I would like to solve for $X$ in the matrix equation
$$
XCX + AX = I
$$
where all the matrices are $n\times n$, have real components, $X$ is positive semidefinite and $C$ is symmetric. My (possibly optimistic) hunch is that $X$ will be unique because of the positive semidefinite requirement; if not I only care about finding a single solution.
My searches have turned up a lot of work on solving non-symmetric riccati equations, unfortunately I don't meet the requirements for any of them. For example, everything I've seen requires that the entries of $C$ all be nonnegative, which is not true in my case.
Best Answer
Both existence and uniqueness can fail spectacularly.
In the special case $A=0$ we get $XCX=I$, which is equivalent to $C = X^{-2}$. Thus we must have $\det C > 0$, else then $C=X^{-2}$ has no real solution at all, even without requiring $X$ to be symmetric, let alone positive-definite. If $X$ is symmetric then $C$ must also be positive-definite. Note that $\det C$ can be zero or negative even when all entries of $C$ are positive.
When a solution does exist then there might even be a positive-dimensional family of positive-definite $X$ satisfying the equation $XCX+AX=I$. For example, let $A=3I$ and let $C=-2I$. Then the equation becomes $2X^2-3X+I=0$, i.e. $(X-I)(2X-I)=0$. So for any orthogonal decomposition ${\bf R}^n = V_1 \oplus V_{1/2}$ we get a solution by making $V_1$ the $1$-eigenspace $V_{1/2}$ the $(1/2)$-eigenspace, i.e. letting $X$ be the transformation that takes any $v_1+v_2$ ($v_i \in V_i$) to $v_1 + \frac12 v_2$.