If $A\in M_{n}(\mathbb{C})$ is a Hermitian matrix  so $I-iA$ is Invertible 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
 A: 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 $A$.
A: $I$ has only eigenvalue 1, and $A$ has only real eigenvalues, thus $iA$ will only have imaginary eigenvalues. Say $A$ has eigenvalues $\lambda_k$, where $\lambda_k$ could possibly be zero, with corresponding eigenvector $v_k$. Then:
$$(I-iA)v_k= v_k - i\lambda_k v_k = (1-i\lambda_k)v_k$$
Thus the eigenvalues for $(I-iA)$ will be on the form $1-i\lambda_k$, $\lambda_k \in \mathbb R$. Since all eigenvalues are non-zero, the matrix will be invertible.
Edit: Hermitian matrices have only real eigenvalues since, if $Av = \lambda v$, then 
$$v^* A v = \lambda v^*v$$
taking the hermitian conjugate of both sides yields: 
$$v^* A^* v = \lambda^* v^* v$$
Since $A = A^*$ we get $\lambda = \lambda^*$ from the two equations above, where $\lambda^*$ is the complex conjugate of $\lambda$. Thus $\lambda$ has to be real. 
