Spanning sets

Vector spaces: Spans

Spanning sets

For each set of vectors of a vector space, there is a smallest linear subspace that contains all the vectors of the set. The linear subspace can be defined as the set of all linear combinations of the vectors from the system and is called the linear span. Below we discuss the concepts of linear combination and span.

Let $n$ be a natural number and let $\vec{a}_1 ,\ldots ,\vec{a}_n$ be $n$ vectors in a vector space $V$ .

A vector $\vec{x}$ of $V$ is called a linear combination of $\vec{a}_1 ,\ldots ,\vec{a}_n$ if there exist scalars $\lambda_1,\ldots ,\lambda_n$ such that $\vec{x} =\lambda_1 \cdot\vec{a}_1 +\lambda_2 \cdot\vec{a}_2 +\cdots +\lambda_n\cdot \vec{a}_n$
We then also say that the vector $\vec{x}$ (linearly) depends on the vectors $\vec{a}_1 ,\ldots ,\vec{a}_n$ .

Examples

The vector $\rv{-1,3,-5}$ is a linear combination of the vectors $\rv{1,1,-1}$ and $\rv{2,0,1}$ in $\mathbb{R}^3$ , for $\rv{-1,3,-5}= 3\cdot \rv{1,1,-1} -2\cdot\rv{ 2,0,1}$

The function $\cos(3x)$ is linearly dependent on the functions $\cos(x)$ and $\cos^3(x)$ in the space of real functions on $\mathbb{R}$ because

$\cos(3x) = 4\cos^3(x)-3\cos(x)$

We are particularly interested in the sets of all linear combinations of a given set of vectors.

Let $n$ be a natural number and let $\vec{a}_1 ,\ldots ,\vec{a}_n$ be $n$ vectors in a vector space.

The set of all linear combinations of $\vec{a}_1 ,\ldots ,\vec{a}_n$ is called the space spanned by the vectors $\vec{a}_1 ,\ldots ,\vec{a}_n$ or their (linear) span, and is denoted as

$\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$

We agree that the span of nothing (the empty set) is equal to $\linspan{}=\{0\}$ .

Notation

Instead of $\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ the following expression can also be found in the literature $\sbspmatrix{ \vec{a}_1 , \ldots, \vec{a}_n}$

The span of a single vector $\vec{v}$ is the line through $\vec{0}$ and $\vec{v}$ . The span of two vectors $\vec{v}$ and $\vec{w}$ that do not lie on a single line through the origin, is the plane through $\vec{0}$ , $\vec{v}$ , and $\vec{w}$ .

The span can also be defined for infinite sets: the span of an arbitrary set of vectors $X$ is the set of all linear combinations of a finite number of vectors from $X$ . Thus, we exclude sums of infinitely many terms.

For example, consider the vector space $P$ of all polynomial functions in $x$ . For every natural number $n$ , any linear combination of $1,x,\ldots,x^n$ is a polynomial of degree at most $n$ , and thus a vector of $P$ . But the infinite sum $\ee^x = \sum_{i=0}^{\infty}\frac{1}{i!}x^i$ is no element of $P$ .

The zero vector always belongs to the linear span: it is the linear combination of $\vec{a}_1 ,\ldots ,\vec{a}_n$ in which all coefficients $\lambda_i$ are equal to $0$ .

The span of a single vector $\vec{a}$ is the straight line through the origin with parametric representation $\lambda\cdot\vec{a}$ and is thus a linear subspace.

The span of two vectors $\vec{a}$ and $\vec{b}$ which are not scalar multiples of each other, is a plane passing through the origin with parametric representation $\lambda\cdot\vec{a}+\mu\cdot\vec{b}$ and is therefore also a linear subspace.

The span of a set of vectors forms a constructive way of determining the smallest subspace containing those vectors:

Spans are linear subspaces

Let $\vec{a}_1 ,\ldots ,\vec{a}_n$ be $n$ vectors of a vector space $V$ . The span $\linspan{\vec{a}_1 ,\ldots ,\vec{a}_n }$ is a linear subspace of $V$ .

The span is contained in all linear subspaces of $V$ which contain the vectors $\vec{a}_1 ,\ldots ,\vec{a}_n$ . In other words, the span is the smallest linear subspace containing $\vec{a}_1 ,\ldots ,\vec{a}_n$ .

We first show that $\linspan{\vec{a}_1 ,\ldots ,\vec{a}_n }$ is a linear subspace of $V$ . This span is not empty, because $\vec{0}$ is the linear combination of $\vec{a}_1 ,\ldots ,\vec{a}_n$ in which all scalars are equal to $0$ .

Let $\vec{p}$ and $\vec{q}$ be vectors in $\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ . Then there are scalars $p_1,\ldots,p_n$ , $q_1,\ldots,q_n$ such that
$\vec{p}=p_1 \cdot\vec{a}_1 +\cdots +p_n \cdot\vec{a}_n \quad \hbox{and} \quad \vec{q}=q_1 \cdot\vec{a}_1 +\cdots +q_n\cdot \vec{a}_n$ Consequently,
$\vec{p}+\vec{q}=(p_1 +q_1 )\cdot \vec{a}_1 +\cdots +(p_n +q_n )\cdot\vec{a}_n \in\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$
Also, for any number $\lambda$ :
$\lambda \cdot\vec{p} =\left(\lambda \cdot p_1 \right)\cdot \vec{a}_1 +\cdots +\left(\lambda \cdot p_n\right)\cdot \vec{a}_n \in \linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ This proves that $\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ is a linear subspace of $V$ .

The span of $\vec{a}_1 ,\ldots ,\vec{a}_n$ is the smallest linear subspace of $V$ containing $\vec{a}_1 ,\ldots ,\vec{a}_n$ : suppose $\vec{u}$ is a vector in $\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ and $U$ is a linear subspace containing $\vec{a}_1 , \ldots, \vec{a}_n$ . The latter property of $U$ and the fact that $U$ is a linear subspace, imply that every linear combination of $\vec{a}_1 ,\ldots ,\vec{a}_n$ also belongs to $U$ . This means that $\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ is a subset of $U$ . Because $\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ itself is a linear subspace which contains $\vec{a}_1 ,\ldots ,\vec{a}_n$ , the intersection of all of linear subspaces of $V$ which contain $\vec{a}_1 ,\ldots ,\vec{a}_n$ , is equal to $\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}$ .

In $\mathbb{R}^3$ , consider the vectors $\vec{a}=\rv{1,1,-2}$ , $\vec{b}=\rv{-1,1,0}$ , and $\vec{c}=\rv{0,1,-1}$ .

Prove that $\linspan{\vec{a},\vec{b},\vec{c}}=\linspan{\vec{a},\vec{c}}$

Write $W=\linspan{\vec{a},\vec{b},\vec{c}}$ .

Every linear combination of $\vec{a}$ and $\vec{c}$ is also a linear combination of $\vec{a}, \vec{b}$ , and $\vec{c}$ (add $0\,\vec{b}$ to it), so $\linspan{\vec{a},\vec{c}}$ is contained in $W$ .

The point is to show that every $\vec{x} \in W$ also belongs to $\linspan{\vec{a},\vec{c}}$ . Such a vector $\vec{x}$ can be written as
$\vec{x} =x_1 \cdot\vec{a} +x_2\cdot \vec{b} +x_3\cdot \vec{c}$ Crucial for the proof is the observation $\vec{b} =2\vec{c} -\vec{a}$ This means that
$\begin{array}{rcl} \vec{x}&=& x_1 \cdot\vec{a} + x_2\cdot (2\vec{c} -\vec{a}) +x_3\cdot \vec{c} \\ & =&\left(x_1 -x_2\right )\cdot \vec{a} +\left(2x_2 +x_3\right )\cdot \vec{c} \end{array}$
Thus, each vector of $W$ is a linear combination of $\vec{a}$ and $\vec{c}$ , so $W$ is contained in $\linspan{\vec{a},\vec{c}}$ . The conclusion is $W=\linspan{\vec{a},\vec{c}}$ , which is what we had to prove.

New example