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The following question is almost the converse of this question.

Let $A$ and $B$, $A \neq B$, be two $2$ by $2$ matrices over a field of characteristic zero. Assume that one of the eigenvalues of $AB$ (and hence also of $BA$) is $1$.

What can be said about the eigenvalues of $A$ and $B$?

Any hints and comments are welcome!

user26857
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user237522
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  • Almost nothing can be said, except that the eigenvalues of $A$ and $B$ are nonzero when $AB$ is nonsingular, or at least one of $A,B$ has a zero eigenvalue when $AB$ is singular. Consider the case where $A$ is a nonsingular complex matrix and $B=A^{-1}$. All nonzero complex numbers are possible eigenvalues of $A$. – user1551 Jun 30 '20 at 10:00
  • @user1551, thank you! I liked the example with $B=A^{-1}$. What about the eigenvectors of $A$, $B$ and $AB$? Any relation between them? – user237522 Jun 30 '20 at 10:03
  • I haven't any nice illustrative examples at hand, but in general, the eigenvectors or eigenvalues of $A$ and $B$ have very little to do with the eigenstructure of $AB$. Conjugation (i.e. $X\mapsto PXP^{-1}$) preserves eigenstructure, but the mere multiplication of $A$ and $B$ does not preserve the eigenstructure of any one of $A$ and $B$. – user1551 Jun 30 '20 at 10:17
  • @user1551, thank you! So only when $A=B$ more can be said, like in the following question https://math.stackexchange.com/questions/241764/eigenvalues-and-power-of-a-matrix?noredirect=1&lq=1 ? – user237522 Jun 30 '20 at 10:31
  • There are other nice scenarios. When $A$ and $B$ commute and the underlying field is algebraically closed, the two matrices can be simultaneously triangularised. In this case, $A,B$ and $AB$ share at least one eigenvector and every eigenvalue of $AB$ is a product of an eigenvalue from $A$ and an eigenvalue from $B$. If both $A$ and $B$ are diagonalisable, $A,B$ and $AB$ share even an eigenbasis. – user1551 Jun 30 '20 at 11:10
  • Thank you very much! – user237522 Jun 30 '20 at 11:41

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