Show that $A^m=I_n$ is diagonalizable

Question

Let $A \in M_n(\mathbb{C})$ such as $A^m=I_n$ or $A^m=-I_n , m\ge1$. Show that A is diagonalizable.

Answer 1

This follows because the polynomials $t^n-1$ and $t^n+1$ both have $n$ distinct roots. See here for example.

Answer 2

Consider the Jordan normal form $J$ of $A$. Then $J^m=\pm\operatorname{Id}_n$ and therefore each block of $J$ will be a $k\times k$ matrix of the type$$\begin{bmatrix}\omega&1&0&0&\ldots&0\\0&\omega&1&0&\ldots&0\\0&0&\omega&1&\ldots&0\\0&0&0&\omega&\cdots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\ldots&\omega\end{bmatrix},$$where $\omega$ is a root of unity. But then, unless $k=1$, no power of this matrix can be $\pm\operatorname{Id}_k$, because the entries of the line above the diagonal will all be equal to $k\omega^{k-1}$. Therefore, $J$ is a diagonal matrix.

Answer 3

Hint: Show that the minimal polynomial of $A$ has distinct roots. Since $A^m-I_n=0$ or $A^m+I_n=0$ , the minimal polynomial divides $x^m-1$ or $x^m+1$ which has distinct roots.