- #1

N00813

- 32

- 0

## Homework Statement

A is an anti-Hermitian matrix.

Show, by diagonalising iA, that:

|det(1+A)|^2 >=1

## Homework Equations

A^H denotes hermitian conjugate of A; A^H = -A\

x = Ox' transforms vector components between 2 basis sets.

## 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.