Intuition for bases having the same orientation Let $V$ be a finite dimensional real vector space.  I'm trying to develop an intuitive understanding of what it means for two ordered bases $\alpha = (u_1,\ldots,u_n)$ and $\beta = (v_1,\ldots,v_n)$ of $V$ to have the same orientation.
A popular definition is that $\alpha$ and $\beta$ have the same orientation if and only if the change of basis matrix from $\alpha$ to $\beta$ has a positive determinant.  (For example, on p. 238 [section 21.1] of An Introduction to Manifolds by Tu, there is a statement, "We say that two ordered bases are equivalent if the change-of-basis matrix $A$ has positive determinant.")
However, this definition does not seem very intuitively clear to me. I can't see directly what this definition means in an intuitive way.
Here is an attempt at a more intuitive definition.  Ordered bases $\alpha$ and $\beta$ have the same orientation if and only if there exist continuous functions $w_i:[0,1] \to V$ (for $i = 1,\ldots, n$) such that:


*

*The set $\{w_1(t),\ldots,w_n(t)\}$ is linearly independent for all $t \in [0,1]$.

*$w_i(0) = u_i$ for $i = 1,\ldots, n$.

*$w_i(1) = v_i$ for $i = 1,\ldots, n$.


Intuitively, this definition means that $\alpha$ can be "continuously deformed" into $\beta$ without passing through a degenerate configuration.  I like this because it matches my intuition from $\mathbb R^3$: when I try to imagine deforming a right-handed basis into a left-handed basis, I always have to pass through a degenerate configuration.  I can also visualize one and two dimensional examples.
I would like to show that this definition is equivalent to the more standard definition that I gave at the beginning.  Here is an attempted proof.  First of all, suppose that $\alpha$ and $\beta$ have the same orientation according to my proposed definition.  Let $R(t)$ be the change of basis matrix from $\alpha$ to the ordered basis $(w_1(t),\ldots, w_n(t))$.  Because $\det R(t)$ is a continuous function that is initially equal to $1$ and is never $0$, it must be true that $\det R(t) > 0$ for all $t \in [0,1]$.  In particular, $\det R(1) > 0$.  This shows that the change of basis matrix from $\alpha$ to $\beta$ has positive determinant.
Conversely, I need to show that if $\alpha$ and $\beta$ have the same orientation according to the standard definition (i.e., the change of basis matrix from $\alpha$ to $\beta$ has positive determinant), then $\alpha$ and $\beta$ have the same orientation according to my proposed definition.  
My question: How do I show this?
I've considered some ideas based on using the SVD of the change of basis matrix, or the QR factorization, but I'm not sure they will work.  (Also, I suspect a simpler approach should be available.)
Additional question: Please provide any comments that you think might be relevant, or tangentially relevant -- anything that you suspect might improve my understanding of this topic.  Maybe a completely different way of looking at orientation would be more intuitive or enlightening.
 A: For the opposite direction, I have a solution which is certainly overkilled but short. Consider $\alpha, \beta$ two basis with the same orientation. We can assume $\alpha = (e_1, \dots, e_n)$. This means that $\beta$ has positive determinant. Let $B$ be the matrix of $\beta$, there is a matrix $C$ with $\exp(C) = B$ since $B$ has positive determinant. Now you can take $B_t = \exp(tC)$. We have $B_0 = \alpha$ and $B_1 = \beta$.
A: Let $B$ be the matrix that represents the change of the bases. We can use the QR-decomposition of $B$. If $B$ has positive determinant, then there is an orthogonal $Q$ with $\det(Q)=1$ and an upper triangular $R$ with positive elements on the diagonal such that $QR=B$. Now we only need to find continuous and non-degenerate functions 
$$
q:[0,1]\rightarrow\mathbb{R}^{n\times n} \;\;\mbox{with}\;\; q(0)=I_n,\;q(1)=Q \\ r:[0,1]\rightarrow\mathbb{R}^{n\times n} \;\;\mbox{with}\;\; r(0)=I_n, \; r(1)=R
$$
and you can set $w(t)=q(t)r(t)$.
The construction of $r$ is trivial, we can use $$r(t) = (1-t)I_n + tR.$$ As the diagonal elements of $R$ are positive and the diagonal elements of $r(t)$ are convex combinations of the diagonal elements of $R$ and $I$, the diagonal elements of $r(t)$ will be positive, too.
For the construction of $q$, we recall that every orthogonal matrix $Q$ with $\det(Q)=1$ is a rotation, i.e. $Q=SD(\alpha_1,\ldots,\alpha_s)S^{-1}$ with
$$
D(\alpha_1,\ldots,\alpha_s) =
\begin{pmatrix}
\cos\alpha_1 & -\sin\alpha_1 & & & & & \\
\sin\alpha_1 & \cos\alpha_1 & & & & & \\
& & \cos\alpha_2 & -\sin\alpha_2 & & & \\
& & \sin\alpha_2 & \cos\alpha_2 & & & \\
& & & & \ddots & & \\
& & & & & \cos\alpha_s & -\sin\alpha_s \\
& & & & & \sin\alpha_s & \cos\alpha_s 
\end{pmatrix}
$$
(for even $n$) or
$$
D(\alpha_1,\ldots,\alpha_s) =
\begin{pmatrix}
\cos\alpha_1 & -\sin\alpha_1 & & & & & & \\
\sin\alpha_1 & \cos\alpha_1 & & & & & & \\
& & \cos\alpha_2 & -\sin\alpha_2 & & & & \\
& & \sin\alpha_2 & \cos\alpha_2 & & & & \\
& & & & \ddots & & &\\
& & & & & \cos\alpha_s & -\sin\alpha_s & \\
& & & & & \sin\alpha_s & \cos\alpha_s & \\ 
& & & & & & & 1
\end{pmatrix}
$$
(for odd $n$).
With these definitions, we can set $$q(t) = SD(t\alpha_1,\ldots,t\alpha_s)S^{-1}.$$
