Newton sign iteration converges off the imaginary axis and is locally quadratic

In a Jordan representation of $A$, [F1]–[F2] give $S=I$ on every right-half-plane block and $S=-I$ on every left-half-plane block. Hence $S^2=I$ and $AS=SA$. Conjugate the exact iterates into this basis. Initially they are block diagonal with upper-triangular Jordan blocks. If $X_k$ has this form, its inverse, when defined, has the same block structure and its diagonal entries are the reciprocals of those of $X_k$. Thus each diagonal entry follows the scalar update ${\lambda }_{k+1}=({\lambda }_k+{\lambda }_k^{-1})\mathrm{/}2$. Whenever $\mathrm{Re}{\lambda }_k\mathrm{\ne }0$, $\mathrm{Re}{\lambda }_{k+1}=\frac{\mathrm{Re}{\lambda }_k}{2}(1+\mathrm{\mid }{\lambda }_k{\mathrm{\mid }}^{-2}),$ so its nonzero real-part sign is unchanged. By induction all diagonal entries remain nonzero and off $i\mathbb{R}$, each $X_k$ is invertible, and the next update is defined.

Since the Jordan-block diagonal values remain in the same half-planes, [F2] gives $\mathrm{sign}(X_k)=S$ for every $k$. Also $X_k$ commutes with $S$: this holds for $X_0=A$, and if it holds for $X_k$, then its inverse also commutes with $S$, so the update [F3] preserves commutation. The eigenvalues of $X_k+S$ are ${\lambda }_k+s$ on the corresponding blocks, with $\mathrm{Re}({\lambda }_k+s)\mathrm{\ne }0$ because $\mathrm{Re}{\lambda }_k$ has sign $s$. Thus $X_k+S$ is invertible. Define $G_k=(X_k-S)(X_k+S{)}^{-1}$.

Using $S^2=I$ and commutation with $X_k$, expansion gives $X_{k+1}\pm S=\frac{1}{2}{X}_k^{-1}(X_k\pm S{)}^2,G_{k+1}=G_k^2\mathrm{.}$ Hence $G_k=G_0^{2^k}$. The eigenvalues of $G_0$ are $(\lambda -s)\mathrm{/}(\lambda +s)$; their moduli are less than one because $\mathrm{\mid }\lambda +s{\mathrm{\mid }}^2-\mathrm{\mid }\lambda -s{\mathrm{\mid }}^2=4s\mathrm{Re}\lambda >0$.

In a Jordan basis of $A$, $G_0=(A-S)(A+S{)}^{-1}$ is block upper triangular; on a block at $\lambda$ its diagonal is $\mu =(\lambda -s)\mathrm{/}(\lambda +s)$ and its strictly upper triangular part is nilpotent, say $N_0^d=0$. Its $q$th power is ${\sum }_{j<d}\left(\genfrac{}{}{0px}{}{q}{j}\right){\mu }^{q-j}{N}_0^j$. Each term tends to zero as $q\to \mathrm{\infty }$ because $\mathrm{\mid }\mu \mathrm{\mid }<1$ and exponential decay dominates the polynomial factor $\left(\genfrac{}{}{0px}{}{q}{j}\right)$; if $\mu =0$, the nilpotent block power is eventually zero. Thus $G_0^{2^k}\to 0$. The eigenvalues of $G_k=G_0^{2^k}$ have modulus less than one, so $I-G_k$ is invertible. Rearranging the definition of $G_k$ gives $X_k=(I-G_k{)}^{-1}(I+G_k)S$, so $X_k\to S$.

Direct expansion, using $X_{k}S=S{X}_k$ and $S^2=I$, gives $X_{k+1}-S=\frac{1}{2}{X}_k^{-1}(X_k-S{)}^2$. Taking any submultiplicative norm proves the stated quadratic inequality. To bound the inverse near the limit, write $X_k=S(I+Y_k)$ with $Y_k=S^{-1}(X_k-S)$. Since $X_k\to S$, eventually $\mathrm{\parallel }{Y}_k\mathrm{\parallel }<1\mathrm{/}2$. The geometric series ${\sum }_{j\ge 0}(-Y_k{)}^j$ is Cauchy in the chosen norm because its tail is bounded by the scalar geometric tail; the matrix space is complete by [F4]. Multiplication of its partial sums by $I+Y_k$ shows its sum is $(I+Y_k{)}^{-1}$, whose norm is at most $\mathrm{\parallel }I\mathrm{\parallel }+\mathrm{\parallel }{Y}_k\mathrm{\parallel }\mathrm{/}(1-\mathrm{\parallel }{Y}_k\mathrm{\parallel })\le \mathrm{\parallel }I\mathrm{\parallel }+1$. Thus $\mathrm{\parallel }{X}_k^{-1}\mathrm{\parallel }\le (\mathrm{\parallel }I\mathrm{\parallel }+1)\mathrm{\parallel }{S}^{-1}\mathrm{\parallel }$ eventually, so the displayed inequality has a fixed quadratic constant near the limit.