Showing that $\langle T(u),T(v)\rangle = \langle u, v \rangle$ implies $T$ is a linear isometry

Question

Let $T$ belong to $\mathcal{L}(H)$ (i.e., the set of linear operators from $H \mapsto H$ where $H$ is a Hilbert space).

I need to show that $T$ is an isometry iff $\langle T(u),T(v) \rangle = \langle u, v \rangle$ for all $u,v \in H$.

I have been successfully able to show that if $T$ is an isometry, then $\langle T(u), T(v) \rangle=\langle u, v \rangle $:

Suppose that $T$ is an isometry; i.e., that $\forall w \in H$, $||T(w)||=||w||$.

Then, of course we have that $||T(w)||^{2} = ||w||^{2}$.

Now, consider $u+v \in H$. Then, since $T$ is an isometry, $||T(u+v)||^{2} = ||u+v||^{2}$ implies that $\langle T(u+v),T(u+v) \rangle = \langle u + v, u+v \rangle \implies \langle T(u), T(u+v)\rangle + \langle T(v), T(u+v)\rangle = \langle u,u+v \rangle + \langle v, u+v \rangle \implies \langle T(u), T(u) \rangle + \langle T(u), T(v) \rangle + \langle T(v), T(u) \rangle + \langle T(v), T(v) \rangle = \langle u, u \rangle + \langle u, v \rangle + \langle v, u \rangle + \langle v,v \rangle \implies \langle T(u), T(u) \rangle + \langle T(u), T(v) \rangle + \langle T(u), T(v) \rangle + \langle T(v), T(v) \rangle = \langle u, u \rangle + \langle u, v \rangle + \langle u, v \rangle + \langle v,v \rangle \implies \langle T(u), T(u) \rangle + 2\langle T(u), T(v) \rangle + \langle T(v), T(v) \rangle = \langle u, u \rangle + 2\langle u, v \rangle + \langle v,v \rangle \implies ||T(u)||^{2} + 2\langle T(u),T(v) \rangle + ||T(v)||^{2} = ||u||^{2} + 2\langle u,v \rangle + ||v||^{2}$.

Then, because $T$ is an isometry, $||T(u)||=||u|| \implies ||T(u)||^{2} = ||u||^{2}$ and $||T(v)||=||v|| \implies ||T(v)||^{2} = ||v||^{2}$, we have that

$2\langle T(u),T(v) \rangle = 2\langle u,v \rangle \implies \langle T(u),T(v) \rangle = \langle u,v \rangle$.

I am having difficulty showing the other direction without using the thing I am trying to prove! Everywhere I've looked has said this direction is "obvious", which, although it might be true to those more knowledgeable than I, is meaningless to me.

Short of telling me this is "obvious" or "directly follows from" something without explaining in detail why, could someone please help me show the $\langle T(u), T(v) \rangle = \langle u, v \rangle \implies \, T\,\text{is an isometry}$ direction? Thank you.

Also, I understand that a lot of books give this as the definition of what it means to be an isometry; however, it is not. It's something that needs to be proven. The definition of isometry I am assuming as a basic principle here is $\forall w \in H$, $||T(w)||=||w||$.

And please refrain from snarkiness. Sometimes I have trouble with very simple things, even though very complicated things come easy to me. Please try to understand that not everybody's brain works the same way.

Answer 1

Note that $\langle Tw, Tw\rangle=\langle w, w\rangle$. Taking $w=u-v$ gives $\langle T(u-v),T(u-v)\rangle=\langle(u-v),(u-v)\rangle\implies \left| u-v\right|^2=\left|Tu-Tv\right|^2$