Rotation matrix (nxn).

1. Jul 30, 2008

MathematicalPhysicist

I know how the rotation matrix looks like in the 2x2 and 3x3 orders, but how does it look in general?

2. Jul 30, 2008

HallsofIvy

I know that, in 2 dimensions, a rotation about (0,0) by angle $\theta$ can be represented as a matrix by
$$\left[\begin{array}{cc}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{array}\right]$$
but what are you saying a rotation in 3 dimensions looks like?

Any matrix with determinant 1 can represent a rotation.

3. Jul 30, 2008

maze

Think of a rotation as taking the x,y,z,w,... axes, and rotating them around to a new set of axes x2,y2,z2,w2,... If you take the unit vectors in the representing of the new axes, and put them in the columns of a matrix in their respective order, then that matrix you get will be the rotation matrix.

In R2, you may notice that the vector (cos(theta),sin(theta)) and (-sin(theta),cos(theta)) are perpendicular unit vectors rotated theta degrees from the original (1,0) and (0,1) vectors.

4. Jul 30, 2008

Peeter

If you are looking for a R^N rotation in a plane, rotating the components of a vector that lies in the plane and leaving the "perpendicular" components (ie: components in all normal directions) untouched, then this can be expressed with the following geometric-algebra/clifford-product expansion:

$$R(x) = e^{-i\theta/2} x e^{i\theta/2} = x \cos^2(\theta/2) - i x i \sin^2(\theta/2) + 2 \cos(\theta/2) \sin(\theta/2) x \cdot i$$

Here the unit bivector i is formed by the product of two non-colinear vectors n_k in the plane:

$$i=\frac{n_1 \wedge n_2}{\lvert n_1 \wedge n_2 \rvert}.$$

It's a bit messy, but you can calculate the rotation matrix from this if you really wanted to (with an orthonormal basis basis e_i and calculation of R(e_i) \cdot e_j gives the matrix).

5. Jul 30, 2008

Peeter

another way to look at this is, just like so:

$$R(x) = x_{\perp} + x_{\parallel}\left( cos\theta + i\sin\theta\right)$$

where $x_{\perp}$ is the component not in the plane of rotation, $x_{\parallel}$ is the component in the plane of rotation, and $x_{\parallel} i$ is a 90 degree rotation in the desired sense of the component of the vector in the plane. Then given any way you are comfortable with (ie: projection and rejection matrixes P, and (I-P)) for finding the components of the vector in and out of the plane, you can use that to compute the rotation matrix.

6. Jul 30, 2008

mathwonk

i think a rotation is an orthogonal matrix with determinant one?

so there is some basis in which the matrix looks like a bunch of 1's on the diagonal, and then a bunch of 2 by 2 matrices representing plane rotations.

7. Jul 31, 2008

MathematicalPhysicist

Yes, thanks mathwonk I also got this answer after solving a question from Gullimin and Pollack:" if k is odd then f:S^k->S^k f(x)=-x is homotpic to the identity function".

The homotopy is a sort of rotation matrix times x, where the arguement is (pi*t), so if k is odd we can find a rotation matrix (k+1)x(k+1) (k+1 is even) which consists of orthogonal matrices on the diagonal representing plane rotations.

P.S
You got to love geometry! (-:

8. Jul 31, 2008

maze

So that raises an interesting question - what is the minimum number of sequential plane rotations required to represent a general rotation in Rn?

In R2, its just 1. In R3 its 2 since you rotate xhat to xhat', and then roll around xhat'. What about higher dimensions?

9. Jul 31, 2008

Ben Niehoff

Every plane rotation can be factored into two reflections. It might be easier to think in terms of reflections.

There's a really elegant derivation of this (i.e., the number of plane rotations needed to give a general rotation) in The Road to Reality using Clifford algebras, but I forget it.

10. Aug 1, 2008

MathematicalPhysicist

well if we have R^n where n is odd, then there must be 1 in the diagonal of the matrix, while in an even n we can handle without it (if we want).