Given that A is anti-hermitian, i.e. $A^\dagger = -A$, show, by diagonalising $iA$, that $$|\det(1+A)|^2 \geqslant 1$$
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So what I thought was that I need to show that $\det(1+A)[\det(1+A)]^* \geqslant 1$, i.e. $\det(1+A)\det(1+A^\dagger) \geqslant 1$.
Since I know the determinant is equal to the product of the eigenvalues, I have found the eigenvalues of $(1+A)$ by $$(1+A)\textbf{x} = k\textbf{x}$$ $$A\textbf{x}=(k-1)\textbf{x}$$ $$A\textbf{x}=\lambda \textbf{x}$$ $$k=\lambda +1$$
where $\lambda$ is purely imaginary because A is anti-hermitian.
Similarly, I can show that the eigenvalue of $(1+A^\dagger)=(1-A)$ to be $(1-\lambda)$. Therefore, $$\det(1+A)\det(1+A^\dagger) = \prod_{i=1}^{n} (1+\lambda_i)(1-\lambda_i)=\prod_{i=1}^{n} (1+|\lambda_i|^2) \geqslant 1$$
However, I am not sure if this actually proves it, and especially I don't know why the question asks to diagonalize $iA$.
Thank you.