Question about rotation $2times 2$ rotation matrices












3












$begingroup$


How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= begin{bmatrix} cos{x}& -sin{x}\ sin{x} & cos{x}end{bmatrix} ,$$
for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.






linear-algebra rotations







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= begin{bmatrix} cos{x}& -sin{x}\ sin{x} & cos{x}end{bmatrix} ,$$
    for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.






    linear-algebra rotations







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= begin{bmatrix} cos{x}& -sin{x}\ sin{x} & cos{x}end{bmatrix} ,$$
      for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.






      linear-algebra rotations







      share|cite|improve this question











      $endgroup$




      How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= begin{bmatrix} cos{x}& -sin{x}\ sin{x} & cos{x}end{bmatrix} ,$$
      for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.






      linear-algebra rotations




      linear-algebra rotations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 22:51









      Namaste

      1




      1










      asked Dec 16 '18 at 15:08









      Tanny SiebenTanny Sieben

      41028




      41028






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Even for the case $A^2=R$ there are many possible roots.



          Look for instance at:



          https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf



          With the additional constraint $det(R)=1$ here you get $$operatorname{tr}(A)A=Rpm I$$



          and have to discuss according values of $x$, whether the trace is zero or not and so on.




          • e.g. $R=I$ and null trace, gives $begin{pmatrix}a&b\c&-aend{pmatrix}$ with $a^2+bc=1$


          [ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]





          For $A^n=R$ you get even more solutions.



          Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...






          share|cite|improve this answer











          $endgroup$





















            4












            $begingroup$

            "Very obvious" or not, it's not true. If $A=begin{bmatrix}0&1\1&0end{bmatrix}$ then $A^2=I$ but $A$ is not a rotation.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
              $endgroup$
              – Tanny Sieben
              Dec 16 '18 at 15:31








            • 2




              $begingroup$
              I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
              $endgroup$
              – David C. Ullrich
              Dec 16 '18 at 15:41












            • $begingroup$
              I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
              $endgroup$
              – Tanny Sieben
              Dec 16 '18 at 15:52





















            1












            $begingroup$

            $DeclareMathOperator{Tr}{Tr}$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:



            If $R=begin{pmatrix}-1\&-1end{pmatrix}$, then $A=begin{pmatrix}alpha&beta\gamma&-alphaend{pmatrix}$ with $alpha^2+betagamma=-1$.



            If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfrac{R+I}{sqrt{2cosphi+2}}$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

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              2












              $begingroup$

              Even for the case $A^2=R$ there are many possible roots.



              Look for instance at:



              https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf



              With the additional constraint $det(R)=1$ here you get $$operatorname{tr}(A)A=Rpm I$$



              and have to discuss according values of $x$, whether the trace is zero or not and so on.




              • e.g. $R=I$ and null trace, gives $begin{pmatrix}a&b\c&-aend{pmatrix}$ with $a^2+bc=1$


              [ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]





              For $A^n=R$ you get even more solutions.



              Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Even for the case $A^2=R$ there are many possible roots.



                Look for instance at:



                https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf



                With the additional constraint $det(R)=1$ here you get $$operatorname{tr}(A)A=Rpm I$$



                and have to discuss according values of $x$, whether the trace is zero or not and so on.




                • e.g. $R=I$ and null trace, gives $begin{pmatrix}a&b\c&-aend{pmatrix}$ with $a^2+bc=1$


                [ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]





                For $A^n=R$ you get even more solutions.



                Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Even for the case $A^2=R$ there are many possible roots.



                  Look for instance at:



                  https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf



                  With the additional constraint $det(R)=1$ here you get $$operatorname{tr}(A)A=Rpm I$$



                  and have to discuss according values of $x$, whether the trace is zero or not and so on.




                  • e.g. $R=I$ and null trace, gives $begin{pmatrix}a&b\c&-aend{pmatrix}$ with $a^2+bc=1$


                  [ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]





                  For $A^n=R$ you get even more solutions.



                  Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...






                  share|cite|improve this answer











                  $endgroup$



                  Even for the case $A^2=R$ there are many possible roots.



                  Look for instance at:



                  https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf



                  With the additional constraint $det(R)=1$ here you get $$operatorname{tr}(A)A=Rpm I$$



                  and have to discuss according values of $x$, whether the trace is zero or not and so on.




                  • e.g. $R=I$ and null trace, gives $begin{pmatrix}a&b\c&-aend{pmatrix}$ with $a^2+bc=1$


                  [ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]





                  For $A^n=R$ you get even more solutions.



                  Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 16 '18 at 16:35

























                  answered Dec 16 '18 at 16:29









                  zwimzwim

                  12.5k831




                  12.5k831























                      4












                      $begingroup$

                      "Very obvious" or not, it's not true. If $A=begin{bmatrix}0&1\1&0end{bmatrix}$ then $A^2=I$ but $A$ is not a rotation.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:31








                      • 2




                        $begingroup$
                        I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                        $endgroup$
                        – David C. Ullrich
                        Dec 16 '18 at 15:41












                      • $begingroup$
                        I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:52


















                      4












                      $begingroup$

                      "Very obvious" or not, it's not true. If $A=begin{bmatrix}0&1\1&0end{bmatrix}$ then $A^2=I$ but $A$ is not a rotation.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:31








                      • 2




                        $begingroup$
                        I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                        $endgroup$
                        – David C. Ullrich
                        Dec 16 '18 at 15:41












                      • $begingroup$
                        I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:52
















                      4












                      4








                      4





                      $begingroup$

                      "Very obvious" or not, it's not true. If $A=begin{bmatrix}0&1\1&0end{bmatrix}$ then $A^2=I$ but $A$ is not a rotation.






                      share|cite|improve this answer









                      $endgroup$



                      "Very obvious" or not, it's not true. If $A=begin{bmatrix}0&1\1&0end{bmatrix}$ then $A^2=I$ but $A$ is not a rotation.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 16 '18 at 15:17









                      David C. UllrichDavid C. Ullrich

                      61.2k43994




                      61.2k43994












                      • $begingroup$
                        I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:31








                      • 2




                        $begingroup$
                        I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                        $endgroup$
                        – David C. Ullrich
                        Dec 16 '18 at 15:41












                      • $begingroup$
                        I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:52




















                      • $begingroup$
                        I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:31








                      • 2




                        $begingroup$
                        I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                        $endgroup$
                        – David C. Ullrich
                        Dec 16 '18 at 15:41












                      • $begingroup$
                        I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
                        $endgroup$
                        – Tanny Sieben
                        Dec 16 '18 at 15:52


















                      $begingroup$
                      I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
                      $endgroup$
                      – Tanny Sieben
                      Dec 16 '18 at 15:31






                      $begingroup$
                      I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
                      $endgroup$
                      – Tanny Sieben
                      Dec 16 '18 at 15:31






                      2




                      2




                      $begingroup$
                      I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                      $endgroup$
                      – David C. Ullrich
                      Dec 16 '18 at 15:41






                      $begingroup$
                      I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                      $endgroup$
                      – David C. Ullrich
                      Dec 16 '18 at 15:41














                      $begingroup$
                      I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
                      $endgroup$
                      – Tanny Sieben
                      Dec 16 '18 at 15:52






                      $begingroup$
                      I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
                      $endgroup$
                      – Tanny Sieben
                      Dec 16 '18 at 15:52













                      1












                      $begingroup$

                      $DeclareMathOperator{Tr}{Tr}$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:



                      If $R=begin{pmatrix}-1\&-1end{pmatrix}$, then $A=begin{pmatrix}alpha&beta\gamma&-alphaend{pmatrix}$ with $alpha^2+betagamma=-1$.



                      If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfrac{R+I}{sqrt{2cosphi+2}}$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        $DeclareMathOperator{Tr}{Tr}$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:



                        If $R=begin{pmatrix}-1\&-1end{pmatrix}$, then $A=begin{pmatrix}alpha&beta\gamma&-alphaend{pmatrix}$ with $alpha^2+betagamma=-1$.



                        If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfrac{R+I}{sqrt{2cosphi+2}}$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          $DeclareMathOperator{Tr}{Tr}$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:



                          If $R=begin{pmatrix}-1\&-1end{pmatrix}$, then $A=begin{pmatrix}alpha&beta\gamma&-alphaend{pmatrix}$ with $alpha^2+betagamma=-1$.



                          If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfrac{R+I}{sqrt{2cosphi+2}}$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.






                          share|cite|improve this answer









                          $endgroup$



                          $DeclareMathOperator{Tr}{Tr}$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:



                          If $R=begin{pmatrix}-1\&-1end{pmatrix}$, then $A=begin{pmatrix}alpha&beta\gamma&-alphaend{pmatrix}$ with $alpha^2+betagamma=-1$.



                          If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfrac{R+I}{sqrt{2cosphi+2}}$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 16 '18 at 21:09









                          I like SerenaI like Serena

                          4,2721722




                          4,2721722






























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