I wanted to show that for a anti-Hermitian matrix of order $4n$ with purely imaginary eigenvalues, the determinant of Matrix is real and positive. Here’s what I’ve done: Let $v$ be the eigenvector of matrix and $ \lambda $ be the purely complex eigenvalue.
$$ Mv=\lambda v $$
Now taking dagger on both sides:
$$ (Mv)^\dagger=(\lambda v)^\dagger $$
$$ v^\dagger M^\dagger=\lambda^* v^\dagger $$
Now multiplying by $v$ from the LHS: $$ vv^\dagger (-M)=\lambda^* v v^\dagger $$
$$ -M=\lambda^* I $$ Now taking determinant on both sides: $$ (-1)^{4n}\det(M)=(\lambda^*)^{4n}\det(I) $$
Now, as for purely complex number,
$$ (\lambda^*)^{4n}\geq 0 $$
$$ \therefore \det(M)\geq 0 $$
Is this method correct? I needed to show this as part of my Mathematical Physics assignment.