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Let $A$ be an $n\times n$ complex matrix. If $A^m=I_n$ for some positive integer $n$. How to show that $A$ is diagonalizable?

Ronald
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1 Answers1

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Hint

What we can say for the polynomial $x^m-1$?

  • $x=\pm 1$.. I think i got it – Ronald Aug 22 '13 at 19:52
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    @Danial A little more! You're in $\Bbb C$, not (algebraically) boring $\Bbb R$! – Pedro Aug 22 '13 at 19:53
  • @Danial What if $m$ is odd? – rurouniwallace Aug 22 '13 at 19:58
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    @Danial so as not to tease you - what you should notice is that the polynomial has no multiple roots – user8268 Aug 22 '13 at 20:02
  • I am trying the following: $A=P^{-1}JP\Rightarrow A^m=P^{-1}J^mP=I_n\Rightarrow J^m=I_n$ but i don't see what can i do further.. I don't know how to use the hint above .. i feel i stupid – Ronald Aug 22 '13 at 20:06
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    @Danial $A$ is diagonalizable iff the minimal polynomial splits into linear factors so with the given hypothesis of your question can we say that we have this condition? –  Aug 22 '13 at 20:12
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    @Peter Tamaroff: Not to be contentious, but I think $\Bbb R$ is at least as interesting as $\Bbb C$. That is to say, $\Bbb C$ is no more boring than $\Bbb R$. – Robert Lewis Aug 22 '13 at 20:15
  • @Peter Tamaroff: algebraically and/or otherwise! ;) – Robert Lewis Aug 22 '13 at 20:15
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    @Danial notice, as $x^m - 1$ has no duplicate roots, it is irreducible, and $0 = A^m - I = (x^m - 1)[A]$ by requirement. From this we have: 1. $\mu(x) = x^m - 1$ is irreducible, 2. $\mu(A) = 0$. $$$$ Conclusion? – AlexR Aug 22 '13 at 20:32
  • Yes thank you all .. I solved it :) – Ronald Aug 22 '13 at 20:35
  • @RobertLewis Sure. The above was merely a joke to hint on what the OP was missing. – Pedro Aug 22 '13 at 21:07
  • @Peter Tamaroff: As was my remark, albeit directed toward the general readership, since I admit it didn't really address the question at hand. Humor and math, great together or separately! – Robert Lewis Aug 22 '13 at 21:15
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    It's interesting to note that using this technique you can actually prove the somewhat stronger result: if $A$ is a complex matrix (or a matrix over any alg closed field with characteristic prime to the integer $m$ about to appear) such that $A^m$ is diagonalizable for some $m$, then $A$ is diagonalizable. This follows because $m_A(x)\mid m_{A^m}(x^m)$ and by assumption $m_{A^m}(x^m)$ is separable and splits in the field. – Alex Youcis Aug 22 '13 at 21:39