Geometric Spans Generated by Vectors in R2

From Combinations to Geometry in the Plane

The span of a list is the set of all vectors that can be formed by choosing coefficients and combining the listed vectors. In \(\mathbb{R}^2\), this set has a direct geometric description. Depending on the list, its span may consist only of the origin, may be a line through the origin, or may be the entire plane. The key question is whether the vectors offer one direction or two genuinely different directions.

All vectors below are columns, and all coefficients are real. A linear combination of \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is a vector \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where each \(c_i\in\mathbb{R}\).

Every span contains \(\mathbf{0}\), because choosing every coefficient to be zero produces \(\mathbf{0}\). A list consisting only of zero vectors cannot produce anything else. At the other extreme, two vectors pointing in different directions can produce any point in the plane by taking a suitable combination. Between these cases, if all the vectors point along one direction, their combinations stay on that line.

What Two Vectors Can Generate

For two vectors, “pointing in different directions” means that neither is a scalar multiple of the other. The next result gives a precise test and shows why that condition is exactly what allows the pair to generate all of \(\mathbb{R}^2\).

Proof. First suppose neither vector is a scalar multiple of the other. Set \(D=ad-bc\). We show that \(D\ne0\). If \(a\ne0\) and \(D=0\), then \(ad=bc\), so \(d=(c/a)b\), while \(c=(c/a)a\). Thus \(\mathbf{v}=(c/a)\mathbf{u}\), a contradiction. If \(a=0\), then \(\mathbf{u}\ne\mathbf{0}\) implies \(b\ne0\). The equation \(D=0\) would give \(-bc=0\), hence \(c=0\) and \(\mathbf{v}=(d/b)\mathbf{u}\), again a contradiction. Therefore \(D\ne0\).

Now take any target \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2\), and choose \(s=(dx-cy)/D\) and \(t=(ay-bx)/D\). Direct substitution gives

and

Consequently \(s\mathbf{u}+t\mathbf{v}=\mathbf{x}\). Since the target was arbitrary, the pair spans \(\mathbb{R}^2\).

For the reverse implication, suppose one vector is a scalar multiple of the other. If both are zero, their span is \(\{\mathbf{0}\}\). Otherwise, choose a nonzero vector among them, say \(\mathbf{u}\). Every combination of the pair is then a scalar multiple of \(\mathbf{u}\), so it lies on the line \(\{r\mathbf{u}:r\in\mathbb{R}\}\). This line is not all of \(\mathbb{R}^2\): if \(\mathbf{u}=\begin{bmatrix}a\\b\end{bmatrix}\), at least one of \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\) is not a multiple of \(\mathbf{u}\). Hence the pair cannot span the plane. This proves both directions. \(\square\)

The Three Geometric Possibilities

The pair test extends to any finite list. A nonzero vector in the list gives a candidate direction. If every listed vector is a multiple of it, all combinations remain on the same line. If some listed vector is not a multiple of it, those two vectors already span the plane, so the full list does as well.

Proof. Consider a finite list. If it is empty or every vector in it is zero, each linear combination is \(\mathbf{0}\), and the zero combination is available, so its span is exactly \(\{\mathbf{0}\}\). Otherwise, choose a nonzero listed vector \(\mathbf{u}\). If every listed vector is a multiple of \(\mathbf{u}\), every combination is a multiple of \(\mathbf{u}\). Conversely, every \(t\mathbf{u}\) is a combination of the list: give \(\mathbf{u}\) coefficient \(t\) and every other vector coefficient zero. Thus the span is exactly the line through \(\mathbf{u}\).

If not every listed vector is a multiple of \(\mathbf{u}\), there is a listed vector \(\mathbf{v}\) that is not a multiple of \(\mathbf{u}\). The pair theorem shows that \(\mathbf{u}\) and \(\mathbf{v}\) span \(\mathbb{R}^2\). Every combination of that pair is also a combination of the full list, using coefficient zero on all other vectors. Thus the full list spans \(\mathbb{R}^2\). These cases are mutually exclusive: the zero set has no nonzero vectors, a line contains nonzero vectors but is not the plane, and the plane is not a line. \(\square\)

Worked Examples in Each Case

Verifying a Claimed Span

To show that a proposed set is the span, verify two inclusions. First show that every combination of the generators belongs to the proposed set. Then show that every vector in the proposed set can be obtained by choosing coefficients. The first direction rules out extra vectors; the second rules out missing ones.

This method is especially useful when checking whether a line description is exact. For instance, merely noticing that each generator is on a line proves that all combinations remain on it, but it does not by itself prove that every point on the line can be reached. Using one nonzero generator with an arbitrary coefficient supplies that missing direction.

Common Pitfall: Two Vectors Do Not Always Span the Plane

A common error is to reason that any two vectors in \(\mathbb{R}^2\) must span the plane simply because there are two of them. For example, \(\begin{bmatrix}1\\2\end{bmatrix}\) and \(\begin{bmatrix}2\\4\end{bmatrix}\) are two listed vectors, but the second is twice the first. A mistaken claim that their span is \(\mathbb{R}^2\) would say, in particular, that \(\begin{bmatrix}1\\0\end{bmatrix}\) is a combination of them.

That would require coefficients satisfying \(c_1+2c_2=1\) and \(2c_1+4c_2=0\). The left side of the second equation is twice the left side of the first, so these equations would force \(0=2\), which is impossible. In fact every combination has the form

The span is the line through \(\begin{bmatrix}1\\2\end{bmatrix}\), not all of \(\mathbb{R}^2\). Having two vectors is not enough; they must provide two different directions.