Re: I thought there was perfect symmetry... Indeed there is. That is the point of my posting a week or so ago about the fibo-quaternion sequence: Given a 'pure' vector quaternion v (scalar part of v =0), we can compute |v|=sqrt(-vv)=sqrt(-v^2), and therefore 'normalize' it: Define U(v) = v/sqrt(-v^2) We can compute the normalized cross product if quaternions q0,q1: Define UX(q0,q1)= U(q0q1-q1q0) Given two arbitrary quaternions q0,q1 s.t., q0q1-q1q0 /= 0, we can form the sequence: q0,q1,q2=UX(q0,q1),q3=UX(q1,q2),q4=UX(q2,q3),... We can now choose I=q2, J=q3, K=q4, and everything 'works'. Given our new 'constants' I,J,K, where II=JJ=KK=IJK=-1, we can define the 'conjugate' of q: q' = -(q+IqI+JqJ+KqK)/2 The scalar and vector parts/components of q can be extracted: s(q) = (q+q')/2 V(q) = (q-q')/2 If we want to express q as a pair of 'complex' numbers, A,B, where A=ar+ai*I, B=br+bi*I, ar,ai,br,bi all real, q=A+BJ, we can extract the A, B components using our constants I,J: A = (q-IqI)/2 B = (q+IqI)/(2J) (be sure to noncommutatively divide on the right) Obviously, the representation A+BJ is relative to the I,J basis. ---- Regarding matrices and eigenvectors, any talk of 'triangularization', 'diagonal matrix', etc., depends upon a particular choice of basis. So if we are to 'extract' the eigenvalues & eigenvectors of a linear transformation, then these eigenvectors have to be expressed in some basis. I find it extremely cool & elegant that we can find a basis where the quaternion q itself can be expressed as its own eigenvalue(!), by choosing the 'complex number' 'I' as the unit vector part of q. The reason why I say 'eigenvalue' rather than 'eigenvalues' is that the quaternion q represents *both* of its (conjugate) eigenvalues at the same time. Yes, this representation is a bit of a pun, because the 'complex numbers' that we are using are a particular injection into the quaternions, but what a lovely pun! At 10:58 AM 10/18/2020, Keith F. Lynch wrote:
Henry Baker <hbaker1@pipeline.com> wrote:
There's a *huge* difference between 'i' and 'j', so the eye might not pick up on this subtlety. In the quaternions, 'i' is perpendicular to 'j', and 'i' doesn't commute with 'j'.
I thought there was perfect symmetry, i.e. that in any quaternion equation you can replace i with j if you also simultaneously replace j with k and k with i. And that you can also swap i with j if you simultaneously stick a minus sign on one (but not both) of them.
This is analogous to how in complex numbers you can swap i with -i.
And also analogous with the symmetry between electricity and magnetism if you make the Maxwell equations symmetric by removing the bald assertion that there are no magnetic monopoles. Sure, they haven't found any *yet*, but they haven't looked everywhere. As far as I know, nobody has checked under my bed.
You can also swap positive for negative electric charges, or north for south magnetic charges.