# Rotations in SO(3)

I've read in a book that every rotation in the $SO(3)$ can be represented by a matrix of the form $\begin{pmatrix} \cos(\theta)&-\sin(\theta)&0 \\ \sin(\theta)&\cos(\theta)&0 \\ 0&0&1 \end{pmatrix}$ for some $\theta$.

Now $\begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&-1 \end{pmatrix}$ is still a rotation and not of the given form. What is wrong with the first representation?

The book means that every rotation can be written in that form in some basis. In the chosen form, the axis of rotation is chosen to be in the third coordinate.

So to change bases to get the desired form, all you have to do is

$$\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}\begin{bmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&-1 \end{bmatrix}\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}$$

What's meant most likely is that every element of $\mathsf{SO}(3,\mathbb R)$ is congugate to a matrix of that form.

Now \begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&-1 \end{pmatrix} is conjugate to \begin{pmatrix} -1&0&0 \\ 0&-1&0 \\ 0&0&1 \end{pmatrix} via a permutation of the basis, which is of the stated form with $\theta=\pi$.