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Prove that $A^n=I$ over an field $\mathbb{F}$ of char 0 iff $A$ is diagonalisable with $n$-th roots of unity on the diagonal.

Remark: One direction is clear. For the other dierection: Assume $A^n=I$ then $x^n-1=0$ is an annihilating polynomial for $A$ and so all eigenvalues are $n$-th roots of unity; but how to prove $A$ diagonalisable? I aim to show its minimal polynomial has no repeated roots by using Jordan normal form..

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    Please edit the title so that it becomes a correct statement: as it is now, it isn 't!2019-05-24

1 Answers 1

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The claim is false without further conditions on the field. For example

$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\mathcal M_2(\overline{\Bbb F}_2)$

has $\,x^2-1=(x-1)^2\,$ as characteristic and minimal polynomial and thus $\,A^2=I\,$ , yet it even isn't diagonalizable...

Added as a result of an added condition: if $\,\operatorname {char}\Bbb F=0\,$ then all the roots of $\,x^n-1\,$ are simple (why?), and since the definition field $\,\Bbb F\,$ is algebraically closed then we can write

$x^n-1=\prod_{i}(x-a_i)$

where $\,a_i^n=1\,$ are $\,n\,$-th roots of unit in the field. Since clearly the characteristic and the minimal polynomial of $\,A\,$ coincide in this case (why?), the last one can be written as a product of different linear factors and thus $\,A\,$ is diagonalizable, with its eigenvalues, i.e. the roots of unit $\,a_i'\,$s, on its main diagonal.