Proving a matrix is unitary given another matrix is skew-adjoint

Question

I know that for a matrix $Q$ to be unitary, then $Q^\star = Q^{-1}$.
If we're given that $(I - A)$ is invertible, where $I$ is the identity matrix and $A$ is an n x n skew-adjoint matrix that is, $A^\star = -A = \bar{A^{T}}$, how would we show that
$$U = (I + A)(I - A)^{-1}$$ is unitary?
I found $U^{-1}$ and $U^{\star}$ and they were equal, but that was under naive assumptions.
These were the assumptions (and I'm not sure if it's true, can anyone confirm) that I had.
$$\text{1.} \qquad \overline{(I-A)^{-1}} = (\bar{I} - \bar{A})^{-1}. \\ \text{2.} \quad (I+A) \quad \text{is invertible.}$$ Can someone confirm if these are true in this case, and in general, and if not, how would I have done this question?

Answer 1

Instead of (1) and (2) do you want $((I-A)^{-1})^*=(I-A^*)^{-1}=(I+A)^{-1}$?

It is true and easy to show that if $B$ is invertible, then so is $B^{*}$, and $(B^{-1})^*=(B^{*})^{-1}$. Just multiply: $B^{*}(B^{-1})^*=(B^{-1}B)^*=I^*=I$ and similarly in the other order. In your case, that means if you already know that $I-A$ is invertible, then so is $I+A=(I-A)^*$, and $(I+A)^{-1}=((I-A)^{-1})^*$.

But why is $I-A$ (and hence $I+A$) invertible in the first place? Because $A$ is skew-adjoint, so all of its eigenvalues are purely imaginary.