User:Prof McCarthy/Rotation matrix

As far as i can tell this article on the rotation matrix was created in 2004 and by 2005, the signs on the off-diagonal terms had already been changed to try to find a consistent definition. The current talk page goes back to 2006 with many editors citing references for the location of the minus sign on both sides of the diagonal. In the past, I have been reluctant to wade into this discussion, but the recent flare up has moved me to offer the following.

Maybe it will help to begin by saying both views are correct, which sounds strange, but hopefully allows us to focus on what are actually two different ways of viewing a rotation: (i) a transformation from coordinates in one frame to coordinates in another frame, and (ii) a transformation between two sets of basis vectors defining coordinates in the same frame. The two formulations are closely related an perhaps it is no surprise that they result in matrices that are inverses of each other, which for rotation matrices means the transpose of each other and differ only by the location of the minus sign above and below the diagonal.

(i) Consider the first case, where a reference frame M is rotated counter-clockwise by the angle θ relative to a reference frame F. A vector in M has the coordinate x=xi + yj, where i=(1,0) and j=(0,1) are the natural basis vectors along the coordinate axes of M, so x,=(x, y) in this reference frame. Now consider the coordinates X=(X, Y) of the same point but now measured in F. This is easily done by considering the vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) that are the images of i and j of M, but now measured in the frame F, so X=xe_r+ye_r, or

${\begin{Bmatrix}X\\Y\end{Bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{Bmatrix}x\\y\end{Bmatrix}}.$

This version of the rotation matrix has the vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) as column vectors, which means the minus sign is located on the upper right sine term.

(ii) Now consider the second case, where the vectors are measured in the same frame F, so i and j are the natural basis vectors along the x and y axis of F, and we have the unit vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) measured in F that define a pair of orthogonal unit vectors rotated by the angle θ in the counter clockwise direction. Now, the transformation from coordinates relative to the basis vectors i and j of F to the new basis vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) in F, is easily defined by the computing i=cosθe_r-sinθe_t j=sinθe_r+cosθe_t. This means a vector X=Xi+Yj in F is transformed to a vector in the rotated basis, by the transformation,

${\begin{Bmatrix}x\\y\end{Bmatrix}}={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}}{\begin{Bmatrix}X\\Y\end{Bmatrix}}$

Now the columns of the rotation matrix are the coordinates of the natural basis vectors i and j as measured in the new basis e_r and e_t. This places the minus sign on the sine term below the diagonal.

Hopefully, it is clear that the two ways to define a rotation matrix are inverses of each other, and simply moves the minus sign. This same situation applies to the rotations about each of the coordinate axes in three dimensions. The first case results in rotation matrices that have the minus sign in the upper diagonal for Rx and Rz and the lower diagonal for Rz, while the second case yields the transpose of these matrices, with the minus sign in the lower diagonal for Rx and Rz and in the upper diagonal for Ry. I hope this is helpful. Prof McCarthy (talk) 04:52, 19 April 2013 (UTC)