Eigenvalues and roots of unity

Question

Let $A \in \mathcal{M}_{n}(\mathbb{C})$ such that $A^{n} = \mathrm{I}_{n}$ and the family $(\mathrm{I}_{n},\ldots,A^{n-1})$ is linearly independent. I would like to prove that $\mathrm{Tr}(A) = 0$.

Here is what I tried : since $X^{n}-1$ is a null polynomial of $A$ and the roots of $X^{n}-1$ have all multiplicity equal to $1$, the minimal polynomial of $A$ has simple roots and $A$ is diagonalizable. If $\mathrm{Sp}(A)$ denotes the set of complex eigenvalues of $A$, then $\mathrm{Sp}(A) \subset \mathbb{U}_{n}$, where :

$$ \mathbb{U}_{n} = \left\{ \exp\Big( \frac{2ik \pi}{n} \Big), \; k = 0,\ldots,n-1 \right\}. $$

If I could prove that $\mathbb{U}_{n} \subset \mathrm{Sp}(A)$, then it would be easy to see that $\mathrm{Tr}(A) = 0$. I don't know how to prove this inclusion.

Answer 1

Hint: Since $A^n - I = 0$, the minimal polynomial of $A$ divides $x^n - 1$. Since $(I,\dots,A^{n-1})$ are linearly independent, the minimal polynomial must have degree $n$.

We conclude that the minimal polynomial of $A$ is $q(x)= x^{n} - 1$.

Answer 2

Hint: for any square matrix $\;n\times n\;,\;\;Tr. A\;$ is the coefficient of $\;x^{n-1}\;$ in the characteristic polynomial of $\;A\;$ .

Another hint in the same spirit as above: the trace of $\;A\;$ is the sum of all its eigenvalues, which are the roots of the charac. polynomial, and then...

Answer 3

Since the family $(I,A,\ldots,A^{n-1})$ is linearly independent then there's no non trivial linear combination of these matrices that gives $0$ so the polynomial $P=x^n-1$ which annihilates $A$ has the minimal degree so it's the minimal polynomial and the characteristic polynomial so their roots: the $n$- roots of the unity are the eigenvalues of $A$ and their sum i.e. the trace of $A$ is the coefficient of $x^{n-1}$ in $P$ so it's $0$.