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I have a quite more general problem that I have reduced to the following question, which I don have any idea how to attack.
Let $A$ and $B$ be square matrices s.t. $AB$ has eigenvalue 1, take $P$ an orthogonal matrix, is 1 still an eigenvalue for $APBP^{-1}$. Taking $B$ a diagonal matrix this is trivially true, so it is not far-fetched to think this is possible, if not I would like a counterexample if possible. Thank in advance.

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    $\begingroup$ Is it trivially true for $B$ diagonal? We can get any permutation of a diagonal $B$ as a $PBP^{-1},$ so we have when $$A=B=\begin{pmatrix}1&0\\0&2\end{pmatrix}$$ and $$P=P^{-1}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$ then $AB$ is diagonal with entries $1,4$ and $APBP^{-1}$ is diagonal with entries $2,2.$ $\endgroup$ Jun 28 at 12:40
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    $\begingroup$ Maybe you meant when $B=\lambda I,$ a constant diagonal matrix? Then it is trivially true. $\endgroup$ Jun 28 at 12:44
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    $\begingroup$ Yeah I expressed myself the wrong way, but thanks very much for your example! I was actually rooting for my question not to have a positive answer and this shows it. $\endgroup$
    – kvicente
    Jun 28 at 12:47

2 Answers 2

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If $$A=B=\begin{pmatrix}1&0\\0&2\end{pmatrix},P=P^{-1}= \begin{pmatrix}0&1\\1&0\end{pmatrix}$$ then $$AB=\begin{pmatrix}1&0\\0&4\end{pmatrix}, APBP^{-1}=\begin{pmatrix}2&0\\0&2\end{pmatrix}$$

So it is not even true for $B$ diagonal, only true for $B=\lambda I,$ or diagonal matrix with a constant on the diagonal.

If $A$ and $B$ are diagonal $n\times n$ matrices, it is true for permutation matrices, $P,$ if there is an eigenvalue $\lambda$ of $A$ repeated $k$ times such that $\lambda^{-1}$ is an eigenvalue of $B$ at least $n-k+1$ times.

But the permutation matrices are a very small part of the set of orthogonal matrices. I’d bet, given $A,B,$ you can always find a counterexample $P$ unless $A=\lambda I$ or $B=\lambda I$ for some $\lambda.$

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  • $\begingroup$ Yes, this is easier (+1)! $\endgroup$ Jun 28 at 17:35
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This is certainly not true already for $2\times 2$-matrices. Here is an example with non-diagonal matrices. $$ A=\begin{pmatrix} 1 & 2 \cr 3 & 7 \end{pmatrix}, \; B=\begin{pmatrix} 7 & -2 \cr -3 & 1 \end{pmatrix}, \; P=\frac{1}{3}\cdot \begin{pmatrix} 1 & \sqrt{8} \cr \sqrt{8} & -1 \end{pmatrix}, \; $$ Then $AB=I$ has both eigenvalues $1$, but $$ APBP^{-1}=\frac{1}{9}\begin{pmatrix} 14\sqrt{2}-11 & 4(8\sqrt{2} + 23) \cr 2(27\sqrt{2} - 23) & 106\sqrt{2} + 333 \end{pmatrix} $$ has no eigenvalue $1$.

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  • $\begingroup$ Thanks @DietrichBurde for taking the time to write a formal answer, I think is good to have it on record for the future, someone may need it $\endgroup$
    – kvicente
    Jun 28 at 12:49
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    $\begingroup$ +1, but that example is way more complicated than necessary. 🤓 $\endgroup$ Jun 28 at 13:10
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    $\begingroup$ @ThomasAndrews Yes, of course. But your example is hard to beat, so I took a complicated one with $AB=I$. It shows that $APBP^{-1}$ is "far away" from the identity matrix. $\endgroup$ Jun 28 at 13:14

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