Vector space/K/Inner product/Linear isometry/Introduction/Section

Definition

Let ${}V,W$ be vector spaces over ${}{\mathbb {K} }$, endowed with inner products, and let

$\varphi \colon V\longrightarrow W$

be s linear mapping. Then ${}\varphi$ is called an isometry if

${}\left\langle \varphi (v),\varphi (w)\right\rangle =\left\langle v,w\right\rangle \,$

holds for all

${}v,w\in V$.

An isometry is always injective. For ${}{\mathbb {K} }=\mathbb {C}$, we also talk about an unitary mapping. As there are also affine isometries, we talk about a linear isometry.

Lemma

Let ${}V$ and ${}W$ be vector spaces over ${}{\mathbb {K} }$, both endowed with an inner product, and let ${}\varphi \colon V\rightarrow W$ be a

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is an isometry.
2. For all ${}u,v\in V$, we have ${}d(\varphi (u),\varphi (v))=d(u,v)$.
3. For all ${}v\in V$, we have ${}\Vert {\varphi (v)}\Vert =\Vert {v}\Vert$.
4. For all ${}v\in V$ fulfilling ${}\Vert {v}\Vert =1$, we have ${}\Vert {\varphi (v)}\Vert =1$.

Proof

The implications ${}(1)\Rightarrow (2)$, ${}(2)\Rightarrow (3)$ and ${}(3)\Rightarrow (4)$ are restrictions. ${}(4)\Rightarrow (3)$. For the zero vecto r,the statement ${}(3)$ is clear; so suppose that ${}v\neq 0$. Then, ${}{\frac {v}{\Vert {v}\Vert }}$ has norm ${}1$, and, because of

${}\varphi (v)=\varphi {\left(\Vert {v}\Vert {\frac {v}{\Vert {v}\Vert }}\right)}=\Vert {v}\Vert \varphi {\left({\frac {v}{\Vert {v}\Vert }}\right)}\,,$

we have

${}\Vert {\varphi (v)}\Vert =\Vert {v}\Vert \,.$

${}(3)\Rightarrow (1)$ follows from fact.

$\Box$

Therefore, an isometry is just a (linear) mapping that preserves distances. The set of all the vectors with norm ${}1$ in a Euclidean vector space is also called the sphere. Hence, an isometry is characterized by the property that it maps the sphere to the sphere.

Lemma

Let ${}V$ and ${}W$ be euclidean vector spaces, and let

$\varphi \colon V\longrightarrow W$

denote a

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is an isometry.
2. For every orthonormal basis ${}u_{i}$, ${}i=1,\ldots ,n$, of ${}V$, ${}\varphi (u_{i})$, ${}i=1,\ldots ,n$, is part of an orthonormal basis of ${}W$.
3. There exists an orthonormal basis ${}u_{i}$, ${}i=1,\ldots ,n$, of ${}V$ such that ${}\varphi (u_{i})$, ${}i=1,\ldots ,n$, is part of an orthonormal basis of ${}W$.

Proof

See exercise.

$\Box$

Theorem

For every euclidean vector space ${}V$, there exists a bijective isometry

$\varphi \colon \mathbb {R} ^{n}\longrightarrow V,$

where ${}\mathbb {R} ^{n}$ carries the

standard inner product.

Proof

Let ${}u_{1},\ldots ,u_{n}$ be an orthonormal basis of ${}V$, and let

$\varphi \colon \mathbb {R} ^{n}\longrightarrow V$

be the linear mapping given by

${}\varphi (e_{i})=u_{i}\,.$

Because of fact  (3), this is an isometry.

$\Box$