Characterisation of isometries

Orthogonal and symmetric maps: Isometries

Characterisation of isometries

Earlier we saw that translations are isometries which are not linear. The following theorem says that every isometry can be written in a unique way as the composition of a linear isometry and a translation. We recall that a property of isometries is that they are injective.

Characterization of isometries Let $V$ and $W$ be inner product spaces.

Every isometry $V\to W$ which fixes the zero vector, is linear.
Each isometry $H:V\to W$ can be written in exactly one way as the composition $H = T_{\vec{a}}\, L$ of a translation $T_{\vec{a}}$ along a vector $\vec{a}$ of $V$ and a linear isometry $L:V\to W$ , where $\vec{a}= H(\vec{0})$ .

Fixing

We say that a map $V\to W$ fixes the zero vector if it maps $\vec{0}_V$ onto $\vec{0}_W$ .

1. Let $H$ be an isometry that fixes the zero vector, that is, $H(\vec{0}_V) =\vec{0}_W$ . Then $H$ preserves the norm, because the norm of each vector $\vec{x}$ is its distance to the vector $\vec{0}_V$ , and therefore equal to the distance from $H(\vec{x})$ to $H(\vec{0}_V)= \vec{0}_W$ .

Because $H$ preserves the norm, $H$ also leaves invariant the inner product. After all, the polarization formula shows

$\begin{array}{rcl} \dotprod{H(\vec{x})}{H(\vec{y})} &=&\displaystyle -\dfrac{1}{2}\left({\parallel H(\vec{x})-H(\vec{y})\parallel} + {\parallel H(\vec{x})\parallel}+{\parallel H(\vec{y})\parallel}\right)\\ &&\phantom{xxx}\color{blue}{\text{polarization formula applied to }H(\vec{x})\text{ and }-H(\vec{y})}\\& =& \displaystyle-\frac12\left({\parallel \vec{x}-\vec{y}\parallel} + {\parallel \vec{x}\parallel}+{\parallel \vec{y}\parallel}\right) \\&&\phantom{xxx}\color{blue}{H \text{ preserves distance and norm}}\\&=& \dotprod{\vec{x}}{\vec{y}}\\&&\phantom{xxx}\color{blue}{\text{polarization formula applied to }\vec{x}\text{ and }-\vec{y}}\end{array}$

First we use the fact that $H$ preserves the inner product in order to prove the summation rule (which states that $H$ respects the addition). Let $\vec{x}$ , $\vec{y}$ be vectors of $V$ . We show that $H(\vec{x}+\vec{y}) = H(\vec{x})+H(\vec{y})$ holds by deriving that the inner product of the difference

$\vec{v}=H(\vec{x}+\vec{y}) - \left(H(\vec{x})+H(\vec{y})\right)$ of the left- and right-hand sides with itself equals $0$ :

$\begin{array}{rcl}\dotprod{\vec{v}}{\vec{v}}&=&\dotprod{\left(H(\vec{x}+\vec{y}) -( H(\vec{x})+H(\vec{y}))\right)}{\left(H(\vec{x}+\vec{y}) -( H(\vec{x})+H(\vec{y}))\right)} \\ &&\phantom{xxx}\color{blue}{\text{definition of }\vec{v}}\\ &=&\dotprod{H(\vec{x}+\vec{y})}{H(\vec{x}+\vec{y})} + \dotprod{H(\vec{x})}{H(\vec{x})}+ \dotprod{H(\vec{y})}{H(\vec{y})}\\ &&+2 \dotprod{H(\vec{x})}{H(\vec{y})}-2 \dotprod{H(\vec{x}+\vec{y})}{H(\vec{x})}-2\dotprod{H(\vec{x}+\vec{y}) }{H(\vec{y})}\\&&\phantom{xxx}\color{blue}{\text{brackets expanded by use of linearity of the inner product}}\\&=&\dotprod{(\vec{x}+\vec{y})}{(\vec{x}+\vec{y})} + \dotprod{\vec{x}}{\vec{x}}+ \dotprod{\vec{y}}{\vec{y}}\\ &&+2 \dotprod{\vec{x}}{\vec{y}}-2 \dotprod{(\vec{x}+\vec{y})}{\vec{x}} -2\dotprod{(\vec{x}+\vec{y}) }{\vec{y}}\\&&\phantom{xxx}\color{blue}{H\text{ preserves the inner product}}\\&=&\dotprod{\vec{x}}{\vec{x}}+\dotprod{\vec{y}}{\vec{y}} +2\dotprod{\vec{x}}{\vec{y}} + \dotprod{\vec{x}}{\vec{x}}+ \dotprod{\vec{y}}{\vec{y}}\\ &&+2 \dotprod{\vec{x}}{\vec{y}}-2 \dotprod{\vec{x}}{\vec{x}}-2\dotprod{\vec{y}}{\vec{x}} -2\dotprod{\vec{x} }{\vec{y}}-2\dotprod{\vec{y} }{\vec{y}}\\&&\phantom{xxx}\color{blue}{\text{brackets expanded by use of linearity of the inner product}}\\&=& 0 \\&&\phantom{xxx}\color{blue}{\text{simplified}}\end{array}$ Because of the positive definiteness of the inner product, it follows that $\vec{v} = \vec{0}$ , so $H(\vec{x}+\vec{y}) = H(\vec{x})+H(\vec{y})$ .

Now we prove the scalar rule, that is, the fact that $H$ preserves scalar multiplication. Let $\vec{x}$ be a vector $V$ and let $\lambda$ be a scalar. We show that $H(\lambda \cdot \vec{x}) = \lambda \cdot H(\vec{x})$ holds by deriving that the inner product of the difference

$\vec{w}=H(\lambda \cdot \vec{x}) - \left(\lambda \cdot H(\vec{x})\right)$ of the left- and right-hand side with itself equals $0$ :

$\begin{array}{rcl}\dotprod{\vec{w}}{\vec{w}}&=&\dotprod{\left(H(\lambda \cdot \vec{x}) -\lambda \cdot H( \vec{x})\right)}{\left(H(\lambda \cdot \vec{x}) -\lambda \cdot H( \vec{x})\right)} \\ &&\phantom{xxx}\color{blue}{\text{definition of }\vec{w}}\\ &=&\dotprod{H(\lambda \cdot \vec{x})}{H(\lambda \cdot \vec{x})} + \dotprod{\lambda \cdot H(\vec{x})}{\left(\lambda \cdot H(\vec{x})\right)}-2 \dotprod{H(\lambda\cdot\vec{x})}{\left(\lambda\cdot H(\vec{x})\right)}\\&&\phantom{xxx}\color{blue}{\text{brackets expanded by use of linearity of the inner product}}\\&=&\dotprod{(\lambda \cdot \vec{x})}{(\lambda \cdot \vec{x})} + \dotprod{(\lambda \cdot \vec{x})}{\left(\lambda \cdot \vec{x}\right)}-2 \dotprod{(\lambda\cdot\vec{x})}{\left(\lambda\cdot \vec{x}\right)}\\&&\phantom{xxx}\color{blue}{H\text{ preserves the inner product}}\\&=&\lambda^2\cdot \dotprod{\vec{x}}{\vec{x}}+\lambda^2\cdot \dotprod{\vec{x}}{\vec{x}} -2\lambda^2\cdot \dotprod{\vec{x}}{\vec{x}} \\ &&\phantom{xxx}\color{blue}{\lambda \text{ pulled outside brackets by use of linearity of the inner product}}\\&=& 0 \\&&\phantom{xxx}\color{blue}{\text{simplified}}\end{array}$ Due to the positive definiteness of the inner product, it follows that $\vec{w} = \vec{0}$ , so $H(\lambda\cdot \vec{x}) = \lambda\cdot H(\vec{x})$ .

According to the definition of a linear map this shows that $H$ is linear.

2. Write $\vec{a}= H(\vec{0})$ and $L =T_{-\vec{a}}\, H$ . Because $L$ is the composition of two isometries (namely $H$ followed by a translation in $W$ ), the map $L$ is an isometry (See properties of isometries). Moreover,

$L(\vec{0}) = T_{-\vec{a}}\, H(\vec{0}) =T_{-\vec{a}} (\vec{a}) = \vec{a}-\vec{a} = \vec{0}$ so $L$ fixes the zero vector. Using statement 1, we conclude that $L$ is linear. By applying the translation $T_{\vec{a}}$ to both sides of the definition, we find $T_{\vec{a}}\,L =T_{\vec{a}}\,T_{-\vec{a}}\, H = H$ .

The expression is unique: Suppose that $H =T_{\vec{b}}M$ , for a vector $\vec{b}$ of $W$ and a linear isometry $M: V\to W$ . Then $\vec{b} =T_{\vec{b}}M(\vec{0}) = H(\vec{0}) = \vec{a}$ , so $\vec{b}=\vec{a}$ . Consequently, we have $T_{\vec{a}}\,M =T_{\vec{b}}\,M = H =T_{\vec{a}}\,L$ . Since translations are bijective, we conclude that $M = L$ .

Every linear map $L:\mathbb{R}\to \mathbb{R}$ is given by the multiplication by a number $\lambda$ . If $L$ is orthogonal, then $|\lambda| = 1$ . According to the theorem, each isometry $H : \mathbb{R}\to \mathbb{R}$ is of the form $L\,T_{a}$ for a real number $a$ , so

$H(x) = \pm(x+a)\phantom{xxx}\text{ for }\phantom{xxx}x\in \mathbb{R}$

In accordance with the first statement, the isometry $L:V\to V$ from an inner product space $V$ to itself is orthogonal if and only if $L$ fixes $\vec{0}$ .

For each vector $\vec{w} \in W$ and each linear isometry $L:V\to W$ , the composition $T_{\vec{w}}\, L$ also is an isometry. According to the above theorem, this isometry is of the form $M\, T_{\vec{a}}$ . The first rules below indicates what $M$ and $\vec{a}$ are.

Rules of calculation with translations

Let $V$ and $W$ be inner product spaces.

1. If $L: V\to W$ is a linear map and $\vec{c}$ is a vector of $V$ , then the following commutation rule for a translation holds:

$L\,T_{\vec{c}} = T_{L(\vec{c})}\, L$

2. If $L$ and $N$ are linear maps $V\to V$ and $\vec{a}$ , $\vec{c}$ are vectors of $V$ , then the composition of $T_{\vec{a}}\, L$ and $T_{\vec{c}}\,N$ is given by

$\left(T_{\vec{a}}\, L\right)\,\left(T_{\vec{c}}\,N \right)= T_{\vec{a}+L(\vec{c})}\,\left(L\,N\right)$

If, in addition, $L$ and $N$ are orthogonal maps $V\to V$ , then the composition of the two isometries $T_{\vec{a}}\, L$ and $T_{\vec{c}}\,N$ is an isometry again.

1. For each $\vec{x}$ in $V$ we have

$\begin{array}{rcl}L \, T_{\vec{c}} (\vec{x}) &=& L(\vec{x}+\vec{c})\\ &&\phantom{xxx}\color{blue}{\text{mapping rule of }T_{\vec{c}}}\\ &=& L(\vec{x})+L(\vec{c})\\&&\phantom{xxx}\color{blue}{\text{linearity of }L} \\&=& T_{L(\vec{c})}\, L(\vec{x})\\ &&\phantom{xxx}\color{blue}{\text{mapping rule of }T_{L(\vec{c})}}\end{array}$

2. According to the proposition, the image $K= (T_{\vec{a}}\, L)\,(T_{\vec{c}}\,N)$ can be written as $T_{\vec{d}}\,P$ , where $\vec{d}$ is a vector, and $P:V\to W$ a linear isometry. The uniqueness tells us how $\vec{d}$ and $P$ can be expressed in $\vec{a}, L, \vec{c},N$ :

$\begin{array}{rcl}\vec{d} &=& K(\vec{0}) = T_{\vec{a}}\, L\,T_{\vec{c}}\,N(\vec{0}) = T_{\vec{a}}\, L\,T_{\vec{c}}(\vec{0}) = T_{\vec{a}}\, L\,(\vec{c}) = \vec{a}+ L\,(\vec{c}) \\ P &=& L\, N\phantom{xx}\color{blue}{\text{(the product of the linear isometries)}} \end{array}$

Also, a direct calculation using rule 1 leads to the entire result:

$\left(T_{\vec{a}}\, L\right)\,\left(T_{\vec{c}}\,N \right)=T_{\vec{a}}\,\left( L\,T_{\vec{c}}\right)\,N =T_{\vec{a}}\,T_{L(\vec{c})}\,L\,N =T_{\vec{a}+L(\vec{c})}\,L\,N$

If $L:V\to V$ is invertible, then, for $\vec{a}$ in $V$ , the commutation rule for a translation can also be written as

$L \, T_{\vec{a}}\,L^{-1} = T_{L(\vec{a})}$ The left-hand side is sometimes referred to as the conjugate of $T_{\vec{a}}$ by $L$ . Conjugation by $L$ thus transforms translations to translations.

Dimension 1

We saw above that each isometry $H : \mathbb{R}\to \mathbb{R}$ is of the form $L\,T_{a}$ for a real number $a$ , so

$H(x) = \pm(x+a)\phantom{xxx}\text{ for }\phantom{xxx}x\in \mathbb{R}$

In accordance with the first rule of calculation with translations, we have, with $\delta = \pm 1$ and $L_\delta$ scalar multiplication by $\delta$ on $\mathbb{R}$ ,

$H = L_\delta \, T_a = T_{\delta\cdot a} L_\delta$

Consider the isometry $H:\mathbb{R}^2\to\mathbb{R}^2$ which assigns to each vector $\vec{x}=\rv{x_1,x_2}$ the vector

$H(\vec{x}) = \rv{-{{27}\over{25}}+\frac{7}{25} x_1 -\,\frac{24}{25} x_2, {{86}\over{25}}+\frac{24}{25} x_1 +\frac{7}{25} x_2}$ Determine the vector $\vec{v}$ for which $H$ can be written in the form $A\, T_{\vec{v}}$ , where $A$ is an orthogonal $(2\times2)$ -matrix.

$\vec{v} =$ $\rv{3,2}$

Rewriting the image of $\vec{x} = \rv{x_1,x_2}$ under $H$ shows that

$\begin{array}{rcl}H\left(\vec{x}\right)&=&\displaystyle \rv{-{{27}\over{25}} +\frac{7}{25} x_1-\,\frac{24}{25} x_2, {{86}\over{25}}+\frac{24}{25} x_1 + \frac{7}{25} x_2}\\ &=&\displaystyle T_{\rv{-{{27}\over{25}},{{86}\over{25}}}}\,\left(\rv{\frac{7}{25} x_1-\,\frac{24}{25} x_2, \frac{24}{25} x_1 + \frac{7}{25} x_2}\right)\\ \\ &=& T_{\rv{-{{27}\over{25}},{{86}\over{25}}}}\,\matrix{\frac{7}{25} & -\,\frac{24}{25}\\ \frac{24}{25} & \frac{7}{25}}\, \vec{x} \end{array}$
so $H = T_{\rv{-{{27}\over{25}},{{86}\over{25}}}}\, A$ , where $A = \matrix{\frac{7}{25} & -\,\frac{24}{25}\\ \frac{24}{25} & \frac{7}{25}}$ .

In order to find $\vec{v}$ , we use the commutation rule for a translation:

$\begin{array}{rcl}T_{\rv{-{{27}\over{25}},{{86}\over{25}}}}\, A &=& A\, A^{\top}\, T_{\rv{-{{27}\over{25}},{{86}\over{25}}}}\, A\\ &&\phantom{xx}\color{blue}{A\text{ is orthogonal}}\\ &=&A\, T_{A^{\top}\,\rv{-{{27}\over{25}},{{86}\over{25}}}} \, A^{\top} A\\&&\phantom{xx}\color{blue}{\text{commutation rule}}\\&=& A\, T_{A^{\top}\,\rv{-{{27}\over{25}},{{86}\over{25}}}} \\&&\phantom{xx}\color{blue}{A\text{ is orthogonal}}\end{array}$ We conclude that

$\vec{v} =A^{\top}\,\rv{-{{27}\over{25}},{{86}\over{25}}} = \matrix{\frac{7}{25} & \frac{24}{25}\\ -\,\frac{24}{25} & \frac{7}{25}}\,\rv{-{{27}\over{25}},{{86}\over{25}}}=\rv{3,2}$

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