TITLE: If $A\in M_{n}(\mathbb{C})$ is a Hermitian matrix  so $I-iA$ is Invertible
QUESTION [6 upvotes]: I'd love your help with this.
Let $A\in M_{n}(\mathbb{C})$ be a Hermitian matrix $(A^{*}=A)$.
How can I prove that $I-iA$ is Invertible?
Thank you

REPLY [7 votes]: Define $$B= (I+A^2)^{-1}(I +i A).$$ Note that $I+A^2$ is invertible, as $\det(I+A^2) = \det(I+A^\dagger A) \geq 1$ (a sum of a positive definite and a positive semidefinite matrix is invertible). Then
$$B(I-iA) = (I+A^2)^{-1}(I +i A)(I-iA) = (I+A^2)^{-1} (I+ A^2) =I,$$
so $B$ is the inverse of $A$.