If T is an endomorphism of finite order, i.e, T^n=1 for some n>=1, give a necessary and sufficient condition on the field F for T to be diagonalizable
If is an endomorphism of finite order, then is diagonalisable if and only if is an algebraically closed field where the characteristic of the field does not divide the order of the endomorphism. This follows from the fact that is diagonalisable if and only if it's minimal polynomial in splits in and has distinct roots (the first condition gives algebraic closure, the second gives the condition for characteristic not dividing the order of endomorphism).
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