6.3 Linear combinations and span

Once a vector space has been defined, the next question is not merely whether its operations make sense. The real question is what those operations can produce.

If you start with a list of vectors, how many new vectors can be built by adding and scaling them? That is the problem of linear combinations and span.

This section is fundamental because it converts geometric language such as "these vectors generate a plane" into exact algebra. It also provides the bridge from vector-space language back to linear systems.

Linear combinations are the basic building blocks

Definition

Linear combination

Let $u_1, u_2, \ldots, u_n$ be vectors in a vector space $V$ , and let $\alpha_1, \alpha_2, \ldots, \alpha_n \in R$ . Then

\alpha_1 u_1 + \alpha_2 u_2 + \cdots + \alpha_n u_n

is called a linear combination of $u_1, u_2, \ldots, u_n$ .

The important point is that the coefficients are completely unrestricted real numbers. They may be positive, negative, fractional, irrational, or zero.

That flexibility is why linear combinations are so expressive.

Worked example

Examples of linear combinations

If $u_1, \ldots, u_n$ are vectors, then all of the following are linear combinations of them:

23u_1 - \sqrt{2}\,u_2 + \pi u_n,

0 = 0u_1 + 0u_2 + \cdots + 0u_n,

and

u_1 = 1u_1 + 0u_2 + \cdots + 0u_n.

There is also a closure fact: if v and w are themselves linear combinations of $u_1, \ldots, u_n$ , then any expression $cv + dw$ is again a linear combination of $u_1, \ldots, u_n$ .

So linear combinations are stable under the very operations that created them in the first place.

Membership in a span is a solvability question

In practice, the question "Is b a linear combination of $u_1, \ldots, u_n$ ?" is not answered by inspection. It is answered by solving a linear system.

Theorem

Linear combinations and linear systems

Let $u_1, u_2, \ldots, u_n$ and b be vectors in $R^m$ , and let

A = \begin{bmatrix} u_1 & u_2 & \cdots & u_n \end{bmatrix}.

Then b can be expressed as a linear combination of $u_1, \ldots, u_n$ if and only if the linear system

Ax = b

has a solution.

This theorem is one of the most useful ideas in the chapter. It means that a generation question in a vector space becomes a consistency question for a matrix equation.

Worked example

Testing whether a vector is a linear combination

Let

u_1 = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}, \qquad u_2 = \begin{bmatrix} 3 \\ -5 \\ 4 \end{bmatrix}.

Consider the vector

w = \begin{bmatrix} 2 \\ -6 \\ -16 \end{bmatrix}.

To ask whether w is a linear combination of $u_1$ and $u_2$ is to ask whether there exist scalars $c_1, c_2$ such that

c_1u_1 + c_2u_2 = w.

Equivalently,

\begin{bmatrix} 1 & 3 \\ -2 & -5 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 2 \\ -6 \\ -16 \end{bmatrix}.

Row reduction shows that the system is consistent and gives

c_1 = 8, \qquad c_2 = -2.

Hence

w = 8u_1 - 2u_2.

So w belongs to the span of $u_1$ and $u_2$ .

The same method can also prove non-membership: if the system is inconsistent, then the target vector is not in the span.

Span collects all possible outputs

Definition

Span

Let $u_1, u_2, \ldots, u_n$ be vectors in a vector space $V$ . The span of these vectors is the set of all linear combinations of them:

\operatorname{Span}\{u_1, u_2, \ldots, u_n\} = \{\alpha_1u_1 + \alpha_2u_2 + \cdots + \alpha_nu_n \mid \alpha_i \in R\}.

So the span is not one vector. It is the whole collection of vectors that can be built from the given list.

This is why span is often read as "the subspace generated by" or "the set spanned by" the given vectors.

Standard geometric examples

Worked example

What span looks like in $R^3$

Let

e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \qquad e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \qquad e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.

Then every vector in $\operatorname{Span}\{e_1, e_2\}$ has the form

\alpha e_1 + \beta e_2 = \begin{bmatrix} \alpha \\ \beta \\ 0 \end{bmatrix}.

\operatorname{Span}\{e_1, e_2\}

is exactly the $x_1x_2$ -plane in $R^3$ .

If we include $e_3$ as well, then every vector in $R^3$ can be written as

\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3,

\operatorname{Span}\{e_1, e_2, e_3\} = R^3.

This example is important because it shows that span can describe:

a line,
a plane,
or the whole ambient space,

depending on the vectors you start with.

Why the span is always a subspace

The span of a set of vectors is not just some arbitrary subset. It is the smallest natural vector-space object generated by those vectors.

Theorem

The span of any set of vectors is a subspace

If $u_1, \ldots, u_n$ are vectors in a vector space $V$ , then

\operatorname{Span}\{u_1, \ldots, u_n\}

is a subspace of $V$ .

Proof

Why the span is a subspace

This theorem is one of the main reasons span matters so much. It gives a systematic way to build subspaces from a generating list.

Use the explorer as a supporting figure

The span explorer below is useful only after the algebraic meaning is clear. It lets you vary the coefficients and watch a vector move through the generated set, but it should be read as a supporting visualization of the definitions and theorem above.

Read and try

Build one vector from a span

The live explorer lets you vary coefficients and watch the resulting vector move inside the span.

(1, 0)

(0, 1)

Result

αu + βv = (1, 0)

Every output vector is built from the horizontal and vertical directions.

Common mistakes

Common mistake

Span is not the set of positive multiples only

The coefficients in a linear combination may be any real numbers. Negative and zero coefficients are just as legitimate as positive ones.

Common mistake

To prove membership in a span, you need coefficients

It is not enough to say that a target vector "looks similar." To prove that a vector lies in a span, you must either produce the coefficients explicitly or solve the associated system and show it is consistent.

Quick checks

Quick check

Is the zero vector always in the span of any finite list of vectors?

Use the definition of linear combination directly.

Solution

Answer

Quick check

Why does $\operatorname{Span}\{e_1, e_2\}$ in $R^3$ contain $(2, -1, 0)$ ?

Write the vector as a combination of $e_1$ and $e_2$ .

Solution

Answer

Exercises

Quick check

Explain why every vector $u_i$ in a generating list automatically lies in $\operatorname{Span}\{u_1, \ldots, u_n\}$ .

Use coefficients.

Solution

Guided solution

Quick check

Why does the theorem $b \in \operatorname{Span}\{u_1, \ldots, u_n\}$ if and only if $Ax = b$ is solvable matter so much in practice?

Answer in terms of what it lets you do computationally.

Solution

Guided solution

Read 6.2 Subspaces for the subspace tests used here, and then continue to 6.4 Linear dependence and independence, where span is paired with the question of redundancy in a generating list.

6.3 Linear combinations and span

MATH1030: Linear algebra I

Linear combinations are the basic building blocks

Linear combination

Examples of linear combinations

Membership in a span is a solvability question

Linear combinations and linear systems

Testing whether a vector is a linear combination

Span collects all possible outputs

Span

Standard geometric examples

What span looks like in $R^3$

Why the span is always a subspace

The span of any set of vectors is a subspace

Why the span is a subspace

Use the explorer as a supporting figure

Build one vector from a span

Common mistakes

Span is not the set of positive multiples only

To prove membership in a span, you need coefficients

Quick checks

Is the zero vector always in the span of any finite list of vectors?

Answer

Why does $\operatorname{Span}\{e_1, e_2\}$ in $R^3$ contain $(2, -1, 0)$ ?

Answer

Exercises

Explain why every vector $u_i$ in a generating list automatically lies in $\operatorname{Span}\{u_1, \ldots, u_n\}$ .

Guided solution

Why does the theorem $b \in \operatorname{Span}\{u_1, \ldots, u_n\}$ if and only if $Ax = b$ is solvable matter so much in practice?

Guided solution

Prerequisites

Key terms in this unit

More notes in this series

6.3 Linear combinations and span

MATH1030: Linear algebra I

Linear combinations are the basic building blocks

Linear combination

Examples of linear combinations

Membership in a span is a solvability question

Linear combinations and linear systems

Testing whether a vector is a linear combination

Span collects all possible outputs

Span

Standard geometric examples

What span looks like in R3R^3R3

Why the span is always a subspace

The span of any set of vectors is a subspace

Why the span is a subspace

Use the explorer as a supporting figure

Build one vector from a span

Common mistakes

Span is not the set of positive multiples only

To prove membership in a span, you need coefficients

Quick checks

Is the zero vector always in the span of any finite list of vectors?

Answer

Why does Span⁡{e1,e2}\operatorname{Span}\{e_1, e_2\}Span{e1​,e2​} in R3R^3R3 contain (2,−1,0)(2, -1, 0)(2,−1,0)?

Answer

Exercises

Explain why every vector uiu_iui​ in a generating list automatically lies in Span⁡{u1,…,un}\operatorname{Span}\{u_1, \ldots, u_n\}Span{u1​,…,un​}.

Guided solution

Why does the theorem b∈Span⁡{u1,…,un}b \in \operatorname{Span}\{u_1, \ldots, u_n\}b∈Span{u1​,…,un​} if and only if Ax=bAx = bAx=b is solvable matter so much in practice?

Guided solution

Related notes

Prerequisites

Key terms in this unit

More notes in this series

What span looks like in $R^3$

Why does $\operatorname{Span}\{e_1, e_2\}$ in $R^3$ contain $(2, -1, 0)$ ?

Explain why every vector $u_i$ in a generating list automatically lies in $\operatorname{Span}\{u_1, \ldots, u_n\}$ .

Why does the theorem $b \in \operatorname{Span}\{u_1, \ldots, u_n\}$ if and only if $Ax = b$ is solvable matter so much in practice?