Linear Combinations and Span

Quick answer

A linear combination of vectors v₁ through v_k is any sum c₁v₁ + ⋯ + c_kv_k with real coefficients. The set of all such combinations is the span of those vectors. The span of one nonzero vector is a line through the origin; the span of two vectors that are not multiples of each other is a plane through the origin. Asking whether a vector b is in a span is the same as asking whether a system of linear equations has a solution, which is how spans get computed.

What you'll learn

Write and evaluate linear combinations of vectors
Describe the span of one, two or more vectors geometrically
Turn the question is b in the span into a system of equations
Recognize when a vector adds nothing to a span

Building from a few vectors

Given some vectors, the two operations available are scaling and adding. Using both together on $\mathbf{v}_1, \ldots, \mathbf{v}_k$ produces a linear combination:

c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k

The numbers $c_i$ are any real numbers, positive, negative or zero. For $\mathbf{u} = (1, 2)$ and $\mathbf{v} = (3, 1)$ ,

2\mathbf{u} + \mathbf{v} = (2, 4) + (3, 1) = (5, 5)

so $(5, 5)$ is one of the vectors reachable from $\mathbf{u}$ and $\mathbf{v}$ . The set of all vectors reachable this way is the span:

\text{Span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} = \left\{c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k \;:\; c_i \in \mathbb{R}\right\}

Taking every coefficient $0$ gives $\mathbf{0}$ , so every span contains the origin.

(5, 5) as 2u + v

What a span looks like

One nonzero vector. Every multiple $c\mathbf{v}$ lies on the line through the origin in the direction of $\mathbf{v}$ , and the span is that whole line.
Two vectors, neither a multiple of the other. Their combinations sweep out a plane through the origin. In ℝ² that plane is all of ℝ²; in ℝ³ it is a flat sheet through the origin.
Two vectors on the same line. The second one adds nothing, and the span is still a line.

the span of (2, 1)

The span of one vector is a line through the origin

Why a span question is a system of equations

Asking whether $\mathbf{b}$ is in $\text{Span}\{\mathbf{u}, \mathbf{v}\}$ means asking whether some coefficients make

c_1\mathbf{u} + c_2\mathbf{v} = \mathbf{b}

Both sides are vectors, and two vectors are equal exactly when every component matches. One component gives one equation, so this single vector equation is a system of linear equations with the coefficients as unknowns. A span question is a solvability question: $\mathbf{b}$ lies in the span exactly when that system has at least one solution. Counting components shows the size of the system: $k$ vectors in ℝⁿ give $n$ equations in $k$ unknowns.

Worked examples

Common mistakes

Practice problems

Compute $3(1, 2) - 2(4, 0)$ .

Answer

$(-5, 6)$

Full solution

$(3, 6) - (8, 0) = (-5, 6)$ .
Is $(7, 4)$ in $\text{Span}\{(1, 2), (3, 1)\}$ ?

Answer

Yes: $(7, 4) = 1(1, 2) + 2(3, 1)$

Full solution

Matching components gives $c_1 + 3c_2 = 7$ and $2c_1 + c_2 = 4$ . Solving, $c_1 = 1$ and $c_2 = 2$ .
Is $(4, 2)$ in $\text{Span}\{(2, 1)\}$ ?

Answer

Yes, with coefficient $2$

Full solution

$(4, 2) = 2(2, 1)$ , so it lies on the line spanned by $(2, 1)$ .
Is $(3, 5)$ in $\text{Span}\{(1, 2)\}$ ?

Answer

No

Full solution

$c(1, 2) = (3, 5)$ needs $c = 3$ and $2c = 5$ at once.
Describe $\text{Span}\{(0, 0)\}$ .

Answer

The origin alone

Full solution

Every multiple of the zero vector is the zero vector, so the span is the single point $\mathbf{0}$ .
Is $(1, 1, 1)$ in $\text{Span}\{(1, 0, 1), (0, 1, 1)\}$ ?

Answer

No

Full solution

The first two components force $c_1 = 1$ and $c_2 = 1$ , and then the third component would have to be $1 + 1 = 2$ , not $1$ .
Describe $\text{Span}\{(1, 0, 0), (0, 0, 1)\}$ .

Answer

The $xz$ -plane: every vector with second component $0$

Full solution

Combinations are $(c_1, 0, c_2)$ , which is a plane through the origin containing the $x$ - and $z$ -axes.
Compare $\text{Span}\{(2, 4)\}$ with $\text{Span}\{(2, 4), (3, 6)\}$ .

Answer

They are the same line.

Full solution

$(3, 6) = 1.5(2, 4)$ , so the second vector is already in the first span and adds nothing.
How many equations and unknowns does the question “is $\mathbf{b}$ in $\text{Span}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ ” produce, for vectors in ℝ⁵?

Answer

Five equations in three unknowns

Full solution

Each of the five components gives one equation, and the unknowns are the three coefficients.
A student says $\text{Span}\{(1, 2), (2, 4), (3, 6)\}$ is all of ℝ², since three vectors is more than enough for a plane. What went wrong?

Hint

Compare the three vectors.

Answer

All three lie on one line, so the span is that line, not ℝ².

Full solution

$(2, 4) = 2(1, 2)$ and $(3, 6) = 3(1, 2)$ , so every combination is a multiple of $(1, 2)$ . What fills a plane is two directions, not a count of vectors.

Frequently asked questions

What is a linear combination?

A sum of scalar multiples: c₁v₁ + c₂v₂ + ⋯ + c_kv_k, where the coefficients c are any real numbers.

What is the span of a set of vectors?

The set of every linear combination of them. It is written Span{v₁, …, v_k} and always contains the zero vector.

What does a span look like?

The span of one nonzero vector is a line through the origin. The span of two vectors that are not multiples is a plane through the origin. In ℝⁿ larger sets can fill more.

How do you check whether b is in a span?

Set up c₁v₁ + ⋯ + c_kv_k = b. That is a system of linear equations in the coefficients, and b is in the span exactly when the system has a solution.

Why does every span contain the origin?

Taking every coefficient to be 0 gives the zero vector, so it belongs to every span.

What to learn next

Row Reduction and Echelon FormSolve any linear system by row reducing its augmented matrix: three row operations, echelon form, pivots, and the reduced form that reads off the answer.

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