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Preparation for a grad school linear algebra exam. This question has been on multiple past exams, so I should be able to do it, but I don't really know where to start. $V$ is a vector space of finite dimension on an algebraically closed field with null characteristic. $f$ is an endomorphism, and $f^m=\operatorname{id}(V)$. I'm supposed to show that $f$ is diagonalizable.

I have that $f^m-\operatorname{id}=0$, so $x^m-1$ is a "polynome annulateur" of $f$ (nullifiying polynomial??). $V$ algebraically closed means it's factorizable, therefore the minimum polynomial is as well and $f$ is triagonalisable. The hint is to use the Jordan normal form.

Any ideas?

Thank you in advance.

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Do you know that $f$ is diagonalizable if the minimal polynomial splits? – user27126 Jan 13 '13 at 21:28
This doctoral dissertation translates polynôme annulateur as annihilator polynomial. – Brian M. Scott Jan 13 '13 at 21:45
@Sanchez, I believe f is diagonalizable if the minimal polynomial splits into distinct factors, but I just have splits, not necessarily distinct factors. – JKH Jan 13 '13 at 21:49
@JKH, if you are talking about complex numbers, then $x^m-1$ splits into distinct factors indeed. – user27126 Jan 13 '13 at 21:52
The field isn't specified as C, but it's algebraically closed and of null characteristic, so maybe that works here as well? – JKH Jan 13 '13 at 21:57

You do not need to use Jordan normal form here.

Since the polynomial $X^m-1=\prod_{j=1}^m(X-\lambda_j)$ has pairwise distinct roots and annihilates $f$, we have: $$ V = \bigoplus_{j=1}^m \mbox{Ker} (f-\lambda_j Id). $$

I am simply using the following lemma: if $P_1$ and $P_2$ are relatively prime polynomials, then $$ \mbox{Ker} (P_1P_2)(f)=\mbox{Ker}P_1(f)\oplus \mbox{Ker}P_2(f). $$

Since you seem to speak french, here is a reference:

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I'm sorry, I must be missing something; how do we know X^m−1 has pairwise distinct roots? Couldn't it have roots of higher degree? – JKH Jan 13 '13 at 21:54
@JKH In $\mathbb{C}$, these roots are the $\exp(2ik/m)$ for $k=0, 1, \ldots, m-1$. These are the $m$ $m$-roots of unity. They form the vertices of a regular $m$-gon. – 1015 Jan 13 '13 at 22:09
@JKH In a general algebraically closed field of characteristic zero, observe that $P(X)=X^m-1$ and $P'(X)=mX^{m-1}$ are relatively prime. It follows that $P$ splits into the product of $m$ pairwise distinct linear factors. – 1015 Jan 13 '13 at 22:13
Thank you, that's the piece of information I was missing. – JKH Jan 14 '13 at 8:38

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