Month: June 2010

3D Rotations

On By JKMSMKJIn Linear Algebra, Mathematics, Physics, TimePassLeave a comment

You should have come from here!

Here’s a quickie: What are the eigenvalues of a 2D rotation matrix?

Here’s a problem: For a bunch of rotations performed one after another on a 3D object, find an equivalent single rotation which would give the same result.

Here’s a solution: First of all, multiply all the rotation matrices (obviously!) such that R_{s}=R_{n}R_{n-1}\cdots R_{2}R_{1}. A single rotation is characterized by an axis and an angle of rotation. Let’s get the angle first. Find the eigenvalues of R_{s}. Two of them will be of the form: e^{\dot{\iota}\theta}\,\&\,e^{-\dot{\iota}\theta}. This θ is the required angle. Let’s get the axis now. So what would be the third eigenvalue? Remember R_{s} \in SO(3) which means the third one is 1! Find the eigenvector corresponding to this eigenvalue which is the required axis!

Here’s an example: Let’s do the problem mentioned in Gravitation.

R_{1}=\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix};

R_{2}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix}.

\Rightarrow R_{s}=R_{2}R_{1}=\begin{pmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{pmatrix}.

Relevant eigenvalues of R_{s} are -\frac{1}{2}\pm\frac{\sqrt{3}}{2}\dot{\iota} which give us the angle: \mathrm{tan}^{-1}\left(-\sqrt{3}\right)=120^{\circ}. The eigenvector corresponding to 1 is \frac{1}{\sqrt{3}}(1,-1,1) so the axis is one of the diagonals! This solution agrees with the one given in the book.