Show that an isometric linear operator $T:H\to H$ satisfies $T^* T=I$, where $I$ is the identity operator on $H$. Show that an isometric linear operator $T:H\to H$ satisfies $T^* T=I$, where $I$ is the identity operator on $H$.
I've been stuck on this for a while and don't really know where to start.

The polarization identity for a complex Hilbert space $H$ is
$$
           (x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2},\;\;\; x,y\in H.
$$
If $T$ is a linear isometry, then
$$
\begin{align}
    (T^{\star}Tx,y) & =(Tx,Ty) \\
        & = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|Tx+i^{n}Ty\|^{2} \\
        & = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|T(x+i^{n}y)\|^{2} \\
        & = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2} = (x,y).
\end{align}
$$
Therefore, $((T^{\star}T-I)x,y)=0$ for all $x,y$. By a judicious choice of $y$, it follows that $(T^{\star}T-I)x=0$ for all $x$ and, hence, $T^{\star}T-I=0$.
Conversely, if $T$ is a bounded linear operator for which $T^{\star}T=I$, then $T$ is isometric because
$$
     \|Tx\|^{2}=(Tx,Tx)=(T^{\star}Tx,x)=(x,x)=\|x\|^{2},\;\;\; x \in X.
$$
