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.

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 $T^*T=I$.
