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I am currently facing a Sylvester equation

$AX+XB = C$

where $A$, $B$, $C$ are all symmetric and a special constraint here is that $X$ should be orthogonal. The Sylvester equation itself may not hold perfectly and hence the least-square solution is also ok.

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    $\begingroup$ Try a numerical solution first, e.g., $\min_{X^TX=I} \|AX+XB-C\|_F$ using manopt.org $\endgroup$
    – Suvrit
    Commented Oct 24, 2019 at 15:19
  • $\begingroup$ If you relax $X^T X = I$ into $X^T X \preceq I$, you have a semidefinite program. $\endgroup$
    – Rodrigo de Azevedo
    Commented Oct 24, 2019 at 18:48
  • $\begingroup$ Thank you for the information. I should have made my expression more clear. That is, I prefer an analytical solution rather than a numerical one. And regarding to relax it into a semidefinite program question, may I say that the solution is still not analytical? $\endgroup$
    – lisi
    Commented Oct 25, 2019 at 3:16
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    $\begingroup$ @lisi Analytical solutions are a luxury and a rarity. $\endgroup$
    – Rodrigo de Azevedo
    Commented Oct 25, 2019 at 6:10

1 Answer 1

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Let me assume that $A$ and $-B$ have no common eigenvalues. Without the orthogonality constraint there is then a unique solution $Y$ of the Sylvester equation $AY+YB=C$, which you can find using known methods. Let $Y$ have the singular value decomposition $Y=U\Sigma V^T$. Then $X=UV^T$ is the orthogonal matrix that minimizes $\sum_{ij}(X_{ij}-Y_{ij})^2$. (For a proof, see here.)

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  • $\begingroup$ I got it! Thank you! $\endgroup$
    – lisi
    Commented Oct 24, 2019 at 8:30

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