Trace conserves singularity for certain case

Question

Suppose $A, B$ are matrices and $A\sim D$.

Then does $\text{tr}(ABA^{-1}B^{-1})=\text{tr}(DBD^{-1}B^{-1})$ hold?

I think this works because $\text{tr}(AB)=\text{tr}(BA)$, but it isn't easy to prove.

Answer 1

No. It should be easy to construct a counterexample. E.g. when $B$ commutes with $A$ but not $D$, the commutator $ABA^{-1}B^{-1}$ is always equal to $I$ but $DBD^{-1}B^{-1}$ is more arbitrary and the equality in question will likely fail. Here is a concrete counterexample on any field of characteristic $\ne2$: \begin{aligned} &A=B=B^{-1}=\pmatrix{0&1\\ 1&0},\ D=\pmatrix{1&0\\ 0&-1},\\ &ABA^{-1}B^{-1}=I,\\ &DBD^{-1}B^{-1}=-I,\\ &\operatorname{tr}(ABA^{-1}B^{-1})=2\ne-2=\operatorname{tr}(DBD^{-1}B^{-1}). \end{aligned} On a field of characteristic $2$, replace the $D$ above by $D_1=\pmatrix{1&1\\ 0&1}$. Then $A\sim D_1$ and $\operatorname{tr}(ABA^{-1}B^{-1})=0\ne1=\operatorname{tr}(D_1BD_1^{-1}B^{-1})$.