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Let $A\in M_{n}(C)$ and $A^{n}=2A$ with $n>1$. I need to show that $A$ is diagonalizable.

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    The title of your question seems misleading to me: the answer to the question asked in the title is clearly no, but in your question there is an additional hypothesis.2012-12-11
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    What happens when you use the Jordan decomposition?2012-12-11

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The polynomial $P=X^n-2X=X(X^{n-1}-2)$ is separable and has $n$ distinct roots over $\mathbb{C}$ and $P(A)=0$ so $A$ is diagonalizable over $\mathbb{C}$.

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    I would say have distinct roots and not separable, for example note that $x^n$ is separable but does not have distinct roots.2012-12-11
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    @Belgi: Opinions differ on whether "$f$ is separable" means "$f$ has distinct roots" or "every irreducible factor of $f$ has distinct roots". I agree that this it would be better to be clearer.2012-12-11
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    @Belgi Sorry for the lack of precision, English is not my mother tongue: I meant "$P$ has $n$ distinct roots".2012-12-11