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As discussed in this other post, $AB$ and $BA$ always have the same characteristic polynomial, but not necessarily the same minimal polynomial.

This means that $AB$ diagonalisable does not imply $BA$ diagonalisable, as shown by the example $$A=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ On the other hand, we know that if both $A,B$ are invertible, then the statement is true.

As mentioned here, it is also the case that the non-zero spectrum of $AB$ and $BA$ is always the same.

I am looking for an answer to the slightly different question: assuming $A,B$ are both diagonalisable, is it true that $AB$ diagonalisable implies $BA$ diagonalisable? The counterexamples I've seen (the post linked above and this one) all involve at least one non-diagonalisable matrix, hence my question.

If the above is not true, what about the more restrictive case of $A,B$ both normal? Or maybe the case with $AB,BA\neq0$?

glS
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1 Answers1

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Counterexample: \begin{aligned} AB&=\pmatrix{1&0&0\\ 0&0&0\\ 0&0&1}\pmatrix{0&0&0\\ 1&1&0\\ 0&0&1}=\pmatrix{0&0&0\\ 0&0&0\\ 0&0&1},\\ BA&=\pmatrix{0&0&0\\ 1&1&0\\ 0&0&1}\pmatrix{1&0&0\\ 0&0&0\\ 0&0&1}=\pmatrix{0&0&0\\ 1&0&0\\ 0&0&1}. \end{aligned}

user1551
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  • well, this covers all bases, but still also kind of follows the same patterns as the other cases. You fix one of the two matrices not being diagonalisable by using $\begin{pmatrix}0&0\1&1\end{pmatrix}$ rather than $\begin{pmatrix}0&0\1&0\end{pmatrix}$. This however still results in $AB=0$, and so you embed the matrices into a larger one acting trivially on a third mode to fix it. I wonder, can you also find an example with two diagonalisable not mutually diagonalisable matrices? This would make the $AB,BA\neq0$ property more "genuine" – glS Feb 22 '20 at 18:25
  • @glS I suspect that this is impossible, but I haven't any proof yet. – user1551 Feb 23 '20 at 09:07