How to solve this quadratic matrix equation for $X$?

Question

Let matrices $A$ and $B$ be symmetric and positive definite, and of the same size. How do we solve the quadratic matrix equation $XAX = B$?

I was wondering about this, because I can't just use $X = B^\frac{1}{2}A^\frac{1}{2}$, as $B^\frac{1}{2}A^\frac{1}{2} \neq A^\frac{1}{2}B^\frac{1}{2}$.

In broader context, I am trying to maximize $2 \operatorname{Tr} (C)$ with the constraint $A - C B C^\top \succeq 0$. So $X$ would be the Lagrange multiplier.

I believe your maximization problem is probably more well-posed than the quadratic matrix equation — whpowell96, Feb 19 '25 at 17:33
How exactly did you go from the LMI $A - X B X^\top \succeq 0$ to the quadratic matrix equation? So, do you want to solve a quadratic matrix equation or a quadratic LMI? — Rodrigo de Azevedo, Feb 21 '25 at 13:54
$X = A^{-1/2} (A^{1/2} B A^{1/2})^{1/2} A^{-1/2};$ as explained in this post. — greg, Apr 05 '25 at 19:18

score 1 · Answer 1 · answered Feb 19 '25 at 17:49

Let $X = A^{-1} Y$ then your equation is $A^{-1} YAA^{-1} Y = A^{-1} Y Y = B$. Hence $Y Y = A B$.

If an eigenvalue decomposition is given as $A B = U^{-1} \Lambda U$ then $Y = U^{-1} \Gamma U$ where $\Gamma$ and $\Lambda$ are diagonal matrices with termwise $\gamma_i^2 = \lambda_i$. So in total, $X = A^{-1} U^{-1} \Gamma U$.

How to solve this quadratic matrix equation for $X$?

1 Answers1