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.
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})#.
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.
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.
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 \[H(\vec{x}) = \rv{{{28}\over{17}}+\frac{8}{17} x_1 -\,\frac{5}{17} x_2, -{{27}\over{17}}+\frac{5}{17} x_1 +\frac{8}{17} x_2}\]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{1,-4}#
Rewriting the image of #\vec{x} = \rv{x_1,x_2}# under #H# shows that
\[\begin{array}{rcl}H\left(\vec{x}\right)&=&\displaystyle \rv{{{28}\over{17}} +\frac{8}{17} x_1-\,\frac{5}{17} x_2, -{{27}\over{17}}+\frac{5}{17} x_1 + \frac{8}{17} x_2}\\ &=&\displaystyle
T_{\rv{{{28}\over{17}},-{{27}\over{17}}}}\,\left(\rv{\frac{8}{17} x_1-\,\frac{5}{17} x_2, \frac{5}{17} x_1 + \frac{8}{17} x_2}\right)\\ \\ &=&
T_{\rv{{{28}\over{17}},-{{27}\over{17}}}}\,\matrix{\frac{8}{17} & -\,\frac{5}{17}\\ \frac{5}{17} & \frac{8}{17}}\, \vec{x}
\end{array}\]
so \(H = T_{\rv{{{28}\over{17}},-{{27}\over{17}}}}\, A\), where #A = \matrix{\frac{8}{17} & -\,\frac{5}{17}\\ \frac{5}{17} & \frac{8}{17}}#.
In order to find #\vec{v}#, we use the
commutation rule for a translation:
\[ \begin{array}{rcl}T_{\rv{{{28}\over{17}},-{{27}\over{17}}}}\, A &=& A\, A^{\top}\, T_{\rv{{{28}\over{17}},-{{27}\over{17}}}}\, A\\
&&\phantom{xx}\color{blue}{A\text{ is orthogonal}}\\
&=&A\, T_{A^{\top}\,\rv{{{28}\over{17}},-{{27}\over{17}}}} \, A^{\top} A\\&&\phantom{xx}\color{blue}{\text{commutation rule}}\\&=& A\, T_{A^{\top}\,\rv{{{28}\over{17}},-{{27}\over{17}}}} \\&&\phantom{xx}\color{blue}{A\text{ is orthogonal}}\end{array}\] We conclude that
\[ \vec{v} =A^{\top}\,\rv{{{28}\over{17}},-{{27}\over{17}}} = \matrix{\frac{8}{17} & \frac{5}{17}\\ -\,\frac{5}{17} & \frac{8}{17}}\,\rv{{{28}\over{17}},-{{27}\over{17}}}=\rv{1,-4}\]
Characterisation of isometries
