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?

  • 1
    $\begingroup$ @Ram A is invertible, but is $A-I$? $\endgroup$ – Thomas Andrews 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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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. $\endgroup$ – Davide Giraudo Sep 27 '13 at 19:37
  • $\begingroup$ 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.