TITLE: Show that if $A^{n}=I$ then $A$ is diagonalizable.
QUESTION [13 upvotes]: Suppose $A$ is an $m \times m$ matrix which satisfies $A^{n}=1$ for some $n$, then why is $A$ necessarily diagonalizable.
Not sure if this is helpful, but here's my thinking so far: We know that $A$ satisfies $p(x)=x^{n}-1=(x-1)(x^{n-1}+\ldots+x+1)$.  If $A=I$ it is clearly diagonalizable so we may assume that $A$ is a root of the other factor. 
Edit: Actually, I'm a bit confused and not even sure if we can say that much.  Since the ring of $m \times m$ matrices is not an integral domain, we can not conclude that if $A-I \not = 0$ then $(A^{n-1}+\ldots+A+I)=0$, correct?

REPLY [6 votes]: Davide showed what happens in an algebraically closed field of characteristic $0$.
If the field is not algebraically closed, the result is not true, for example
$$A=\begin{pmatrix}
0 & 1\\
-1 & 0
\end{pmatrix}$$
Satisfies $A^4=I$ but is not diagonalizable over $\mathbb R$, as it has complex eigenvalues. 
Also 
$$A=\begin{pmatrix}
1& 1\\
0 & 1
\end{pmatrix}$$
is not diagonalizable in an algebraically closed field of characteristic $2$, but $A^2=I_2$. Note that the reason why $A$ is not diagonalizable is simple: both eigenvalues are $1$, thus if $A$ is diagonalizable, $D=I$ and thus $A=PDP^{-1}=I$ contradiction.