Proof that, given $U$ unitary there is a unique $A$ such that $U = (I-A)(I+A)^{-1}$

Question

a) If A is an anti-Hermitian matrix, show that $$U = (I − A)(I + A)^{-1} \qquad(*)$$ is a unitary matrix

b) If U is a unitary matrix such that 1 + U is invertible, show that there is a unique matrix A satisfying (∗).

---

I have managed the first part of the question using: $$U^\dagger = (I+A^\dagger)^{-1}(I-A^\dagger)=(I-A)^{-1}(I+A)$$ $$ \rightarrow U^\dagger U = I$$

But I am stuck on the second part. How do I prove that A is unique? By simply using (*) I have managed to show that we must have $A^\dagger = -A$, ie that it is antihermitian, but not that it is unique.

Answer 1

We can solve for $A$: $$U=(I-A)(I+A)^{-1} \iff U + UA = I - A \iff A = (I+U)^{-1}(I-U)$$

To show unqiueness assume that $(I-A)(I+A)^{-1} = (I-B)(I+B)^{-1}$. Note that $I-A$ and $(I+A)^{-1}$ commute, hence:

\begin{align*} (I-A)(I+A)^{-1} = (I-B)(I+B)^{-1} &\iff (I-A)(I+B) = (I+A)(I-B)\\ \iff I-A+B-AB = I+A-B-AB&\iff A=B\end{align*}

Answer 2

This is a consequence of what is known as the Cayley Transform. See "Matrix maps" in (https://en.wikipedia.org/wiki/Cayley_transform)