The Span Operator — Linear Algebra

Span as a closure operator. A function maps each input to the output. Viewing it as an operator emphasizes how it transforms mathematical objects. When an operator maps a collection of objects to itself, we can compose it with itself and apply it repeatedly. This leads us to study how an object changes under one application, what happens under successive applications, and which objects remain unchanged. Taking the span of a set can be viewed as an operator, specifically as a closure operator on the power set of a vector space. We begin by studying a closure operator through its relation to inclusion and its behavior under iteration, and then use the general properties of this transformation to describe the span operator.

Definition 1.5.1. An operator $\operatorname{cl}$ on the collection of subsets of a set $$X$$ is called a closure operator on $$X$$ if it satisfies the following axioms: for all $$S,T\subseteq X$,$

$S\subseteq \operatorname{cl}(S)$ (extensive);
if $$S\subseteq T$,$ then $\operatorname{cl}(S)\subseteq\operatorname{cl}(T)$ (monotone);
$\operatorname{cl}(\operatorname{cl}(S))=\operatorname{cl}(S)$ (idempotent).

Definition 1.5.2. Let $\operatorname{cl}$ be a closure operator on a set $$X$,$ and let $$S\subseteq X$.$ The set $\operatorname{cl}(S)$ is called the closure of $$S$.$ If $$S=\operatorname{cl}(S)$,$ then $$S$$ is called a closed set.

Proposition 1.5.3. Let $\operatorname{cl}$ be a closure operator on a set $$X$,$ and let $$S,T\subseteq X$.$ Then,

$$\operatorname{cl}(S)\cup \operatorname{cl}(T) \subseteq \operatorname{cl}(S\cup T)$;$
$$\operatorname{cl}(S\cup T) = \operatorname{cl}(\operatorname{cl}(S)\cup \operatorname{cl}(T))$.$

Proof. (1) Since $$S,T\subseteq S\cup T$,$ monotonicity gives $\operatorname{cl}(S)\subseteq \operatorname{cl}(S\cup T)$ and $$\operatorname{cl}(T)\subseteq \operatorname{cl}(S\cup T)$.$ Hence $$\operatorname{cl}(S)\cup \operatorname{cl}(T) \subseteq \operatorname{cl}(S\cup T)$.$ (2) Since $S\subseteq \operatorname{cl}(S)$ and $$T\subseteq \operatorname{cl}(T)$,$ we have $$S\cup T\subseteq \operatorname{cl}(S)\cup \operatorname{cl}(T)$.$ The inclusion $\operatorname{cl}(S\cup T) \subseteq \operatorname{cl}(\operatorname{cl}(S)\cup \operatorname{cl}(T))$ follows from monotonicity. Since $\operatorname{cl}(S) \cup \operatorname{cl}(T) \subseteq \operatorname{cl}(S\cup T)$ by (1), monotonicity and idempotence give the reverse inclusion. $\square$

The closure axioms can be expressed as a single inclusion criterion: replacing a set by its closure does not change whether it lies inside a closed set.

Theorem 1.5.4. Let $$X$$ be a set. An operator $\operatorname{cl}\colon \mathcal P(X)\to \mathcal P(X)$ is a closure operator on $$X$$ if and only if

\[S\subseteq \operatorname{cl}(T) \iff \operatorname{cl}(S)\subseteq \operatorname{cl}(T) \tag{$\ast$}\]

for any $$S,T\subseteq X$.$

Proof. Suppose $\operatorname{cl}$ is a closure operator on $$X$.$ Since $\operatorname{cl}$ is monotone and idempotent, if $$S\subseteq \operatorname{cl}(T)$,$ then $$\operatorname{cl}(S)\subseteq \operatorname{cl}(\operatorname{cl}(T))=\operatorname{cl}(T)$.$ Since $\operatorname{cl}$ is extensive, if $$\operatorname{cl}(S)\subseteq \operatorname{cl}(T)$,$ then $$S\subseteq\operatorname{cl}(S)\subseteq \operatorname{cl}(T)$.$ Conversely, suppose $(\ast)$ holds for any $$S,T\subseteq X$.$ Since $$\operatorname{cl}(S)\subseteq \operatorname{cl}(S)$,$ we have $$S\subseteq \operatorname{cl}(S)$;$ hence $\operatorname{cl}$ is extensive. Since $\operatorname{cl}$ is extensive, if $$S\subseteq T$,$ then $$S\subseteq \operatorname{cl}(T)$.$ Then $\operatorname{cl}(S)\subseteq \operatorname{cl}(T)$ follows from $$(\ast)$;$ $\operatorname{cl}$ is monotone. The inclusion $\operatorname{cl}(\operatorname{cl}(S))\subseteq \operatorname{cl}(S)$ follows from $\operatorname{cl}(S)\subseteq \operatorname{cl}(S)$ by $$(\ast)$.$ Since extensivity gives the reverse inclusion, $$\operatorname{cl}(\operatorname{cl}(S))=\operatorname{cl}(S)$;$ $\operatorname{cl}$ is idempotent. $\square$

We now turn from the behavior of the operator to the property of the collection $\mathcal C$ of all its closed sets. A union of closed sets need not be closed; applying the closure operator produces a closed set containing that union, as in Proposition 1.5.3(2). An arbitrary intersection of closed sets, however, is already closed. The following theorem establishes this intersection property of $$\mathcal C$.$

Theorem 1.5.5. Let $\operatorname{cl}$ be a closure operator on a set $$X$.$ The collection $\mathcal C$ of all closed subsets of $$X$$ is a Moore collection.

Proof. Let $\{S_i\}_{i\in I}$ be any collection of elements of $$\mathcal C$.$ If $$I=\varnothing$,$ its intersection is $$X$.$ Since $\operatorname{cl}(X)\subseteq X$ and extensivity gives the reverse inclusion, $$\operatorname{cl}(X)=X$.$ Thus the empty intersection belongs to $$\mathcal C$.$ Now assume $$I\ne\varnothing$,$ and put $$S=\bigcap_{i\in I}S_i$.$ For every $$i\in I$,$ we have $$S\subseteq S_i$.$ By monotonicity and the fact that $$S_i$$ is closed, $$\operatorname{cl}(S)\subseteq\operatorname{cl}(S_i)=S_i$.$ It follows that $$\operatorname{cl}(S)\subseteq S$.$ Since $\operatorname{cl}$ is extensive, $$S\subseteq\operatorname{cl}(S)$,$ so $$\operatorname{cl}(S)=S$.$ Hence $$S$$ is closed and belongs to $$\mathcal C$.$ Therefore $\mathcal C$ is closed under arbitrary intersections, including the empty intersection, and is a Moore collection by Definition 1.3.8. $\square$

We now reverse the point of view: once $\mathcal C$ is known, the value $\operatorname{cl}(S)$ can be recovered for every $S\subseteq X$ by intersecting all members of $\mathcal C$ that contain $$S$.$ Thus the collection of closed sets completely determines the closure operator.

Theorem 1.5.6. Let $\operatorname{cl}$ be a closure operator on a set $$X$,$ and let $\mathcal C$ be the collection of all closed subsets of $$X$.$ Then, for all $$S\subseteq X$,$

\[\operatorname{cl}(S) = \bigcap \{C\in \mathcal C: C\supseteq S\}.\]

Proof. Let $$\mathcal C_{S}=\{C\in \mathcal C: C\supseteq S\}$.$ Since $$\operatorname{cl}(S)\in \mathcal C_S$,$ we have the first inclusion $$\bigcap \mathcal C_S\subseteq \operatorname{cl}(S)$.$ By Theorem 1.5.4, for all $$C\in \mathcal C_S$,$ $S\subseteq C$ implies $$\operatorname{cl}(S) \subseteq C$.$ Therefore we have the second inclusion $$\operatorname{cl}(S)\subseteq \bigcap \mathcal C_S$.$ $\square$

For a fixed $$S\subseteq X$,$ Theorem 1.5.6 also has a pointwise interpretation. The closure $\operatorname{cl}(S)$ is itself closed, contains $$S$,$ and is contained in every closed set containing $$S$.$ Hence $\operatorname{cl}(S)$ is the minimal closed set containing $$S$.$ We now show that span is a closure operator on a vector space.

Theorem 1.5.7. Let $$V$$ be a vector space. The self-map

\[\operatorname{span}\colon \mathcal P(V)\to \mathcal P(V), \qquad S\mapsto \operatorname{span}(S)\]

is a closure operator.

Proof. Since Definition 1.4.9 guarantees that $\operatorname{span}(S)\in \mathcal P(V)$ for all $$S\in \mathcal P(V)$,$ the operation is well-defined. Firstly, if $$v\in S$,$ then

\[v=\sum_{i=1}^1 1v\in \operatorname{span}(S).\]

Hence $S\subseteq \operatorname{span}(S)$ and $\operatorname{span}$ is extensive. Secondly, suppose $S,T\in \mathcal P(V)$ and $$S\subseteq T$.$ If $$v\in \operatorname{span}(S)$,$ then there exist a finite index set $$I$,$ an indexed family $\{u_i\}_{i\in I}$ in $$S$,$ and an indexed family $\{a_i\}_{i\in I}$ in $$F$$ such that

\[v=\sum_{i\in I}a_iu_i.\]

Since $$S\subseteq T$,$ the family $\{u_i\}_{i\in I}$ is in $$T$.$ By existential introduction, $$v\in \operatorname{span}(T)$.$ Hence, $\operatorname{span}$ is monotone. Lastly, we check idempotence. If $$v\in\operatorname{span}(\operatorname{span}(S))$,$ then there exist a finite index set $$I$,$ an indexed family $\{u_i\}_{i\in I}$ in $$\operatorname{span}(S)$,$ and an indexed family $\{a_i\}_{i\in I}$ in $$F$$ such that

\[v=\sum_{i\in I}a_iu_i.\]

Since $$u_i\in \operatorname{span}(S)$,$ there exist a finite index set $$J_i$,$ an indexed family $\{w_{ij}\}_{j\in J_i}$ in $$S$,$ and an indexed family $\{b_{ij}\}_{j\in J_i}$ in $$F$$ such that

\[u_i=\sum_{j\in J_i}b_{ij}w_{ij}\]

for each $$i\in I$.$ Put $$K=\{(i,j):i\in I,\ j\in J_i\}$.$ By Proposition 1.4.7(2) and Theorem 1.4.4,

\[\begin{align} v &=\sum_{i\in I}a_i\lowparen{\sum_{j\in J_i}b_{ij}w_{ij}}\\ &=\sum_{i\in I}\sum_{j\in J_i}a_i(b_{ij}w_{ij})\\ &=\sum_{(i,j)\in K}(a_ib_{ij})w_{ij}. \end{align}\]

Therefore, by existential introduction, $$v$$ is a linear combination of vectors in $$S$,$ so $$v\in\operatorname{span}(S)$.$ Hence $$\operatorname{span}(\operatorname{span}(S))\subseteq\operatorname{span}(S)$.$ Extensivity gives the reverse inclusion, so $$\operatorname{span}(\operatorname{span}(S))=\operatorname{span}(S)$;$ the span operator is idempotent. $\square$

Theorem 1.5.8. Let $$V$$ be a vector space over $$F$,$ and let $$S\subseteq V$.$ Then $$S$$ is closed if and only if $$S$$ is a subspace of $$V$.$

Proof. Suppose first that $$S$$ is closed. First, choosing $$I=\varnothing$,$ we have $$0_V\in\operatorname{span}(S)=S$.$ Next, let $u,v\in S$ and $$a\in F$.$ Since both $$u+v$$ and $$au$$ are linear combinations of vectors in $$S$,$ we have

\[u+v\in\operatorname{span}(S)=S \quad\text{and}\quad au\in\operatorname{span}(S)=S.\]

By Theorem 1.3.2, $$S$$ is a subspace of $$V$.$ Conversely, suppose that $$S$$ is a subspace of $$V$.$ Since $\operatorname{span}$ is extensive, we have $$S\subseteq\operatorname{span}(S)$.$ It remains to show that $$\operatorname{span}(S)\subseteq S$.$ Let $$v\in\operatorname{span}(S)$.$ By Definitions 1.4.8 and 1.4.9, there exist a finite index set $$I$,$ $\{u_i\}_{i\in I}$ in $$S$,$ and $\{a_i\}_{i\in I}$ in $$F$$ such that

\[v=\sum_{i\in I}a_i u_i.\]

We show by mathematical induction on $$|I|$$ that $$v\in S$.$ For the base case $$|I|=0$,$ we have $$I=\varnothing$,$ so $$v=0_V\in S$.$ Now suppose that every linear combination indexed by a set of cardinality $$n$$ belongs to $$S$.$ Let $$|I|=n+1$,$ and choose $$j\in I$.$ Then

\[v=a_j u_j + \sum_{i\in I\setminus\{j\}}a_i u_i.\]

Since $$S$$ is a subspace, the first summand belongs to $$S$.$ Since $$|I\setminus\{j\}|=n$,$ the second summand belongs to $$S$$ by the induction hypothesis. Hence $$v\in S$.$ It follows that $$\operatorname{span}(S)\subseteq S$.$ Consequently $S=\operatorname{span}(S)$ and $$S$$ is closed. $\square$

Since span is a closure operator whose closed sets are precisely the subspaces, Theorem 1.5.6 specializes immediately to the following description.

Corollary 1.5.9. Let $$S$$ be a subset of a vector space $$V$.$ Then,

\[\operatorname{span}(S) =\bigcap\{W\subseteq V: W\text{ is a subspace containing }S\}.\]

In general closure-operator theory, to combine two closed sets is to take the closure of the union of the two sets. In the setting of vector spaces, this construction corresponds to taking the span of two subspaces $$U$$ and $$W$.$ Theorem 1.3.12 and Corollary 1.5.9 characterize $\operatorname{span}(U\cup W)$ and $$U+W$$ by the same universal property: each is the minimal subspace containing both $$U$$ and $$W$.$ Therefore,

\[\operatorname{span}(U\cup W)=U+W.\]

Thus, the sum of subspaces is the specific realization of the general construction of combining two closed sets. The next corollary extends this connection from subspaces to arbitrary subsets $$S$$ and $$T$.$

Corollary 1.5.10. For any subsets $$S$$ and $$T$$ of a vector space $$V$,$

\[\operatorname{span}(S\cup T)=\operatorname{span}(S)+\operatorname{span}(T).\]

Proof. By Theorems 1.5.7 and 1.5.8, $\operatorname{span}(S\cup T)$ is a subspace. It contains both $\operatorname{span}(S)$ and $\operatorname{span}(T)$ by Proposition 1.5.3(1). Theorem 1.3.12(2) then gives the inclusion $$\operatorname{span}(S)+\operatorname{span}(T) \subseteq \operatorname{span}(S\cup T)$.$ On the other hand, $\operatorname{span}(S)+\operatorname{span}(T)$ is a subspace by Theorem 1.3.12(1). It contains $S\cup T$ by the same theorem and extensivity. Corollary 1.5.9 therefore gives the reverse inclusion. $\square$

Further properties of the span operator. The span operator has further properties that distinguish it from an arbitrary closure operator. They will be especially important in the study of linear dependence, bases, and dimension developed in the following sections. One of them is finite character: although a generating set may be infinite, membership in its span is always witnessed by finitely many generators.

Proposition 1.5.11. Let $$S$$ be a subset of a vector space $$V$,$ and let $$v\in V$.$ Then $v\in\operatorname{span}(S)$ if and only if there exists a finite subset $S_0\subseteq S$ such that $$v\in\operatorname{span}(S_0)$.$

Proof. If $$v\in\operatorname{span}(S)$,$ then there exists a finitely supported function $a\colon S\to F$ such that

\[v=\sum_{s\in\operatorname{supp}(a)}a(s)s\]

by Proposition 1.4.14. Put $S_0 = \operatorname{supp}(a)$ and define the restriction $a|_{S_0}$ of $$a$.$ Then $$S_0\subseteq S$,$ and $v\in\operatorname{span}(S_0)$ by Proposition 1.4.14. The converse is trivial: since $$S_0\subseteq S$,$ extensivity gives $$v\in \operatorname{span}(S_0)\subseteq \operatorname{span}(S)$.$ $\square$

Example 1.5.12. The vector space $$F[x]$$ of all polynomials has an infinite generating set:

\[F[x]=\operatorname{span}\{x^k:k\in\mathbb Z_{\ge0}\}.\]

Any polynomial $p(x)=a_0+a_1x+\cdots+a_nx^n$ in $$F[x]$,$ however, uses only finitely many of these generators:

\[p(x) \in \operatorname{span}\{1,x,\ldots,x^n\}.\]

Fix a subset $$S$$ of $$V$.$ We say that the $$u$-$ adding effect with respect to $$v$$ occurs if adjoining $$u$$ makes $$S$$ newly generate $$v$.$ Then, saying that no such effect occurs means that either $$v$$ was already generated by $$S$$ or it remains ungenerated after adjoining $$u$.$ The span operator has two notable properties in connection with this adding effects. First, it exhibits a certain symmetry: The $$u$-$ adding effect with respect to $$v$$ occurs if and only if the $$v$-$ adding effect with respect to $$u$$ occurs. Equivalently, the absence of such an effect is also reciprocal. Second, whenever an adding effect occurs, the two vectors can be exchanged as additional generators without changing the resulting span. In particular, this property is called the exchange property of the span operator.

Theorem 1.5.13. Let $$S$$ be a subset of a vector space $$V$,$ and let $$u,v\in V$.$ Then the following hold:

We have $u\in\operatorname{span}(S\cup\{v\})\setminus\operatorname{span}(S)$ if and only if $$v\in\operatorname{span}(S\cup\{u\})\setminus \operatorname{span}(S)$.$
If either of the equivalent conditions in (1) holds, then $$\operatorname{span}(S\cup\{u\}) = \operatorname{span}(S\cup\{v\})$.$

Proof. (1) Suppose

\[u\in\operatorname{span}(S\cup\{v\})\setminus\operatorname{span}(S).\]

By Corollary 1.5.10, there exist $s\in \operatorname{span}(S)$ and $a\in F$ such that

\[u = s + av.\]

Since $\operatorname{span}(S)$ is a subspace and $$u\notin\operatorname{span}(S)$,$ we must have $a\ne0$ and $$v\notin \operatorname{span}(S)$.$ It follows that

\[v=a^{-1}u - a^{-1}s \in \operatorname{span}(S\cup \{u\}),\]

which proves the forward implication. Since $$u$$ and $$v$$ were arbitrary, the same argument with their roles interchanged proves the backward implication. (2) Since $$u\in\operatorname{span}(S\cup\{v\})$,$

\[S\cup \{u\}\subseteq \operatorname{span}(S\cup \{v\}).\]

Since the right hand side is closed, Theorem 1.5.4 gives the inclusion

\[\operatorname{span}(S\cup\{u\}) \subseteq \operatorname{span}(S\cup\{v\}).\]

The reverse inclusion follows similarly from $$v\in\operatorname{span}(S\cup\{u\})$.$ Therefore the equality holds. $\square$

Exercises

1. Let $$S$$ and $$T$$ be subsets of a vector space $$V$.$

Prove that $\operatorname{span}(S\cap T) \subseteq \operatorname{span}(S)\cap\operatorname{span}(T).$
Give an example in $\mathbb R^2$ for which the inclusion is strict.

2. Let $$W$$ be a subspace of a vector space $$V$$ over $$F$,$ and let $$S\subseteq V$.$

Prove that $\{s+W:s\in S\}$ generates $$V/W$$ if and only if
\[W+\operatorname{span}(S)=V.\]
Apply this criterion to the quotient space in Example 1.3.20, with $$S=\{(0,1)\}$.$
Determine whether $\{s+W:s\in S\}$ generates $$V/W$$ and whether $$S$$ generates $$V$.$
Explain how the two questions in (3) differ when changes by elements of $$W$$ are regarded as insignificant.