complex numbers and rotation matrices












1












$begingroup$


I have just learned that complex numbers and rotation matrices in the plane are the same thing (up to isomorphism). Is there any deep reason for this? Is it anything more than the fact that complex numbers encode rotations in $mathbb{R}^2$?



I am looking for intuition and insight into the bigger picture if it exists.



Thanks






linear-algebra matrices complex-numbers linear-transformations intuition







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I suppose $Bbb C$ is the even part of the Clifford algebra over $Bbb R$ associated to the quadratic form $x^2+y^2$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 23 at 18:53






  • 1




    $begingroup$
    It is the same on the real line. Real numbers correspond to scalings and rotations (which are fairly limited on the line).
    $endgroup$
    – copper.hat
    Jan 23 at 18:57






  • 1




    $begingroup$
    $$e^{i alpha} cdot e^{i beta} = e^{i(alpha + beta)}$$
    $endgroup$
    – user635162
    Jan 23 at 18:57








  • 1




    $begingroup$
    You should rather say that the unit circle in the complex plane is in correspondence with (orientation preserving) rotations. The set of nonzero complex numbers describes a much larger group generated by rotations and scalings.
    $endgroup$
    – rschwieb
    Jan 23 at 19:29
















1












$begingroup$


I have just learned that complex numbers and rotation matrices in the plane are the same thing (up to isomorphism). Is there any deep reason for this? Is it anything more than the fact that complex numbers encode rotations in $mathbb{R}^2$?



I am looking for intuition and insight into the bigger picture if it exists.



Thanks






linear-algebra matrices complex-numbers linear-transformations intuition







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I suppose $Bbb C$ is the even part of the Clifford algebra over $Bbb R$ associated to the quadratic form $x^2+y^2$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 23 at 18:53






  • 1




    $begingroup$
    It is the same on the real line. Real numbers correspond to scalings and rotations (which are fairly limited on the line).
    $endgroup$
    – copper.hat
    Jan 23 at 18:57






  • 1




    $begingroup$
    $$e^{i alpha} cdot e^{i beta} = e^{i(alpha + beta)}$$
    $endgroup$
    – user635162
    Jan 23 at 18:57








  • 1




    $begingroup$
    You should rather say that the unit circle in the complex plane is in correspondence with (orientation preserving) rotations. The set of nonzero complex numbers describes a much larger group generated by rotations and scalings.
    $endgroup$
    – rschwieb
    Jan 23 at 19:29














1












1








1





$begingroup$


I have just learned that complex numbers and rotation matrices in the plane are the same thing (up to isomorphism). Is there any deep reason for this? Is it anything more than the fact that complex numbers encode rotations in $mathbb{R}^2$?



I am looking for intuition and insight into the bigger picture if it exists.



Thanks






linear-algebra matrices complex-numbers linear-transformations intuition







share|cite|improve this question









$endgroup$




I have just learned that complex numbers and rotation matrices in the plane are the same thing (up to isomorphism). Is there any deep reason for this? Is it anything more than the fact that complex numbers encode rotations in $mathbb{R}^2$?



I am looking for intuition and insight into the bigger picture if it exists.



Thanks






linear-algebra matrices complex-numbers linear-transformations intuition




linear-algebra matrices complex-numbers linear-transformations intuition






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 23 at 18:48









Ac711Ac711

345




345








  • 2




    $begingroup$
    I suppose $Bbb C$ is the even part of the Clifford algebra over $Bbb R$ associated to the quadratic form $x^2+y^2$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 23 at 18:53






  • 1




    $begingroup$
    It is the same on the real line. Real numbers correspond to scalings and rotations (which are fairly limited on the line).
    $endgroup$
    – copper.hat
    Jan 23 at 18:57






  • 1




    $begingroup$
    $$e^{i alpha} cdot e^{i beta} = e^{i(alpha + beta)}$$
    $endgroup$
    – user635162
    Jan 23 at 18:57








  • 1




    $begingroup$
    You should rather say that the unit circle in the complex plane is in correspondence with (orientation preserving) rotations. The set of nonzero complex numbers describes a much larger group generated by rotations and scalings.
    $endgroup$
    – rschwieb
    Jan 23 at 19:29














  • 2




    $begingroup$
    I suppose $Bbb C$ is the even part of the Clifford algebra over $Bbb R$ associated to the quadratic form $x^2+y^2$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 23 at 18:53






  • 1




    $begingroup$
    It is the same on the real line. Real numbers correspond to scalings and rotations (which are fairly limited on the line).
    $endgroup$
    – copper.hat
    Jan 23 at 18:57






  • 1




    $begingroup$
    $$e^{i alpha} cdot e^{i beta} = e^{i(alpha + beta)}$$
    $endgroup$
    – user635162
    Jan 23 at 18:57








  • 1




    $begingroup$
    You should rather say that the unit circle in the complex plane is in correspondence with (orientation preserving) rotations. The set of nonzero complex numbers describes a much larger group generated by rotations and scalings.
    $endgroup$
    – rschwieb
    Jan 23 at 19:29








2




2




$begingroup$
I suppose $Bbb C$ is the even part of the Clifford algebra over $Bbb R$ associated to the quadratic form $x^2+y^2$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 18:53




$begingroup$
I suppose $Bbb C$ is the even part of the Clifford algebra over $Bbb R$ associated to the quadratic form $x^2+y^2$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 18:53




1




1




$begingroup$
It is the same on the real line. Real numbers correspond to scalings and rotations (which are fairly limited on the line).
$endgroup$
– copper.hat
Jan 23 at 18:57




$begingroup$
It is the same on the real line. Real numbers correspond to scalings and rotations (which are fairly limited on the line).
$endgroup$
– copper.hat
Jan 23 at 18:57




1




1




$begingroup$
$$e^{i alpha} cdot e^{i beta} = e^{i(alpha + beta)}$$
$endgroup$
– user635162
Jan 23 at 18:57






$begingroup$
$$e^{i alpha} cdot e^{i beta} = e^{i(alpha + beta)}$$
$endgroup$
– user635162
Jan 23 at 18:57






1




1




$begingroup$
You should rather say that the unit circle in the complex plane is in correspondence with (orientation preserving) rotations. The set of nonzero complex numbers describes a much larger group generated by rotations and scalings.
$endgroup$
– rschwieb
Jan 23 at 19:29




$begingroup$
You should rather say that the unit circle in the complex plane is in correspondence with (orientation preserving) rotations. The set of nonzero complex numbers describes a much larger group generated by rotations and scalings.
$endgroup$
– rschwieb
Jan 23 at 19:29










1 Answer
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2












$begingroup$

What you are probably looking for is a discussion on Clifford algebras or maybe one tailored to so-called geometric algebras which is pretty much under the same umbrella.



Clifford algebra explains the connection between elements of a special algebra and rotations in a vector space (and more.) In particular, it generalizes what you see for the reals, complexes, and quaternions on $mathbb R$, $mathbb R^2$ and $mathbb R^3$ respectively.






share|cite|improve this answer









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    1 Answer
    1






    active

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    active

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    active

    oldest

    votes









    2












    $begingroup$

    What you are probably looking for is a discussion on Clifford algebras or maybe one tailored to so-called geometric algebras which is pretty much under the same umbrella.



    Clifford algebra explains the connection between elements of a special algebra and rotations in a vector space (and more.) In particular, it generalizes what you see for the reals, complexes, and quaternions on $mathbb R$, $mathbb R^2$ and $mathbb R^3$ respectively.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      What you are probably looking for is a discussion on Clifford algebras or maybe one tailored to so-called geometric algebras which is pretty much under the same umbrella.



      Clifford algebra explains the connection between elements of a special algebra and rotations in a vector space (and more.) In particular, it generalizes what you see for the reals, complexes, and quaternions on $mathbb R$, $mathbb R^2$ and $mathbb R^3$ respectively.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        What you are probably looking for is a discussion on Clifford algebras or maybe one tailored to so-called geometric algebras which is pretty much under the same umbrella.



        Clifford algebra explains the connection between elements of a special algebra and rotations in a vector space (and more.) In particular, it generalizes what you see for the reals, complexes, and quaternions on $mathbb R$, $mathbb R^2$ and $mathbb R^3$ respectively.






        share|cite|improve this answer









        $endgroup$



        What you are probably looking for is a discussion on Clifford algebras or maybe one tailored to so-called geometric algebras which is pretty much under the same umbrella.



        Clifford algebra explains the connection between elements of a special algebra and rotations in a vector space (and more.) In particular, it generalizes what you see for the reals, complexes, and quaternions on $mathbb R$, $mathbb R^2$ and $mathbb R^3$ respectively.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 23 at 19:35









        rschwiebrschwieb

        107k12102251




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