Consider the group $SO(3)$. The rotation around the $x$-axis is represented by the matrix

$$R_x = \left ( \begin{array}{ ccc } 1 & 0 & 0 \\ 0 & \cos \Theta & - \sin \Theta \\ 0 & \sin \Theta & \cos \Theta \end{array}\right ) $$

Similarly, for the rotations around the other axes. It is very easy to verify that these three are linearly independent.

I wanted to show that they form a basis for $SO(3)$. But this is where I got stuck. The idea should be to consider an arbitrary rotation and then show that it is a linear combination of these $3$ matrices. But I don't know what a generic rotation matrix looks like. So maybe there is a different way of showing this or maybe it is even false.

Please can someone help me?

  • 4
    $\begingroup$ $SO(3)$ is not a group under addition, so you're not going to get very far with showing something is a basis in the sense of linear algebra. The relevant notion is of a generating set of a group. $\endgroup$
    – curious
    Nov 9 '14 at 1:36
  • $\begingroup$ Consider the action of $SO(3)$ on $S^2$. $\endgroup$
    – anomaly
    Nov 9 '14 at 1:55
  • $\begingroup$ @curious But it is a group under multiplication. Why does it matter what we call the group operation? $\endgroup$ Nov 10 '14 at 1:11
  • $\begingroup$ Uhm, maybe I get it. What we call addition would be linear with scalar multiplcation but multiplication is not linear with scalar multiplication. $\endgroup$ Nov 10 '14 at 1:13

Basis is the wrong word. However, you are correct; an oriented (orthogonal) basis can be rotated to the standard $x,y,z$ directions in no more than 3 standard rotations. One to put the $x$ vector in the $xz$ plane. a second rotation to send it to the positive $x$ axis. One final rotation in the $yz$ plane to take those vectors to their proper positions.


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