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

Let $$A\in M_n(\mathbb{Q})$$ 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.

## closed as off-topic by Saad, Namaste, Holo, Arnaud D., RRLJan 10 at 17:28

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## 2 Answers

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 $$\{\lambda_i\}\subset \mu_k =\{\zeta\in\mathbb{C}\;|\;\zeta^k=1\}$$. Let $$\omega$$ be the $$k$$-th primitive root of unity. Since $$(j,k)=1$$, we can define $$\sigma\in \text{Aut}(\mathbb{Q}(\mu_k)/\mathbb{Q})$$ by letting $$\sigma(\omega)= \omega^j.$$ Then, it holds $$p(t) = \sigma(p(t)) = \sigma\left[\prod_{i=1}^n (t-\lambda_i)\right]=\prod_{i=1}^n (t-\sigma(\lambda_i))=\prod_{i=1}^n (t-\lambda_i^j).$$ Now, it follows $$\text{tr}(A) = \sum_{i=1}^n \lambda_i = \sum_{i=1}^n \lambda_i^j = \text{tr}(A^j).$$

• This is a nice answer that also nicely shows why we do need to do this over $\mathbb{Q}$, since it will fail in suitable extensions. – Tobias Kildetoft Jan 10 at 11:46
• I think this answer is wonderful..... I sincerely appreciate you! – w.sdka Jan 10 at 12:00
• why does an eigenvalue $\lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $\lambda_i^j$ of $A^j$? – M. Van Jan 10 at 13:19
• @M.Van Because $\{\lambda_i\} = \{\lambda_i^j\}$ ..? Equality not as a set, but counted with multiplicity. – Song Jan 10 at 13:32
• Umm... Could you explain why $\lambda_i \in \mu_k$?.... why $\prod_{i=1}^n \lambda_i^k =1$ implies $\lambda_i \in \mu_k$?? – w.sdka Jan 10 at 13:43

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

$$\textbf{Proposition.}$$ When $$A\in M_n(\mathbb{Q})$$ 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 $$j. Clearly, $$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 $$\phi_a(x)=\Pi_{(u,a)=1}(x-e^{2i\pi u/a})\in\mathbb{Q}[x]$$ where $$a|k$$. Since $$(j,a)=1$$, it is easy to see that $$\phi_a(x)=\Pi_{(u,a)=1}(x-(e^{2i\pi u/a})^j)$$ and we are done. $$\square$$