Orientations of $n$-dimensional vector spaces Let $V$ be a $n$-dimensional vector space. An $n$-tuple $(\mathbf{a_1},...,\mathbf{a_n})$ of independent vectors in $V$ is called an $n$-frame in $V$. In $\mathbb{R}^n$ we call such a frame right-handed if:$$\det[\mathbf{a_1},...,\mathbf{a_n}]>0;$$
we call left-handed otherwise. Now choose a liner isomorphism $T:\mathbb{R}^n\to V$, and define one orientation of $V$ to consist of all frames of the form $(T(\mathbf{a_1}),...,T(\mathbf{a_n}))$ for which $(\mathbf{a_1},...,\mathbf{a_n})$ is a right-handed frame in $\mathbb{R}^n$, and the other orientation of $V$ to consist of all such frames for which $(\mathbf{a_1},...,\mathbf{a_n})$ is left-handed.
How can I show that this notion is well defined, that is independent of the choice of $T$? Any hint?
 A: My comment was a good hint...for a slightly different question. 
First, on $\mathbb R^n$, we have two classes $R$ and $L$, of orientations, right? You understand that any two frames in the same class differ by a transformation of positive determinant, and because determinants multiply nicely, that makes the classes well-defined. 
Now you want to define orientations on $V$. You pick one linear isomorphism of $\mathbb R^n$ with $V$, say $T$, and create a class $R'$ of frames in $V$ consisting of $\{ T(F) \mid F \in R \}$, where by $T(F)$, I mean "T applied to each element of the frame F", OK? 
You define $L' = \{ T(F) \mid F \in L \}$. 
And presumably you can see that $L'$ and $R'$ are disjoint classes of frames on $V$, and that every frame lies in one of the two. 
Now you're perhaps asking "But what if I picked a different isomorphism $S$ instead of $T$, and used $S$ to define the two classes? How would I know I got the same thing?"
Answer: the two classes would be the same, but their names might be opposite: the class $R'$, as defined by $S$, might be the class $L'$ as defined by $T$. In fact, if we let 
$$
H: R^n \to R^n : (x_1, x_2, x_3 \ldots, x_n) \mapsto (x_2, x_1, x_3, \ldots, x_n)
$$
and define $S = T \circ H$, then they certainly will swap names. 
Note that the person who wrote the stuff you quoted was quite careful not to say "then we define the right-handed coordinates in $V$ to be $T$ applied to right-handed frames in $\mathbb R^n$; instead, the author said "we can defined two classes of orientations." 
If I had been the author, I might have taken the opportunity to emphasize that point to avoid confusion in the reader's mind. 
Clarification/Expansion
Suppose that $T$ and $S$ have the property that $K = S^{-1} \circ T$ has positive determinant. I'll then show that $R'_S = R'_T$. (In the case where the map has negative determinant, it'll turn out that $R'_S = L'_T$, and I'll bet you can show that once you've seen this proof. 
Let's suppose that $T(A)$ is an element of $R'_T$. Then I want to show that it's also in $R'_S$, that is, that there's a frame $B \in R$ with $S(B) = T(A)$. 
Brief digression: Now suppose for a moment that we had found such a frame $B$. Then by inverting, we'd have that 
$$
B = S^{-1}(T(A)),
$$
right? Well, that suggests a pretty good way to proceed. End of digression.
Consider the frame $B = K(A)$. Since $K$ is positive determinant, we know that $B$, too, is in $R$. But what's $S(B)$? It's
\begin{align}
S(B) &= S(K(A)) \\
&= S(S^{-1} \circ T (A))\\
&= (S \circ S^{-1})( T (A))\\
&= T (A))
\end{align}
which shows that any element of $R'_T$ (such as $T(A)$) is also an element of $R'_S$. Hence $R'_T \subset R'_S$; but the same argument in reverse shows that $R'_S \subset R'_T$, so they're equal as sets. 
A: Suppose:
S, T are linear isomorphisms from R^n to V.
Define:
T(X) = (T(x1),...,T(xn)), S(X) = (S(x1),...,S(xn)), where X = (x1, ..., xn) is any n-frame in R^n;
W = { T(X) | X is a right-handed n-frame in R^n }, U = { S(X) | X is a right-handed n-frame in R^n }.
We want to show that W = U (i.e., that the orientation of V is well-defined, independent of the choice of linear isomorphism from R^n to V).
Suppose A = (a1,...,an) is a right-handed n-frame in R^n. I need to show that there is a right-handed n-frame B in R^n such that
(1) T(A) = S(B).
Because S is onto, there is an n-frame C such that
(2) S(C) = T(A).
From (2) it follows that
(3) C = S^−1 ∘ T(A).
Note that H = S^−1 ∘ T is a linear isomorphism (from R^n x ... x R^n to R^n x ... x R^n) which can be represented by a non-singular n x n matrix, call it H'. If det H' > 0, then C and A belong to the same orientation and we're done.
