I have the following problem:

$$ q_2 = q_aq_bq_1{q_b}^{-1} $$

All the $q$'s are quaternions and I want to solve for $q_a$ and $q_b$, given more than one $[q_1, q_2]$ pairs, the last term is the inverse of $q_b$.

I tried to solve them like a typical simultaneous equation problem, but the inverse and the non-commutative nature is giving me a hard time.

  • $\begingroup$ Do you mean you want to find possible $q_1,\,q_2$? If so, it would be clearer to say "solve for given $q_a$ and $q_b$, since "solve for" often otherwise implying you're staying what is sought and unknown. $\endgroup$ – J.G. Mar 14 at 7:43
  • $\begingroup$ Actually no, I want to find $q_a$ and $q_b$, let me clarify that $\endgroup$ – klWu Mar 14 at 7:46

Choose any $q_b\ne 0$ you want; then for given $q_1,\,q_2$, we can find a unique $q_a$ that works, viz. $$q_2=q_aq_bq_1q_b^{-1}\iff q_2q_b=q_aq_bq_1\iff q_2q_bq_1^{-1}=q_aq_b\iff q_a=q_2q_bq_1^{-1}q_b^{-1}.$$ Replacing $q_1,\,q_2$ with $q_3,\,q_4$ allows the same $q_a,\,q_b$ as a solution iff $q_4q_bq_3^{-1}=q_2q_bq_1^{-1}$, or equivalently $q_b=q_4^{-1}q_2q_bq_1^{-1}q_3$. So this gets into how one solves $q=pqr$ with $|pr|=1$. For $p:=q_4^{-1}q_2,\,r:=q_1^{-1}q_3$, I recommend solving this by simultaneous equations. Once you have $q_b$ up to real scaling, you get $q_a$. If there are three or more pairs $q_a,\,q_b$ have to work for, you'll get even more constraints that will still keep the same scaling redundancy, if there are still nonzero solutions.

  • $\begingroup$ Thanks, but solving for $q=pqr$ is exactly where I'm stuck on. Isn't simultaneous equation solving exactly how we get to this step? I don't know how to isolate q to one side of the equal sign. $\endgroup$ – klWu Mar 14 at 8:46
  • $\begingroup$ @KlWu I mean solve for the four parts of $q$ as real variables; each of the four parts of $q=pqr$ gives an equation in these variables, so in the end it's equivalent to solving $Mv=v$ for $v\in\Bbb R^4$, for some matrix $M\in\Bbb R^{4\times 4}$ whose entries are expressible in terms of the components of $p,\,r$. Note $M$ won't be invertible, but you can get $v$ up to scaling. $\endgroup$ – J.G. Mar 14 at 8:49
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    $\begingroup$ Oh I see what you mean! $\endgroup$ – klWu Mar 14 at 8:51
  • $\begingroup$ I have been solving it with the $Mv=v$ way by looking for the eigenvector of M with eigenvalue of 1. I ended up with 4 complex eigen values all with modus of 1 and 4 sets of eigen vectors with complex coefficients. What went wrong? $\endgroup$ – klWu Mar 15 at 8:29
  • $\begingroup$ @klwu I recommend you ask a new question in which you work through such calculations so I or others can say. $\endgroup$ – J.G. Mar 15 at 9:51

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