Span

Definition: Span

The set of all linear combinations of a list of vectors $v_1,\dots ,v_m$ in $V$ is called the span of $v_1,\dots ,v_m$, denoted by $\text{span}(v_1,\dots ,v_m)$. In other words,

$$\text{span}(v_1,\dots ,v_m)=\{a_1{v}_1+\cdots +a_m{v}_m:a_1,\dots ,a_m\in \mathbb{F}\}$$

The span of the empty list $()$ is defined to be $\{0\}$.

Following the example from Linear Combinations, we saw that

• $(17,-4,2)$ can be written as $(17,-4,2)=6(2,1,3)+5(1,-2,4)$. Therefore, $(17,-4,2)\in \text{span}((2,1,-3),(1,-2,4))$.
• $(17,-4,5)$ cannot be written as any linear combination $(17,-4,5)=a_1(2,1,-3)+a_2(1,-2,4)$. Therefore, $(17,-4,5)\notin \text{span}((2,1,-3),(1,-2,4))$.

Span = Smallest Containing Subpace

Theorem: Span is the smallest containing subspace

The span of a list of vectors in $V$ is the smallest subspace of $V$ containing all vectors in the list.

Proof. Suppose $v_1,\dots ,v_m$ is a list of vectors in $V$.

First, we show that $\text{span}(v_1,\dots ,v_m)$ is a subspace of $V$. The additive identity is in $\text{span}(v_1,\dots ,v_m)$ because we always have

$$0=0v_1+\cdots +0v_m$$

Also, $\text{span}(v_1,\dots v_m)$ is closed under addition because

$$(a_1{v}_1+\cdots +a_m{v}_m)+(c_1{v}_1+\cdots +c_m{v}_m)=(a_1+c_1)v_1+\cdots +(a_m+c_m)v_m$$

Furthermore, $\text{span}(v_1,\dots v_m)$ is closed under addition because

$$\lambda (a_1{v}_1+\cdots +a_m{v}_m)=\lambda a_1{v}_1+\cdots +\lambda a_m{v}_m$$

Thus, $\text{span}(v_1,\dots v_m)$ is a subspace of $V$.

Each $v_k$ can be written as a linear combination of $v_1,\dots ,v_m$, as we can set $a_k=1$ and let all other $a$‘s in $a_1{v}_1+\cdots +a_m{v}_m$ be zero. Thus, $\text{span}(v_1,\dots v_m)$ contains each $v_k$.

Because subspaces are closed under scalar multiplication and addition, any subspace of $V$ that contains all $v_1,\dots ,v_k,\dots ,v_m$ must also contain every linear combination of these vectors, or $\text{span}(v_1,\dots v_m)$. Thus, $\text{span}(v_1,\dots ,v_m)$ is the smallest subspace of $V$ containing all vectors $v_1,\dots ,v_m$, since it’s the minimum set that guarantees closure under addition and scalar multiplication.

Spanning a Subspace

Definition: Spans

If $\text{span}(v_1,\dots ,v_m)$ equals $V$, we say that the list $v_1,\dots ,v_m$ spans $V$.

Example: Suppose $n$ is a positive integer. We want to show that

$$(1,0,\dots ,0),(0,1,0,\dots ,0),\dots ,(0,\dots ,0,1)$$

spans ${\mathbb{F}}^n$. Here, the $k$-th vector in the list above has $1$ in the $k$th slot and $0$ in all other slots.

Suppose $(x_1,\dots ,x_n)\in {\mathbb{F}}^n$. Then

$$(x_1,\dots ,x_n)=x_1(1,0,\dots ,0)+x_2(0,1,0,\dots ,0)+\cdots +x_n(0,\dots ,0,1)$$

Thus $(x_1,\dots ,x_n)\in \text{span}((1,0,\dots ,0),(0,1,0,\dots ,0),\dots ,(0,\dots ,0,1))$.