Element of finite order is semisimple and eigenvalues are roots of unity
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Suppose is a field, is a finite-dimensional vector space over and is an element in such that there exists with equal to the identity matrix. Then:
- All the eigenvalues of over the algebraic closure of are roots of unity.
- Suppose that either has characteristic zero or is relatively prime to the characteristic of . is semisimple, i.e. it is diagonalizable over the algebraic closure of .
satisfies the polynomial , hence the minimal polynomial of must divide this polynomial. So, every eigenvalue of must satisfy the polynomial , hence must be a root of unity.
If is relatively prime to the characteristic, then the polynomial has no repeated roots, hence the minimal polynomial of has no repeated roots. So, is semisimple, i.e. it is diagonalizable over the algebraic closure.