Linear span

The span of a set of vectors, also called linear span, is the linear space formed by all the vectors that can be written as linear combinations of the vectors belonging to the given set.

Table of contents

Definition
A linear span is a linear space
Solved exercises
1. Exercise 1
2. Exercise 2

Definition

Let us start with a formal definition of span.

Definition Let $S$ be a linear space. Let $x_{1}, \dots, x_{n} \in S$ be $n$ vectors. The linear span of $x_{1}, \dots, x_{n}$ , denoted by

span (x_{1}, \dots, x_{n})

is the set of all the linear combinations

x = α_{1} x_{1} + \dots + α_{n} x_{n}

that can be obtained by arbitrarily choosing $n$ scalars $α_{1}$ , ... , $α_{n}$ .

A very simple example of a linear span follows.

Example Let $x_{1}$ and $x_{2}$ be $2 \times 1$ column vectors defined as follows:

x_{1} = 10 x_{2} = 01

Let $x$ be a linear combination of $x_{1}$ and $x_{2}$ with coefficients $α_{1}$ and $α_{2}$ . Then,

x = α_{1} x_{1} + α_{2} x_{2} = α_{1} 10 + α_{2} 01 = α_{1} \cdot 1 α_{1} \cdot 0 + α_{2} \cdot 0 α_{2} \cdot 1 = α_{1} 0 + 0 α_{2} = α_{1} + 0 0 + α_{2} = α_{1} α_{2}

Thus, the linear span is the set of all vectors $x$ that can be written as

x = α_{1} α_{2}

where $α_{1}$ and $α_{2}$ are two arbitrary scalars.

A linear span is a linear space

The following proposition, although elementary, is extremely important.

Proposition The linear span of a set of vectors is a linear space.

Proof

Let $S$ be the linear span of $n$ vectors $x_{1}, \dots, x_{n}$ . Then, $S$ is the set of all vectors $x$ that can be represented as linear combinations

x = α_{1} x_{1} + \dots + α_{n} x_{n}

Take two vectors $s_{1}$ and $s_{2}$ belonging to $S$ . Then, there exist coefficients $σ_{11}, \dots, σ_{1 n}$ and $σ_{21}, \dots, σ_{2 n}$ such that

s_{1} s_{2} = σ_{11} x_{1} + \dots + σ_{1 n} x_{n} = σ_{21} x_{1} + \dots + σ_{2 n} x_{n}

The span $S$ is a linear space if and only if, for any two coefficients $β_{1}$ and $β_{2}$ , the linear combination

β_{1} s_{1} + β_{2} s_{2}

also belongs to $S$ . But,

β_{1} s_{1} + β_{2} s_{2} = β_{1} (σ_{11} x_{1} + \dots + σ_{1 n} x_{n}) + β_{2} (σ_{21} x_{1} + \dots + σ_{2 n} x_{n}) = (β_{1} σ_{11} + β_{2} σ_{21}) x_{1} + \dots + (β_{1} σ_{1 n} + β_{2} σ_{2 n}) x_{n}

Thus, the linear combination

β_{1} s_{1} + β_{2} s_{2}

can itself be expressed as a linear combination of the vectors $x_{1}, \dots, x_{n}$ with coefficients $β_{1} σ_{11} + β_{2} σ_{21}$ , ... , $β_{1} σ_{1 n} + β_{2} σ_{2 n}$ . As a consequence, it belongs to the span $S$ . In summary, we have proved that any linear combination of vectors belonging to the span $S$ also belongs to the span $S$ . This means that $S$ is a linear space.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Define the following $2 \times 1$ vectors:

x_{1} = 11 x_{2} = 22 x_{3} = 01

Does $x_{3}$ belong to the linear span of $x_{1}$ and $x_{2}$ ?

Solution

The linear span of $x_{1}$ and $x_{2}$ is the set of all vectors $s$ that can be written as linear combinations of $x_{1}$ and $x_{2}$ with scalar coefficients $α_{1}$ and $α_{2}$ :

s = α_{1} x_{1} + α_{2} x_{2} = α_{1} 11 + α_{2} 22 = α_{1} + 2 α_{2} α_{1} + 2 α_{2} = (α_{1} + 2 α_{2}) 11

In other words, $span (x_{1}, x_{2})$ contains all the scalar multiples of the vector

x_{1} = 11

But $x_{3}$ is not a scalar multiple of $x_{1}$ . Therefore, $x_{3}$ does not belong to $span (x_{1}, x_{2})$ .

Exercise 2

Does the zero vector

00

belong to the span of the vectors $x_{1}$ and $x_{3}$ defined above?

Solution

We have proved that the span is a linear space, and the zero vector always belongs to a linear space (by the very definition of linear space).

How to cite

Please cite as:

Taboga, Marco (2021). "Linear span", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/linear-span.