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Let $A, B \in \mathbb{R}^{n \times n}$ be symmetric, positive-semidefinite, full-rank matrices. I would like to understand the set of $X \in \mathbb{R}^{n \times n}$ which are themselves symmetric and positive-semidefinite and verify $X A X^T = B$.

As a first step, I have decomposed $A = A^{1/2} A^{1/2}$ and $B = B^{1/2} B^{1/2}$ and thereby observed that any solution must verify $X A^{1/2} = B^{1/2} O$, and thus $X = B^{1/2} O A^{-1/2}$, for some orthogonal matrix $O$. However, I am stumped as for how to find the set of orthogonal matrices $O$ for which the result is symmetric and PSD.

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  • $\begingroup$ there is no closed-form solution, you will need to solve some $n^2$ quadratic equations; if $\Lambda_{A,B}$ is the diagonal matrix of eigenvalues of $A$ or $B$, you will need to find a matrix $Y$ such that $Y\Lambda_AY^\top=\Lambda_B$; the desired $X$ then follows by multiplication of $Y$ with matrices that diagonalise $A,B$. $\endgroup$
    – Carlo Beenakker
    Commented Mar 15, 2023 at 12:05
  • $\begingroup$ I do not understand what the problem is, since for any invertible $n \times n$ real matrix $X$, the matrix $XAX^\top$ is congruent to $A$, hence is still symmetric positive-definite (and invertible). Therefore any orthogonal matrix $O$ works. $\endgroup$
    – Christophe Leuridan
    Commented Mar 15, 2023 at 18:59
  • $\begingroup$ Just a note: If a positive-semidefinite symmetric square matrix is full-rank, then it is positive-definite. $\endgroup$
    – Wei Wang
    Commented Mar 21, 2023 at 15:17

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