$A$ is skew hermitian, prove $(I-A)^{-1} (I+A)$ is unitary Given $A$ is a skew-hermitian, (i.e $A^H=−A$), the Cayley transform of $A$ is defined as: $W=(I-A)^{-1} (I+A)$. How can be proved that $W$ is unitary (i.e. $W^H W = W W^H = I$)?

First of all, 
$$
I^2-A^2=(I-A)(I+A)=(I+A)(I-A)
$$ 
Multiply by $(I-A)^{-1}$ pre and post, to get
$$
(I+A)(I-A)^{-1}=(I-A)^{-1}(I+A) \tag{1}
$$
So order really doesn't matter: these matrices commute.
The Hermitian transpose of the product in (1) is
$$ (I+A^H)(I-A^H)^{-1} = (I-A)(I+A)^{-1} \tag{2}$$
The inverse of this product is
$$
[(I-A)^{-1}(I+A)]^{-1} =(I+A)^{-1}(I-A)\tag{3}
$$
By (1), the matrices in (2) and (3) are equal: that is, the inverse of $I^2-A^2$ is its Hermitian transpose. Hence, that product matrix is unitary. 
