# Hint for proving that there are exactly two equivalence classes under the relation being consistently orientable

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?

• Take two ordered bases you know are in different equivalence classes, and show that every ordered basis is equivalent to one of these. For example, you could take one to be $(v_1 ,v_2,... ,v_n)$ and the second to be $(v_2 ,v_1, ..., v_n)$. – астон вілла олоф мэллбэрг Apr 27 '17 at 8:48
• @астонвіллаолофмэллбэрг That was exactly what I thought, but the case $n = 1$ is then not possible to prove. Oh, I could just use $$(v_1,v_2,\dots,v_n) \qquad \text{and} \qquad (-v_1,v_2,\dots,v_n)$$ – TheGeekGreek Apr 27 '17 at 8:53
• Yes, that is correct. You could try to do this. – астон вілла олоф мэллбэрг Apr 27 '17 at 8:55
• @астонвіллаолофмэллбэрг Thanks for your help. – TheGeekGreek Apr 27 '17 at 8:57
• You are welcome! But since the question isn't solved yet, you can come back to me if there are more difficulties. – астон вілла олоф мэллбэрг Apr 27 '17 at 8:58

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 $$(w_1,\dots,w_n) \sim (\tilde{v}_1,\dots,\tilde{v}_n)$$