This question relates to Solving a transformation equation involving vectors and quaternions, however my equation was a bit different. After a bit of solving it arrives at ( I am trying to solve for $Q_x$ )

$$Q_{x}=Q_{known1}\times Q_{x}\times Q_{known2} $$

Now if we were not talking about quaternions, things would be simple. Here however I do not know how to go on and solve for $Q_x$

I have looked around a bit, but am still unsure on how to proceed, apart from maybe the (obvious but complicated?) way of spliting every quaternion to its elements and trying to derive a system of equations... Any help?

  • $\begingroup$ Explain a little more what your equation means, please. (The sign * is multiplication maybe?) $\endgroup$
    – Piquito
    Aug 17, 2016 at 12:38
  • $\begingroup$ Yes, sorry, here * is multiplication, basically i need to find $Q_x$ for which two rotations should have the same effect e.g. I start from $Q_A *Q_x=Q_x*Q_B$ which is $Q_x=Q_A^-1 *Q_x* Q_B$ $\endgroup$
    – ntg
    Aug 17, 2016 at 12:48

1 Answer 1


I'm going to abbreviate the original equation to $ax=xb$ where all three symbols are quaternions. Instead of putting $a$ and $b$ on the same side, you might learn more by putting the $x$'s on the same side like this:


That is, if $x$ is nonzero. We can already see that $0$ is always a solution to $ax=xb$ for any $a,b$. By normalizing $x$, we can also assume, without loss of generality, that $x$ is a unit quaternion (has norm $1$).

It's well-known that $\mathrm{Re}( x^{-1}qx)=\mathrm{Re}(q)$ so the above equation has no solutions if the real parts of $a$ and $b$ are unequal.

Furthermore, the subspace of quaternions with real part $0$ is invariant under conjugation. Rotation quaternions act transitively on the sets of quaternions with real part $0$ which share the same length. So we have a further constraint that $|\mathrm{Im}(a)|=|\mathrm{Im}(b)|$ for there to be any more nonzero solutions.

Under these conditions, you can find a rotation quaternion $x$ such that $x^{-1}\mathrm{Im}(a)x=\mathrm{Im}(b)$. Since $x^{-1}\mathrm{Re}(a)x=\mathrm{Re}(a)=\mathrm{Re}(b)$ already, we get that $x^{-1}ax=b$.

But this $x$ you found is not unique: if $u$ is any rotation around the axis determined by $b$'s complex part, $u^{-1}\mathrm{Im}(b)u=\mathrm{Im}(b)$, and there is such a $u$ for each angle of rotation you can think of. Then $(xu)^{-1}a(xu)=b$ for every such $u$.

So in summary, the solution set will be $0$ if the real parts of $a$ and $b$ don't match or if the pure quaternion parts of $a$ and $b$ have different length. Otherwise, you will also get one particular $x$ with the strategy I mentioned above, and further solutions will be $\{xu\mid u\in \mathrm{Stab}(\mathrm{Im}(b))\}$ where $\mathrm{Stab}(y)=\{q\in \mathbb H\mid q^{-1}yq=y\}$.

I think this is a complete set of solutions.

  • 1
    $\begingroup$ Don't you also need that $a$ and $b$ have the same norm somewhere? Surely $2i$ and $i$ aren't conjugated? $\endgroup$
    – Myself
    Aug 17, 2016 at 19:01
  • $\begingroup$ @Myself Should be good now. Thanks for the close reading. $\endgroup$
    – rschwieb
    Aug 18, 2016 at 13:08

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