Isometries on $\mathbb{E}^{2}$

Question

A function $T: \mathbb{E}^{2} \longrightarrow \mathbb{E}^{2}$ is an isometry if $\langle T(\vec{x}), T(\vec{y}) \rangle = \langle \vec{x}, \vec{y} \rangle$ for all $\vec{x}, \vec{y} \in \mathbb{E}^{2}$.

Prove that $T$ is linear, i. e.,

1. $T(\vec{x}+ \vec{y}) = T(\vec{x}) + T(\vec{y})$;
2. $T(\alpha \cdot \vec{x}) = \alpha \cdot T(\vec{x})$.

It is clear that $T$ is injecive (since $\ker \ T = \{0\}$) and $T$ is an isometry if and only if, $||\vec{x}|| = ||T(\vec{x})|| \ \forall \ \vec{x} \in \mathbb{E}^{2}$, but I'm having a little bit trouble with the proof.

Thanks for the help x)

Answer 1

To prove 1. I suggest you to compute $<T(x+y)-T(x)-T(y),T(x+y)-T(x)-T(y)>$ and find that it is zero.

To prove 2. I suggest the same as before but to the expression $<T(\alpha x)- \alpha T(x),T(\alpha x)- \alpha T(x)>$.