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1

$begingroup$

I have the following problem:

$$
q_2 = q_aq_bq_1q_b^-1
$$

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.

quaternions

share|cite|improve this question

edited Mar 14 at 7:47

klWu

asked Mar 14 at 7:33

klWuklWu

83

New contributor

klWu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

1

$begingroup$

I have the following problem:

$$
q_2 = q_aq_bq_1q_b^-1
$$

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.

quaternions

share|cite|improve this question

edited Mar 14 at 7:47

klWu

asked Mar 14 at 7:33

klWuklWu

83

New contributor

klWu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

I have the following problem:

$$
q_2 = q_aq_bq_1q_b^-1
$$

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.

quaternions

share|cite|improve this question

edited Mar 14 at 7:47

klWu

asked Mar 14 at 7:33

klWuklWu

83

New contributor

klWu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 14 at 7:47

klWu

asked Mar 14 at 7:33

klWuklWu

83

New contributor

klWu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 14 at 7:47

klWu

asked Mar 14 at 7:33

klWuklWu

83

New contributor

klWu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 14 at 7:33

klWuklWu

83

New contributor

klWu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 14 at 7:33

klWuklWu

83

1 Answer
1

active

oldest

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0

$begingroup$

Choose any $q_bne 0$ you want; then for given $q_1,,q_2$, we can find a unique $q_a$ that works, viz. $$q_2=q_aq_bq_1q_b^-1iff q_2q_b=q_aq_bq_1iff q_2q_bq_1^-1=q_aq_biff q_a=q_2q_bq_1^-1q_b^-1.$$ 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 $q_b=q_4^-1q_2q_bq_1^-1q_3$. So this gets into how one solves $q=pqr$ with $|pr|=1$. For $p:=q_4^-1q_2,,r:=q_1^-1q_3$, 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.

share|cite|improve this answer

answered Mar 14 at 8:32

J.G.J.G.

30.9k23149

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $vinBbb R^4$, for some matrix $MinBbb R^4times 4$ 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
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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0

$begingroup$

Choose any $q_bne 0$ you want; then for given $q_1,,q_2$, we can find a unique $q_a$ that works, viz. $$q_2=q_aq_bq_1q_b^-1iff q_2q_b=q_aq_bq_1iff q_2q_bq_1^-1=q_aq_biff q_a=q_2q_bq_1^-1q_b^-1.$$ 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 $q_b=q_4^-1q_2q_bq_1^-1q_3$. So this gets into how one solves $q=pqr$ with $|pr|=1$. For $p:=q_4^-1q_2,,r:=q_1^-1q_3$, 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.

share|cite|improve this answer

answered Mar 14 at 8:32

J.G.J.G.

30.9k23149

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $vinBbb R^4$, for some matrix $MinBbb R^4times 4$ 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
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

0

$begingroup$

Choose any $q_bne 0$ you want; then for given $q_1,,q_2$, we can find a unique $q_a$ that works, viz. $$q_2=q_aq_bq_1q_b^-1iff q_2q_b=q_aq_bq_1iff q_2q_bq_1^-1=q_aq_biff q_a=q_2q_bq_1^-1q_b^-1.$$ 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 $q_b=q_4^-1q_2q_bq_1^-1q_3$. So this gets into how one solves $q=pqr$ with $|pr|=1$. For $p:=q_4^-1q_2,,r:=q_1^-1q_3$, 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.

share|cite|improve this answer

answered Mar 14 at 8:32

J.G.J.G.

30.9k23149

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $vinBbb R^4$, for some matrix $MinBbb R^4times 4$ 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
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

Choose any $q_bne 0$ you want; then for given $q_1,,q_2$, we can find a unique $q_a$ that works, viz. $$q_2=q_aq_bq_1q_b^-1iff q_2q_b=q_aq_bq_1iff q_2q_bq_1^-1=q_aq_biff q_a=q_2q_bq_1^-1q_b^-1.$$ 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 $q_b=q_4^-1q_2q_bq_1^-1q_3$. So this gets into how one solves $q=pqr$ with $|pr|=1$. For $p:=q_4^-1q_2,,r:=q_1^-1q_3$, 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.

share|cite|improve this answer

answered Mar 14 at 8:32

J.G.J.G.

30.9k23149

$endgroup$

share|cite|improve this answer

answered Mar 14 at 8:32

J.G.J.G.

30.9k23149

answered Mar 14 at 8:32

J.G.J.G.

30.9k23149

answered Mar 14 at 8:32

J.G.J.G.

30.9k23149