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Let $Ain M_n(mathbb{Q})$ with $A^k=I_n$. If $j$ is a positive
integer with $gcd(j,k)=1$, show that $ operatorname{tr}(A)=
operatorname{tr}(A^j)$.

I don't know how to start to prove that. I tried to find the matrix $B$ similar with $A$, but I am stuck...

Thank you.

Note that eigenvalues $lambda_i, i=1,ldots, n$ of $A$ are roots of the rational polynomial
$$
p(t) = det(tI-A)=prod_{i=1}^n (t-lambda_i).
$$ Since $Ax=lambda_i x$ implies $A^kx = x = lambda_i^k x$, we have ${lambda_i}subset mu_k ={zetainmathbb{C};|;zeta^k=1}$. Let $omega$ be the $k$-th primitive root of unity. Since $(j,k)=1$, we can define $sigmain text{Aut}(mathbb{Q}(mu_k)/mathbb{Q})$ by letting
$$
sigma(omega)= omega^j.
$$ Then, it holds
$$
p(t) = sigma(p(t)) = sigmaleft[prod_{i=1}^n (t-lambda_i)right]=prod_{i=1}^n (t-sigma(lambda_i))=prod_{i=1}^n (t-lambda_i^j).
$$ Now, it follows
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$

We must use the fact that $Ain M_n(mathbb{Q})$ because the result is false over $M_n(mathbb{C})$. Indeed, consider $A=diag(e^{2ipi/3},i)$ where $k=12$; then $trace(A)not= trace(A^5)$.

$textbf{Proposition.}$ When $Ain M_n(mathbb{Q})$ and $(j,k)=1$, $spectrum(A)=spectrum(A^j)$.

$textbf{Proof}.$
We may assume that $order(A)=k$ ($k=min{l;A^l=I_2}$) and that $j<k$. Clearly, $order(A^j)=l$. $A$ is diagonalizable over $mathbb{C}$ and its minimal polynomial divides $x^k-1$.

Therefore, $chi_A$, the characteristic polynomial of $A$, is a product of (irreducible) cyclotomic polynomials $phi_a(x)=Pi_{(u,a)=1}(x-e^{2ipi u/a})inmathbb{Q}[x]$ where $a|k$. Since $(j,a)=1$, it is easy to see that $phi_a(x)=Pi_{(u,a)=1}(x-(e^{2ipi u/a})^j)$ and we are done. $square$