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I've been looking up Quaternion multiplication and many resources have stated that
$j*k=i$
and also in other sources I've found
$k*j=-i$


But I have not found any sources stating
$j*k=-i$
and I also have not found any sources stating
$k*j=i$


I know this sounds like a dumb question, but does the commutative property of multiplication not apply when dealing with higher dimensional complex number systems?
$j*k≠k*j$       ?

Albert Renshaw
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    Quaternion multiplication (what sources did you end up looking up...?) –  Jan 27 '16 at 05:05
  • @T.Bongers So even in the article you linked it says jk=i, kj=-i :o Is this really the case? jk≠kj ? – Albert Renshaw Jan 27 '16 at 05:07
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    Yes, and the next boldface-titled section is "Noncommutativity of multiplication." –  Jan 27 '16 at 05:08
  • @T.Bongers Fascinating, I'm surprised I was never taught this before haha! – Albert Renshaw Jan 27 '16 at 05:09
  • @user26857 There was a downvote, it has since then been removed. I'm not sure why it's pathetic, I simply wanted to know what I had done wrong so that I could avoid getting more downvotes in the future? I'm sorry that I've offended you sir – Albert Renshaw Jan 27 '16 at 07:02
  • Nope, quaternions and matrix multiplication are THE two prime examples of non-commutivity. Absolutely does not hold. – fleablood Jan 27 '16 at 07:12
  • @fleablood Thanks, now that I think about it quaternion multiplication is matrix multiplication. Since quaternions are points on a cartesian 4D hyper-field, they can be represented as vectors on [r, i, j, k]. e.g. "j" could be seen as [0,0,1,0] – Albert Renshaw Jan 27 '16 at 07:18
  • I'm not sure why it's pathetic, I simply wanted to know what I had done wrong so that I could avoid getting more downvotes in the future? I think the deal was probably that you asked a question that is answered in every single introduction to the quaternions that exists, which would be impossible to overlook. This egregious oversight probably exasperated the user, and probably others. – rschwieb Feb 12 '16 at 18:40

1 Answers1

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Yes, quaternions multiplication is not commutative.

https://en.wikipedia.org/wiki/Quaternion

$ i \times j = k $

$ j \times i = -k $

Andrew Au
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  • Thank you! That's fascinating. I'd like to understand why this is the case, in an intuitive sense. – Albert Renshaw Jan 27 '16 at 05:08
  • Quaternion is used in computer graphics and robotics to interpolate 3D rotations through conjugate multiplication

    https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation

    3D rotations are not commutative, for example, rotate left and then rotate up is in general not the same as rotate up than rotate left, so in order for quaternion to be useful, it cannot be commutative, in some sense, it is designed that way.

     – Andrew Au Jan 27 '16 at 05:12
  • That makes perfect sense! Thank you Andrew, I'll accept this answer once StackExchange's time limit hits 0 – Albert Renshaw Jan 27 '16 at 05:14