1. The problem statement, all variables and given/known data A is an anti-Hermitian matrix. Show, by diagonalising iA, that: |det(1+A)|^2 >=1 2. Relevant equations A^H denotes hermitian conjugate of A; A^H = -A\ x = Ox' transforms vector components between 2 basis sets. 3. The attempt at a solution I know that iA is a Hermitian matrix, so it is diagonalisable. But I'm not sure what the point of diagonalising iA is. D = O^(-1)(iA)O for diagonal matrix of eigenvalues D. det(iA) = det(D). I'm floundering around. Any help is appreciated.