Determinant of rotation matrices

[loom91]

Hi,

After obtaining the 2D rotation matrix (as a function of rotation angle) once by geometry and once by complex algebra, I tried to obtain it by invariance of the Euclidean metric. By this approach, the four elements of the 2D rotation matrix can be determined in terms of a single adjustable parameter, which is all right as rotation in 2D is characterised by a single parameter.

However when I try this with the 3D rotation matrices, I get 3 adjustable parameters. But 3D rotation is characterised by only 2 independent parameters! What's happening?

Further, I can show that the 2D rotation matrices must have determinant equal to 1/-1. Is the same true for 3 or higher dimensions?

Thanks.

Molu

 

[HallsofIvy]

No, 3D rotation is determined by 3 parameters: the angle of rotation and two angles to determine the direction of the axis. Imagining the axis of rotation going through the origin, those angles might be the angle the axis makes with the z-axis and with the x-axis. The angle it makes with the y-axis could then be calculated.

Equivalently, you can think of the "rotation" as a vector pointing in the direction of the axis of rotation with length equal to the angle of rotation- 3 parameters.

Finally, the rotation is equivalent to an anti-symmetric matrix and so has 3 independent values: all 0s along the diagonal and the lower triangle is just the negative of the three values in the upper triangle.

 

[loom91]

But aren't you talking about rotation of rigid bodies? A vector is characterised by three coordinates. If we take, for example, the spherical polar coordinate system, there is one scale coordinate and two angle coordinates. That means a pure rotation can be described completely by giving the changes in the two angle coordinates. I can't find anything wrong with this line of reasoning.

Molu

 

[D H]

If we take, for example, the spherical polar coordinate system, there is one scale coordinate and two angle coordinates. That means a pure rotation can be described completely by giving the changes in the two angle coordinates. I can't find anything wrong with this line of reasoning.

Molu

The problem is that line of reasoning is a non sequitur. Here's a better way of looking at it: Those two angles are the equation of a line passing through the origin in spherical coordinates. Look at that line as one of the axes of another coordinate system. The other two axes form a plane, and as you know, rotation in a plane can be described in terms of one angle.

Here's another way to look at things. A 3x3 matrix has nine elements. In other words, the 3x3 matrices are nine dimensional entities. Nine numbers picked at random do not form a rotation matrix. Those nine numbers must obey some constraints. The rows (or columns) must be unit vectors (three constraints), and the rows (or columns) must be normal to each other (three more constraints). Nine degrees of freedom less six constraints => three degrees of freedom.

 

[loom91]

I've thought long and hard but I really don't get this. How can the rotation of a vector *not* be described by 2 independent parameters? If I specify the change in angle with the z axis and the change in angle with the x-axis of the projection on the x-y plane, I've completely specified the new direction of the unit vector. What else is there to specify? If it's a rigid body, I completely understand why it has to be 3 parameters, but it really doesn't make sense for vectors.

Also, what will be the determinant of the 3D rotation matrix? Will it be 1/-1? Then how can I prove this?

Thank you.

Molu

 

[loom91]

A little help please?

Molu

 

[D H]

I've thought long and hard but I really don't get this. How can the rotation of a vector *not* be described by 2 independent parameters?

This is still a non sequitur. You are correct in the sense that it does indeed take two parameters to describe the rotation of a single vector in 3 space. However, a transformation matrix (or a rotation matix) describes the transformation (or rotation) of an entire space, not just one vector.

It does take just 2 independent parameters to describe the rotation of a single vector in three space. This means that it takes 2 independent parameters to describe the rotation of a plane (sans coordinates on the plane) in three space. Rotating a normal to the plane will do it. However, this does not give the requisite information for a coordinate transform in three space. The rotated plane needs to be rotated about its normal. Thus, three rotations are required: Two to properly orient a reference plane, and one more to properly align the reference plane.

 

[CrazyIvan]

Think about an airplane. There are 3 different rotation parameters for that: yaw, pitch, and roll.

 

[marcusl]

Look at the description of Euler angles (especially the diagrams) in any mechanics book (Marion, Goldstein, Landau), or look here

You can see clearly why 3 angles are required to rotate a 3D coordinate basis.

EDIT: What happened to the link? It showed up in the preview...
Well, here it is in text:
http://mathworld.wolfram.com/EulerAngles.html

 

[loom91]

This is still a non sequitur. You are correct in the sense that it does indeed take two parameters to describe the rotation of a single vector in 3 space. However, a transformation matrix (or a rotation matix) describes the transformation (or rotation) of an entire space, not just one vector.

It does take just 2 independent parameters to describe the rotation of a single vector in three space. This means that it takes 2 independent parameters to describe the rotation of a plane (sans coordinates on the plane) in three space. Rotating a normal to the plane will do it. However, this does not give the requisite information for a coordinate transform in three space. The rotated plane needs to be rotated about its normal. Thus, three rotations are required: Two to properly orient a reference plane, and one more to properly align the reference plane.

Finally, a ray of light! I think I see this now. The rotational matrix does not describe the rotation of one specific vector, but the entire vector space, which introduces an extra DoF. Right? Thanks a lot. It feels great when the necker cube suddenly falls into place.

Molu