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For an orthonormal basis $v_1, ..., v_n$ of $(V, \cdot )$

$\mathbb{R}^n \ni (x_1, ..., x_n) \rightarrow \sum x_jv_j \in V$

is an isomorphism preserving dot product.

I've already proven that it preserves dot product, but I have problems showing that it is isomorphic.

Could you tell me how to do it?

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    $\begingroup$ I do not understand what you mean with "preserving dot product". $\endgroup$ – Avitus Jun 16 '13 at 16:12
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The linear map $\phi_v:\mathbb R^n\rightarrow V$ is s.t. $\phi_v(x_1,\dots,x_n):=\sum_{i=1}^n x_iv_i$, denoting by $v$ the orthonormal basis $\{v_1,\dots, v_n\}$ in $V$. It admits inverse given by $\psi:V\rightarrow \mathbb R^n$, with $\psi(w)=(w_1,\dots,w_n)$, for any $w\in V$, and $w=\sum_{i=1}^n w_iv_i$.

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  • $\begingroup$ Thank you very much. How stupid of me, I didn't think ogf that. By preserving dot product I mean that if $f: V \rightarrow W$ is linear and preserves dot product then $\forall v_1, v_2 \in V \ : \ v_1 \cdot v_2 = f(v_1) \cdot f(v_2)$ $\endgroup$ – Hagrid Jun 16 '13 at 16:15
  • $\begingroup$ You are welcome. My concern is about dot product: what is, for example the "dot product" on $\mathbb R^n$ you are referring to? If it is the scalar product, then its result is an element of $\mathbb R$, and not a vector itself. $\endgroup$ – Avitus Jun 16 '13 at 16:21
  • $\begingroup$ By dot product I mean scalar product, but I didn't say that its result is an element of $V$. $\endgroup$ – Hagrid Jun 16 '13 at 16:41
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    $\begingroup$ Ok, now I understand: the relation you proved is $(x_1,\dots,x_n)\cdot(y_1,\dots,y_n)=\phi_v(x_1,\dots,x_n)*\phi_v(y_1,\dots,y_n)$. A similar relation holds for the inverse $\psi$, and now the proof is complete :-) Thanks! $\endgroup$ – Avitus Jun 16 '13 at 17:07
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    $\begingroup$ Thank you very much for your help! $\endgroup$ – Hagrid Jun 16 '13 at 17:12

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