# Rotation matrices and $SO(3)$

Let $R$ denote the set of matrices that rotate $\mathbb R^3$ around an axis. For the $x,y,z$ axes the matrices are given here.

Let $SO(3)$ denote the set of orthogonal $3\times 3$ matrices with determinant $1$.

It is clear to me that $R \subseteq SO(3)$.

How can I show that $SO(3) \subseteq R$?

• I found a complete proof here. Nov 13 '14 at 1:51

Let $A\in SO(3)$, and we show that $1$ is an eigenvalue of $A$. The characteristic polynomial is of degree $=3$, and so there are two possible cases:

a) All eigenvalues are real. Since $A$ preserves lengths, all eigenvalues are $\pm1$, and since $\det A=1$, the product of all three eigenvalues is $1$, hence $1$ has to be an eigenvalue.

b) $A$ has one real eigenvalue and two non-real eigenvalues, which are conjugate to one another, and so their product is positive. Hence the real eigenvalue has to be positive as well, and (again) as $A$ preserves lengths, the real eigenvalue must be $1$.

Having verified that $1$ is an eigenvalue, let $v\in\mathbb{R}^3$ such that $Av=v$, and let $U$ denote the orthogonal complement of $v$. Then $U$ is of dimension $2$, it is clearly invariant under $A$, and the restriction $A|_U$ is orthogonal with positive determinant.

If you believe that any matrix in $SO(2)$ represents a rotation (and this is not hard to see), you're done.

• How do you argue that $A_U$ is orthogonal? Nov 13 '14 at 3:07
• An operator is orthogonal if it preserves the inner product (by definition), hence it is obvious that the restriction of an orthogonal operator to an invariant subspace is again orthogonal. Nov 13 '14 at 10:54

Essentially, you need to prove that every orientation-preserving orthonormal matrix is a rotation. First of all, it's obvious that the matrices preserve lengths and angles. You can also show there is always an eigenvalue equal to $1$ (obvious in 3D, because eigenvalues of orthonormal matrices are in complex conjugate pairs). The associated eigenvector is the axis. You can then show that when you go into a coordinate frame with the axis aligned along $z$, there is only one remaining degree of freedom (the matrix in this frame looks like a rotation matrix around $z$).

Or you could use (derive) the Rodriguez formula which is a direct bijection between rotations and axis-angle representation. Essentially, an orthogonal matrix $R$ is a rotation by an angle $${\rm Tr\,}(R)=1+2\cos\theta$$ and $$R-R^T=2\sin\theta\begin{bmatrix}0&a_z & -a_y \\ -a_z & 0 & a_x \\ a_y & -a_x & 0 \end{bmatrix}$$

You can do this for any matrix... but you can show, that for orthogonal matrices, if you do this transformation and then go back with Rodriguez formula, you get the original matrix (therefore the mapping is bijective).