two definitions of isometry

Question

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.

Answer 1

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.