or sufficient for that interpretation.