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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.

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    $\begingroup$ Same trace and determinant for $2\times 2$ means same characteristic polynomial, i.e. same eigenvalues. So $A^k$ and $B^k$ have the same eigenvalues too ($k^\text{th}$ powers of the original ones). Thus the traces (and deteminants) are the same again. $\endgroup$ – A.Γ. Jul 27 '15 at 11:08
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    $\begingroup$ @A.G. Does this break down then if I go up to $n \times n$ matrices since we need more information to get the eigenvalues? $\endgroup$ – Matt B Jul 27 '15 at 11:11
  • $\begingroup$ Correct. Then the characteristic polynomial has more coefficients than the trace and the determinant. $\endgroup$ – A.Γ. Jul 27 '15 at 11:13
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Yes, 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.

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  • $\begingroup$ And what about the case, when they are not diagnoziable? $\endgroup$ – Lurco Jul 27 '15 at 12:12
  • $\begingroup$ It bears mentioning that this approach requires that we make use of the algebraic closure of a finite field $\endgroup$ – Omnomnomnom Jul 27 '15 at 13:04
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Help: (please try proving these yourself)

  • $\operatorname{trace}(A)=\lambda_1 + \dots + \lambda_N$
  • $\lambda(A^k)=\{\lambda_1^k,\dots,\lambda_N^k\}$
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