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Let {$e_i$} be a basis of n dimensional inner product space $V$. Let $f:V \to V$ be a function such that $f(0)=0$ and $||f(x)-f(y)||=||x-y||$. then how can I able to prove that {$f(e_i)$} is also an orthonormal basis of $V$

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At least the answer is true when $f$ is surjective. In this case, the Mazur-Ulam theorem implies that the isometry map $f$ is affine and since $f(0)=0$ we conclude that it is in fact linear. The we can use polarization identities. For example, for a real vector space:

$$ <f(e_i),f(e_j)> = \frac{1}{4} \left( \lVert f(e_i)+f(e_j) \rVert ^2 - \lVert f(e_i)- f(e_j) \rVert ^2 \right) = \frac{1}{4}\left( \lVert e_i+e_j \rVert ^2 - \lVert e_i- e_j \rVert ^2\right) = <e_i,e_j>=\delta_{ij}$$

The Polarization identity has a different form for a complex vector space, but the proof is equally trivial.

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    $\begingroup$ Note that $f$ is not assumed to be linear, So it is better to use $$\langle f(e_i),f(e_j)\rangle=\frac{1}{2}\left(\Vert f(e_i)\Vert^2+\Vert f(e_j)\Vert^2-\Vert f(e_i-e_j)\Vert^2\right).$$ the Complex case needs more attention. $\endgroup$ Apr 10, 2014 at 7:13
  • $\begingroup$ @OmranKouba You are right, only if $f$ is surjective we can use them the way I stated. Thanks for the observation. $\endgroup$ Apr 14, 2014 at 4:11

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