TITLE: $U = (I-iT) (I + iT) ^ {-1}$ is a unitary operator when T is self-adjoint
QUESTION [5 upvotes]: Let $V$ complex inner product space of finite dimension and $T$ an operator over $V$. Show that the transformation
$$U = (I-iT) (I + iT) ^ {-1}$$
is a unitary operator
My Attempt: 
$$\langle U \alpha, U\alpha \rangle = \langle (I−iT)(I+iT)^{-1} \alpha , (I−iT)(I+iT)^{-1} \alpha \rangle = 
\langle \alpha , (I−iT)^{-1}(I+iT)(I−iT)(I+iT)^{-1} \alpha \rangle $$
and I need proof this  $(I−iT)^{-1}(I+iT)(I−iT)(I+iT)^{-1} = I $

REPLY [15 votes]: This is the operator-valued version of the fact that the map
$$x\mapsto \frac{1-ix}{1+ix}$$
transforms the real line onto the unit circle. 
In operator terms, the real line becomes the set of self-adjoint operators. And the unit circle is the group of unitary operators. 
So, let  $T$ be self-adjoint. Then $(I\pm iT)^*=(I\mp iT)$, which implies $U^*=(I-iT)^{-1}(I+iT)$. Since all operators of the form $aI+bT$ commute with each other, we have
$$U^*U = (I-iT)^{-1}(I+iT) (I-iT)(I+iT)^{-1} =(I-iT)^{-1}(I-iT)(I+iT) (I+iT)^{-1} =I $$
and 
$$UU^* =  (I-iT)(I+iT)^{-1} (I-iT)^{-1}(I+iT) = (I-iT) (I-iT)^{-1}(I+iT)^{-1}(I+iT) =I $$