Inversion of rotation matrix
$begingroup$
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
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
asked Aug 27 '18 at 6:13
aleksvaleksv
396
$endgroup$
add a comment |
$begingroup$
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
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
asked Aug 27 '18 at 6:13
aleksvaleksv
396
$endgroup$
add a comment |
$begingroup$
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
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
asked Aug 27 '18 at 6:13
aleksvaleksv
396
$endgroup$
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
matrices rotations
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
asked Aug 27 '18 at 6:13
aleksvaleksv
396
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
asked Aug 27 '18 at 6:13
aleksvaleksv
396
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
edited Aug 27 '18 at 10:34
Jendrik Stelzner
7,88121339
7,88121339
asked Aug 27 '18 at 6:13
aleksvaleksv
396
asked Aug 27 '18 at 6:13
aleksvaleksv
396
asked Aug 27 '18 at 6:13
aleksvaleksv
396
396
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Recall that rotation matrices are orthogonal therefore
$$A^-1=A^T$$
indeed note that
$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Perfect, thanks!
$endgroup$
– aleksv
Aug 27 '18 at 6:35
1
$begingroup$
@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
$endgroup$
– gimusi
Aug 27 '18 at 6:36
add a comment |
$begingroup$
That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
$endgroup$
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
$begingroup$
Recall that rotation matrices are orthogonal therefore
$$A^-1=A^T$$
indeed note that
$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Perfect, thanks!
$endgroup$
– aleksv
Aug 27 '18 at 6:35
1
$begingroup$
@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
$endgroup$
– gimusi
Aug 27 '18 at 6:36
add a comment |
$begingroup$
Recall that rotation matrices are orthogonal therefore
$$A^-1=A^T$$
indeed note that
$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Perfect, thanks!
$endgroup$
– aleksv
Aug 27 '18 at 6:35
1
$begingroup$
@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
$endgroup$
– gimusi
Aug 27 '18 at 6:36
add a comment |
$begingroup$
Recall that rotation matrices are orthogonal therefore
$$A^-1=A^T$$
indeed note that
$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
$endgroup$
Recall that rotation matrices are orthogonal therefore
$$A^-1=A^T$$
indeed note that
$$A^-1=beginbmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendbmatrix^-1
=beginbmatrixcos(-alpha) & -sin(-alpha)\ sin(-alpha) & cos(-alpha)endbmatrix=beginbmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendbmatrix=A^T$$
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
edited Aug 27 '18 at 6:35
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
answered Aug 27 '18 at 6:29
gimusigimusi
92.9k84494
92.9k84494
$begingroup$
Perfect, thanks!
$endgroup$
– aleksv
Aug 27 '18 at 6:35
1
$begingroup$
@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
$endgroup$
– gimusi
Aug 27 '18 at 6:36
add a comment |
$begingroup$
Perfect, thanks!
$endgroup$
– aleksv
Aug 27 '18 at 6:35
1
$begingroup$
@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
$endgroup$
– gimusi
Aug 27 '18 at 6:36
$begingroup$
Perfect, thanks!
$endgroup$
– aleksv
Aug 27 '18 at 6:35
$begingroup$
Perfect, thanks!
$endgroup$
– aleksv
Aug 27 '18 at 6:35
1
1
$begingroup$
@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
$endgroup$
– gimusi
Aug 27 '18 at 6:36
$begingroup$
@aleksv I've also added a simple way to see that assuming the opposite angle. You are welcome! Bye
$endgroup$
– gimusi
Aug 27 '18 at 6:36
add a comment |
$begingroup$
That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
$endgroup$
add a comment |
$begingroup$
That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
$endgroup$
add a comment |
$begingroup$
That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
$endgroup$
That's easy:
$$beginpmatrixcosalpha & -sinalpha \ sinalpha & cosalphaendpmatrix^-1
=beginpmatrixcosalpha & sinalpha \ -sinalpha & cosalphaendpmatrix$$
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
answered Aug 27 '18 at 6:30
md2perpemd2perpe
8,05111028
8,05111028
add a comment |
add a comment |
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