Given $u,v\in \mathbb{S}^n$, then there exists a orthogonal matrix s.t. $u=O v$.

Question

Is the following intuitive statement true:

For given $u,v \in \mathbb{S}^n$, there exists a matrix $R\in O(n)$ s.t. $u=Rv$.

Answer 1

There exist orthonormal basis $(e_1=u,e_2,...,e_n)$ and $(f_1=v,...,f_n)$ define $A$ by $A(e_i)=f_i$.

Answer 2

Yes it's true. We can take $S^2$ as model. Without loosing generality we can change the basis and get v e u as: \begin{equation} u=\left(\begin{array}{c} \cos\varphi_{1}\cos\theta_{1}\\ \cos\varphi_{1}\sin\theta_{1}\\ \sin\varphi_{1} \end{array}\right),\quad v=\left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right) \end{equation} Then for example we can consider the following rotation R which has determinant $1$ because is triangular and transform $v$ in $u$: \begin{equation} R=\left(\begin{array}{c} \begin{array}{ccc} \cos\varphi_{1}\cos\theta_{1} & 0 & 0\\ \cos\varphi_{1}\sin\theta_{1} & \left(\cos\varphi_{1}\cos\theta_{1}\right)^{-1} & 0\\ \sin\varphi_{1} & 0 & 1 \end{array}\end{array}\right) \end{equation} The $S^n$ is just the same with $n$ coordinates instead of $3$.