Span as a Set of Linear Combinations

Span Collects All Linear Combinations

A list of vectors does not generate only the vectors written in the list. It also generates every vector that can be made by multiplying those vectors by scalars and adding the results. For example, starting with \((1,2)\), we can form \((3,6)\) by multiplying by \(3\), or \((-1,-2)\) by multiplying by \(-1\). The span records all such possibilities at once.

Throughout this tutorial the vectors lie in \(\mathbb{R}^n\), and the scalars are real numbers. A linear combination of a finite list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is an expression \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where each \(c_i\in\mathbb{R}\). The coefficients may be chosen independently. In particular, they are not required to be \(1\), and they may be zero or negative.

The braces in the span notation name the generators; the definition is about a list of generators and the combinations formed from it. The result is a set, not a list: if different coefficient choices produce the same vector, that vector appears only once in the set. To describe a span in practice, we let the coefficients vary over all real values and simplify the resulting vector expression.

Computing Spans with Parameters

The Empty List and Zero Vectors

The empty-list convention is important because it makes the definition precise even when there are no generators. There is exactly one sum with no terms, and by convention its value is the zero vector. Thus the span of the empty list is the set containing the zero vector—not the empty set.

More generally, the span of any list consisting only of zero vectors is \(\{\mathbf{0}\}\). Every linear combination of such a list is zero. If the list is nonempty, choosing all coefficients zero also gives zero; if it is empty, the empty-sum convention gives zero. In either case, zero is the only possible result.

Basic Properties from the Definition

Two useful facts follow directly from what span means: every listed vector belongs to the span, and combining vectors already in a span produces another vector in that span. The second fact says that once a set contains all the combinations of a list, further linear combinations of those results do not produce anything outside that set.

Proof. If \(p=0\), the definition gives \(\operatorname{span}\{\}=\{\mathbf{0}\}\), so the span contains \(\mathbf{0}\), and there are no listed vectors to check. Now suppose \(p\geq 1\). Choosing every coefficient equal to zero gives \(0\mathbf{v}_1+\cdots+0\mathbf{v}_p=\mathbf{0}\), so \(\mathbf{0}\) is in the span. For any index \(k\) with \(1\leq k\leq p\), choose coefficient \(1\) on \(\mathbf{v}_k\) and \(0\) on every other vector. The resulting linear combination is \(\mathbf{v}_k\), so \(\mathbf{v}_k\) belongs to the span. \(\square\)

Proof. If \(q=0\), the combination is \(\mathbf{0}\), which belongs to \(S\) by the preceding theorem. Suppose \(q\geq1\). Because each \(\mathbf{x}_j\) lies in \(S\), the definition of span gives real numbers \(c_{1j},\dots,c_{pj}\) such that \(\mathbf{x}_j=\sum_{i=1}^{p}c_{ij}\mathbf{v}_i\). Substitute these expressions into the combination and collect the coefficient of each \(\mathbf{v}_i\): $$ \sum_{j=1}^{q}d_j\mathbf{x}_j = \sum_{j=1}^{q}d_j\left(\sum_{i=1}^{p}c_{ij}\mathbf{v}_i\right) = \sum_{i=1}^{p}\left(\sum_{j=1}^{q}d_jc_{ij}\right)\mathbf{v}_i. $$ Each coefficient \(\sum_{j=1}^{q}d_jc_{ij}\) is real, so the final expression is a linear combination of the original list \(\mathbf{v}_1,\dots,\mathbf{v}_p\). It therefore belongs to \(S\). If \(p=0\), then \(S=\{\mathbf{0}\}\); every \(\mathbf{x}_j\) equals zero, and the combination is zero, so the conclusion still holds. \(\square\)

This closure property explains why it is enough to describe a span using arbitrary coefficients on the original generators. First making some combinations and then combining those results cannot enlarge the set: the final vector can always be written directly as a combination of the original list.

A Smallest-Set Characterization

There is another way to understand the word “span.” Consider a set \(U\subseteq\mathbb{R}^n\) that contains each generator and is closed under finite linear combinations, including the empty combination. Then \(U\) must contain every vector in the span, because every such vector is a combination of the generators.

Proof. Take any \(\mathbf{x}\in\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\). By the definition of span, \(\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\) for real coefficients \(c_1,\dots,c_p\). Each \(\mathbf{v}_i\) lies in \(U\) by assumption, so this finite linear combination of elements of \(U\) lies in \(U\). Therefore \(\mathbf{x}\in U\). If \(p=0\), the span contains only \(\mathbf{0}\); the empty-combination condition gives \(\mathbf{0}\in U\), so the same inclusion holds. \(\square\)

The theorem does not say that every set containing the generators must contain their span. It needs the additional condition that the set is closed under finite linear combinations. Under that condition, the span is contained in every such set: it includes exactly the vectors forced by allowing all combinations of the given generators.

Common Mistake: Listing Generators Instead of Combinations

A frequent error is to report the generators themselves as the span. For example, a wrong calculation is \(\operatorname{span}\{(1,0),(0,1)\}=\{(1,0),(0,1)\}\). That right-hand side contains only two vectors and leaves out combinations such as \(2(1,0)-3(0,1)=(2,-3)\). The correction is to allow arbitrary coefficients:

The mistake is to confuse the list of inputs with the set of all outputs permitted by the definition. A reliable check is to write a general coefficient for every listed vector before simplifying. For a list of \(p\) generators, use \(p\) real coefficients; for the empty list, use the empty-sum convention.

Span is useful because it gives a precise description of what a collection of vectors can produce through addition and scalar multiplication. In later work, questions about whether a vector can be assembled from given columns will be expressed using this definition. For now, the essential skill is to state the full set with freely varying coefficients and to check that the parameter description represents exactly the allowed combinations.