For skew-symmetric $U$, prove that $U(1-U^\dagger U)^{-1}$ is, too

Question

Let $U$ be a complex skew-symmetric matrix ($U = -U^\mathsf{T}$, where $U^\mathsf{T}$ means the transpose of $U$). I want to show that $U(1 - U^\dagger U)^{-1}$ is also skew-symmetric. Here $U^\dagger$ is the conjugate transpose (Hermitian conjugate) of $U$.

I have numerically tested this statement, and it seems to be true as long as $1 - U^\dagger U$ is invertible. I attempted to use the eigen-decomposition of $U^\dagger U$ to prove it, but it does not seem to help much.

Answer 1

Let $M = U\left(I - U^{\dagger}U\right)^{-1}$ since $U^{\dagger} = -\overline{U}$, then

$$M = U\left(I + \overline U U\right)^{-1}$$ and

\begin{align} M^\intercal &= \left(I + U\overline U\right)^{-1}U^{\intercal}\\ &= -\left(I + U\overline U\right)^{-1}U\\ &= -\left(I + U\overline U\right)^{-1}U\left(I + \overline U U\right)\left(I + \overline U U\right)^{-1}\\ &= -\left(I + U\overline U\right)^{-1}\left(U + U\overline U U\right)\left(I + \overline U U\right)^{-1}\\ &= -\left(I + U\overline U\right)^{-1}\left(I + U\overline U \right)U\left(I + \overline U U\right)^{-1}\\ &= -U\left(I + \overline U U\right)^{-1} = -M \end{align}