Let $A,B,C$ be $n\times n$ complex Hermitian matrices with non-degenerate spectrum and such that they do not share any eigenstates. Consider now the following two matrices: $S_{AB}=A+B$ and $S_{AC}=A+C$. In addition, it is assumed that $S_{AB}, S_{AC}$ are such that their eigenvalues are strictly positive and their null space is $\{0\}$.
Does it follow that the null space of their commutator $C=[S_{AB},S_{AC}]$ is also $\{0\}$? If not, what condition would guarantee such property?
Note that one could use the inequality $$ \dim \ker (S_{AB}S_{AC}) \leq \dim \ker (S_{AB}) + \dim \ker (S_{AC}), $$ to show that the null space of both $S_{AB}S_{AC}$ and $S_{AC}S_{AB}$ is $\{0\}$ but that is not sufficient to tell us anything about the null space of the commutator (their difference). Right?
What more can be said about the null space of the commutator of matrices with the above properties?
Moreover, can $S_{AB}$ and $S_{AC}$ share any eigenstates?
You can observe that
$[S_{AB},S_{AC}]=(A+B)(A+C)-(A+C)(A+B)=$
$=A^2+AC+BA+BC-(A^2+AB+CA+CB)=$
$=[A,C]+[B,A]+[B,C]=$
$=[A,C]+[B,C]-[A,B]$
so if
$[A,B]$ and $[A,C]+[B,C]$ are the null space $0$ then the null space of $[S_{AB},S_{AC}]$ will be $0$