Isometry on inner product space

Question

$V$ is an inner product vector space. If a transformation $T\colon V\to V$ satisfies $\langle T(x), T(y)\rangle = \langle x, y\rangle$ for every vector $x, y \in V$, prove or disprove that $T$ is linear.

Seems true, but can't prove it. Tried plugging $x+y$ into $x,y$ and got $\langle T(x+y), T(x+y)\rangle = \langle T(x)+T(y),T(x)+T(y)\rangle$

but this do not lead to the conclusion. Also I got that $T$ is one-to-one. Does anyone know the answer? Any help is appreciated!

Answer 1

It's true. Take an orthonormal basis $(e_i)$ for $V$. Then $(Te_i)$ is an orthonormal set, hence a basis. By orthonormal expansion, we have $$v = \sum_i \langle v,e_i\rangle e_i$$for all $v\in V$. Similarly we have $$Tv = \sum_i \langle Tv,Te_i\rangle Te_i.$$But $\langle v,e_i\rangle = \langle Tv,Te_i\rangle$. So $T$ is linear.

Answer 2

I found answer. The point is using that $\langle x,x\rangle=0$ implies $x=0$. Consider $||T(u+v)-T(u)-T(v)||^2$ then the condition directly gives $||T(u+v)-T(u)-T(v)||^2=||u+v-u-v||^2=0$.