Let $A,B$ be two $2 \times 2$ matrices over some finite field $\mathbb{F}_q$, such that they have the same trace and determinant. Does this imply that tr $A^k$ = tr $B^k$ for any integer $k$?
I've checked the case $k=2$, which works and makes use of the fact that they have the same determinant as well as the same trace, but can't seem to generalise this to an induction proof. Do I need stronger hypotheses for this?
Note that the answer is trivially true if the matrices are similar but I am more interested whether it holds even if they are not similar.
Best Answer
Yes, it is not difficult to see. By hypothesis, your matrices have same determinant and same trace; since the order of $A,B$ is $2$, then they have the same characteristic polynomial; this means they have the same eigenvalues.
It is a standard fact in linear algebra that the trace is the sum of eigenvalus; moreover the power of a generic matrix $M$ has eigenvalues which are the same power of the eigenvalues of $M$. So $A^k$ and $B^k$ have the same eigenvalues; then they have the same trace.