Question: Let T be a diagonalizable linear operator on a finite-dimensional vector space, and let m be any positive integer. Prove that T and T^m are simultaneously diagonalizable. Definition: Two linear operators T and U are finite-dimensional vector space V are called simultaneously diagonalizable if there exists an ordered basis B for V such that both [T]B and B are diagonal matrices. Similarly, A and B are called simultaneously diagonalizable if there exists and invertible matrix Q such that Q-1AQ and Q-1BQ are diagonal matrices. Attempt at a solution: If T is a linear operator then there is a basis B for V such that [T]B = A is a matrix representation of T. Now, T is diagonalizable and therefore there is a invertible matrix Q such that C = Q-1AQ is a diagonal matrix. Recall that A = QCQ-1 and A^m = QC^m Q-1 = B which is a diagonal matrix. So, A*A^m = AB = (QCQ-1)(QC^m Q-1) = QC(Q-1 Q)C^m Q-1 = QC C^m Q-1) = QC^m CQ-1 [since diagonal matrices commute] = QC^m Q-1 QCQ-1 = (QC^m Q-1)( QCQ-1) = A^m A = BA. Therefore T and T^m commute and are simultaneously diagonalizable. I think I am missing something. Any suggestions?