This is an extension of another math.SE question. Let $A,B$ be two (complex) square matrices which are both diagonalizable but, unlike in the aforementioned question, not simultaneously diagonalizable (particularly, they do not commute). The following facts are known:
(see e.g. here) If $A,B$ are Hermitian and (say) $A$ is positive definite, then $AB$ ($=A^{\frac{1}{2}}(A^{\frac{1}{2}}BA^{\frac{1}{2}})A^{-\frac{1}{2}}$) is diagonalizable;
If (say) $A$ is nonsingular, then $AB$ ($=A(BA)A^{-1}$) is diagonalizable if and only if $BA$ ($=A^{-1}(AB)A$) is also diagonalizable.
(see e.g. here) On the other hand, there are examples where $A,B$ are both nonsingular and have only real eigenvalues, yet $AB$ is not diagonalizable - moreover, in the example just cited the eigenvalues of $AB$ also happen to be all real.
I am looking for conditions somewhat in between (1) and (3) which guarantee that $AB$ is diagonalizable.
More precisely, suppose that $A,B$ are Hermitian, noncommuting (complex) square matrices with $A$ nonsingular but neither $A$ nor $B$ necessarily positive definite. As (3) shows, this alone does not guarantee diagonalizability of $AB$, even if $AB$ has only real eigenvalues as well. Nonetheless,
Question: Are there other conditions that can be added to the ones listed in the last paragraph above which do guarantee together that $AB$ is diagonalizable?
