# The minimal polynomial of $A$ divides $x^{2013} -1$. Prove that $A$ is diagonalizable over the complex field.

$A$ 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?

• 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). – Jyrki Lahtonen Feb 11 '14 at 8:10
• Alternatively, over $\mathbb C$ one can just say what the roots of $x^{2013} - 1$ are and observe that there are $2013$ different roots. – Jim Feb 11 '14 at 8:17
• @Jyrki, I've never thought of this gcd characterization for roots of polynomials! Quite interesting! – Christopher A. Wong Feb 11 '14 at 9:02
• @Jim how can I know that there are 2013 different roots? – CnR Feb 11 '14 at 9:41
• 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. – LutzL Feb 11 '14 at 16:52

(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 $A=I$.