# Eigenvalues of a composition of matrices

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.

• 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 $2,2.$ Jun 28 at 12:40
• Maybe you meant when $B=\lambda I,$ a constant diagonal matrix? Then it is trivially true. Jun 28 at 12:44
• 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. Jun 28 at 12:47

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 $$\lambda.$$

• Yes, this is easier (+1)! Jun 28 at 17:35

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 $$1$$.

• 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 Jun 28 at 12:49
• +1, but that example is way more complicated than necessary. 🤓 Jun 28 at 13:10
• @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. Jun 28 at 13:14