Let $\alpha: I\rightarrow \mathbb R^3$ arc length parametric curve with positive curvature. Show $\exists \ \omega: I\rightarrow \mathbb R^3$ a curve such that $$T'=\omega\times T\quad N'=\omega\times N \quad B'=\omega\times T$$ with $\{ T,N,B\}$ Frenet Frame of $\alpha$.
I'm not sure how to approach this problem. Any hints are appreciated!
The Frenet–Serret formulae give you three equations that such a $\omega$ has to satisfy: $$ \kappa N = \omega \times T, \\ -\kappa T + \tau B = \omega \times N, \\ -\tau N = \omega \times B. $$ If we write $\omega = aT+bN+cB$, as we may since $\{T,N,B\}$ form a basis, these give three equations for $a,b,c$ using $N \times T = -B$, $B \times T = N$, $T \times N = B$, $B \times N = -T$, $ T \times B = -N $, and $N \times B = T$: $$ \kappa N = -bB + c N \\ -\kappa T+ \tau B = aB - cT \\ -\tau N = -aN + bT. $$ Equating coefficients then gives $c=\kappa$, $b=0$ and $a=\tau$. We verify $$ T' = (\tau T + \kappa B) \times T = \kappa (B \times T) = \kappa N $$ and so on, so $\omega = \tau T + \kappa B$ will work.
(The condition $\kappa \neq 0$ is required for the normal and binormal to be well-defined: otherwise you don't know which direction to take for $N$.)