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












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












    $begingroup$


    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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 28 at 19:28









      greedoid

      46.1k1160117




      46.1k1160117










      asked Feb 14 '17 at 20:14









      Si.0788Si.0788

      1,302627




      1,302627






















          2 Answers
          2






          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



















          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











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2144549%2fisometry-of-mathbb-r2-that-preserves-orientation-is-either-a-rotation-or-a-t%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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

                1,631219




                1,631219






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2144549%2fisometry-of-mathbb-r2-that-preserves-orientation-is-either-a-rotation-or-a-t%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more