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?

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    $\begingroup$ @Ram A is invertible, but is $A-I$? $\endgroup$ Sep 27 '13 at 19:16
  • $\begingroup$ @ThomasAndrews, yes yes I got it. $\endgroup$
    – Ram
    Sep 27 '13 at 19:18

Let $P(X):=X^n-1$. Assume we work on an algebraically closed field $\mathbb K$ of characteristic $0$. We have, since $P$ kills $A$, that the minimal polynomial of $A$ splits on $\mathbb K[X]$ and has distinct roots. We conclude by Theorem 4.11.

  • $\begingroup$ But is this still true if the field is not algebraically closed? like here if $A \in M_m(\mathbb R)$ $\endgroup$
    – Ram
    Sep 27 '13 at 19:17
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    $\begingroup$ I think your proof only works in characteristic 0 ;) $\endgroup$
    – N. S.
    Sep 27 '13 at 19:20
  • $\begingroup$ @N.S. Yes, I've added that. $\endgroup$ Sep 27 '13 at 19:37
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    $\begingroup$ My algebra is very very rusty, so I am not sure about this: I think in positive characteristic $X^n-1$ splits if and only if $p$ doesn't divide $n$. So in that situation the proof would still work, wouldn't it? $\endgroup$
    – N. S.
    Sep 27 '13 at 20:01
  • $\begingroup$ Question: how does this solution change if the vs is not finite dimensional? $\endgroup$
    – HJ32
    Dec 18 '13 at 18:08

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.


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