Do the trace and determinant of an operator remain the sum/product of its eigenvalues if the underlying field is finite?

Question

For fields of characteristic $0$ one can derive from the Jordan normal form that the trace/determinant of an operator are equal to the sum/product of its eigenvalues. Does this fact remain true when the characteristic is $p>0$? I feel like this statement should be derivable from the definition of the characteristic polynomial, but I'm unsure whether the positive characteristic can muck things up.