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How to stack rotation matrices to operate on a matrix of samples?
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I can operate on a $3times 1$ vector, $a$, with a $3times 3$ rotation matrix, $r$. Such that $r a=b$, another $3times 1$ vector.
If I have $N$ of these vectors and rotation matrices I can stack them to create matrix $A$ with dimension $3times N$ and matrix $R$ with dimension $3times 3times N$. Is there a way to make $R$ a $Ntimes N$ matrix that preserves its information? So that I can operate on $A$ from the right to find $B$? or $A^T$ from the left to find $B^T$
linear-algebra linear-transformations rotations
edited Aug 1 at 21:32
Mauricio Cele Lopez Belon
54728
asked Jul 31 at 19:23
Ian Campbell Moore
34
add a comment |Â
up vote
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down vote
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I can operate on a $3times 1$ vector, $a$, with a $3times 3$ rotation matrix, $r$. Such that $r a=b$, another $3times 1$ vector.
If I have $N$ of these vectors and rotation matrices I can stack them to create matrix $A$ with dimension $3times N$ and matrix $R$ with dimension $3times 3times N$. Is there a way to make $R$ a $Ntimes N$ matrix that preserves its information? So that I can operate on $A$ from the right to find $B$? or $A^T$ from the left to find $B^T$
linear-algebra linear-transformations rotations
edited Aug 1 at 21:32
Mauricio Cele Lopez Belon
54728
asked Jul 31 at 19:23
Ian Campbell Moore
34
Please use MathJax to format your questions. You'll get a lot more readership if you do, I think.
– saulspatz
Jul 31 at 20:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I can operate on a $3times 1$ vector, $a$, with a $3times 3$ rotation matrix, $r$. Such that $r a=b$, another $3times 1$ vector.
If I have $N$ of these vectors and rotation matrices I can stack them to create matrix $A$ with dimension $3times N$ and matrix $R$ with dimension $3times 3times N$. Is there a way to make $R$ a $Ntimes N$ matrix that preserves its information? So that I can operate on $A$ from the right to find $B$? or $A^T$ from the left to find $B^T$
linear-algebra linear-transformations rotations
edited Aug 1 at 21:32
Mauricio Cele Lopez Belon
54728
asked Jul 31 at 19:23
Ian Campbell Moore
34
I can operate on a $3times 1$ vector, $a$, with a $3times 3$ rotation matrix, $r$. Such that $r a=b$, another $3times 1$ vector.
If I have $N$ of these vectors and rotation matrices I can stack them to create matrix $A$ with dimension $3times N$ and matrix $R$ with dimension $3times 3times N$. Is there a way to make $R$ a $Ntimes N$ matrix that preserves its information? So that I can operate on $A$ from the right to find $B$? or $A^T$ from the left to find $B^T$
linear-algebra linear-transformations rotations
edited Aug 1 at 21:32
Mauricio Cele Lopez Belon
54728
asked Jul 31 at 19:23
Ian Campbell Moore
34
edited Aug 1 at 21:32
Mauricio Cele Lopez Belon
54728
edited Aug 1 at 21:32
Mauricio Cele Lopez Belon
54728
edited Aug 1 at 21:32
Mauricio Cele Lopez Belon
54728
54728
asked Jul 31 at 19:23
Ian Campbell Moore
34
asked Jul 31 at 19:23
Ian Campbell Moore
34
asked Jul 31 at 19:23
Ian Campbell Moore
34
34
Please use MathJax to format your questions. You'll get a lot more readership if you do, I think.
– saulspatz
Jul 31 at 20:14
add a comment |Â
Please use MathJax to format your questions. You'll get a lot more readership if you do, I think.
– saulspatz
Jul 31 at 20:14
Please use MathJax to format your questions. You'll get a lot more readership if you do, I think.
– saulspatz
Jul 31 at 20:14
Please use MathJax to format your questions. You'll get a lot more readership if you do, I think.
– saulspatz
Jul 31 at 20:14
add a comment |Â
1 Answer
1
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0
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Short answer is almost. The long answer requires to do some precisions:
First, if you "stack" a $3 times 1$ column vector $N$ times you get a $3N times 1$ column vector, not a $3 times N$ matrix.
Second, the matrix $R$ should be $3N times 3N$ square matrix, as it has $N$ number of $3 times 3$ matrices on its diagonal and zeros everywhere esle.
So if you stack your $N$ vectors in a $3N times 1$ vectot $A$ and arrange your matrix $R$ of dimension $3N times 3N$ as described above you will get $ R A = B$ and $ A = R^T B$ as you want.
TL;DR
Now that will be extremely slow in a computer. In order to do better you will need to use sparse matrix multiplication since the matrix $R$ is sparse. However, I think doing the individual $3 times 3$ matrix multiplication with a $3 times 1$ column vector will be much faster since modern linear algebra libraries use vectorized instructions for those tiny matrices which would reduce the multiplication time by a factor of four.
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Short answer is almost. The long answer requires to do some precisions:
First, if you "stack" a $3 times 1$ column vector $N$ times you get a $3N times 1$ column vector, not a $3 times N$ matrix.
Second, the matrix $R$ should be $3N times 3N$ square matrix, as it has $N$ number of $3 times 3$ matrices on its diagonal and zeros everywhere esle.
So if you stack your $N$ vectors in a $3N times 1$ vectot $A$ and arrange your matrix $R$ of dimension $3N times 3N$ as described above you will get $ R A = B$ and $ A = R^T B$ as you want.
TL;DR
Now that will be extremely slow in a computer. In order to do better you will need to use sparse matrix multiplication since the matrix $R$ is sparse. However, I think doing the individual $3 times 3$ matrix multiplication with a $3 times 1$ column vector will be much faster since modern linear algebra libraries use vectorized instructions for those tiny matrices which would reduce the multiplication time by a factor of four.
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
add a comment |Â
up vote
0
down vote
accepted
Short answer is almost. The long answer requires to do some precisions:
First, if you "stack" a $3 times 1$ column vector $N$ times you get a $3N times 1$ column vector, not a $3 times N$ matrix.
Second, the matrix $R$ should be $3N times 3N$ square matrix, as it has $N$ number of $3 times 3$ matrices on its diagonal and zeros everywhere esle.
So if you stack your $N$ vectors in a $3N times 1$ vectot $A$ and arrange your matrix $R$ of dimension $3N times 3N$ as described above you will get $ R A = B$ and $ A = R^T B$ as you want.
TL;DR
Now that will be extremely slow in a computer. In order to do better you will need to use sparse matrix multiplication since the matrix $R$ is sparse. However, I think doing the individual $3 times 3$ matrix multiplication with a $3 times 1$ column vector will be much faster since modern linear algebra libraries use vectorized instructions for those tiny matrices which would reduce the multiplication time by a factor of four.
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Short answer is almost. The long answer requires to do some precisions:
First, if you "stack" a $3 times 1$ column vector $N$ times you get a $3N times 1$ column vector, not a $3 times N$ matrix.
Second, the matrix $R$ should be $3N times 3N$ square matrix, as it has $N$ number of $3 times 3$ matrices on its diagonal and zeros everywhere esle.
So if you stack your $N$ vectors in a $3N times 1$ vectot $A$ and arrange your matrix $R$ of dimension $3N times 3N$ as described above you will get $ R A = B$ and $ A = R^T B$ as you want.
TL;DR
Now that will be extremely slow in a computer. In order to do better you will need to use sparse matrix multiplication since the matrix $R$ is sparse. However, I think doing the individual $3 times 3$ matrix multiplication with a $3 times 1$ column vector will be much faster since modern linear algebra libraries use vectorized instructions for those tiny matrices which would reduce the multiplication time by a factor of four.
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
Short answer is almost. The long answer requires to do some precisions:
First, if you "stack" a $3 times 1$ column vector $N$ times you get a $3N times 1$ column vector, not a $3 times N$ matrix.
Second, the matrix $R$ should be $3N times 3N$ square matrix, as it has $N$ number of $3 times 3$ matrices on its diagonal and zeros everywhere esle.
So if you stack your $N$ vectors in a $3N times 1$ vectot $A$ and arrange your matrix $R$ of dimension $3N times 3N$ as described above you will get $ R A = B$ and $ A = R^T B$ as you want.
TL;DR
Now that will be extremely slow in a computer. In order to do better you will need to use sparse matrix multiplication since the matrix $R$ is sparse. However, I think doing the individual $3 times 3$ matrix multiplication with a $3 times 1$ column vector will be much faster since modern linear algebra libraries use vectorized instructions for those tiny matrices which would reduce the multiplication time by a factor of four.
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
answered Aug 1 at 21:51
Mauricio Cele Lopez Belon
54728
54728
add a comment |Â
add a comment |Â
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Please use MathJax to format your questions. You'll get a lot more readership if you do, I think.
– saulspatz
Jul 31 at 20:14