Show that if $H$ is an Hermitian matrix, then $U=(iI-H)(iI+H)^{-1}$ is unitary How would you prove, that when $H$ is a hermitian matrix, then:
$$U=(iI-H)(iI+H)^{-1},$$
where $U$ is unitary, assuming that $(iI+H)$ is invertible.
I thought that because $U$ is unitary, $UU^*=U^*U=I$.
So I tried to take the conjugate transpose of $U$ where:
$$U^*=(-iI-H)(H-iI)^{-1}.$$
Essentially this is where I am stuck:
1) How do I take the transpose of $U$, so I don't just have the conjugate -- or does it not matter?
2) How do I get rid of the inverse because I don't know what to do with it and how to eventually prove $UU^*=I$?
Thank you!
 A: Let $H$ be Hermitian and $U=(iI-H)(iI+H)^{-1}$. Then
$$
U^*=(-iI+H)^{-1}(-iI-H) = (iI-H)^{-1}(iI+H) .
$$
Therefore
$$
\begin{split}
UU^*
&= (iI-H)(iI+H)^{-1}(iI-H)^{-1}(iI+H) \\
&= (iI-H)(iI+H)^{-1}(iI+H)(iI-H)^{-1} \\
&=(iI-H)(iI-H)^{-1} \\
&= I.
\end{split}
$$

They key fact is that for every $x,y\in\mathbb C$ and every matrix $A$ you have
$$
(xI+A)(yI+A)=(yI+A)(xI+A).
$$
Multiplying by $(yI+A)^{-1}$ both on the left and the right, you get also
$$
(yI+A)^{-1}(xI+A) = (xI+A)(yI+A)^{-1}.
$$
A: The point is that $(AB)^* = B^* A^*$, which implies $(A^{-1})^* = (A^*)^{-1}$.
Thus if $U = (iI-H)(iI+H)^{-1}$, $$U^* = ((iI + H)^{-1})^* (iI - H)^* = 
(-iI+H)^{-1} (-iI-H) = (iI-H)^{-1}(iI+H)$$
and $U U^* = U^* U = I$ (note that $iI+H$ and $iI-H$ and their inverses commute).
A: The matrices $iI+H$ and $iI-H$ are invertible, because $H$ only has real eigenvalues, being Hermitian.
Your equality can also be written as $U(iI-H)=(iI+H)$. Taking the Hermitian transpose,
$$
(iI-H)^*U^*=(iI+H)^*
$$
so that
$$
(-iI-H)U^*=-iI+H
$$
which is equivalent to $(iI+H)U^*=iI-H$. Multiply the left-hand side by $U(iI-H)$ and the right hand side by $(iI+H)$:
$$
(iI+H)U^*U(iI-H)=(iI-H)(iI+H)
$$
However, $(iI-H)(iI+H)=(iI+H)(iI-H)$ by direct computation, so
$$
(iI+H)U^*U(iI-H)=(iI+H)(iI-H)
$$
and we can cancel on both sides the invertible matrices, leaving $UU^*=I$.
