Span

If we take a set of vectors and work out all of their linear combinations, we get the span of that set of vectors.

Definition of the span

Let x₁, …, x_n be vectors. We already know that we can build a new vector from them using a linear combination. We choose real numbers a₁, …, a_n and get the new vector y by multiplying and adding:

$$\mathbf{y}=a_1\cdot {\mathbf{x}}_1+\dots +a_n\cdot {\mathbf{x}}_n$$

In this way, as long as at least one of the vectors x_i is not the zero vector, we can make infinitely many “new” vectors from the set x₁, …, x_n – we just have to choose different coefficients a₁, …, a_n.

If we keep generating vectors, we end up with all the vectors that can be obtained as a linear combination of x₁, …, x_n. We call this set the span of the vectors x₁, …, x_n. The span of a set of vectors X is written span(X). For a finite number of vectors we can write the span formally as

$$\mathrm{span}({\mathbf{x}}_1,\dots ,{\mathbf{x}}_n)=\{a_1\cdot {\mathbf{x}}_1+\dots +a_n\cdot {\mathbf{x}}_n\mid a_1,\dots ,a_n\in \mathbb{R}\}.$$

If we have an infinite set of vectors X, we can take all the finite subsets Y_i ⊆ X, that is |Y_i| = r for some r ∈ ℤ⁺, and the span of X is then the union of all the sets span(Y_i).

Example

Let’s stay in the familiar vector space ℝ³. Take the one-element set X₁ = {(1, 2, 1)}. What is the span of this set? It is all the linear combinations of the vector (1, 2, 1). The span has the form

$$\mathrm{span}(X_1)=\{(a,2a,a)\mid a\in \mathbb{R}\}$$

So it consists of vectors such as (3, 6, 3),(8, 16, 8),(−1, −2, −1) and so on. We can check that the sum of any two of these vectors gives a new vector of the same form:

$$\begin{array}{rcl}(3,6,3)+(3,6,3)& =& (6,12,6)\\ (3,6,3)+(-1,-2,-1)& =& (2,4,2)\\ (8,16,8)+(2,4,2)& =& (10,20,10)\end{array}$$

For a second example, take the standard vectors X₂ = {(1, 0, 0), (0, 1, 0)}. When we work out all the linear combinations, we find that they have the form (a, b, 0), where a, b ∈ ℝ. This is easy to show, because

$$(a,b,0)=a\cdot (1,0,0)+b\cdot (0,1,0).$$

Vectors in this span include (0, 8, 0) and (14, 15, 0). Formally we would write

$$\mathrm{span}(X_2)=\{(a,b,0)\mid a,b\in \mathbb{R}\}$$

The span as the smallest subspace

Let V be a vector space and X ⊆ V a set of vectors. The span span(X) is then a subspace of V, that is, the span span(X) is itself a vector space.

We only need to show that span(X) is closed under addition and scalar multiplication. It certainly is, because any vector x ∈ span(X) arose as a linear combination of vectors from X.

The span of X is also the smallest subspace that contains all the vectors of X. Why? The smallest subspace containing the vectors of X must also contain all linear combinations of vectors from X – otherwise it would not be a vector space. We can assume for contradiction that there is a smaller subspace, call it Y, that contains all the vectors of X, that is X ⊆ Y. Since Y is, by assumption, smaller than the span span(X), there must be a vector x that is in span(X) but not in Y, that is x ∈ span(X) ∧ x ∉ Y.

But the vector x must be buildable as a linear combination of vectors from X, so there must be vectors x₁, …, x_n ∈ X and coefficients a₁, …, a_n such that

$$\mathbf{x}=a_1\cdot {\mathbf{x}}_1+\dots +a_n\cdot {\mathbf{x}}_n$$

But if Y does not contain this vector x while containing the vectors x₁, …, x_n, then Y cannot be a vector space, and so it cannot be a subspace either. The span span(X) is therefore the smallest subspace that contains all the vectors of X.

Basic properties of the span

• It is always true that W ⊆ span(W). This should be clear. The span of the set W is always the same as, or larger than, the set W. The span span(W) contains all linear combinations of vectors from W, so span(W) must also contain all the vectors of W: if we take a vector x ∈ W, then for a = 1 we get the combination a · x = x.

  Example: the span of the set W = {(1, 0, 0)} is the set

  $$\begin{array}{rcl}\mathrm{span}(W)& =& \{(a,0,0)\mid a\in \mathbb{R}\}.\end{array}$$

  Clearly {(1, 0, 0)} ⊆ {(a, 0, 0)  ∣  a ∈ ℝ}.

• It is always true that span(W) = span(span(W)). Once we have worked out a span, the span of that span is the same set. span(W) contains all the linear combinations from W. If we worked out all the linear combinations from span(W) again, we would not get any new ones.

• Let V be a vector space and let W₁ and W₂ be two subsets of it (not necessarily subspaces), that is W₁, W₂ ⊆ V. If W₁ ⊆ W₂, then span(W₁) ⊆ span(W₂) too.

  In other words: if we have two sets of vectors and one of them is smaller, then the smaller set of vectors has a span that is smaller than or equal to the span of the larger set.

  Example: let W₁ = {(1, 0, 0)} and W₂ = {(1, 0, 0), (0, 1, 0)}. We can see that W₁ ⊆ W₂. What would the spans be?

  $$\begin{array}{rcl}\mathrm{span}(W_1)& =& \{(a,0,0)\mid a\in \mathbb{R}\}\\ \mathrm{span}(W_2)& =& \{(a,b,0)\mid a,b\in \mathbb{R}\}\end{array}$$

  The set W₂ “generated a larger span” than W₁, as we expected. The set W₂ contains all the vectors of W₁, so all the linear combinations of vectors from W₁ can also be generated from the vectors in W₂.

  Another example: W₃ = {(1, 2, 0)} and W₄ = {(1, 2, 0), (5, 10, 0)}. At first glance you can see that the vectors in W₄ are dependent. So both sets generate the same span. That is, although W₃ ⊊ W₄, we have span(W₃) = span(W₄).