prove that an operator is unitary I have $V$ an inner product space above $\mathbb{C}$ and a linear operator $T$ such as $T=T^{*}$ on $V$ (surjective).
I need to prove that $U=\left(I-iT\right)\left(I+iT\right)^{-1}$ satisfies the condition $U^{*}=U^{-1}$, by $*$ I mean adjoint operator.
I know that:
$$\left\|v+iT(v)\right\|=\left\|v-iT(v)\right\|$$
$$v=0 \iff v+iT(v)=0$$
$$(I+iT)^*=I-iT$$
I also know that $(I+iT),=I-iT$ both injective and surjective, so they are invertible.
I shouldn't use the rule $(ST)^*=T^*S^*$ or any other rules ($(S+T)^*=...$ and e.t.c) without proving them.
I tried to do it by the definition $\langle T(u),v \rangle = \langle u,T^*(v) \rangle$, but I got stuck:
$$\left \langle \left(I-iT\right)\left(I+iT\right)^{-1}(u),v \right \rangle = \left \langle \left(I+iT\right)^{-1}(u),\left(I+iT\right)(v) \right \rangle$$
What is the easiest way to show that $U$ is unitary ?
Edit:
it is easy to prove that $(ST)^*=T^*S^*$.
How do I prove that $\left(S^{-1}\right)^{*}=\left(S^{*}\right)^{-1}$
 A: Unfortunately, I don't see an elementary way to solve the problem without making use of some of the rules that you have mentioned you are not allowed to use, but one way to see it is as follows. We need a few lemmas; I will state some with proof and some with just explanation.
Lemma 1. We have that $(T^*)^* = T.$
Proof. By definition, the adjoint is the unique linear operator satisfying $\langle T(v), w \rangle = \langle v, T^*(w) \rangle$ for all vectors $v$ and $w$ in $V.$ We have therefore that $$\langle (T^*)^*(v), w \rangle = \langle v, T^*(w) \rangle = \langle T(v), w \rangle$$ for all vectors $v$ and $w$ in $V.$ We conclude as desired that $(T^*)^* = T.$ QED.
Lemma 2. Given that $T$ is invertible, we have that $(T^{-1})^* = (T^*)^{-1}.$
Proof. We will prove that $T^*(T^{-1})^* = I$ and $(T^{-1})^* T^* = I.$ By definition, the adjoint is the unique linear operator satisfying $\langle T(v), w \rangle = \langle v, T^*(w) \rangle$ for all vectors $v$ and $w$ in $V.$ We have therefore that $$\langle T^* (T^{-1})^*(v), w \rangle = \langle (T^{-1})^*(v), (T^*)^*(w) \rangle = \langle (T^{-1})^*(v), T(w) \rangle = \langle v, T^{-1} T(w) \rangle = \langle v, w \rangle$$ for all vectors $v$ and $w$ of $V,$ where the second equality holds by Lemma 1. Consequently, we have that $T^*(T^{-1})^* = I.$ One can prove the second statement analogously using the fact that $TT^{-1}(w) = w$ for all vectors $w$ of $V.$ We conclude as desired that $(T^{-1})^* = (T^*)^{-1}.$ QED.
Lemma 3. Given linear operators $T$ and $S,$ we have that $(TS)^* = S^* T^*.$
Proof. By definition, the adjoint is the unique linear operator satisfying $\langle T(v), w \rangle = \langle v, T^*(w) \rangle$ for all vectors $v$ and $w$ in $V.$ We have therefore that $$\langle (TS)^*(v), w \rangle = \langle v, TS(w) \rangle = \langle T^*(v), S(w) \rangle = \langle S^* T^*(v), w \rangle$$ for all vectors $v$ and $w$ in $V.$ We conclude as desired that $(TS)^* = S^* T^*.$ QED.
Lemma 4. Given linear operators $T$ and $S,$ we have that $(T + S)^* = T^* + S^*.$
Proof. Use the definition of the adjoint and the linearity of the inner product. QED.
Lemma 5. Given polynomials $f(x)$ and $g(x)$ in $\mathbb C[x],$ we have that $f(T) g(T) = g(T) f(T),$ i.e., polynomials in a linear operator commute with one another.
Proof. Use the fact that $T$ commutes with itself and all scalars commute.

We must assume that $V$ is finite-dimensional.
Proof. Using the lemmas, we have that $$\begin{align*} U^* &= [(I - iT)(I + iT)^{-1}]^* \\ \\ &= [(I + iT)^{-1}]^* (I - iT)^* \tag{by Lemma 3} \\ \\ &= [(I + iT)^*]^{-1} (I - iT)^* \tag{by Lemma 2} \\ \\ &= (I - iT)^{-1}(I + iT) \tag{by Lemma 4} \end{align*}$$ so that $$U^* U = (I - iT)^{-1}(I + iT)(I - iT)(I + iT)^{-1} = (I - iT)^{-1} (I - iT)(I + iT)(I + iT)^{-1} = I,$$ where the second equality holds by Lemma 5. Considering that $V$ is finite-dimensional, we have that $U$ is invertible if and only if $U$ is injective if and only if $U$ has a left-inverse. Certainly, $U^*$ is a left-inverse of $U,$ hence $U^*$ is a right-inverse of $U,$ i.e., $U^* = U^{-1}.$ QED.
A: Set
$U = (I - iT)(I + iT)^{-1}; \tag 1$
for any operators $A$, $B$ and any $x, y \in V$,
$\langle (AB)^\dagger x, y \rangle = \langle x, ABy \rangle = \langle A^\dagger x, By \rangle = \langle B^\dagger A^\dagger x, y  \rangle, \tag 2$
whence
$(AB)^\dagger x = B^\dagger A^\dagger x, \; \forall x \in V, \tag 3$
and thus
$(AB)^\dagger = B^\dagger A^\dagger; \tag 4$
also,
$\langle (A + B)^\dagger x, y \rangle = \langle x, (A + B)y \rangle = \langle x, Ay + By \rangle = \langle x, Ay \rangle + \langle x, By \rangle$
$= \langle A^\dagger x, y \rangle + \langle B^\dagger y, x \rangle = \langle A^\dagger x + B^\dagger x, y \rangle = \langle (A^\dagger + B^\dagger)x, y \rangle, \tag 5$
from which
$(A + B)^\dagger x = (A^\dagger + B^\dagger)x, \; \forall x \in V, \tag 6$
and hence
$(A + B)^\dagger = A^\dagger + B^\dagger; \tag 7$
applying (4) and (7) to (1):
$U^\dagger = ((I + iT)^{-1})^\dagger (I - iT)^\dagger = ((I + iT)^{-1})^\dagger (I - (iT)^\dagger); \tag 8$
we have
$I - (iT)^\dagger = I + iT^\dagger = I + iT, \tag 9$
since
$T^\dagger = T; \tag{10}$
furthermore,
$(I + iT)(I + iT)^{-1} = I, \tag{11}$
from which
$((I + iT)^{-1})^\dagger (I + iT)^\dagger = I^\dagger = I, \tag{12}$
so that
$((I + iT)^{-1})^\dagger = ((I + iT)^\dagger)^{-1} = (I - iT)^{-1}; \tag{13}$
then
$U^\dagger = ((I + iT)^{-1})^\dagger (I - iT)^\dagger$
$= (I - iT)^{-1} (I - (iT)^\dagger) = (I - iT)^{-1} (I + iT);  \tag{14}$
finally,
$U^\dagger U = (I - iT)^{-1} (I + iT)(I - iT)(I + iT)^{-1} = (I - iT)^{-1} (I - iT)(I + iT)(I + iT)^{-1} = II = I, \tag{15}$
and now right multiplying by $U^{-1}$ yields
$U^\dagger = U^\dagger I = U^\dagger (UU^{-1}) = (U^\dagger U)U^{-1} = IU^{-1} = U^{-1}, \tag{16}$
$OE\Delta$.
We note that the argument surrounding (11)-(13) extends to show that
$(S^\dagger)^{-1} = (S^{-1})^\dagger \tag{17}$
for general invertible operators $S$.
