Is $f(x) = T(x) + b$ with $x \in \mathbb{R}^n$, $b \in \mathbb{R}^n$ and $T_i(x) = \sum_j a_{ji}x_j $ an global isometry?

Question

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 global isometry. It's easy to prova that $d(x,y) = d(T(x),T(y))$ and $T$ its bijective because $ker(T) = {0}$. Therefore it's easy to prove that $d(x,y) = d(T(x) + b, T(y) + b)$, but $T(x) + b$ it's not an linear map. How can i prove that $T(x) + b$ is surjective?

And how can i prove that every global isometry $f$ from $\mathbb R^n$ to $\mathbb R^n$ have form $f(x) = T(x) + b$?

Answer 1

Google provided me several links. For instance

• Mikael Forsberg "Isometries of the plane", Theorem 1.4

• Keith Conrad "Isometries of $\Bbb R^n$",Theorems 2.1 and 4.1

• "Isometries in $\Bbb R^n$", Theorem 16.6

• "Isometries of the Euclidean Space", Theorem 1

• Salil Vadhan "Isometries" (only a claim without a proof).

Answer 2

Hint: $f(x)-b= T(x)$ is a linear map that can be expressed as a matrix with determinant 1 or -1.