3
$\begingroup$

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 $A\bar{P}+\bar{P}A^\top =-\bar{X}$?

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.