Let $V$ be a vector space over $\mathbb{C}$ and $T$ a function from $V$ to $V$ such that for every $u,v\in V$, $||T(u)-T(v)||=||u-v||$ and $T(iv)=iT(v)$. Prove that $T$ is linear and $||T(v)||=||v||$ for every $v\in V$.
My attempt:
Since $T(iv)=iT(v)$ we can conclude that $T(-v)=-T(v)$ because if $T(iv)=iT(v)$ then $T(-v)=T(i^2v)=iT(iv)=i^2T(v)=-T(v)$.
With this in mind, we can use the polarization identities like:
$4\left< u,v\right>=||u+v||^2-||u-v||^2$
$4\left< T(u),T(v)\right>=||T(u)+T(v)||^2-||T(u)-T(v)||^2$
And we have that
$4\left< u,v\right>=||u+v||^2-||u-v||^2=||T(u)+T(v)||^2-||T(u)-T(v)||^2=4\left< T(u),T(v)\right>$
And then $\left< T(u),T(v)\right>=\left< u,v\right>$ for every $u,v\in V$
From this, we can inmediatily conclude that $||T(v)||=||v||$ for every $v\in V$ (doing $u=v$ above).
For the linearity: If $r\in\mathbb{C}$ then
$\left< T(ru+v),T(w)\right>=\left< ru+v,w\right>=r\left< u,w\right>+\left< v,w\right>=r\left< T(u),T(w)\right>+\left< T(v),T(w)\right>=\left< rT(u)+T(v),T(w)\right>$
So here is where I have dubs, I want to prove $\left< T(ru+v),T(w)\right>=\left< rT(u)+T(v),x\right>$ for every $x\in V$ to conlude that $T(ru+v)=rT(u)+T(v)$, but I proved that $\left< T(ru+v),T(w)\right>=\left< rT(u)+T(v),T(w)\right>$, what can I do then?
