Suppose $A$ a real matrix of dimension $2$, having two not real conjugate eigenvalues. I want to prove if $B$ is another real $2\times 2$ matrix commuting with $A$, then the eigenvalues of $B$ are conjugate (same norm).
Maybe the proof relies on a manipulation of entries $a_{i,j}$ and $b_{i,j}$ but the conditions are not easy to extract. Thanks for ideas.
If $A$ has conjugate eigenvalues then it has in some basis a representation $PAP^{-1}=A_a=\begin{bmatrix} a & -b \\ b & a \end{bmatrix} $.
Let $B_a$ be any matrix $\begin{bmatrix} c & d \\ e & f \end{bmatrix} $ which is commutative with $A_a$ in this basis
i.e. $A_aB_a=B_aA_a$.
Compare $\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} c & d \\ e & f \end{bmatrix} = \begin{bmatrix} c & d \\ e & f \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$
and you should obtain conditions for $c,d,e,f$
If the representation is of the form like $A_a$ ( $c=f$ and $d=-e$) then you'll achieve the desired result because eigenvalues are constant under a change of basis (so their property of conjugation) and commutation for both matrices holds also under a change of basis.