This is from the book of Guillemin and Pollack, p. 96.
If $A: V\to W$ is an isomorphism of vector spaces, then whenever two ordered bases $\beta$ and $\beta'$ on $V$ belong to the same equivalence class, so do the two ordered bases $A\beta$ and $A\beta'$ on $W$. Thus if both $V$ and $W$ are oriented, meaning that an orientation is specified for both, the sign of $A\beta$ is always the same as the sign of $\beta$ or always opposite. That is, $A$ either preserves or reverses orientation. (Note that if $\beta=(v_1,\dots,v_n)$, then $A\beta$ means $(Av_1,\dots, Av_n)$.)
- What is meant by "if both $V$ and $W$ are oriented, meaning that an orientation is specified for both, the sign of $A\beta$ is always the same as the sign of $\beta$ or always opposite"? I mean that without any additional assumptions there is no a third option -- there are only two possibilities for the sign of $A\beta$. Why is "thus" present in this sentence, and how is it connected with the previous sentence?
- So, according to the above definition, an isomorphism $A:V\to W$ is orientation preserving if for any basis $\beta$ of $V$, the basis $A\beta$ of $W$ has the same sign as $\beta$. Is this the same as requiring that $\det A > 0$? If so, why?
- I can't seem to organize neatly the notation to establish the claim in the very first sentence. Dealing with $n^2$ indices is a pain. How do I prove this assertion easily?