what is the relationship between the eigenvalues of $ABC$ and $BAC$?

Question

I have a question: Let $A, B, C$ be $n$-order square matrices. It is easy to prove that $AB$ and $BA$ have the same eigenvalues. Therefore, we can deduce that $ABC$ has the same eigenvalues as $BCA$ and $CAB$. My question is, if the order is not cyclic, will there be a similar result? Or are there other conclusions? Specifically, what is the relationship between the eigenvalues of $ABC$ and $BAC$? Or can we add some conditions to obtain some interesting conclusions?

Answer 1

You may want to check this answer out:

https://math.stackexchange.com/a/2163596/1270050

In my opinion, looking into other permutations other than the cyclic ones poses a problem because for general $n \times n$ matrices, $\mathcal{A}$ and $\mathcal{B}$, the relationship between their eigenvalues to $\mathcal{AB}$ still remains unknown according to my research:

• https://mathoverflow.net/a/137692

In most cases where a certain relationship exists, matrices $\mathcal{A}$ or $\mathcal{B}$ are assumed Hermitian, or positive (semi)definite, with additional restrictions if needed (first one is not really useful for this problem):

• https://arxiv.org/abs/1601.05525
• Showing that every eigenvalue of $AB$ is the product of an eigenvalue of $A$ and eigenvalue of $B$, for any Hermitian matrices $A$, $B$ and $AB$,

Then there are additional cases when matrices $\mathcal{A}$ and $\mathcal{B}$ commute:

• https://en.wikipedia.org/wiki/Commuting_matrices?wprov=sfla1

But nothing can be said about $n \times n$ matrices in general. So nothing really can be deduced for the case $ABC$ and $BAC$ in general.

If it were given $A,B,AB$ are Hermitian, then $A,B$ commute and hence $ABC = BAC$

To generalise this further (but not to any $n \times n$ matrices), if $A,B$ are simultaneously diagonalisable (of which Hermitian matrices $A,B,AB$ is a special case), then $A,B$ commute (mentioned in the Wikipedia link)