# Statement(s) > [!NOTE] Statement 1 (Necessary Conditions for Map on Real n-Space to be an Isometry) > Let $L:\mathbb{R}^n \to \mathbb{R}^n$ be a [[Linear maps|linear operator]]. Then the following are equivalent: > > (1) $L$ is an [[Isometry|isometry]]. > > (2) $L$ is [[Euclidean Norm|norm]] preserving: for all $\underline{x}\in \mathbb{R}^n,$ $\Vert L(\underline{x})\Vert = \Vert \underline{x} \Vert$ > > (3) $L$ preserves [[Inner products|inner product]]: for all $\underline{x},\underline{y}\in \mathbb{R}^n,$ $\langle L(\underline{x}),L(\underline{y})\rangle =\langle \underline{x}, \underline{y}\rangle$ > > (4) The [[Matrix representations of linear map|matrix]] of $A$ is [[Orthogonal endomorphisms of Euclidean spaces|orthogonal]]. # Proof(s) **Proof of statement 1:** (1) $\Rightarrow(2)$ : Using definition of the metric, defining property of isometries, and linearity of $L$ : $\|\mathbf{x}\|=d(\mathbf{x}, \mathbf{0})=d(L(\mathbf{x}), L(\mathbf{0}))=d(L(\mathbf{x}), \mathbf{0})=\|L(\mathbf{x})\|$. $(2) \Rightarrow(3)$ : This follows from [[Euclidean Norm Satisfies Polarization Identity]]. $(3) \Rightarrow(4)$ : Now compute the elements of $A^T A$. By the formula for matrix multiplication$\left(A^T A\right)_{i j}=\sum_{k=1}^n\left(A^T\right)_{i k} A_{k j}=\sum_{k=1}^n A_{k i} A_{k j}=\sum_{k=1}^n L\left(\mathbf{e}_i\right)_k L\left(e_j\right)_k=\left\langle L\left(\mathbf{e}_i\right), L\left(e_j\right)\right\rangle$ So the entries of $A^T A$ are the inner products of the columns of $A$. Since the $\mathbf{e}_i$ are the standard basis, and by hypothesis $(3)$, $L$ preserves the inner product, we have $\left(A^T A\right)_{i j}=\left\langle L\left(\mathbf{e}_i\right), L\left(e_j\right)\right\rangle=\left\langle\mathbf{e}_i, e_j\right\rangle= \begin{cases}1 & \text { if } i=j \\ 0 & \text { if } i \neq j\end{cases}$ These are the elements of the identity matrix, so $A$ is orthogonal. $(4) \Rightarrow(1)$ : Assuming $(4)$, $A$ is invertible with inverse $A^T$, so $L$ is a bijection. So we just need to check $L$ is distance preserving. First check that $L$ is norm preserving: $\begin{aligned} \|L(\mathbf{x})\|^2 & =\langle L(\mathbf{x}), L(\mathbf{x})\rangle \\ & =(L(\mathbf{x}))^T L(\mathbf{x}) \text { using matrix notation for dot product } \\ & =(A \mathbf{x})^T(A \mathbf{x}) \text { writing } L(\mathbf{x})=A \mathbf{x} \\ & =\mathbf{x}^T A^T A \mathbf{x} \text { property of transpose } \\ & =\mathbf{x}^T \mathbf{x} \text { since } A \text { is orthogonal } \\ & =\langle\mathbf{x}, \mathbf{x}\rangle, \text { by definition of inner product } \\ & =\|\mathbf{x}\|^2, \text { by definition of norm. } \end{aligned}$ Now distance preservation follows from the linearity of $L$: $d(L(\mathbf{x}), L(\mathbf{y}))=\|L(\mathbf{x})-L(\mathbf{y})\|=\|L(\mathbf{x}-\mathbf{y})\|=\|\mathbf{x}-\mathbf{y}\|=d(\mathbf{x}, \mathbf{y})$ Since $A$ has inverse $A^T, L$ is invertible, so $L$ is an isometry. $\blacksquare$ # Application(s) **Consequences**: **Examples**: # Bibliography