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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. $W^H W = W W^H = I$)?

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I present a slight modification on the answer to a very similar question. I assume we are working over the complex numbers.

Recall the eigenvalues of a skew-Hermitian matrix are imaginary. So the set of eigenvalues of $J=I-A$ cannot contain zero, and the matrix is invertible. We then have $W=J^{-1} J^H$. it remains to compute:

$$J^{-1} J^H (J^{-1} J^H)^H= J^{-1} J^H J (J^{-1})^H = J^{-1} J^H J (J^{H})^{-1}$$

Note now that $J$ and $J^H$ commute.

$$J^{-1} J^H J (J^{H})^{-1}=J^{-1} J J^H (J^{H})^{-1}=I.$$

The other computation is almost exactly the same.

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.

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