two definitions of isometry Let $V$ be an inner product space, and $F :V\to V$ a linear map. 
Some say
that $F$ is an isometry when $\langle F(u),F(v)\rangle = \langle u,v\rangle \forall u,v\in V$. Others say that $F$ is an isometry when $||F(u)||=||u||$ $\forall u\in V$.

Show that these two meanings are equivalent.


Note that $||x||={\langle x ,x\rangle }^{\frac{1}{2}}$
$F$ is said to be an isometry if ${\langle Fu ,Fv\rangle }={\langle u ,v\rangle }$
So when $F$ is an isometry then $||Fu||^2={\langle Fu ,Fu\rangle }={\langle u ,u\rangle }=||u||^2$
Moreover, when  $||Fu||^2=||u||^2$, then 
$\langle Fu, Fv\rangle$ 
$=\frac{1}{4}  \{\{||F(u)+Fv||^2-||Fu-Fv||^2\}+i\{||Fu+iFv||^2-||Fu-iFv||^2\}\}=\frac{1}{4}  \{\{||(u)+v||^2-||u-v||^2\}+i\{||u+iv||^2-||u-iv||^2\}\}$ 
$= \langle u,v\rangle$ 
by the Polarization Identity.
Hence, we have proved the equivalence of definitions.
