TITLE: If $A$ is a complex matrix of size $n$ of finite order then is $A$ diagonalizable ?
QUESTION [12 upvotes]: Let $A$ be a complex matrix of size $n$ if for some positive integer $k$ , $A^k=I_n$ , then is $A$ diagonalizable ?

REPLY [3 votes]: Nothing wrong with the other answers. I simply want to record the following simple argument specific to this particular case. Also, this can be seen as an application of the representation theory of cyclic groups as well as discrete Fourier analysis.
Let $\omega=e^{2\pi i/k}$ be primitive $k$th root of unity in $\Bbb{C}$. Diagonalizability of a matrix $A$ is equivalent to every vector being a sum of eigenvectors of $A$. So let $x\in\Bbb{C}^n$ be arbitrary.
For all indices $\ell=0,1,\ldots,k-1$, consider the vector
$$
x_\ell=\frac1k(x+\omega^\ell Ax+\omega^{2\ell}A^2x+\cdots+\omega^{(k-1)\ell} A^{k-1}x).\tag{1}
$$
As $A^k=I$, applying $A$ cyclically permutes the vectors $x, Ax, A^2x,\ldots,A^{k-1}x$, so we easily see that
$$
Ax_\ell=\omega^{-\ell} x_\ell
$$
for all $\ell$. In other words, whenever $x_\ell\neq0$ it is an eigenvector of $A$ belonging to the eigenvalue $\omega^{-\ell}$.
Furthermore, if $0<\ell<k$ we have
$$
1+\omega^\ell+\omega^{2\ell}+\cdots+\omega^{(k-1)\ell}=0.\tag{2}
$$
This can be seen with the aid of the formula for a geometric sum. Or by using the fact that $\omega^\ell\neq1$ is a zero of $(z^k-1)/(z-1)=1+z+z^2+\cdots+z^{k-1}$.
Using the equation $(2)$ when summing $\sum_\ell x_\ell$ of $(1)$ thus yields
$$
\begin{aligned}
x_0+x_1+x_2+\cdots+x_{k-1}&=
\frac1k \sum_{j=0}^{k-1}\left(\sum_{\ell=0}^{k-1}\omega^{\ell j}\right)A^jx\\
&=\frac kk x+\sum_{j=1}^{k-1}\left(\sum_{\ell=0}^{k-1}\omega^{\ell j}\right)A^jx\\
&=x.
\end{aligned}
$$
This shows that $x$ is a sum of eigenvectors of $A$, and proves the main claim.

This is an extension of the common trick of getting eigenvectors of an order two operator. The formula
$$
f(x)=\frac12(f(x)+f(-x))+\frac12(f(x)-f(-x))
$$
writing an arbitrary function as a sum of an even function and an odd function is, perhaps, the earliest encounter of this. In that case the operator is $T:f(x)\mapsto f(-x)$, we have $T^2=id$ and $-1$ is the second root of unity. The even (respectively odd) functions being eigenvectors of $T$ belonging to $\lambda=+1$ and $\lambda =-1$ respectively.