Is the action of $SO(3, \mathbb R)$ on $S^2:=\{(x,y,z) \in \mathbb R^3 |x^2+y^2+z^2=1\}$ defined as $(A,x) \to Ax$ transitive?

Question

Consider the action of $SO(3, \mathbb R)$ on $S^2:=\{(x,y,z) \in \mathbb R^3 |x^2+y^2+z^2=1\}$ as $(A,x) \to Ax$ . Is this action transitive ? i.e. is it true that for every $x,y \in S^2$ , $\exists A \in SO(3,\mathbb R)$ such that $Ax=y$ ?

Answer 1

Here's an approach using linear algebra: given a point $q_1=(x,y,z)\in S^2$, expand $q_1$ to an orthonormal basis $\{q_1,q_2,q_3\}$ for $\mathbb{R}^3$, and let $Q$ be the $3\times 3$ matrix whose rows are $q_1,q_2,q_3$.

By interchanging $q_2$ and $q_3$ if necessary we may assume that $\det(Q)=1$, hence $Q\in\mathrm{SO}(3)$. And by construction $$ Q\cdot\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$$

Therefore every point in $S^2$ can be rotated to $(1,0,0)$, hence the action of $\mathrm{SO}(3)$ is transitive.