The notion of isometry

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.

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 other words, a map $L:V\to W$ is an isometry when the distance between each pair of vectors $\vec{x}$ and $\vec{y}$ is preserved under the map $L$ . An isometry does not necessarily preserve the length of a vector. If $L$ is a linear isometry, then it will preserve the length, as will become apparent from the theorem below.

Orthogonal maps If $W=V$ and $L$ is a linear map, then $L$ is an orthogonal map. Also, every orthogonal map is an isometry.

Translation Earlier we saw that a translation preserves length and distance, but is not linear. Thus, translations are examples of non-linear isometries.

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.

Let $V$ 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 $\dotprod{L(\vec{x})}{L(\vec{y})}=\dotprod{\vec{x}}{\vec{y}}$ .

We first prove the implication $1\Rightarrow 2$ . Let $V$ , $W$ , and $L$ be as in the statement of the theorem. We now assume that $\norm{L(\vec{x})}=\norm{\vec{x}}$ for each $\vec{x}$ in $V$ . The following equations show that $L$ preserves distance: $\norm{L(\vec{x})-L(\vec{y})}=\norm{L(\vec{x}-\vec{y})}=\norm{\vec{x}-\vec{y}}$

The proof $2\Rightarrow 1$ follows from the following equations: $\norm{L(\vec{x})}=\norm{L(\vec{x})-L(\vec{0})}=\norm{\vec{x}-\vec{0}}=\norm{\vec{x}}$ In both proofs we used that $L$ is linear.

To establish the equivalence of $1$ and $3$ we repeat the proof given for the case of orthogonal maps.

Here are some general 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.

1. For all $\vec{x},\vec{y}\in U$ we have $\norm{L\,M(\vec{x})-L\,M(\vec{y} )}=\norm{M(\vec{x})-M(\vec{y})}=\norm{\vec{x}-\vec{y}}$ because $L$ and $M$ are isometries, so also $L\,M$ is an isometry.

2. Set $L(\vec{x}) = L(\vec{y})$ . Then $L(\vec{x})-L(\vec{y})=\vec{0}$ , so $\norm{L(\vec{x}) - L(\vec{y})}=0$ . Because $L$ is an isometry, it follows that $\norm{\vec{x} - \vec{y}}=0$ , and, because of the positivity of the norm, $\vec{x} -\vec{y}=\vec{0}$ , so $\vec{x}=\vec{y}$ . This proves that $L$ is injective.

3. First we show that $L$ is invertible. Statement 2 shows that $L$ is injective. Because $V$ is finite-dimensional and $L$ is linear, it follows from Invertiblility with the same dimensions for domain and codomain that $L$ is invertible. All $\vec{x}\in V$ satisfy $\vec{x}=L\, L^{-1}(\vec{x})$ . Because $L$ is linear, the previous theorem gives that $\norm{L(\vec{x})}=\norm{\vec{x}}$ . This implies that $\norm{\vec{x}}=\norm{L\,L^{-1}\vec{x}}=\norm{L^{-1}\vec{x}}$ so $L^{-1}$ also is a linear isometry.

Let $a$ be a non-negative real number and $L :\mathbb{R}\to \mathbb{R}^3$ the linear map given by
$L (x)=\frac{x}{19}\,\rv{a, -15, -6 }$ We work with the standard inner product on $\mathbb{R}$ and on $\mathbb{R}^3$ .

For what value of $a$ is $L$ an isometry?

$a=$ $10$

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: $\begin{array}{rcl}\norm{\dfrac{x}{19}\,\rv{a, -15, -6 }}&=&\abs{x}\\ &&\phantom{xx}\color{blue}{\text{mapping rule for }L\text{ used}}\\ \abs{\dfrac{x}{19}}\cdot \norm{\rv{a, -15, -6}}&=&\abs{x}\\ &&\phantom{xx}\color{blue}{\text{multiplicativity of the norm }}\\ \norm{\rv{a, -15, -6}}&=&19\\ &&\phantom{xx}\color{blue}{x = 1\text{ substituted and multiplied by }19}\\ a^2+(-15)^2+(-6)^2&=&{19}^2\\ &&\phantom{xx}\color{blue}{\text{length calculated and both sides squared}}\\ a&=&\pm 10\\&&\phantom{xx}\color{blue}{\text{quadratic equation solved}}\\ a &=& 10\\&&\phantom{xx}\color{blue}{a\ge 0\text{ used}} \end{array}$

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