Quadratic equation of matrices Let $A$ be positive definite and $I$ denote the identity matrix. Does the following matrix quadratic equation have a solution (not necessarily unique) when we solve for $X$?
$X^T A X + X^T B + C = I$
If not, perhaps it does when $C$ is also positive definite?
 A: Suppose the system yields a solution, then we have
$$
X^{\top}AX+X^{\top}B+C=I
$$
and its transpose
$$
X^{\top}AX+B^{\top}X+C^{\top}=I.
$$
Add up these two equations, and
$$
X^{\top}AX+\frac{1}{2}\left(X^{\top}B+B^{\top}X\right)+\frac{1}{2}\left(C+C^{\top}\right)=I.
$$
This can be rewritten as
$$
\left(X+\frac{1}{2}A^{-1}B\right)^{\top}A\left(X+\frac{1}{2}A^{-1}B\right)=I-\frac{1}{2}\left(C+C^{\top}\right)+\frac{1}{4}B^{\top}A^{-1}B,
$$
or
$$
Y^{\top}AY=F
$$
for short, where $A$ is positive definite, while $F$ is symmetric.
Note that the LHS is always positive semi-definite. So if $F$ is not positive semi-definite, there would be no solution.
When $F$ is positive semi-definite, $F^{1/2}$ is well defined by using orthogonal diagonalization. So is $A^{1/2}$. As such, the equation reads (note that $A^{1/2}$ and $F^{1/2}$ are symmetric as well)
$$
\left(A^{1/2}Y\right)^{\top}\left(A^{1/2}Y\right)=F^{\top/2}F^{1/2}.
$$
Therefore,
$$
A^{1/2}Y=UF^{1/2}\iff Y=A^{-1/2}UF^{1/2}\iff X=A^{-1/2}UF^{1/2}-\frac{1}{2}A^{-1}B,
$$
where $U$ is an arbitrary orthogonal matrix.
This is a necessary condition for $X$ being a solution. However, even if the above form makes sense, i.e., $F^{1/2}$ is well defined, it does not mean that the original system has a solution, because it merely corresponds to the symmetrized system
$$
X^{\top}AX+\frac{1}{2}\left(X^{\top}B+B^{\top}X\right)+\frac{1}{2}\left(C+C^{\top}\right)=I.
$$
A: @hypernova , your candidate $X_0=f(U)$ is interesting when $rank(F)=r$ is small; indeed, $X_0$ does not depend on $n(n-1)/2=dim(O(n))$ parameters but only on $r(n-(r+1)/2)$ parameters; cf. my post in
Solution form of a certain matrix relation 
