Let $V$ be a real vector space of dimension $n \geq 1$. We say that two ordered bases $(v_1,\dots,v_n)$ and $(\widehat{v}_1,\dots,\widehat{v}_n)$ are consistently oriented if $\det B > 0$ where $B$ is the transition matrix defined by $$v_j = B^i_j \widehat{v}_i$$ Show that being consistently oriented constitutes an equivalence relation on the set of all ordered bases for $V$ and that there are exactly two equivalence classes.
I have shown that above relation is an equivalence relation. But I am not sure, how to prove that there are exactly two equivalence classes. I mean, somehow it is clear, that in the second equivalence class we have $\det B < 0$ ($\det B = 0$ is impossible since $B$ is an isomorphism). Has someone a hint for the structure of the proof? Proof by contradiction?
Let $V$ be a real vector space with dimension $n \geq 1$. Since every vector space has a basis, say $$(v_1,\dots,v_n)$$ and clearly $$(\tilde{v}_1,\dots,\tilde{v}_n):=(-v_1,\dots,v_n)$$ is also one by considering the transition matrix $$B = \begin{pmatrix} -1\\ & 1 \\ && \ddots\\ &&&1\end{pmatrix}$$ We show that any other oriented basis $(w_1,\dots,w_n)$ is equivalent to one of those. Let us consider the transition matrix $\widehat{B}$ defined by $$w_j = \widehat{B}^i_j v_i$$ If $\det \widehat{B} > 0$, we have that $(w_1,\dots,w_n) \sim (v_1,\dots,v_n)$ and if $\det \widehat{B} < 0$, we have that $$w_j = \widehat{B}^i_jv_i = \widehat{B}^i_j B_i^k \tilde{v}_k$$ and by $$\det(\widehat{B}B) = \det(\widehat{B})\det(B) > 0$$ we get that $$(w_1,\dots,w_n) \sim (\tilde{v}_1,\dots,\tilde{v}_n)$$