From Zorich II, pag. 177, exercise 4a:
A subspace $\mathbb{R}^{n-1}$ has been fixed, a vector $\mathbf{v}\in\mathbb{R}^n\setminus\mathbb{R}^{n-1}$ has been chosen, along with two frames $\{\mathbf{\xi}_1,...,\mathbf{\xi}_{n-1}\}$ and $\{\eta{\xi}_1,...,\eta{\xi}_{n-1}\}$ of the subspace $\mathbb{R}^{n-1}$. Verify that these frames belong to the same orientation class of frames of $\mathbb{R}^{n-1}$ if and only if the frames $\{\mathbf{v},\mathbf{\xi}_1,...,\mathbf{\xi}_{n-1}\}$ and $\{\mathbf{v},\eta{\xi}_1,...,\eta{\xi}_{n-1}\}$ define the same orientation on $\mathbb{R}^n$.
I can't understand how the "if and only if" could apply. In fact, I think that the proof is simply in this logic chain:
$$\left\{\begin{matrix} \left | M_{\{\mathbf{e}_1,...,\mathbf{e}_{n}\} \leftarrow\{\mathbf{v},\mathbf{\xi}_1,...,\mathbf{\xi}_{n-1}\}} \right |>0\\ \left | M_{ \{\mathbf{v},\mathbf{\eta}_1,...,\mathbf{\eta}_{n-1}\}\leftarrow \{\mathbf{e}_1,...,\mathbf{e}_{n}\}} \right |>0 \end{matrix}\right.\Rightarrow $$
$$\Rightarrow \left | M_{\{\mathbf{v},\mathbf{\eta}_1,...,\mathbf{\eta}_{n-1}\} \leftarrow\{\mathbf{v},\mathbf{\xi}_1,...,\mathbf{\xi}_{n-1}\}} \right |>0\iff \left | M_{\{\mathbf{\eta}_1,...,\mathbf{\eta}_{n-1}\} \leftarrow\{\mathbf{\xi}_1,...,\mathbf{\xi}_{n-1}\}} \right |>0$$
where $M_{F_1 \leftarrow F_2}$ is the transition matrix from the frame $F_2$ to $F_1$. How the arrow $\Rightarrow$ in my proof could become a $\iff$?
I will explain my answer here.
Two frames $\mathcal{F}_1, \mathcal{F}_2$ define the same orientation iff the Transition matrix of these two has a positive determinant.
Now suppose that $|M_{\{ \xi_1,...\xi_{n-1}\} \leftarrow \{ \eta_1,..., \eta_{n-1}\}}| > 0.$
Call this matrix $A$. then Transition matrix $M_{\{ \mathbf{v}, \xi_1,...x_{n-1}\} \leftarrow \{ \mathbf{v}, \eta_1,..., \eta_{n-1}\}} $ is of the form :
$$ \begin{bmatrix} 1 & 0 \\\ 0 & A \end{bmatrix} \hspace{1 cm} (i) $$ So the determinant of this function would be the same as $A$, and we are done. The reason is that $\mathbf{v}$ was chosen outside of $\Bbb{R}^{n-1}$.
For the converse, the transition matrix is of the form $(i)$. and because it has a positive determinant thus $A$ has positive determinant. But $A$ is the transition matrix of the frames $(\xi_1,...,\xi_{n-1})$ and $(\eta_1,...,\eta_{n-1})$.