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.
 A: Isometric means
$$
\|Tx\|=\|x\|,
$$
for all $x\in H$. Equivalently, for all $x,y\in H$
$$
\|T(x+y)\|^2=\|(x+y)\|^2.
$$
But
$$
\|(x+y)\|^2=\langle x+y,x+y\rangle=\|x\|^2+2\langle x,y\rangle+\|y\|^2,
$$
while
$$
\|T(x+y)\|^2=\langle T(x+y),T(x+y)\rangle=\|Tx\|^2+2\langle Tx,Ty\rangle+\|Ty\|^2.
$$
Thus, for all $x,y\in H$
$$
\langle Tx,Ty\rangle=\langle x,y\rangle.
$$
Note that $\langle x,Ty\rangle=\langle T^*,y\rangle$, and the above becomes
$$
\langle T^*Tx,y\rangle=\langle x,y\rangle\quad\text{or}\quad \langle (T^*T-I)x,y\rangle=0
$$
for all $x,y\in H$. In particular, for $y=(T^*T-I)x$, it becomes
$$
0=\langle (T^*T-I)x,(T^*T-I)x\rangle=\|(T^*T-I)x\|^2,
$$
for all $x\in H$, and thus $T^*T=I$.
A: 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
$$
     \|Tx\|^{2}=(Tx,Tx)=(T^{\star}Tx,x)=(x,x)=\|x\|^{2},\;\;\; x \in X.
$$
A: Assuming $H$ is a Hiblert space:
Let $\{e_i\}$ be an orthonormal Schauder basis for the Hilbert space.  We note that 
$$
T^*T = I \iff 
\forall x,y: \langle x,T^*Ty\rangle = \langle x,y\rangle\\
\iff \forall x,y: \langle Tx, Ty \rangle = \langle x,y \rangle
$$
Try to deduce this last statement using the fact that
$$
\forall x,y: \langle Tx, Tx \rangle = \langle x,x \rangle
$$
