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$A$ is a $n\times n$ real matrix. The minimal polynomial of A divides $x^{2013} -1$.

I need to prove that:

(1) A is diagonalizable over the complex field.

(2) If A is diagonalizable over the reals, then it must be the identity matrix.

At first I want to show that the minimal polynomial is consisted of different linear factors. I know that over the complex field every polynomial is decomposed to linear factors, but how can I prove that they are different?

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    $\begingroup$ Hint: The zeros of a polynomial $f(x)\in\Bbb{C}[x]$ are distinct, if and only if $$\gcd(f(x),f'(x))=1.$$ The same fact holds for all fields (when you think of the zeros in an algebraic closure of the field of coefficients). $\endgroup$ – Jyrki Lahtonen Feb 11 '14 at 8:10
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    $\begingroup$ Alternatively, over $\mathbb C$ one can just say what the roots of $x^{2013} - 1$ are and observe that there are $2013$ different roots. $\endgroup$ – Jim Feb 11 '14 at 8:17
  • $\begingroup$ @Jyrki, I've never thought of this gcd characterization for roots of polynomials! Quite interesting! $\endgroup$ – Christopher A. Wong Feb 11 '14 at 9:02
  • $\begingroup$ @Jim how can I know that there are 2013 different roots? $\endgroup$ – CnR Feb 11 '14 at 9:41
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    $\begingroup$ You can enumerate the roots explicitly as $z_k=\exp(2\pi i k/2013)$. Any factor of the polynomial can only have each root at most once. $\endgroup$ – LutzL Feb 11 '14 at 16:52
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(1) $x^{2013} -1$ has only simple roots in $\mathbb C$ and so must the minimal polynomial if $A$.

(2) The only real root of $x^{2013} -1$ is $1$ and so $A$ is similar to the identity matrix, and this can only happen when $A=I$.

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