Linear Span

Given a handful of vectors, what can you build from them using only addition and scalar multiplication? The answer — the set of all possible outputs — is the span. Span is the central construction of linear algebra: every subspace, every basis, every row space and column space is ultimately described as the span of something.

Definition

Let $V$ be a vector space over a field $F$, and let $S \subseteq V$ be any subset. The span of $S$ is the set of all linear combinations of elements of $S$:

\text{span}(S) \coloneqq \bigl\{ c_1 v_1 + \cdots + c_k v_k : k \ge 0,\ v_1, \ldots, v_k \in S,\ c_1, \ldots, c_k \in F \bigr\}. \tag{1}

Two boundary cases:

• $\text{span}(\emptyset) = \{\mathbf{0}\}$ by convention: a sum of zero vectors is the empty sum, which equals $\mathbf{0}$.
• $\text{span}(\{v\}) = \{cv : c \in F\}$, the line through the origin in the direction of $v$.

Examples

Span in $\mathbb{R}^2$. Let $e_1 = (1, 0)$ and $e_2 = (0, 1)$.

• $\text{span}(\{e_1\}) = \{(c, 0) : c \in \mathbb{R}\}$ — the $x$-axis.
• $\text{span}(\{e_1, e_2\}) = \{(c_1, c_2) : c_1, c_2 \in \mathbb{R}\} = \mathbb{R}^2$ — the whole plane.
• $\text{span}(\{(1,0), (2,0)\}) = \{(c,0) : c \in \mathbb{R}\}$ — still just the $x$-axis, because $(2,0)$ is already a multiple of $(1,0)$ and contributes nothing new.

Span in $\mathbb{R}^3$. Let $u = (1, 0, 0)$ and $v = (0, 1, 0)$.

• $\text{span}(\{u, v\}) = \{(c_1, c_2, 0) : c_1, c_2 \in \mathbb{R}\}$ — the $xy$-plane.
• Adding $w = (0, 0, 1)$: $\text{span}(\{u, v, w\}) = \mathbb{R}^3$.
• Adding $(1, 1, 0)$ instead: $\text{span}(\{u, v, (1,1,0)\}) = \text{span}(\{u,v\})$ — still just the $xy$-plane, since $(1,1,0) = u + v$.

The pattern: a redundant vector (one that is already a linear combination of the others) does not enlarge the span.

The span is always a subspace

Theorem: For any $S \subseteq V$, the set $\text{span}(S)$ is a subspace of $V$.

Proof. $\text{span}(S)$ is non-empty ($\mathbf{0} \in \text{span}(S)$ via the empty combination). If $x = c_1 v_1 + \cdots + c_k v_k$ and $y = d_1 w_1 + \cdots + d_l w_l$ are two elements of $\text{span}(S)$, and $a, b \in F$, then

$ax + by = ac_1 v_1 + \cdots + ac_k v_k + bd_1 w_1 + \cdots + bd_l w_l$

is again a linear combination of elements of $S$, so $ax + by \in \text{span}(S)$. $\square$

Moreover, $\text{span}(S)$ is the smallest subspace of $V$ containing $S$: any subspace $W$ that contains $S$ must be closed under linear combinations, so it must contain every element of $\text{span}(S)$. In symbols:

\text{span}(S) = \bigcap \{W \subseteq V : W \text{ is a subspace and } S \subseteq W\}. \tag{2}

Spanning sets

A subset $S \subseteq V$ spans $V$ (or is a spanning set for $V$) if $\text{span}(S) = V$ — that is, every vector in $V$ can be expressed as a linear combination of vectors in $S$.

Example: $\{(1,0),(0,1)\}$ spans $\mathbb{R}^2$. So does $\{(1,0),(0,1),(1,1)\}$ — but the third vector is redundant. And $\{(1,0)\}$ alone does not span $\mathbb{R}^2$.

A spanning set can be reduced: if any vector in $S$ is a linear combination of the others, it can be removed without changing $\text{span}(S)$. Repeating this process yields a minimal spanning set — one in which no vector is redundant — which is exactly what Linear Subspace calls a basis.

Why span is the right notion

Span answers the reachability question: which vectors are in the “reach” of a given set? This question appears in every corner of linear algebra:

• Is a vector $b$ in the column space of $A$? Equivalently, is $b \in \text{span}(\text{columns of } A)$?
• Does a set of vectors cover all of $V$? Equivalently, does it span $V$?
• What is the smallest subspace containing a given set? Its span.

Summary

• $\text{span}(S)$ is the set of all linear combinations of elements of $S$; by convention, $\text{span}(\emptyset) = \{\mathbf{0}\}$.
• $\text{span}(S)$ is always a subspace of $V$, and it is the smallest subspace containing $S$.
• A redundant vector — one that is already a linear combination of the others — does not change the span.
• $S$ spans $V$ if $\text{span}(S) = V$. Removing all redundancies from a spanning set yields a basis.