# Prove that T and T^m are simultaneously diagonalizable.

1. Feb 4, 2012

### amanda_ou812

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?

2. Feb 4, 2012

### Deveno

question: why do you care if T and Tm commute? (i mean, they obviously do, but how is that relevant)?

if T is diagonalizable, then for any basis B, if [T]B = A, there is some invertible matrix P

(why?) so that:

P-1AP = D, where D is diagonal. furthermore Dm is diagonal with:

Dm = P-1AmP.