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How to solve the Sylvester equation $$ax + xb = c$$ over quaternion? I tried to consider operator $$D = a^2 + a\cdot(b+\overline{b}) + b \cdot \overline{b} $$ and calculate $Dx$. But it didn't help.

P.S: What is the general meaning of this equation?

Thank you in advance!

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  • $\begingroup$ What do you mean by general meaning? It means Sylvester's equation, not in $M_n(K)$, but in the quaternion algebra. $\endgroup$ – Dietrich Burde Jan 6 at 19:27
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See this paper. I'd interpret $ax+xb=c$ as an equation asking for a quaternion $x=x_0+ x_1i+x_2j+x_3 k$, given quaternions $a$, $b$, and $c$ in a similar way. Write out this equation component wise, and obtain linear equations for the $x_i$. For the existence of a unique solution there will be some assumptions on the parameters $a$, $b$, and $c$.

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