Isometry of $mathbb R^2$ that preserves orientation is either a rotation or a translation












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I want to show that if $gamma$ is an isometry of $mathbb{R}^2$ that preserves orientation, then it is either a translation or a rotation about a point.



If $gamma$ is an isometry of $mathbb{R}^2$ then it is of the form $vec{gamma}(vec{x})=Avec{x}+vec{b}$ for some orthogonal matrix $A$ and $vec{b} in mathbb{R}^2$. Since $gamma$ preserves orientation we have that $det A=1$.



If $A=I$, then clearly $gamma$ is a translation $vec{x} mapsto vec{x} + vec{b}$.



I'm stuck with the case $A neq I$, though. How do I proceed?






geometry geometric-transformation







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    0












    $begingroup$


    I want to show that if $gamma$ is an isometry of $mathbb{R}^2$ that preserves orientation, then it is either a translation or a rotation about a point.



    If $gamma$ is an isometry of $mathbb{R}^2$ then it is of the form $vec{gamma}(vec{x})=Avec{x}+vec{b}$ for some orthogonal matrix $A$ and $vec{b} in mathbb{R}^2$. Since $gamma$ preserves orientation we have that $det A=1$.



    If $A=I$, then clearly $gamma$ is a translation $vec{x} mapsto vec{x} + vec{b}$.



    I'm stuck with the case $A neq I$, though. How do I proceed?






    geometry geometric-transformation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I want to show that if $gamma$ is an isometry of $mathbb{R}^2$ that preserves orientation, then it is either a translation or a rotation about a point.



      If $gamma$ is an isometry of $mathbb{R}^2$ then it is of the form $vec{gamma}(vec{x})=Avec{x}+vec{b}$ for some orthogonal matrix $A$ and $vec{b} in mathbb{R}^2$. Since $gamma$ preserves orientation we have that $det A=1$.



      If $A=I$, then clearly $gamma$ is a translation $vec{x} mapsto vec{x} + vec{b}$.



      I'm stuck with the case $A neq I$, though. How do I proceed?






      geometry geometric-transformation







      share|cite|improve this question











      $endgroup$




      I want to show that if $gamma$ is an isometry of $mathbb{R}^2$ that preserves orientation, then it is either a translation or a rotation about a point.



      If $gamma$ is an isometry of $mathbb{R}^2$ then it is of the form $vec{gamma}(vec{x})=Avec{x}+vec{b}$ for some orthogonal matrix $A$ and $vec{b} in mathbb{R}^2$. Since $gamma$ preserves orientation we have that $det A=1$.



      If $A=I$, then clearly $gamma$ is a translation $vec{x} mapsto vec{x} + vec{b}$.



      I'm stuck with the case $A neq I$, though. How do I proceed?






      geometry geometric-transformation




      geometry geometric-transformation






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      edited Jan 28 at 19:28









      greedoid

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      46.1k1160117










      asked Feb 14 '17 at 20:14









      Si.0788Si.0788

      1,302627




      1,302627






















          2 Answers
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          $begingroup$

          Orthogonal matrices with determinant 1 are exactly rotation matrices (about the origin). Hence $vec{gamma}(vec{x})=Avec{x}+vec{b}$ is a rotation plus a translation, i.e. a rotation about the point which has the coordinates of the vector $vec{b}$.






          share|cite|improve this answer









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          • 1




            $begingroup$
            It's rather a rotation that takes $0$ into $b$.
            $endgroup$
            – Berci
            Feb 14 '17 at 20:27










          • $begingroup$
            How would one justify that a rotation plus a translation is indeed a rotation about the point $vec{b}$?
            $endgroup$
            – Si.0788
            Feb 15 '17 at 17:20



















          0












          $begingroup$

          First observe that translations are rigid motoins or isometry and any rigid motion which fixes the origin preserves the inner product i.e. any rigid motion which fixes the origin is an orthogonal transformation. Now suppose $m$ is an isometry. Then $m$ sends the origin $O$ to somewhere let's say $b$ i.e. $m(O) = b,$ where $b in Bbb R^n$. Let $t_{-b}$ be the translation by the vector $-b$ i.e. $t_{-b} (v) = v-b, v in Bbb R^n$. Let $A=t_{-b} circ m$. Then clearly $A$ is an isometry since so are $t_{-b}$ and $v$. Also $A(O) = O$ i.e. $A$ fixes the origin. So by our previous observation it follows that $A$ is an orthogonal transformation. Since translations are invertible and $t_{-b} = (t_b)^{-1}$ for all $b in Bbb R^n$. So we have $m = t_b circ A$. So $m(v) = Av + b, text {for all} v in Bbb R^n$ where $A in O(n)$.






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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Orthogonal matrices with determinant 1 are exactly rotation matrices (about the origin). Hence $vec{gamma}(vec{x})=Avec{x}+vec{b}$ is a rotation plus a translation, i.e. a rotation about the point which has the coordinates of the vector $vec{b}$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              It's rather a rotation that takes $0$ into $b$.
              $endgroup$
              – Berci
              Feb 14 '17 at 20:27










            • $begingroup$
              How would one justify that a rotation plus a translation is indeed a rotation about the point $vec{b}$?
              $endgroup$
              – Si.0788
              Feb 15 '17 at 17:20
















            1












            $begingroup$

            Orthogonal matrices with determinant 1 are exactly rotation matrices (about the origin). Hence $vec{gamma}(vec{x})=Avec{x}+vec{b}$ is a rotation plus a translation, i.e. a rotation about the point which has the coordinates of the vector $vec{b}$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              It's rather a rotation that takes $0$ into $b$.
              $endgroup$
              – Berci
              Feb 14 '17 at 20:27










            • $begingroup$
              How would one justify that a rotation plus a translation is indeed a rotation about the point $vec{b}$?
              $endgroup$
              – Si.0788
              Feb 15 '17 at 17:20














            1












            1








            1





            $begingroup$

            Orthogonal matrices with determinant 1 are exactly rotation matrices (about the origin). Hence $vec{gamma}(vec{x})=Avec{x}+vec{b}$ is a rotation plus a translation, i.e. a rotation about the point which has the coordinates of the vector $vec{b}$.






            share|cite|improve this answer









            $endgroup$



            Orthogonal matrices with determinant 1 are exactly rotation matrices (about the origin). Hence $vec{gamma}(vec{x})=Avec{x}+vec{b}$ is a rotation plus a translation, i.e. a rotation about the point which has the coordinates of the vector $vec{b}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 14 '17 at 20:25









            MuziMuzi

            424320




            424320








            • 1




              $begingroup$
              It's rather a rotation that takes $0$ into $b$.
              $endgroup$
              – Berci
              Feb 14 '17 at 20:27










            • $begingroup$
              How would one justify that a rotation plus a translation is indeed a rotation about the point $vec{b}$?
              $endgroup$
              – Si.0788
              Feb 15 '17 at 17:20














            • 1




              $begingroup$
              It's rather a rotation that takes $0$ into $b$.
              $endgroup$
              – Berci
              Feb 14 '17 at 20:27










            • $begingroup$
              How would one justify that a rotation plus a translation is indeed a rotation about the point $vec{b}$?
              $endgroup$
              – Si.0788
              Feb 15 '17 at 17:20








            1




            1




            $begingroup$
            It's rather a rotation that takes $0$ into $b$.
            $endgroup$
            – Berci
            Feb 14 '17 at 20:27




            $begingroup$
            It's rather a rotation that takes $0$ into $b$.
            $endgroup$
            – Berci
            Feb 14 '17 at 20:27












            $begingroup$
            How would one justify that a rotation plus a translation is indeed a rotation about the point $vec{b}$?
            $endgroup$
            – Si.0788
            Feb 15 '17 at 17:20




            $begingroup$
            How would one justify that a rotation plus a translation is indeed a rotation about the point $vec{b}$?
            $endgroup$
            – Si.0788
            Feb 15 '17 at 17:20











            0












            $begingroup$

            First observe that translations are rigid motoins or isometry and any rigid motion which fixes the origin preserves the inner product i.e. any rigid motion which fixes the origin is an orthogonal transformation. Now suppose $m$ is an isometry. Then $m$ sends the origin $O$ to somewhere let's say $b$ i.e. $m(O) = b,$ where $b in Bbb R^n$. Let $t_{-b}$ be the translation by the vector $-b$ i.e. $t_{-b} (v) = v-b, v in Bbb R^n$. Let $A=t_{-b} circ m$. Then clearly $A$ is an isometry since so are $t_{-b}$ and $v$. Also $A(O) = O$ i.e. $A$ fixes the origin. So by our previous observation it follows that $A$ is an orthogonal transformation. Since translations are invertible and $t_{-b} = (t_b)^{-1}$ for all $b in Bbb R^n$. So we have $m = t_b circ A$. So $m(v) = Av + b, text {for all} v in Bbb R^n$ where $A in O(n)$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              First observe that translations are rigid motoins or isometry and any rigid motion which fixes the origin preserves the inner product i.e. any rigid motion which fixes the origin is an orthogonal transformation. Now suppose $m$ is an isometry. Then $m$ sends the origin $O$ to somewhere let's say $b$ i.e. $m(O) = b,$ where $b in Bbb R^n$. Let $t_{-b}$ be the translation by the vector $-b$ i.e. $t_{-b} (v) = v-b, v in Bbb R^n$. Let $A=t_{-b} circ m$. Then clearly $A$ is an isometry since so are $t_{-b}$ and $v$. Also $A(O) = O$ i.e. $A$ fixes the origin. So by our previous observation it follows that $A$ is an orthogonal transformation. Since translations are invertible and $t_{-b} = (t_b)^{-1}$ for all $b in Bbb R^n$. So we have $m = t_b circ A$. So $m(v) = Av + b, text {for all} v in Bbb R^n$ where $A in O(n)$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                First observe that translations are rigid motoins or isometry and any rigid motion which fixes the origin preserves the inner product i.e. any rigid motion which fixes the origin is an orthogonal transformation. Now suppose $m$ is an isometry. Then $m$ sends the origin $O$ to somewhere let's say $b$ i.e. $m(O) = b,$ where $b in Bbb R^n$. Let $t_{-b}$ be the translation by the vector $-b$ i.e. $t_{-b} (v) = v-b, v in Bbb R^n$. Let $A=t_{-b} circ m$. Then clearly $A$ is an isometry since so are $t_{-b}$ and $v$. Also $A(O) = O$ i.e. $A$ fixes the origin. So by our previous observation it follows that $A$ is an orthogonal transformation. Since translations are invertible and $t_{-b} = (t_b)^{-1}$ for all $b in Bbb R^n$. So we have $m = t_b circ A$. So $m(v) = Av + b, text {for all} v in Bbb R^n$ where $A in O(n)$.






                share|cite|improve this answer











                $endgroup$



                First observe that translations are rigid motoins or isometry and any rigid motion which fixes the origin preserves the inner product i.e. any rigid motion which fixes the origin is an orthogonal transformation. Now suppose $m$ is an isometry. Then $m$ sends the origin $O$ to somewhere let's say $b$ i.e. $m(O) = b,$ where $b in Bbb R^n$. Let $t_{-b}$ be the translation by the vector $-b$ i.e. $t_{-b} (v) = v-b, v in Bbb R^n$. Let $A=t_{-b} circ m$. Then clearly $A$ is an isometry since so are $t_{-b}$ and $v$. Also $A(O) = O$ i.e. $A$ fixes the origin. So by our previous observation it follows that $A$ is an orthogonal transformation. Since translations are invertible and $t_{-b} = (t_b)^{-1}$ for all $b in Bbb R^n$. So we have $m = t_b circ A$. So $m(v) = Av + b, text {for all} v in Bbb R^n$ where $A in O(n)$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 28 at 19:18

























                answered Dec 22 '18 at 5:56









                Dbchatto67Dbchatto67

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