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.)

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