# Simultaneous similarity of matrices over finite fields

Suppose $$A,B\in SL(3,F_q)$$, where $$F_q$$ is the finite field of order $$q$$ and $$SL(3,F_q)$$, the group of matrices with determinant one and entries from $$F_q$$ , are such that $$A$$ has eigenvalues in $$F_q$$ and $$B$$ has eigenvalues in $$\overline{F_q}\setminus F_q$$. Also, $$A$$ is diagonalizable over $$F_q$$ and $$B$$ is diagonalizable over $$\overline{F_q}$$, the closure of $$F_q$$. I am trying to show that $$A$$ and $$B$$ are not simultaneously diagonalizable by a matrix $$P\in SL(3,\overline{F_q})$$ (i.e., $$\nexists P\in SL(3,\overline{F_q})$$ such that $$(PAP^{-1},PBP^{-1})$$ are diagonal). I am considering the approach of looking at the $$F_q$$ algebra generated by $$A$$ and $$B$$ and trying to show it is not isomorphic to the algebra generated by $$F_q[PAP^{-1},PBP^{-1}]$$. I am looking for some reference which might be helpful in proving the above. Most of the results I saw had been about irreducible representations which is not helpful for my case. I would appreciate it if you could suggest some reference that could be helpful. Thanks in advance for your time.

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• What about the case where $A$ is the identity matrix? Then they clearly are simultaneously diagonalizable. If $A$ has distinct eigenvalues then you should be OK. – Will Sawin 2 days ago

I think there is a simpler way to see what is going on: first of all, the hypotheses force the characteristic polynomial of $$B$$ to be irreducible of degree $$3$$ over $$F_{q}.$$ On the other hand, if $$PAP^{-1}$$ and $$PBP^{-1}$$ are both diagonal, then $$PAP^{-1}$$ and $$PBP^{-1}$$ certainly commute. Hence $$A$$ and $$B$$ already commute. Since $$B$$ has irreducible characteristic polynomial, (which is also its minimum polynomial in this situation), and since $$B$$ leaves every eigenspace of $$A$$ invariant (as $$A$$ and $$B$$ commute), this forces (given the hypotheses) $$A$$ to have the form $$\lambda I$$ for some $$\lambda \in F_{q}$$.
• So, if you exclude $A$ being a scalar matrix, what you want to be true indeed is. – Geoff Robinson 2 days ago
Your intuition is somewhat correct. The characteristic polynomial of $$B$$ is cubic over $$\mathbf F_q$$, and has no roots in $$\mathbf F_q$$ by assumption, hence is irreducible. Thus, the roots all live in $$\mathbf F_{q^3}$$, and $$\mathbf F_q[B] \cong \mathbf F_{q^3}$$. If $$A$$ and $$B$$ are simultaneously diagonalisable over $$\bar {\mathbf F}_q$$, then $$\mathbf F_q[A,B] \underset{\mathbf F_q}\otimes \bar{\mathbf F}_q \cong \bar{\mathbf F}_q[PAP^{-1},PBP^{-1}]$$ is a subalgebra of the diagonal matrices $$\bar{\mathbf F}_q^3$$, so a dimension count shows that the inclusion $$\mathbf F_q[B] \subseteq \mathbf F_q[A,B]$$ must be an equality. Since $$A$$ has eigenvalues in $$\mathbf F_q$$ and $$\mathbf F_q[B] \cong \mathbf F_{q^3}$$, this forces $$A$$ constant (think about which elements of $$\mathbf F_{q^3}$$ have characteristic polynomial totally split over $$\mathbf F_q$$).
(Conversely, if $$A$$ is constant, as Will Sawin noted, then clearly $$A$$ and $$B$$ are simultaneously diagonalisable over $$\bar{\mathbf F}_q$$).
• Thank you so much for answering. It was very helpful. I was considering $A$ to be the matrices that are not conjugate to the central elements. Sorry I forgot to mention that. – C.T. 2 days ago