Question about rotation matrices











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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







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    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







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      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1





      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















      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






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      share|cite|improve this question













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      edited 6 hours ago









      GNUSupporter 8964民主女神 地下教會

      12.8k72445




      12.8k72445










      asked 6 hours ago









      Tanny Sieben

      30117




      30117






















          3 Answers
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          up vote
          4
          down vote













          "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





















          • 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?
            – Tanny Sieben
            6 hours ago








          • 2




            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 $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
            – David C. Ullrich
            5 hours ago












          • 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?
            – Tanny Sieben
            5 hours ago




















          up vote
          2
          down vote













          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






























            up vote
            0
            down vote













            $DeclareMathOperator{Tr}{Tr}$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





















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              3 Answers
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              3 Answers
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              up vote
              4
              down vote













              "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





















              • 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?
                – Tanny Sieben
                6 hours ago








              • 2




                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 $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                – David C. Ullrich
                5 hours ago












              • 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?
                – Tanny Sieben
                5 hours ago

















              up vote
              4
              down vote













              "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





















              • 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?
                – Tanny Sieben
                6 hours ago








              • 2




                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 $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                – David C. Ullrich
                5 hours ago












              • 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?
                – Tanny Sieben
                5 hours ago















              up vote
              4
              down vote










              up vote
              4
              down vote









              "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












              "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 6 hours ago









              David C. Ullrich

              57.6k43891




              57.6k43891












              • 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?
                – Tanny Sieben
                6 hours ago








              • 2




                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 $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                – David C. Ullrich
                5 hours ago












              • 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?
                – Tanny Sieben
                5 hours ago




















              • 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?
                – Tanny Sieben
                6 hours ago








              • 2




                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 $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
                – David C. Ullrich
                5 hours ago












              • 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?
                – Tanny Sieben
                5 hours ago


















              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?
              – Tanny Sieben
              6 hours ago






              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?
              – Tanny Sieben
              6 hours ago






              2




              2




              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 $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
              – David C. Ullrich
              5 hours ago






              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 $B=A_{(theta+2pi k)/n}$, $k=1,dots,n$.
              – David C. Ullrich
              5 hours ago














              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?
              – Tanny Sieben
              5 hours ago






              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?
              – Tanny Sieben
              5 hours ago












              up vote
              2
              down vote













              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



























                up vote
                2
                down vote













                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

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  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














                  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 5 hours ago

























                  answered 5 hours ago









                  zwim

                  11.3k728




                  11.3k728






















                      up vote
                      0
                      down vote













                      $DeclareMathOperator{Tr}{Tr}$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

























                        up vote
                        0
                        down vote













                        $DeclareMathOperator{Tr}{Tr}$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























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          $DeclareMathOperator{Tr}{Tr}$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












                          $DeclareMathOperator{Tr}{Tr}$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 28 mins ago









                          I like Serena

                          3,4521718




                          3,4521718






























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