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Let $a_1, a_2, a_3,\ldots,a_n \in\mathbb R^n$, $a_i = (a_{i1},a_{i2},\ldots,a_{in})$ , $\langle a_i,a_i\rangle = \sum_j a_{ij}a_{ij} = 1$ and if $i \neq k$ then $\langle a_i,a_k\rangle = \sum_j a_{ij}a_{kj} = 0$. If $T$ is a function from $\mathbb R^n$ to $\mathbb R^n$ defined by $T(x_1,x_2,\ldots,x_n) = (y_1, y_2,\ldots,y_n)$ where $y_i = \sum_j a_{ji}x_j$. Then $T$ is an isometry.

\begin{align} d(T(x),T(y)) & = \sqrt{\sum_i \left(\sum_j a_{ji}x_j-a_{ji}y_j \right)^2} \\[10pt] = & \sqrt{\sum_i \left( \sum_j \sum_k a_{ji}a_{ki}(x_j-y_j)(x_k-y_k)\right)} = \sqrt{\sum_i^{} (x_i-y_i)^2} = d(x,y). \end{align}

I did prove that $d(x,y) = d(T(x),(T(y))$ for any $x,y \in\mathbb R^n$. How can i prove that $T$ is surjective?

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  • $\begingroup$ Proper notation is $\langle a_i,a_k\rangle = \sum_j \cdots,$ not $<a_i,a_k> = \sum_j \cdots.$ I edited accordingly, along with some other copy-editing. $\endgroup$ – Michael Hardy Oct 9 '17 at 4:23
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    $\begingroup$ You have already shown that $T$ preserves norms. Thus, its null space is trivial. $\endgroup$ – littleO Oct 9 '17 at 4:27
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Since $n$ is finite, if $T$ is not surjective, then $\ker(T)\ne\{0\}.$ If $x,y\in \ker(T)$ and $x\ne y,$ then $T(x)=T(y).$ Since $x\ne y,$ then $d(x,y)\ne 0,$ but $T(x)=T(y)=0,$ then $d(T(x),T(y))=0,$ and so $d(x,y) \ne d(T(x),T(y)),$ contradicting our assumption that $T$ is an isometry.

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