Showing that 2 pairs of vectors span the same subspace and that their frames belong to opposite orientations of that subspace

Question

The following question is an exercise from Munkres' Analyis on Manifolds (Chapter 4 - Section 20):

Consider the vectors $a_i$ in $R^3$ such that:

$[a_1\ a_2\ a_3\ a_4] = \begin{bmatrix} 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 2 & 0 \end{bmatrix}$

Let $V$ be the subspace of $R^3$ spanned by $a_1$ and $a_2$. Show that $a_3$ and $a_4$ also span $V$, and that the frames $(a_1,a_2)$ and $(a_3,a_4)$ belong to opposite orientations of $V$.

My initial approach was to span $a_1$ and $a_2$ to determine V. Since two vectors can span at most $R^2$, the third term of any general vector within span must be 0. So, $(x,x,x+y)=0\Longrightarrow x=0,y=0$ And so, V is simply the origin $(0,0)$. Spanning $a_3$ and $a_4$ also yields this, showing that they both span V. This seems like a very odd answer for the first part in my opinion.

However, in trying to answer the second part I am completely lost. In Munkres, we see that the orientation of some $n$-tuple $(x_1,\ldots,x_n)$ is determined by the sign of the determinant of the matrix they form. But surely the determinant of a $3\times2$ matrix is not defined? One thought I had was to simply take the determinant of the upper $\frac{2}{3}$ of the matrix (so determinant is defined), but the determinant of this is 0, which isn't covered by Munkres' definition, so the frame has no orientation?

Any help / clarification / solutions would be massively appreciated.

Answer 1

Recall that two frames $(a_1,\dots,a_n)$ and $(v_1,\dots,v_n)$ have the same orientation if the linear isomorphism which sends $a_i\mapsto v_i$ (which exists uniquely) has positive determinant.

The subspace $V$ has two bases $a=(1,1,1),b=(0,0,1)$ and $v=(1,1,2),w=(1,1,0)$. If you change from one basis to the other, you use the following formulas: $v=a+b$ and $w=a-b$. This means that the linear isomorphism $a\mapsto v$ and $b\mapsto w$ is described (in the basis $\{a,b\}$) by the matrix $$\begin{bmatrix}1&1\\1&-1\end{bmatrix}$$ which has determinant $-2$ which is negative. This means that the two bases (frames) have opposite orientations.

Answer 2

$\mathbb R^2$ is not a subspace of $\mathbb R^3$, so there’s no way for any of these vectors to combine to span the former. On the other hand, two of them could span a two-dimensional subspace of $\mathbb R^3$. There are many such subspaces besides the one that consists of vectors with last coordinate equal to zero (which is certainly isomorphic to, but not equal to, $\mathbb R^2$). These misconceptions appear to have led you to the absurd conclusion that neither $a_1$ nor $a_2$ belong to their own span—after all, neither of their third coordinates vanish.

The span of $a_1$ and $a_2$ consists of all linear combinations $\lambda a_1+\mu a_2$. They are obviously linearly independent, so this is indeed a two-dimensional subspace of $\mathbb R^3$. Similarly, $a_3$ and $a_4$ are obviously linearly-independent, too. We can find by inspection that $a_3 = a_1+a_2$ and $a_4=a_1-a_2$, so the span of the latter pair is contained in the span of the former. You can solve these equations for $a_1$ and $a_2$ to show that the inclusion goes in the other direction as well, and hence the two subspaces are identical, but it’s enough to note that they, too, are linearly independent without explicitly inverting the equations.

If you didn’t happen to spot these relationships among the vectors, you could instead proceed systematically by computing the row-reduced echelon form of the matrix in your question, which is $$\begin{bmatrix}1&0&1&1 \\ 0&1&1&-1 \\ 0&0&0&0 \end{bmatrix}.$$ The first two columns verify that $a_1$ and $a_2$ are linearly independent, while the second two columns show that the other two vectors are both elements of their span and in fact give the coordinates of $a_3$ and $a_4$ relative to the $(a_1,a_2)$ basis of $V$. This last fact will come in handy for the second part.

These two bases of $V$ have opposite orientations iff the matrix that maps between them has a negative determinant. This will be a $2\times2$ matrix since you’re mapping from a two-dimensional space to another two-dimensional space. Recalling that the columns of a transformation matrix are the images of the basis vectors, this means that the matrix that maps coordinates of elements of $V$ from the $(a_3,a_4)$ basis to the $(a_1,a_2)$ basis is just the upper-right $2\times2$ submatrix of the above rref matrix. Its determinant is $-2$, therefore this change of basis is orientation-reversing.