Inversion of rotation matrix

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up vote
3
down vote

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For example, I have a two-dimensional rotation matrix
$$
beginbmatrix
0.5091 & -0.8607 \
0.8607 & phantom-0.5091
endbmatrix
$$
and I have a vector I'd like to rotate, e.g. $(1, -0.5)$.

My problem is to find an inverse of the rotation matrix so that I can later “undo” the rotation performed on the vector so that I get back the original vector.

The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix.

matrices rotations

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edited Aug 27 at 10:34

Jendrik Stelzner

7,63121037

asked Aug 27 at 6:13

aleksv

396

add a comment | 

up vote
3
down vote

favorite

For example, I have a two-dimensional rotation matrix
$$
beginbmatrix
0.5091 & -0.8607 \
0.8607 & phantom-0.5091
endbmatrix
$$
and I have a vector I'd like to rotate, e.g. $(1, -0.5)$.

My problem is to find an inverse of the rotation matrix so that I can later “undo” the rotation performed on the vector so that I get back the original vector.

The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix.

matrices rotations

share|cite|improve this question

edited Aug 27 at 10:34

Jendrik Stelzner

7,63121037

asked Aug 27 at 6:13

aleksv

396

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

For example, I have a two-dimensional rotation matrix
$$
beginbmatrix
0.5091 & -0.8607 \
0.8607 & phantom-0.5091
endbmatrix
$$
and I have a vector I'd like to rotate, e.g. $(1, -0.5)$.

My problem is to find an inverse of the rotation matrix so that I can later “undo” the rotation performed on the vector so that I get back the original vector.

The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix.

matrices rotations

share|cite|improve this question

edited Aug 27 at 10:34

Jendrik Stelzner

7,63121037

asked Aug 27 at 6:13

aleksv

396

For example, I have a two-dimensional rotation matrix
$$
beginbmatrix
0.5091 & -0.8607 \
0.8607 & phantom-0.5091
endbmatrix
$$
and I have a vector I'd like to rotate, e.g. $(1, -0.5)$.

My problem is to find an inverse of the rotation matrix so that I can later “undo” the rotation performed on the vector so that I get back the original vector.

The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix.

matrices rotations

share|cite|improve this question

edited Aug 27 at 10:34

Jendrik Stelzner

7,63121037

asked Aug 27 at 6:13

aleksv

396

share|cite|improve this question

share|cite|improve this question

edited Aug 27 at 10:34

Jendrik Stelzner

7,63121037

edited Aug 27 at 10:34

Jendrik Stelzner

7,63121037

edited Aug 27 at 10:34

Jendrik Stelzner

7,63121037

7,63121037

asked Aug 27 at 6:13

aleksv

396

asked Aug 27 at 6:13

aleksv

396

asked Aug 27 at 6:13

aleksv

396

396

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
5
down vote



accepted

Recall that rotation matrices are orthogonal therefore

$$A^-1=A^T$$

indeed note that

$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$

share|cite|improve this answer

edited Aug 27 at 6:35

answered Aug 27 at 6:29

gimusi

70.7k73786

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect, thanks!
  – aleksv
  Aug 27 at 6:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
  – gimusi
  Aug 27 at 6:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$

share|cite|improve this answer

answered Aug 27 at 6:30

md2perpe

6,53811023

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Recall that rotation matrices are orthogonal therefore

$$A^-1=A^T$$

indeed note that

$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$

share|cite|improve this answer

edited Aug 27 at 6:35

answered Aug 27 at 6:29

gimusi

70.7k73786

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect, thanks!
  – aleksv
  Aug 27 at 6:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
  – gimusi
  Aug 27 at 6:36
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

Recall that rotation matrices are orthogonal therefore

$$A^-1=A^T$$

indeed note that

$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$

share|cite|improve this answer

edited Aug 27 at 6:35

answered Aug 27 at 6:29

gimusi

70.7k73786

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect, thanks!
  – aleksv
  Aug 27 at 6:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
  – gimusi
  Aug 27 at 6:36
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Recall that rotation matrices are orthogonal therefore

$$A^-1=A^T$$

indeed note that

$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$

share|cite|improve this answer

edited Aug 27 at 6:35

answered Aug 27 at 6:29

gimusi

70.7k73786

Recall that rotation matrices are orthogonal therefore

$$A^-1=A^T$$

indeed note that

$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$

share|cite|improve this answer

edited Aug 27 at 6:35

answered Aug 27 at 6:29

gimusi

70.7k73786

share|cite|improve this answer

share|cite|improve this answer

edited Aug 27 at 6:35

edited Aug 27 at 6:35

edited Aug 27 at 6:35

answered Aug 27 at 6:29

gimusi

70.7k73786

answered Aug 27 at 6:29

gimusi

70.7k73786

answered Aug 27 at 6:29

gimusi

70.7k73786

70.7k73786

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect, thanks!
  – aleksv
  Aug 27 at 6:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
  – gimusi
  Aug 27 at 6:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect, thanks!
  – aleksv
  Aug 27 at 6:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
  – gimusi
  Aug 27 at 6:36
  

  
  
  
  

Perfect, thanks!
– aleksv
Aug 27 at 6:35

Perfect, thanks!
– aleksv
Aug 27 at 6:35

1

1

@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
– gimusi
Aug 27 at 6:36

@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
– gimusi
Aug 27 at 6:36

add a comment | 

up vote
1
down vote

That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$

share|cite|improve this answer

answered Aug 27 at 6:30

md2perpe

6,53811023

add a comment | 

up vote
1
down vote

That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$

share|cite|improve this answer

answered Aug 27 at 6:30

md2perpe

6,53811023

add a comment | 

up vote
1
down vote

up vote
1
down vote

That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$

share|cite|improve this answer

answered Aug 27 at 6:30

md2perpe

6,53811023

That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$

share|cite|improve this answer

answered Aug 27 at 6:30

md2perpe

6,53811023

share|cite|improve this answer

share|cite|improve this answer

answered Aug 27 at 6:30

md2perpe

6,53811023

answered Aug 27 at 6:30

md2perpe

6,53811023

answered Aug 27 at 6:30

md2perpe

6,53811023

6,53811023

add a comment | 

add a comment | 

 

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