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Let $d \times d$ matrices $A, B$ be positive definite. Is there a closed form solution for the following quadratic equation in $X$?

$$X A X^{T} = B$$

Thank you.

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  • $\begingroup$ Just for completeness: this is usually called a Ricatti equation, sometimes also a Lyapunov equation. This might help to lookup references if needed. $\endgroup$ Jul 20, 2020 at 17:10
  • $\begingroup$ Related: mathoverflow.net/q/78106 $\endgroup$ Jul 20, 2020 at 17:28
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    $\begingroup$ Does this answer your question? Solving a quadratic matrix equation $\endgroup$
    – vidyarthi
    Jul 20, 2020 at 17:59
  • $\begingroup$ Thank you for the answers, I think Federico's answer is very neat, at least for the type of problem I specified. $\endgroup$
    – Fabio
    Jul 20, 2020 at 20:29

1 Answer 1

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$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a complete parametrization of the solutions.

Here $A^{1/2}$ is the symmetric square root of $A$ (if you prefer you can work with the Cholesky factor and obtain similar results).

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  • $\begingroup$ @federico-poloni thank you, this was very helpful, could you elaborate on how the Cholesky factor approch would looke like? $\endgroup$
    – user_na
    May 29, 2021 at 6:46
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    $\begingroup$ @user_na If $B=MM^T$ and $A=LL^T$, then following the same steps one gets that $M^{-1}XL=Q$ satisfies $QQ^T=I$. $\endgroup$ May 29, 2021 at 7:15
  • $\begingroup$ And with that, $X=MQL^{-1}$, right? $\endgroup$
    – user_na
    May 29, 2021 at 7:29
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    $\begingroup$ The question asks about positive definite matrices, not symmetric matrices. However, if $A^{1/2}$ is a symmetric square root, $A^T = (A^{1/2}A^{1/2})^T=A$, so $A$ is also symmetric. The given solution only works if $A$ and $B$ are symmetric. Unfortunately Federico Poloni's answer doesn't answer the question in general as I thought. If the question intended to ask about symmetric matrices, maybe it should be edited. $\endgroup$
    – wikiwert
    Sep 15, 2022 at 14:13
  • $\begingroup$ @wikiwert AFAIK the standard definition of "positive definite" includes "symmetric"; if you mean "Hurwitz anti-stable" (all eigenvalues in the right half-plane) please specify it. $\endgroup$ Sep 16, 2022 at 7:28

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