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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