I want to prove this: If $(a_{ij})$ is skew-symmetric, $\{v_i(s)\}_{i=1}^3$ are smooth functions which satisfy, $$v_i'(s)=\sum_{k=1}^3 a_{ik}v_k(s),$$ and that $v_i(s_0)$ are orthonormal for some $s_0$ then $v_i(s)$ are orthonormal for all $s$.
I started like this:
Let $\lambda_{ij}(s)=v_i(s)\cdot v_j(s)$, then we must show that $\lambda_{ij}=\delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta. So we calculate the derivative: $$ \lambda_{ij}'=v_i'\cdot v_j+v_i\cdot v_j'$$ $$ \lambda_{ij}'=\sum_{k=1}^3 a_{ik}v_k\cdot v_j+v_i\cdot \sum_{k=1}^3 a_{jk}v_k$$ $$ \lambda_{ij}'=\sum_{k=1}^3 (a_{ik}v_k\cdot v_j)+\sum_{k=1}^3 (a_{jk}v_i\cdot v_k)$$ From where $$ \lambda_{ij}'=\sum_{k=1}^3 (a_{ik}\lambda_{kj}+a_{jk}\lambda_{ik})$$
I haven't been able to go further. What can I do in this part?
Let $B$ be an $n\times n$ matrix such that the $i$-th row are $v_i$ (treated as row vectors). Then the differential equation is the same as $$B'(s) = A B(s).$$ To check that they are orthogonal, it suffices to check $B^T(s) B(s) = I$ for all $s$. Now we differentiate:
$$ \begin{split} (B^T B)' &= (B^T)' B + B^T B' \\ &= (AB)^T + B^T AB \\ &= B^T A^T B + B^T AB \\ &= 0 \end{split} $$
as $A$ is skew symmetric ($A^T + A = 0$). As $B^T(0) B(0) = I$, we have $B^T(s) B(s) = I$.