### Orthogonal and symmetric maps: Isometries

### The notion of isometry

As we saw earlier, orthogonal maps are linear transformations of an inner product space to itself which preserve length and therefore also distance. Here, we look at the more general case of maps between inner product spaces that maintain distance.

Isometry Let #V# and #W# be real inner product spaces. A map #L :V\rightarrow W# is called an **isometry** if #\norm{L(\vec{x})-L(\vec{y})}=\norm{\vec{x}-\vec{y}}# for all #\vec{x},\vec{y}\in V#. If, moreover, #L# is linear, then #L# is a **linear isometry**.

In some literature the term isometry is used for a linear map that preserves the length. The following theorem will show that the above definition is equivalent as far as linear maps are concerned.

Characterisations of linear isometriesLet #V# and #W# be real inner product spaces and let #L:V\to W# be a linear map. The following statements are equivalent:

- For each #\vec{x}# in #V# we have #\norm{L(\vec{x})}=\norm{\vec{x}}#.
- The map #L# is a linear isometry.
- For each #\vec{x}# and #\vec{y}# in #V# we have #\dotprod{L(\vec{x})}{L(\vec{y})}=\dotprod{\vec{x}}{\vec{y}}#.

Here are some general properties of isometries.

Properties of isometries Let #U#, #V#, #W# be real inner product spaces.

- If #L:V\rightarrow W# and #M :U\rightarrow V# are isometries, then the composition #L\,M:U\rightarrow W# also is an isometry.
- If #L:V\rightarrow W# is an isometry, then #L# is injective.
- If #L:V\rightarrow W# is a linear isometry and #V# and #W# have equal finite dimension, then #L# is invertible and also #L^{-1}# is a linear isometry.

Because #L# is linear, it is an isometry if and only if #\norm{L(x)}=\abs{x} # for all real #x#. This leads to an equation with unknown #a# which we can solve: \[\begin{array}{rcl}\norm{\dfrac{x}{7}\,\rv{a, -3, 6 }}&=&\abs{x}\\ &&\phantom{xx}\color{blue}{\text{mapping rule for }L\text{ used}}\\ \abs{\dfrac{x}{7}}\cdot \norm{\rv{a, -3, 6}}&=&\abs{x}\\ &&\phantom{xx}\color{blue}{\text{multiplicativity of the norm }}\\

\norm{\rv{a, -3, 6}}&=&7\\ &&\phantom{xx}\color{blue}{x = 1\text{ substituted and multiplied by }7}\\

a^2+(-3)^2+{6}^2&=&{7}^2\\ &&\phantom{xx}\color{blue}{\text{length calculated and both sides squared}}\\

a&=&\pm 2\\&&\phantom{xx}\color{blue}{\text{quadratic equation solved}}\\

a &=& 2\\&&\phantom{xx}\color{blue}{a\ge 0\text{ used}}

\end{array}\]

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