Linear Algebra: orientations of vector spaces (problem) This is an exercise from J.Munkres's Analysis on Manifolds:
Consider the vectors $\mathbf{a_i}$ in $\mathbb{R}^3$ such that:$$[\mathbf{a_1},\mathbf{a_2},\mathbf{a_3},\mathbf{a_4}]=\begin{bmatrix} 1 &0&1&1 \\ 1&0&1&1\\1&1&2&0\end{bmatrix}$$
Let $V$ the subspace of $\mathbb{R}^3$ spanned by $\mathbf{a_1}$ and $\mathbf{a_2}$. Show that 1) $\mathbf{a_3}$ and $\mathbf{a_4}$ also span $V$, and that 2) the frames $(\mathbf{a_1},\mathbf{a_2})$ and $(\mathbf{a_3},\mathbf{a_4})$ belong to opposite orientation of $V$.
I'm having troubles showing part 2): J.Munkres gives the following definitions concerning frames and orientations of vector spaces: Orientations of $n$-dimensional vector spaces. Do those definitions apply in this problem?

I'll only explain part 2. 
$\{a_1,a_2\}$ forms a basis for $V$. I want to express $a_3, a_4$ as linear combination of the basis. $a_3=a_1+a_2$, $a_4=a_1-a_2$. Therefore, in the basis $\{a_1,a_2\}$, $a_3, a_4$ can be expressed in a component form: $(1,1), (1,-1)$. 
Observe that $\det \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}=-2$ is negative. By the definition of orientation, the frames $\{a_1, a_2\}$, $\{a_3, a_4\}$ have different orientation. 

Well, if you like, you can do it in another way, without choosing basis. That is, think of the cross product. If $a_1 \times a_2$ has the same direction as $a_3\times a_4$, then they have the same orientation. If the cross products are in the same direction, then they are in the opposite direction. 
