How to prove $\rm{tr}(A)=\rm{tr}(B)$ for real $2018\times2018$-matrices such that $A^{2018}=I=B^{2018}$ and $AB=BA$ and $\rm{tr}(AB)=2018$

Question

Let $A,B$ are two real square matrices of order 2018 such that $A^{2018}=I=B^{2018}$ and $AB=BA$ and $\operatorname{tr}(AB)=2018$. Prove that $\operatorname{tr}(A)=\operatorname{tr}(B)$.

Answer 1

We give a counter-example

Let $\theta=2\pi/2018,U=\begin{pmatrix}\cos(3\theta)&\sin(3\theta)\\-\sin(3\theta)&\cos(3\theta)\end{pmatrix},V=\begin{pmatrix}2\cos(5\theta)-1&c\\b&1\end{pmatrix}$, with the conditions $\det(V)=1,Trace(UV)=2$, that is

$2\cos(5\theta)-2-bc=0,2\cos(5\theta)\cos(3\theta)+b\sin(3\theta)-c\sin(3\theta)-2=0$.

The condition $Trace(U)\not= Trace(V)$ is fulfilled.

We find $b\approx 0.025944824003781216032,c\approx -0.0093410978477481362394$.

Now, take $A=diag(U,I_{2016}),B=diag(V,I_{2016})$.