Let $A$ and $B$ be two $n\times n$ matrices that have no eigenvalues in common. Let $T$ be the transformation $$ T(S):=AS-SB $$ that maps the $n\times n$ matrices, $M_n$, to the $M_n$.
Can we assume that $AS=SB$?
2026-03-30 23:19:49.1774912789
Transformation Definition
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No. If you assume $AS=SB$, then $T(S)$ is a zero map for all $S$. This cannot happen if $A, B$ have no common eigenvalues.