determination of axis of rotation from rotation matrices
Clash Royale CLAN TAG#URR8PPP
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I am given a problem to find the mapping corresponding to the following matrix
$left[beginarraycc
-frac12 & fracsqrt32\
-fracsqrt32 & -frac12
endarrayright]$
Now using the eigenvalue equation $M(r)=r$, I found two equations
$y=sqrt3x$ and
$y=-frac1sqrt3x$
and these are two straight lines crossing at the origin.
I am a bit confused about how to find the axis of rotation from these information.
linear-algebra
edited Aug 13 at 5:41
Anik Bhowmick
432317
asked Aug 13 at 4:48
STGoswami
142
add a comment |Â
up vote
0
down vote
favorite
I am given a problem to find the mapping corresponding to the following matrix
$left[beginarraycc
-frac12 & fracsqrt32\
-fracsqrt32 & -frac12
endarrayright]$
Now using the eigenvalue equation $M(r)=r$, I found two equations
$y=sqrt3x$ and
$y=-frac1sqrt3x$
and these are two straight lines crossing at the origin.
I am a bit confused about how to find the axis of rotation from these information.
linear-algebra
edited Aug 13 at 5:41
Anik Bhowmick
432317
asked Aug 13 at 4:48
STGoswami
142
1
This is a rotation of the plane. There is no axis. Or, the rotation is about the origin. Rotations in $BbbR^3$ have an axis. In higher dimensions it depends on your definition.
– Jyrki Lahtonen
Aug 13 at 4:55
1
I’d be curious to see how you came up with these eigenvectors, since $1$ isn’t even an eigenvalue of this matrix. The only solution to $Mr=r$ is the zero vector.
– amd
Aug 13 at 5:02
1
@amd I think that, in solving for eigenvectors for non-eigenvalue $1$, the asker found themselves solving those two equations simultaneously, and also arrived at the conclusion that the only solution was the $0$ vector.
– Theo Bendit
Aug 13 at 5:21
@amd, I simply thought that there should be some rotation axis the vectors along which would remain the same under the matrix operation. I took this idea from a 3 by 3 matrix which may correspond to a rotation about some axis. Now as has been commented above, there is no axis for this 2 by 2 matrix since this corresponds to a rotation in plane devoid of a third axis. Is there something wrong here?
– STGoswami
Aug 13 at 15:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given a problem to find the mapping corresponding to the following matrix
$left[beginarraycc
-frac12 & fracsqrt32\
-fracsqrt32 & -frac12
endarrayright]$
Now using the eigenvalue equation $M(r)=r$, I found two equations
$y=sqrt3x$ and
$y=-frac1sqrt3x$
and these are two straight lines crossing at the origin.
I am a bit confused about how to find the axis of rotation from these information.
linear-algebra
edited Aug 13 at 5:41
Anik Bhowmick
432317
asked Aug 13 at 4:48
STGoswami
142
I am given a problem to find the mapping corresponding to the following matrix
$left[beginarraycc
-frac12 & fracsqrt32\
-fracsqrt32 & -frac12
endarrayright]$
Now using the eigenvalue equation $M(r)=r$, I found two equations
$y=sqrt3x$ and
$y=-frac1sqrt3x$
and these are two straight lines crossing at the origin.
I am a bit confused about how to find the axis of rotation from these information.
linear-algebra
edited Aug 13 at 5:41
Anik Bhowmick
432317
asked Aug 13 at 4:48
STGoswami
142
edited Aug 13 at 5:41
Anik Bhowmick
432317
edited Aug 13 at 5:41
Anik Bhowmick
432317
edited Aug 13 at 5:41
Anik Bhowmick
432317
432317
asked Aug 13 at 4:48
STGoswami
142
asked Aug 13 at 4:48
STGoswami
142
asked Aug 13 at 4:48
STGoswami
142
142
1
This is a rotation of the plane. There is no axis. Or, the rotation is about the origin. Rotations in $BbbR^3$ have an axis. In higher dimensions it depends on your definition.
– Jyrki Lahtonen
Aug 13 at 4:55
1
I’d be curious to see how you came up with these eigenvectors, since $1$ isn’t even an eigenvalue of this matrix. The only solution to $Mr=r$ is the zero vector.
– amd
Aug 13 at 5:02
1
@amd I think that, in solving for eigenvectors for non-eigenvalue $1$, the asker found themselves solving those two equations simultaneously, and also arrived at the conclusion that the only solution was the $0$ vector.
– Theo Bendit
Aug 13 at 5:21
@amd, I simply thought that there should be some rotation axis the vectors along which would remain the same under the matrix operation. I took this idea from a 3 by 3 matrix which may correspond to a rotation about some axis. Now as has been commented above, there is no axis for this 2 by 2 matrix since this corresponds to a rotation in plane devoid of a third axis. Is there something wrong here?
– STGoswami
Aug 13 at 15:09
add a comment |Â
1
This is a rotation of the plane. There is no axis. Or, the rotation is about the origin. Rotations in $BbbR^3$ have an axis. In higher dimensions it depends on your definition.
– Jyrki Lahtonen
Aug 13 at 4:55
1
I’d be curious to see how you came up with these eigenvectors, since $1$ isn’t even an eigenvalue of this matrix. The only solution to $Mr=r$ is the zero vector.
– amd
Aug 13 at 5:02
1
@amd I think that, in solving for eigenvectors for non-eigenvalue $1$, the asker found themselves solving those two equations simultaneously, and also arrived at the conclusion that the only solution was the $0$ vector.
– Theo Bendit
Aug 13 at 5:21
@amd, I simply thought that there should be some rotation axis the vectors along which would remain the same under the matrix operation. I took this idea from a 3 by 3 matrix which may correspond to a rotation about some axis. Now as has been commented above, there is no axis for this 2 by 2 matrix since this corresponds to a rotation in plane devoid of a third axis. Is there something wrong here?
– STGoswami
Aug 13 at 15:09
1
1
This is a rotation of the plane. There is no axis. Or, the rotation is about the origin. Rotations in $BbbR^3$ have an axis. In higher dimensions it depends on your definition.
– Jyrki Lahtonen
Aug 13 at 4:55
This is a rotation of the plane. There is no axis. Or, the rotation is about the origin. Rotations in $BbbR^3$ have an axis. In higher dimensions it depends on your definition.
– Jyrki Lahtonen
Aug 13 at 4:55
1
1
I’d be curious to see how you came up with these eigenvectors, since $1$ isn’t even an eigenvalue of this matrix. The only solution to $Mr=r$ is the zero vector.
– amd
Aug 13 at 5:02
I’d be curious to see how you came up with these eigenvectors, since $1$ isn’t even an eigenvalue of this matrix. The only solution to $Mr=r$ is the zero vector.
– amd
Aug 13 at 5:02
1
1
@amd I think that, in solving for eigenvectors for non-eigenvalue $1$, the asker found themselves solving those two equations simultaneously, and also arrived at the conclusion that the only solution was the $0$ vector.
– Theo Bendit
Aug 13 at 5:21
@amd I think that, in solving for eigenvectors for non-eigenvalue $1$, the asker found themselves solving those two equations simultaneously, and also arrived at the conclusion that the only solution was the $0$ vector.
– Theo Bendit
Aug 13 at 5:21
@amd, I simply thought that there should be some rotation axis the vectors along which would remain the same under the matrix operation. I took this idea from a 3 by 3 matrix which may correspond to a rotation about some axis. Now as has been commented above, there is no axis for this 2 by 2 matrix since this corresponds to a rotation in plane devoid of a third axis. Is there something wrong here?
– STGoswami
Aug 13 at 15:09
@amd, I simply thought that there should be some rotation axis the vectors along which would remain the same under the matrix operation. I took this idea from a 3 by 3 matrix which may correspond to a rotation about some axis. Now as has been commented above, there is no axis for this 2 by 2 matrix since this corresponds to a rotation in plane devoid of a third axis. Is there something wrong here?
– STGoswami
Aug 13 at 15:09
add a comment |Â
1 Answer
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votes
up vote
1
down vote
The idea of an axis of rotation is pretty much unique to three-dimensional spaces. In general, a rotation takes place in some plane. The orthogonal complement of this plane remains unaffected by the rotation. In three dimensions, this space is one-dimensional, so you can identify a single line through the center of rotation as its axis. In higher-dimensional spaces, this complementary space has more than one dimension, so there’s no unique line that is left fixed by it. (One could by extension call this space the “axis,†but I’ve not found that very useful.) In a two-dimensional space, there’s nothing outside of the plane of rotation, which is the entire space, so it doesn’t really make sense to speak of a rotation axis there, either. You often see this done, anyway, but doing so views the plane as embedded in a three-dimensional space instead of standing on its own.
You can identify a fixed point in the plane of rotation, though. This is the center of rotation within the rotation plane and makes sense in any number of dimensions. For a rotation that’s a linear transformation like your example, this will always be the origin.
Now, you can certainly find the axis of a 3-D rotation matrix by computing eigenvectors of $1$, but that works because $1$ is always an eigenvalue of a 3-D rotation and as I noted above, the subspace that’s left unchanged by a nontrivial 3-D rotation is one-dimensional. As you discovered, the only solution of the equation $Mv=v$ for your 2-D rotation is $(0,0)$, but that’s to be expected because $1$ isn’t an eigenvalue of $M$—the only point that gets mapped to itself is the origin.
answered Aug 13 at 19:53
amd
26.1k2944
Wouldn't there be an axis of rotation for any $n-1$ dimensional hyperplane in $BbbR^n?$
– Chickenmancer
Aug 13 at 20:47
@Chickenmancer Rotations occur in *planes*—two-dimensional flats—regardless of the dimensionality of the space they’re embedded in.
– amd
Aug 13 at 22:38
1
+1 of course. The description of a rotation as moving only points of a plane, and leaving the $(n-2)$-dimensional orthogonal complement fixed pointwise, is fine. But it has the shortcoming that, unless $nle3$, they don't form a group. To remedy that we can also declare that linear transformations with matrices $A$ satisfying $A^TA=I_n, det A=1$, could all be called rotations. In that case there may be several angles of rotation. See here for more discussion about rotations in nD.
– Jyrki Lahtonen
Aug 14 at 4:29
Thank you @amd for this nice explanation
– STGoswami
Aug 14 at 6:48
@JyrkiLahtonen Indeed. Things can get quite interesting once you start combining these “basic†rotations in various ways.
– amd
Aug 14 at 20:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The idea of an axis of rotation is pretty much unique to three-dimensional spaces. In general, a rotation takes place in some plane. The orthogonal complement of this plane remains unaffected by the rotation. In three dimensions, this space is one-dimensional, so you can identify a single line through the center of rotation as its axis. In higher-dimensional spaces, this complementary space has more than one dimension, so there’s no unique line that is left fixed by it. (One could by extension call this space the “axis,†but I’ve not found that very useful.) In a two-dimensional space, there’s nothing outside of the plane of rotation, which is the entire space, so it doesn’t really make sense to speak of a rotation axis there, either. You often see this done, anyway, but doing so views the plane as embedded in a three-dimensional space instead of standing on its own.
You can identify a fixed point in the plane of rotation, though. This is the center of rotation within the rotation plane and makes sense in any number of dimensions. For a rotation that’s a linear transformation like your example, this will always be the origin.
Now, you can certainly find the axis of a 3-D rotation matrix by computing eigenvectors of $1$, but that works because $1$ is always an eigenvalue of a 3-D rotation and as I noted above, the subspace that’s left unchanged by a nontrivial 3-D rotation is one-dimensional. As you discovered, the only solution of the equation $Mv=v$ for your 2-D rotation is $(0,0)$, but that’s to be expected because $1$ isn’t an eigenvalue of $M$—the only point that gets mapped to itself is the origin.
answered Aug 13 at 19:53
amd
26.1k2944
Wouldn't there be an axis of rotation for any $n-1$ dimensional hyperplane in $BbbR^n?$
– Chickenmancer
Aug 13 at 20:47
@Chickenmancer Rotations occur in *planes*—two-dimensional flats—regardless of the dimensionality of the space they’re embedded in.
– amd
Aug 13 at 22:38
1
+1 of course. The description of a rotation as moving only points of a plane, and leaving the $(n-2)$-dimensional orthogonal complement fixed pointwise, is fine. But it has the shortcoming that, unless $nle3$, they don't form a group. To remedy that we can also declare that linear transformations with matrices $A$ satisfying $A^TA=I_n, det A=1$, could all be called rotations. In that case there may be several angles of rotation. See here for more discussion about rotations in nD.
– Jyrki Lahtonen
Aug 14 at 4:29
Thank you @amd for this nice explanation
– STGoswami
Aug 14 at 6:48
@JyrkiLahtonen Indeed. Things can get quite interesting once you start combining these “basic†rotations in various ways.
– amd
Aug 14 at 20:16
add a comment |Â
up vote
1
down vote
The idea of an axis of rotation is pretty much unique to three-dimensional spaces. In general, a rotation takes place in some plane. The orthogonal complement of this plane remains unaffected by the rotation. In three dimensions, this space is one-dimensional, so you can identify a single line through the center of rotation as its axis. In higher-dimensional spaces, this complementary space has more than one dimension, so there’s no unique line that is left fixed by it. (One could by extension call this space the “axis,†but I’ve not found that very useful.) In a two-dimensional space, there’s nothing outside of the plane of rotation, which is the entire space, so it doesn’t really make sense to speak of a rotation axis there, either. You often see this done, anyway, but doing so views the plane as embedded in a three-dimensional space instead of standing on its own.
You can identify a fixed point in the plane of rotation, though. This is the center of rotation within the rotation plane and makes sense in any number of dimensions. For a rotation that’s a linear transformation like your example, this will always be the origin.
Now, you can certainly find the axis of a 3-D rotation matrix by computing eigenvectors of $1$, but that works because $1$ is always an eigenvalue of a 3-D rotation and as I noted above, the subspace that’s left unchanged by a nontrivial 3-D rotation is one-dimensional. As you discovered, the only solution of the equation $Mv=v$ for your 2-D rotation is $(0,0)$, but that’s to be expected because $1$ isn’t an eigenvalue of $M$—the only point that gets mapped to itself is the origin.
answered Aug 13 at 19:53
amd
26.1k2944
Wouldn't there be an axis of rotation for any $n-1$ dimensional hyperplane in $BbbR^n?$
– Chickenmancer
Aug 13 at 20:47
@Chickenmancer Rotations occur in *planes*—two-dimensional flats—regardless of the dimensionality of the space they’re embedded in.
– amd
Aug 13 at 22:38
1
+1 of course. The description of a rotation as moving only points of a plane, and leaving the $(n-2)$-dimensional orthogonal complement fixed pointwise, is fine. But it has the shortcoming that, unless $nle3$, they don't form a group. To remedy that we can also declare that linear transformations with matrices $A$ satisfying $A^TA=I_n, det A=1$, could all be called rotations. In that case there may be several angles of rotation. See here for more discussion about rotations in nD.
– Jyrki Lahtonen
Aug 14 at 4:29
Thank you @amd for this nice explanation
– STGoswami
Aug 14 at 6:48
@JyrkiLahtonen Indeed. Things can get quite interesting once you start combining these “basic†rotations in various ways.
– amd
Aug 14 at 20:16
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The idea of an axis of rotation is pretty much unique to three-dimensional spaces. In general, a rotation takes place in some plane. The orthogonal complement of this plane remains unaffected by the rotation. In three dimensions, this space is one-dimensional, so you can identify a single line through the center of rotation as its axis. In higher-dimensional spaces, this complementary space has more than one dimension, so there’s no unique line that is left fixed by it. (One could by extension call this space the “axis,†but I’ve not found that very useful.) In a two-dimensional space, there’s nothing outside of the plane of rotation, which is the entire space, so it doesn’t really make sense to speak of a rotation axis there, either. You often see this done, anyway, but doing so views the plane as embedded in a three-dimensional space instead of standing on its own.
You can identify a fixed point in the plane of rotation, though. This is the center of rotation within the rotation plane and makes sense in any number of dimensions. For a rotation that’s a linear transformation like your example, this will always be the origin.
Now, you can certainly find the axis of a 3-D rotation matrix by computing eigenvectors of $1$, but that works because $1$ is always an eigenvalue of a 3-D rotation and as I noted above, the subspace that’s left unchanged by a nontrivial 3-D rotation is one-dimensional. As you discovered, the only solution of the equation $Mv=v$ for your 2-D rotation is $(0,0)$, but that’s to be expected because $1$ isn’t an eigenvalue of $M$—the only point that gets mapped to itself is the origin.
answered Aug 13 at 19:53
amd
26.1k2944
The idea of an axis of rotation is pretty much unique to three-dimensional spaces. In general, a rotation takes place in some plane. The orthogonal complement of this plane remains unaffected by the rotation. In three dimensions, this space is one-dimensional, so you can identify a single line through the center of rotation as its axis. In higher-dimensional spaces, this complementary space has more than one dimension, so there’s no unique line that is left fixed by it. (One could by extension call this space the “axis,†but I’ve not found that very useful.) In a two-dimensional space, there’s nothing outside of the plane of rotation, which is the entire space, so it doesn’t really make sense to speak of a rotation axis there, either. You often see this done, anyway, but doing so views the plane as embedded in a three-dimensional space instead of standing on its own.
You can identify a fixed point in the plane of rotation, though. This is the center of rotation within the rotation plane and makes sense in any number of dimensions. For a rotation that’s a linear transformation like your example, this will always be the origin.
Now, you can certainly find the axis of a 3-D rotation matrix by computing eigenvectors of $1$, but that works because $1$ is always an eigenvalue of a 3-D rotation and as I noted above, the subspace that’s left unchanged by a nontrivial 3-D rotation is one-dimensional. As you discovered, the only solution of the equation $Mv=v$ for your 2-D rotation is $(0,0)$, but that’s to be expected because $1$ isn’t an eigenvalue of $M$—the only point that gets mapped to itself is the origin.
answered Aug 13 at 19:53
amd
26.1k2944
answered Aug 13 at 19:53
amd
26.1k2944
answered Aug 13 at 19:53
amd
26.1k2944
answered Aug 13 at 19:53
amd
26.1k2944
26.1k2944
Wouldn't there be an axis of rotation for any $n-1$ dimensional hyperplane in $BbbR^n?$
– Chickenmancer
Aug 13 at 20:47
@Chickenmancer Rotations occur in *planes*—two-dimensional flats—regardless of the dimensionality of the space they’re embedded in.
– amd
Aug 13 at 22:38
1
+1 of course. The description of a rotation as moving only points of a plane, and leaving the $(n-2)$-dimensional orthogonal complement fixed pointwise, is fine. But it has the shortcoming that, unless $nle3$, they don't form a group. To remedy that we can also declare that linear transformations with matrices $A$ satisfying $A^TA=I_n, det A=1$, could all be called rotations. In that case there may be several angles of rotation. See here for more discussion about rotations in nD.
– Jyrki Lahtonen
Aug 14 at 4:29
Thank you @amd for this nice explanation
– STGoswami
Aug 14 at 6:48
@JyrkiLahtonen Indeed. Things can get quite interesting once you start combining these “basic†rotations in various ways.
– amd
Aug 14 at 20:16
add a comment |Â
Wouldn't there be an axis of rotation for any $n-1$ dimensional hyperplane in $BbbR^n?$
– Chickenmancer
Aug 13 at 20:47
@Chickenmancer Rotations occur in *planes*—two-dimensional flats—regardless of the dimensionality of the space they’re embedded in.
– amd
Aug 13 at 22:38
1
+1 of course. The description of a rotation as moving only points of a plane, and leaving the $(n-2)$-dimensional orthogonal complement fixed pointwise, is fine. But it has the shortcoming that, unless $nle3$, they don't form a group. To remedy that we can also declare that linear transformations with matrices $A$ satisfying $A^TA=I_n, det A=1$, could all be called rotations. In that case there may be several angles of rotation. See here for more discussion about rotations in nD.
– Jyrki Lahtonen
Aug 14 at 4:29
Thank you @amd for this nice explanation
– STGoswami
Aug 14 at 6:48
@JyrkiLahtonen Indeed. Things can get quite interesting once you start combining these “basic†rotations in various ways.
– amd
Aug 14 at 20:16
Wouldn't there be an axis of rotation for any $n-1$ dimensional hyperplane in $BbbR^n?$
– Chickenmancer
Aug 13 at 20:47
Wouldn't there be an axis of rotation for any $n-1$ dimensional hyperplane in $BbbR^n?$
– Chickenmancer
Aug 13 at 20:47
@Chickenmancer Rotations occur in *planes*—two-dimensional flats—regardless of the dimensionality of the space they’re embedded in.
– amd
Aug 13 at 22:38
@Chickenmancer Rotations occur in *planes*—two-dimensional flats—regardless of the dimensionality of the space they’re embedded in.
– amd
Aug 13 at 22:38
1
1
+1 of course. The description of a rotation as moving only points of a plane, and leaving the $(n-2)$-dimensional orthogonal complement fixed pointwise, is fine. But it has the shortcoming that, unless $nle3$, they don't form a group. To remedy that we can also declare that linear transformations with matrices $A$ satisfying $A^TA=I_n, det A=1$, could all be called rotations. In that case there may be several angles of rotation. See here for more discussion about rotations in nD.
– Jyrki Lahtonen
Aug 14 at 4:29
+1 of course. The description of a rotation as moving only points of a plane, and leaving the $(n-2)$-dimensional orthogonal complement fixed pointwise, is fine. But it has the shortcoming that, unless $nle3$, they don't form a group. To remedy that we can also declare that linear transformations with matrices $A$ satisfying $A^TA=I_n, det A=1$, could all be called rotations. In that case there may be several angles of rotation. See here for more discussion about rotations in nD.
– Jyrki Lahtonen
Aug 14 at 4:29
Thank you @amd for this nice explanation
– STGoswami
Aug 14 at 6:48
Thank you @amd for this nice explanation
– STGoswami
Aug 14 at 6:48
@JyrkiLahtonen Indeed. Things can get quite interesting once you start combining these “basic†rotations in various ways.
– amd
Aug 14 at 20:16
@JyrkiLahtonen Indeed. Things can get quite interesting once you start combining these “basic†rotations in various ways.
– amd
Aug 14 at 20:16
add a comment |Â
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1
This is a rotation of the plane. There is no axis. Or, the rotation is about the origin. Rotations in $BbbR^3$ have an axis. In higher dimensions it depends on your definition.
– Jyrki Lahtonen
Aug 13 at 4:55
1
I’d be curious to see how you came up with these eigenvectors, since $1$ isn’t even an eigenvalue of this matrix. The only solution to $Mr=r$ is the zero vector.
– amd
Aug 13 at 5:02
1
@amd I think that, in solving for eigenvectors for non-eigenvalue $1$, the asker found themselves solving those two equations simultaneously, and also arrived at the conclusion that the only solution was the $0$ vector.
– Theo Bendit
Aug 13 at 5:21
@amd, I simply thought that there should be some rotation axis the vectors along which would remain the same under the matrix operation. I took this idea from a 3 by 3 matrix which may correspond to a rotation about some axis. Now as has been commented above, there is no axis for this 2 by 2 matrix since this corresponds to a rotation in plane devoid of a third axis. Is there something wrong here?
– STGoswami
Aug 13 at 15:09