# If A is anti-hermitian, show that $|\det(1+A)|^2 \geqslant 1$

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.

• Well, $iA$ is Hermitian, so its eigenvalues are real, so those of $A$ are purely imaginary. – Lord Shark the Unknown Jan 1 at 11:51
• @Lord Shark the Unknown So the questions asks us to diagonalise $iA$ so that we can show that eigenvalues of A are purely imaginary? Does it have no other purposes? What about the rest of my working, are they valid? – Student 1 Jan 1 at 11:58