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 $ tr(A)= tr(A^j)$.

Question

I saw this question and answer at following site.

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The answer at this site is here.

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 my question : Why $\{\sigma(\lambda_i)\;|\;p(\lambda_i) =0 \} =\{\lambda_i\in\mathbb{C}\;|\;p(\lambda_i) =0\}$ is correct?

I think this needs that $p(t)$ has multiplicity $1$ and $n=k$.

For example, $n=k=5$, $ H=\{\zeta\in\mathbb{C}\;|\;\zeta^k=1\}=\{\;\zeta^1,\;\zeta^2,\;\zeta^3,\;\zeta^4,\;1\;\} $ and let $\sigma(\omega)= \omega^3$.

Then $\sigma(H) = \{\;\zeta^1,\;\zeta^2,\;\zeta^3,\;\zeta^4,\;1\;\}$

But when $n < k$ and has not multiplicity $1$, following is possible.

For example , If eigen value $\lambda_i = \zeta^1, \zeta^1,\zeta^2$ , then $\sigma(\lambda_i) = \zeta^3, \zeta^3,\zeta^1$. That is, $\{\sigma(\lambda_i)\;|\;p(\lambda_i) =0 \} \neq \{\lambda_i\in\mathbb{C}\;|\;p(\lambda_i) =0\}$

How to prove this?

Answer 1

We need following observations:

• The polynomial $p(t)$ has a unique factorization $p(t)=\prod_{i=1}^n(t-\lambda_i)$. So the set of roots of $p(t)$ is unique counted with multiplicity.

• Since $\sigma \in\text{Aut}(\Bbb Q(\mu_k)/\Bbb Q)$ fixes $\Bbb Q$ and $p(t)$ is a rational polynomial, say $p(t)=t^n+a_{n-1}t^{n-1}+$$\cdots +a_0$ where all $a_i\in \Bbb Q$, $p(t)$ is fixed by $\sigma$. That is, $\sigma(p(t))=t^n+\sigma(a_{n-1})t^{n-1}+\cdots +\sigma(a_0)=p(t)$.

• If $p(t)$ is factorized into $\prod_{i=1}^n(t-\lambda_i)$, then $\sigma(p(t))$ is factorized into $$\sigma\left[\prod_{i=1}^n(t-\lambda_i)\right]=\prod_{i=1}^n(t-\sigma(\lambda_i))=\prod_{i=1}^n(t-\lambda_i^j).$$
  

Thus it should be that $p(t)=\prod_{i=1}^n(t-\lambda_i)=\prod_{i=1}^n(t-\lambda_i^j)$ and $$ \{\lambda_i:i=1,2,\ldots,n\}=\{\lambda_i^j:i=1,2,\ldots,n\} $$ counted with multiplicity.

Note: Multiplicity of $\lambda_i$'s needs not be $1$, it can be any integer $m\ge 1$ and $n=mk$. Your scenario in which $\{\lambda_i\}=\{\zeta,\zeta,\zeta^3\}$ is not plausible because $$ (t-\zeta)^2(t-\zeta^3) $$ is not a rational polynomial.