Find a feasible solution for matrix inequation $AXA^T\succ B$ Suppose that $A,X,B$ are all $n\times n$ symmetrical matrices. Also suppose that $\det(A)=0$(otherwise the problem is trivial). Then how to find a feasible solution $X$ for the matrix inequation $AXA^T\succ B$ or judging whether there exists a solution.Here for matrices $M\succ N$ denotes that $M-N$ is positive. I try to use some properties about generalized inverse matrix but still don't know how to solve it.
 A: Let $U$ be an orthogonal matrix that diagonalizes $A$, so that
$$
U^TAU = \pmatrix{D & 0\\0 & 0},
$$
where $D$ is diagonal with non-zero diagonal entries. Partition $U$ into $\pmatrix{U_1 & U_2}$ so that $U_1$ has the same number of columns as $D$ (note: it follows that $D = U_1^TAU_1$). We have
$$
AXA^T \succ B \iff (U^TAU)(U^TXU)(U^TAU)^T \succ U^TBU\\
\iff \pmatrix{D & 0\\0 & 0} \pmatrix{U_1^T XU_1 & U_1^TX U_2\\ U_2^T XU_1 & U_2^TXU_2} \pmatrix{D & 0\\0 & 0} \succ \pmatrix{U_1^TBU_1 & U_1^TBU_2\\ U_2^TBU_1 
& U_2^T BU_2}\\
\iff 
\pmatrix{D[U_1^T XU_1]D & 0\\ 0 & 0} \succ \pmatrix{U_1^TBU_1 & U_1^TBU_2\\ U_2^TBU_1 
& U_2^T BU_2}
\\ \iff
\pmatrix{D[U_1^T XU_1]D - U_1^TBU_1 & -U_1^TBU_2\\ -U_2^TBU_1 
& -U_2^T BU_2} \succ 0.
$$
We can now see that a solution can only exist if $U_2^T BU_2 \prec 0$. If $U_2^T BU_2 \prec 0$ and we merely want at least one feasible solution, we can always find a solution of the form $X = \lambda I$ with $\lambda > 0$. Indeed: plugging in this form of $X$ yields the inequality
$$
\pmatrix{\lambda D^2 - U_1^TBU_1 & -U_1^TBU_2\\ -U_2^TBU_1 
& -U_2^T BU_2} \succ 0.
$$
Using the Schur complement, we see that this matrix is positive if and only if
$$
(\lambda D^2 - U_1^TBU_1) - (-U_1^TBU_2)(-U_2BU_2)^{-1}(-U_2^TBU_1) =\\
\lambda D^2 - \underbrace{[U_1^TBU_1 + U_1^TBU_2(-U_2BU_2)^{-1}U_2^TBU_1]}_M \succ 0.
$$
Because $D^2 \succ 0$, there exists a value of $\lambda$ such that $\lambda D^2 \succ M$, or equivalently $D^{-1}MD^{-1} \prec \lambda I$.
