If $A$ is hermitian matrix s.t. $A^{99}=I_n$, show that $A=I_n$.

Question

Let $A \in M_n(\mathbb{C})$ be a hermitian matrix that satisfies the property $A^{99}=I_n$. I want to show that $A=I_n$.

By definition, $A$ is a hermitian matrix if $a_{ij}=\overline{a_{ji}}$.

From the fact that $A^{99}=I_n$, we deduce that $A$ is invertible and its inverse is $A^{98}$. Does that help to get that $A=I_n$ ? Or do we use somehow the definition of a hermitian matrix?

Answer 1

$A$ is hermitian so $\sigma(A) \subseteq \mathbb{R}$. On the other hand, the polynomial $z^{99} - 1$ annihilates $A$ so

$$\sigma(A) \subseteq \{\text{zeroes of }z^{99}-1\} \cap \mathbb{R} = \left\{e^{i\frac{2k\pi}{99}} : k \in \{0, \ldots, 98\}\right\} \cap \mathbb{R} = \{1\}$$

so $\sigma(A) = \{1\}$.

Since $A$ is diagonalizable, it diagonalizes to the indentity matrix so $A = I$.