Originally Posted by

**Magnite** Couldn't you just prove that they are similar like this:

PA(P^-1) = D

((P^-1)^T)(A^T)(P^T) = D^T, transpose of both sides

(Q^-1)(A^T)Q = D, By letting Q = P^T, and transpose of diagonal matrix is still diagonal (also note: (P^-1)^T = (P^T)^-1 = Q^-1)

(P^-1)(Q^-1)(A^T)(Q)(P) = A, so they are similar, share the same eigenvalues and therefore have the same characteristic polynomial.

(Note: (QP)^-1 = (P^-1)(Q^-1), in case you were wondering why the previous equations suggests similar matrices)