Linearly dependent vector fields that are not spanned by fewer (continuous) vector fields. Consider the following continuous vector fields in $\mathbb{C}^3$:
\begin{equation*}
w_1(x,y,z):=(x,y,0), \; \; w_2(x,y,z):=(-z,0,y), \;\; w_3(x,y,z):= (0,z,x).
\end{equation*}
At every point $(x,y,z) \in \mathbb{C}^3$, the vectors are linearly dependent, as the matrix
\begin{pmatrix}
x & y & 0 \\ -z & 0 & y \\ 0 & z & x
\end{pmatrix}
has determinant zero. In particular, this means that at each point there are two vectors $v_1, v_2$ that linearly generate $w_1,w_2,w_3$ at that point. Choosing such a pair of vectors at each point we end up with two vector fields $v_1(x,y,z),v_2(x,y,z)$ which span the other three as $w_i(x,y,z) = \alpha_i(x,y,z)v_1(x,y,z) + \beta_i(x,y,z) v_2(x,y,z)$ for some functions $\alpha_i, \beta_i : \mathbb{C}^3 \rightarrow\mathbb{C}$.
The question is, can there be two continuous vector fields $v_1, v_2$ with that property? (that they span $w_1,w_2,w_3$ at all points). I strongly suspect that this is not the case, but I am not sure how to approach this. Am I missing some obvious condition the vector fields should satisfy for this to be true? I am actually interested in a higher dimensional example, so I am looking for a general proof strategy or hints on how to attack this problem in general rather than a trick that works for this particular example.
 A: So the following strategy seems to work (in some situations) to prove that such vector fields do not exist. I'll explain it for the vectors that I gave as an example.
If they did, we can first restrict to the space $\mathbb{R}^3$. Then, note that at every point in $\mathbb{R}^3 \setminus\{(0,0,0)\}$ the fields $w_1, w_2, w_3$ span a two dimensional subspace. Therefore, this gives a rank $2$ vector bundle over the sphere $S^2$. In the example that I give, it turns out to be isomorphic to the tangent bundle, and so it doesn't have two linearly independent sections $v_1, v_2$.
Unfortunately, this does not seem to work the other way around: after removing the "singularities" (the points at which the vector fields do not span a vector space of generic dimension) you could in principle be left with a trivial bundle in the remaining space but still no vector fields in the whole space that span the original ones.
A: One way is to find an easy map directly and check. As you write, the matrix is linearly dependent, so we can take first and third vectors and generate the second one:
$$
(-z,0,y) = \frac{y}{x} (0,z,x) - \frac{z}{x} (x,y,0)
$$
which seems to be continuous away from $(0,0,0)$
