# On the existence of fixed points of a matrix iteration

Let $$A\in\mathbb{R}^{n\times n}$$ be a Hurwitz stable matrix (i.e. all eigenvalues of $$A$$ have negative real part). Let $$\succeq$$ denote the standard partial order in the cone of positive semidefinite matrices, and consider the following iteration, for $$k\in\mathbb{Z}$$, $$k\ge 0$$, $$\tag{1}\label{eq:1} X_{k+1} = \frac{P_k^{1/2}X_k P_k^{1/2}}{\mathrm{tr}(P_k^{1/2}X_k P_k^{1/2})}, \ \ \ X_0\succeq 0,\ \mathrm{tr}(X_0)=1,$$ where $$P_k\succeq 0$$ is the solution of $$AP_k+P_kA^\top =-X_k$$, and $$\cdot^{1/2}$$ denotes the principal matrix square root.

First, note that a fixed point of \eqref{eq:1} always exists by Brouwer's fixed point theorem (indeed, \eqref{eq:1} is continuous and maps the compact set of unit trace positive semidefinite matrices to itself). However, the fixed point is, in general, not unique.

Second, if $$A+A^\top \prec 0$$ then there exists a fixed point $$\bar{X}\succeq 0$$ of \eqref{eq:1} such that $$\bar{P}\succ 0$$, where $$\bar{P}$$ is the solution of $$A\bar{P}+\bar{P}A^\top =-\bar{X}$$ (to see this pick for instance $$\bar{P}=-\frac{1}{2\mathrm{tr}(A)}I$$).

Hence, my question: Does there always (i.e., for any Hurwitz stable $$A$$) exist a fixed point $$\bar{X}\succeq 0$$ of \eqref{eq:1} such that $$\bar{P}\succ 0$$, where $$A\bar{P}+\bar{P}A^\top =-\bar{X}$$?