Given $U \in \mathbb{C^{n \times n}}$ a unitary Matrix that doesn't have $-1$ as an Eigenvalue.
I am trying show that $U + I$ is invertiable and that $−i(U − E)(U + E)^{-1}$ is a Hermitian matrix.
Let $\lambda$ be an Eigenvalue of $U$ It follows that the Eigenvalues of $U + I$ have the form $\lambda+1$. Since $\lambda \neq -1$ then $U + I$ doesn't have zero as an Eigenvalue and therefore invertiable. (Is this correct?)
Let $B = −i(U − I)(U + I)^{-1} = (-iU + iI)(U + I)^{-1}$
We can show that $B^{H} = B$ or $B^{T} = \overline{B}$
$$\overline{B} = \overline{(-iU + iI)(U + I)^{-1}} = (-i \overline{U} - i I)(\overline{U} + I)^{-1}$$ $$B^{T} = (U^{T} + I)^{-1}(-i U^{T} + i I)$$
There is where I get stuck. Not sure how to relate $U^{T}$ and $\overline{U}$
Since a unitary matrix is normal, it is unitarily diagonalizable. So all that needs to be shown is that if $|\lambda|=1$, $\lambda\ne-1$, then $$ -i\,\frac{\lambda-1}{\lambda+1}\in \mathbb R. $$ For this, $$ -i\,\frac{\lambda-1}{\lambda+1} =-i\,\frac{(\lambda-1)(\bar\lambda+1)}{|\lambda+1|^2}, $$ so we only care about the numerator now. And $$ -i(\lambda-1)(\bar\lambda+1)=-i(|\lambda|^2-1+\lambda-\bar\lambda)=-i(\lambda-\bar\lambda)=2\operatorname{Im}\lambda\in\mathbb R. $$