## Over CC, matrices of finite multiplicative order are diagonalizable

Let $A$ be an $n \times n$ matrix over $\mathbb{C}$. Show that if $A^k=I$ for some $k$, then $A$ is diagonalizable.

Let $F$ be a field of characteristic $p$. Show that $A = \begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix}$ has finite order but cannot be diagonalized over $F$ unless $\alpha = 0$.

Since $A^k = I$, the minimal polynomial $m$ of $A$ over $\mathbb{C}$ divides $x^k-1$. In particular, the roots of $m$ are distinct. Since $\mathbb{C}$ contains all the roots of unity, by Corollary 25 on page 494 of D&F, $A$ is diagonalizable over $\mathbb{C}$.

Note that $\begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \alpha+\beta \\ 0 & 1 \end{bmatrix}$. By an easy inductive argument, then, $\begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix}^t = \begin{bmatrix} 1 & t\alpha \\ 0 & 1 \end{bmatrix}$, and in particular, $A^p = I$.

Suppose $\alpha \neq 0$. Now $\frac{1}{\alpha}A$ is in Jordan canonical form, and is not diagonalizable. (See Corollary 24 on page 493 of D&F.) So $A$ cannot be diagonalizable, for if it were, then so would $\frac{1}{\alpha}A$. (If $P^{-1}AP = D$ is diagonal, then so is $P^{-1}\frac{1}{\alpha}AP = \frac{1}{\alpha}D$.)