Matrix equation $X+AX^{-1}A^T=B$

Question

Given non-singular matrices $A, B \in\mathbb{R}^{n \times n}$, is there a way to solve the following matrix equation in matrix unknown $X \in\mathbb{R}^{n \times n}$?

$$X+AX^{-1}A^T=B$$

Answer 1

The equation can be rewritten as $Y+Y^{-1}P=Q$ where $Y=A^{-1}X,\,P=A^{-1}A^T$ and $Q=A^{-1}B$. Therefore, the problem boils down to solving the quadratic matrix equation (sometimes referred to as a non-symmetric algebraic Ricatti equation) $$ Y^2-YQ+P=0\tag{1} $$ for $Y$. (Note that when $P$ is nonsingular, any solution $Y$ is automatically nonsingular.) This equation is not always solvable. E.g. when $Q=I$ (i.e., when $A=B$), we have $Y^2-Y+P=0$. Hence $(Y-\frac12I)^2=\frac14I-P$ and solutions exist only if $\frac14I-P$ has a real square root.

I am not familiar with quadratic matrix equations. A quick internet search reveals that for some classes of matrices $P$ and $Q$, necessary and sufficient conditions for the existence of solutions are available. See the paper Quadratic Matrix Equations by Shurbet et al. (1974). Numerical methods for solving $(1)$ are studied in the Ph.D. thesis of Hyun-Min Kim (2000).