Proof explanation: T is unitary $ \iff $ $ T^*T = I $

Theorem: T is unitary $ \iff $ $ T^*T = I $
Proof:
$ ( \rightarrow ) $ If $ T $ is unitary, then for all $ u,v \in V$ we have that $ \langle u,T^* T v \rangle = \langle Tu, Tv \rangle = \langle u , v \rangle $ hence $ T^* Tv = v $ for all $ v \in V $, meaning $ T^* T = I $.
$ ( \leftarrow ) $ If $ T^* T = I $, then for all $ u,v \in V $ we have that $ \langle u,v \rangle = \langle u , T^* T v \rangle = \langle Tu , Tv \rangle $, hence $ T $ is unitary.

My question: I haven't understood in the " $ ( \rightarrow ) $ " part where we said $ T^* Tv = v $.
We arrived to $ \langle u,T^* T v \rangle = \langle u , v \rangle $, how from here I deduce that $ T^* Tv = v $? I tried using linearity property of inner-product as follows:
$ \langle u,T^* T v \rangle - \langle u , v \rangle = 0 \rightarrow \langle u,T^* T v - v \rangle = 0 $ , however now I got stuck and couldn't proceed.
In general I know that:
if $ V $ is an arbitrary vector space over some field $ \mathbb{F} $, suppose an inner-product $ \langle \cdot ,\cdot \rangle : V^2 \to \mathbb{F} $ is defined on the vector space. Then it is not necessarily true that if $ \langle u ,v \rangle = \langle u ,w \rangle $ for arbitrary $ u,v,w \in V $, then $ v = w $.
Also, it is not necessarily true that if $ \langle u, v \rangle = 0 $ then $ u = v $.

Therefore, by having $ \langle u,T^* T v - v \rangle = 0 $ , how does one deduce $ T^* Tv = v $ ?

Thanks in advance for any help!

Solution 1:

The point is that $u$ is arbitrary. You can choose $u=\lambda(T^*Tv-v)=\lambda w\neq0$, and by axioms of inner product $\langle\lambda w,w\rangle\neq0,\,w,\lambda\neq0$. As the product is in fact equal to zero, either $\lambda$ or $w$ is zero, but as $\lambda$ can be arbitrarily chosen from the scalar field $w=0$, that is: $(T^*T-I)v=0$ for all $v$. If $T^*T-I\neq\bf 0$, then there exist $v$ for which the product is not zero. Again, as $v$ is arbitrary, we have that $T^*T-I=\bf0$ assuming a non empty vector space.

Solution 2:

Hint: You have $\langle u, (T^*T - I)v \rangle = 0$ for all $u,v$. Take $u = (T^*T - I)v$, to get $$\|(T^*T-I)v \| = 0$$ for all $v$. Can you conclude?