I'm trying to prove the problem below:
Let $\Omega = (\omega_{ij})$ be a skew-symmetric $3×3$ matrix, i.e., $\Omega^T=-\Omega$.
Let $e_1(s)$, $e_2(s)$ and $e_3(s)$ be smooth vector-valued functions of a parametre $s$ satisfying the differential equations $\frac{d}{ds}e_i=\sum^3_{k=1} \omega_{ik} e_k$, where $i=1,2,3$.
Suppose that for some parametre value $s_0$, the vectors $e_1(s_0)$, $e_2(s_0)$ and $e_3(s_0)$ are orthonormal.
Show that the vectors $e_1(s)$, $e_2(s)$ and $e_3(s)$ are orthonormal for all $s$.
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But what I have been doing is just expanding out the differential equations $\frac{d}{ds}e_i$ and translating the condition of "orthonormal" for $e_1(s_0)$, $e_2(s_0)$ and $e_3(s_0)$.
I have also defined a function $f_{ik}(s)=e_i(s) \cdot e_k(s)$, where $i,k=1,2,3$, and found its derivative by the product rule: $\frac{d}{ds}f_{ik}(s)=\frac{d}{ds}e_i(s) \cdot e_k(s) + e_i(s) \cdot \frac{d}{ds}e_k(s)$.
Any ideas on how to carrying on this proof? Thanks!
You can write the set of differential equations as \begin{equation} \frac{\partial}{\partial s}\begin{pmatrix} e_1(s) & e_2(s) & e_3(s) \end{pmatrix} = \Omega \begin{pmatrix} e_1(s) & e_2(s) & e_3(s) \end{pmatrix} \end{equation} with the solution \begin{align} \begin{pmatrix} e_1(s) & e_2(s) & e_3(s) \end{pmatrix} &= e^{\Omega s}\begin{pmatrix} e_1(s_0) & e_2(s_0) & e_3(s_0) \end{pmatrix}\\ &= e^{-\Omega^{\top} s}\begin{pmatrix} e_1(s_0) & e_2(s_0) & e_3(s_0) \end{pmatrix}. \end{align} It turns out that $e^{\Omega s}$ is a rotation matrix (formally you can look into the special orthogonal group), which means that for any $s$ the frame $\begin{pmatrix} e_1(s_0) & e_2(s_0) & e_3(s_0) \end{pmatrix}$ is rotated, which preserves relative orientation, e.g. pick up a box, you can rotate it all day long, the sides remain fixed with respect to each other.