Prove that linear operator that has the power of M and equals the unit operator is diagonalizable.

Question

So I have linear operator in the power of M (integer, >1) that equals the unit operator, how exactly can I prove that it is diagonalizable.

Answer 1

This property is true over any field that contains $M$ distinct $M$-th roots of unity (which requires that its characteristic does not divide$~M$); in particular it always holds over an algebraically closed field of characteristic$~0$, like$~\Bbb C$. The reason is that $X^M-1$ is an annihilating polynomial of the operator, and by assumption it is split with simple roots (any operator annihilated by a polynomial that is split with simple roots is diagonalisable).