[SOLVED] permutation of eigenvalues

Hello Kitty

Suppose H is a subgroup of O(n,R) (orthogonal matrices) and that I have an unambiguous way of ordering the eigenvalues for all matrices in H. Let A \in H have n distinct eigenvalues. Now suppose I have an invertible matrix P from the normalizer of H in O(n,R) such that PAP^{-1} gives me a matrix where exactly two of my eigenvalues have switched places. I would like to establish whether P must have determinant -1.

I think there may be a connection between how a permutation matrix (in the standard sense) has determinant -1 if it corresponds to a transposition (or indeed any odd permutation).