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I have this: $$ x = q_1 \cdot A \cdot \overline{q_1} - q_2 \cdot A \cdot \overline{q_2} $$ $\cdot$ is a standard quaternion multiplication.

Is it somehow possible to express $A$ from it?

Harald Hanche-Olsen
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Martin Perry
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  • What are the conditions on the $q $s? For example, if they're equal, you're out of luck. – Mark S. Mar 21 '17 at 17:30
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    More generally, if they have the same absolute value, there is trouble, for then the RHS vanishes when $A$ is real, which in turn means there may not be a solution in general, or if there is, it is not unique. (Standard linear algebra fare, this.) – Harald Hanche-Olsen Mar 21 '17 at 17:34
  • qs are unit-length quaternions. A is also a quaternion – Martin Perry Mar 22 '17 at 06:46
  • Then my argument shows that, at least, the real part of $A$ cannot be determined uniquely, as it doesn't even enter into the equation. So there are, in effect, only three real unknowns, and four equations. – Harald Hanche-Olsen Mar 23 '17 at 10:05

2 Answers2

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I have only a partial answer, but here goes:

First, find the kernel (null space, if you prefer) of the map $T\colon\mathbb{H}\to\mathbb{H}$ given by $TA=q_1A\bar q_1-q_2A\bar q_2$. We readily find that $A\in\operatorname{ker}T$ iff $A$ commutes with $\bar q_2q_1$. Thus $\operatorname{ker}T=\operatorname{lin}(\bar q_2q_1,1)$. This is two-dimensional, except in the degenerate case when $\bar q_2q_1$ is real, which you probably want to avoid anyhow.

Next, note that $\mathbb{H}$ is a real inner product space, with inner product $\langle x,y\rangle=\operatorname{Re}(x\bar y)$.

A straightforward calculation shows that the adjoint of $T$ is given by $T*A=\bar q_1Aq_1-\bar q_2Aq_2$. This has the same form as $T$, so we readily find $\operatorname{ker}T^*=\operatorname{lin}(q_2\bar q_1,1)$.

Findally we find the range of $T$ to be $\operatorname{im}T=(\operatorname{ker}T^*)^\perp$, which is the two-dimensional space spanned by the unit quaternions with zero real part and orthogonal to $q_2\bar q_1$.

In other words, $x$ had better belong to that space, or there is no solution. And when there is a solution, is only unique if you constrain it further, to have zero real part and be orthogonal to $q_2\bar q_1$.

Harald Hanche-Olsen
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  • Found a typo: $TA$ should be $T^A$ in the fourth paragraph. Not correcting it now, to avoid gratuitously bumping up to the front page. – Harald Hanche-Olsen Mar 23 '17 at 20:34
  • You can make this a complete solution. Rewrite as $$ \overline{q_1}xq_1=A-\overline{q_1}q_2A\overline{q_2}q_1.$$ Use the substitutions $y=\overline{q_1}xq_1$ and $p=\overline{q_1}q_2$. It becomes $$ y=A-pA,\overline{p}.$$ Write $A=A_{|}+A_\perp$, substutite in, obtain $A_\perp=y/2$ and $A_{|}$ is arbitrary. Now you have conditions on when a solution exists, and a characterization of all solutions. – anon Mar 23 '17 at 22:51
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Since the quaternion multiplication is not commutative, there is no simple way to extract $A$ from the LHS using the properties of operations on$ \mathbb{H}$.

We can find $A=a_1+a_2\hat i+a_3 \hat j +a_4 \hat k$ performing the operations in LHS until we arrive at a result of the form $$ x_1+x_2\hat i+x_3 \hat j +x_4 \hat k=m_1+m_2\hat i+m_3 \hat j +m_4 \hat k $$

where the terms $m_i$ are linear functions of the unknowns $a_i$, so that, using the identity of quaternion, we find a linear systems with $4$ equations in the $4$ unknowns $ a_i$: $$ \begin {cases} m_1(a_1,a_2,a_3,a_4)=x_1\\m_2(a_1,a_2,a_3,a_4)=x_2\\m_3(a_1,a_2,a_3,a_4)=x_3\\m_4(a_1,a_2,a_3,a_4)=x_4 \end {cases} $$

solving this system (if it has a solution) we find $A$.

Emilio Novati
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