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Let $A$ be a $n\times n$ matrix over $\mathbb{C}$. If $A^n$ = $I$ for some $n>0$, then how to show that $A$ can be diagonalized? The hint I have is to show that the characteristic polynomial has no repeated root.

3x89g2
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  • The characteristic polynomial can have multiple roots in this case, think about $I_2.$ I think the hint is about the minimal polynomial, as Mariano answered you. – Balloon Sep 05 '16 at 22:55
  • I am almost sure that I have written the same answer in some other question… Have you searched before asking? – Mariano Suárez-Álvarez Sep 05 '16 at 22:57
  • @MarianoSuárez-Álvarez Yes I do. The thing with searching though is, say I searched $A^n$ or with MathJax, I usually get zero result, not sure why. – 3x89g2 Sep 05 '16 at 23:00
  • http://math.stackexchange.com/questions/1013429/if-a-is-a-complex-matrix-of-size-n-of-finite-order-then-is-a-diagonalizabl and a few others. – Mariano Suárez-Álvarez Sep 05 '16 at 23:05

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If you evaluate the polynomial $X^n-1$ at $A$ you get zero. It follows that the minimal polynomial of $A$ divides $X^n-1$. Since the latter has all its roots simple —because, for example, it is coprime with its derivative— the same is true of the minimal polynomial.

Now it is a standard fact that a matrix is diagonalizable iff its minimal polynomial has all its roots simple, so...

Mariano Suárez-Álvarez
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