# General Sylvester's linear matrix equation

For what conditions on $$A$$, $$B$$ and $$C$$ (square matrices of size $$n$$) would there be a unique solution to $$ABX + AXC + XBC = D,$$ for any $$D$$? Can one expect a characterization similar to the Sylvester Theorem, which states that there always exists a unique solution to $$AX + XB = C$$, for any $$C$$, if and only if $$A$$ and $$-B$$ do not share an eigenvalue? And then, can this be extended to equations of the form $$ABCX + ABXD + AXCD + XBCD = E$$, and beyond?

• I didn't even know Sylvester was in the Army, much less that he held the rank of a General. Jan 26 at 3:57
• @Gerry Myerson may be he meant the concept of generalization in mathematics. Jan 26 at 6:48
• @GerryMyerson: of course, he was: en.wikipedia.org/wiki/John_B._Sylvester
– M.G.
Jan 26 at 11:50
• @AhmadJamilAhmadMasad: no need to apologize, it was a joke :) The general is a different Sylvester. Or is he?...
– M.G.
Jan 26 at 12:14
• @GerryMyerson. I heard this joke 20 years ago, from Efimov, à propos of his proof of the General Burnside conjecture. Jan 26 at 14:35

As far as I know, no, apart from very special cases where the coefficients can be triangularized simultaneously. There is a big gap in difficulty between the 2-term case (where there is a canonical form for pairs of matrices, an $$O(n^3)$$ algorithm, etc.) and the 3-term case, when there is basically nothing apart from turning it into a $$n^2 \times n^2$$ linear system.
• Maybe a more definite negative answer (for fixed $n\ge 2$) would be to show that the characterization does not only depend on the conjugacy classes of $A$, $B$, and $C$.