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Let $(H, <\cdot ,\cdot>) $ be a Hilbert space and let $T: H \to H$ be an isometry so that $||T(x)-T(y)|| = ||x - y||$ for all $x, y \in H$ and $T(0)=0$. Prove $T$ is linear.

To solve this, I need the equation $<T(x),T(y)>=<x,y>$ but I can only show that the real components of the left and right side are equal. How do I show that their imaginary components are equal as well?

This question is very similar to Map of Pre-Hilbert Space is affine if it is an isometry with the only difference being that our space is not necessarily real. Is there an easy trick I am missing? Thanks in advance.

Update: The statement is false. Take for instance the complex conjugation as a counterexample. The information, that H is real was simply missing.

Florian Ente
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    Use the https://en.wikipedia.org/wiki/Parallelogram_law for complex inner product space. – Just a user Oct 22 '21 at 18:43
  • Note that $<x,iy>+<iy,x>=i(<y,x>-\overline {<y,x>})=-2\Im <y,x>)$ so apply the same reasoning (ie parallelogram rule) to $x,iy$ and conclude – Conrad Oct 22 '21 at 20:28
  • The problem here for me is that I don't know $T(iy)=iT(y)$. With that I could solve using your argument. – Florian Ente Oct 23 '21 at 07:40

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