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If V is a finite dimensional real inner product space with an orthonormal basis $e_1,...,e_n$, and there exists a linear operator $T$ such that $\|T(e_i+e_j)\| = \|e_i+e_j\|$ then $T$ is an isometry.

My attempts so far have been to write $v=\langle v,e_1\rangle e_1+...$ from which it follows $Tv=\langle v,e_1\rangle T e_1+...$ and thus $\|T(v)\|^2=\|\langle v,e_1\rangle T e_1+...\|^2$ but I cannot break this down further since we don't know if the $Te_i$'s aren't orthonormal. I also attempt to use the fact given that $\|T(e_i+e_j)\| = \|e_i+e_j\|$ but cannot find a way to incorporate that into the expansion.

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Hint: For $i \neq j$ we get $\|Te_i+Te_j\|^{2}=\|e_i+e_j||^{2}=2$. So $\|Te_i\|^{2}+\|Te_j\|^{2}+2 \langle Te_i, Te_j \rangle=2$. But $\|Te_i\|^{2}=\|Te_j\|^{2}=1$ so we get $ \langle Te_i, Te_j \rangle=0$.

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  • $\begingroup$ @DavidSkarsgaard Just put $i=j$ in the hypothesis. $\endgroup$ Feb 19, 2021 at 8:40

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