Source: Friedberg, Insel & Spence, Linear Algebra 4th Ed., Ch. 1.4
Tags: linear combination, span, spanning set, generates, system of linear equations, Gaussian elimination, row reduction, vector space, subspace, Friedberg chapter 1
Difficulty: Foundational to Intermediate Prerequisites: Vector space definition and subspace test (Sections 1.1–1.3 study notes).
This section answers two linked questions: "Can I build a target vector from a given collection of vectors?" and "What is the set of everything I can build?" The first question reduces to solving a system of linear equations. The second introduces the span, which is always a subspace. Together, these ideas connect abstract vector space theory to the concrete computational tool of row reduction. If you are behind, note that the mechanics of solving linear systems here are a preview of the full Gaussian elimination method in Chapter 3.
A linear combination of vectors u₁,...,uₙ is any expression a₁u₁ + ... + aₙuₙ. The span of a set S is the collection of all linear combinations of vectors in S, and it is always a subspace. Determining whether a vector is a linear combination of others amounts to solving a system of linear equations using elimination.
Linear combination
A vector v is a linear combination of vectors u₁, u₂, ..., uₙ if there exist scalars a₁, a₂, ..., aₙ such that v = a₁u₁ + a₂u₂ + ... + aₙuₙ. The scalars are called the coefficients.
In simple terms, v can be built by scaling each uᵢ and adding the results.
Span (span(S))
The set of all linear combinations of vectors in a nonempty set S. By convention, span(∅) = {0}.
Think of it as: all the vectors you can "reach" by combining elements of S with any scalars.
Generates (spans)
A subset S of a vector space V generates (or spans) V if span(S) = V, meaning every vector in V can be written as a linear combination of vectors in S.
System of linear equations
A collection of equations, each linear in the unknowns. Solving it means finding all tuples of scalars satisfying every equation simultaneously.
The zero vector is always a linear combination of any nonempty set (take all coefficients to be zero). Determining whether a specific vector v is a linear combination of u₁,...,uₙ amounts to asking whether there exist scalars a₁,...,aₙ satisfying:
v = a₁u₁ + a₂u₂ + ... + aₙuₙ
Equating coordinates (or coefficients, for polynomials/matrices) produces a system of linear equations in the unknowns a₁,...,aₙ.
Three operations that do not change the solution set:
Interchange two equations
Multiply an equation by a nonzero constant
Add a constant multiple of one equation to another
The goal is to reach a system in "reduced" form where:
The first nonzero coefficient in each equation is 1 (leading 1)
A variable that is the leading variable in one equation does not appear (with nonzero coefficient) in any other equation
Leading variables appear in increasing order down the rows
Once reduced, you can read off solutions directly. Variables corresponding to leading 1s are expressed in terms of the remaining "free" variables.
To check if (2, 6, 8) is a linear combination of u₁ = (1,2,1), u₂ = (−2,−4,−2), u₃ = (0,2,3), u₄ = (2,0,−3), u₅ = (−3,8,16):
Setting up a₁u₁ + ... + a₅u₅ = (2,6,8) and equating coordinates gives three equations in five unknowns. After elimination, the reduced system is:
a₁ − 2a₂ + a₅ = −4
a₃ + 3a₅ = 7
a₄ − 2a₅ = 3
Setting free variables a₂ = 0, a₅ = 0 gives a₁ = −4, a₃ = 7, a₄ = 3. So (2,6,8) = −4u₁ + 0u₂ + 7u₃ + 3u₄ + 0u₅.
If during elimination you obtain an equation of the form 0 = c where c ≠ 0, the system has no solution. This means the target vector is not a linear combination of the given vectors.
Example: 3x³ − 2x² + 7x + 8 is not a linear combination of x³ − 2x² − 5x − 3 and 3x³ − 5x² − 4x − 9, because the corresponding system yields 0 = 17.
Theorem 1.5: The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace that contains S must also contain span(S).
This means span(S) is the smallest subspace containing S.
Proof sketch:
If S = ∅, span(∅) = {0}, a subspace.
If S ≠ ∅, pick z ∈ S; then 0 = 0z ∈ span(S).
If x, y ∈ span(S), both are linear combinations of vectors in S, so x + y and cx are also linear combinations. Closure holds.
Any subspace W containing S must contain all linear combinations of elements of S (by closure), so span(S) ⊆ W.
{(1,1,0), (1,0,1), (0,1,1)} generates R³. For any (a₁,a₂,a₃), the coefficients r, s, t can be found explicitly:
r = ½(a₁ + a₂ − a₃), s = ½(a₁ − a₂ + a₃), t = ½(−a₁ + a₂ + a₃)
{e₁, e₂, ..., eₙ} generates Fⁿ, where eⱼ has 1 in position j and 0 elsewhere.
{1, x, x², ..., xⁿ} generates Pₙ(F).
The four matrices E¹¹, E¹², E²¹, E²² generate M₂ₓ₂(F).
span({x}) = {ax : a ∈ F} for any vector x. Geometrically in R³, this is a line through the origin.
If S₁ ⊆ S₂, then span(S₁) ⊆ span(S₂). In particular, if span(S₁) = V and S₁ ⊆ S₂, then span(S₂) = V.
span(S₁ ∪ S₂) = span(S₁) + span(S₂).
W is a subspace if and only if span(W) = W.
Linear combination:
v = a₁u₁ + a₂u₂ + ... + aₙuₙ
Plane through origin (Section 1.1 connection):
x = su + tv (all linear combinations of u and v)
This set equals span({u, v}), a subspace of R³.
Expressing a target as a linear combination of available resources is the mathematical core of mixture problems. The text's vitamin example shows this directly: the vitamin content of wild rice can be replicated by combining specific amounts of cupcake, custard pie, brown rice, and soy sauce, because the wild rice vitamin vector is a linear combination of the others.
Students often think span(∅) is ∅. It is not. By convention, span(∅) = {0}.
"Span" is both a noun (the set) and a verb (to generate). A set S spans V means span(S) = V.
Not every system of linear equations has a solution. When checking whether v is in span(S), you must be prepared for the system to be inconsistent.
Students sometimes forget that there may be infinitely many ways to write v as a linear combination. The existence of free variables in the reduced system means the representation is not unique (unless the set is linearly independent, covered in Section 1.5).
⚠️ "Express v as a linear combination of ..." is a standard exam question format. You need to set up and solve the system correctly.
⚠️ Know Theorem 1.5 and be able to prove that span(S) is a subspace.
⚠️ Recognising when a system is inconsistent (0 = nonzero) versus when it has free variables is crucial.
⚠️ You will need to show that a given set generates a specific vector space. This means showing that every vector in the space can be written as a linear combination.
⚠️ The three elementary row operations are the computational backbone. They reappear formally in Chapter 3 with row echelon form.
True or false: the zero vector is a linear combination of any nonempty set of vectors.
True or false: span(∅) = ∅.
Fill in the blank: if S is a subset of V, then span(S) is the ______ subspace of V containing S.
True or false: every system of linear equations has a solution.
True or false: if span(S₁) = V and S₁ ⊆ S₂, then span(S₂) = V.
Answers: 1. True (all coefficients zero). 2. False (span(∅) = {0}). 3. Smallest. 4. False. 5. True.
Q: Determine whether (−2, 0, 3) is a linear combination of (1, 3, 0) and (2, 4, −1).
A: Solve a(1,3,0) + b(2,4,−1) = (−2,0,3). The system is: a + 2b = −2, 3a + 4b = 0, −b = 3. From the third equation, b = −3. Then a = −2 − 2(−3) = 4. Check: 3(4) + 4(−3) = 0 ✓. So yes, (−2,0,3) = 4(1,3,0) − 3(2,4,−1).
Q: Show that {1, x, x², ..., xⁿ} generates Pₙ(F).
A: Any polynomial f(x) = aₙxⁿ + ... + a₁x + a₀ in Pₙ(F) can be written as a₀·1 + a₁·x + ... + aₙ·xⁿ, which is a linear combination of {1, x, ..., xⁿ}. Hence span({1, x, ..., xⁿ}) = Pₙ(F).
Q: Let S₁ and S₂ be subsets of V with S₁ ⊆ S₂. Prove that span(S₁) ⊆ span(S₂).
A: If v ∈ span(S₁), then v = a₁u₁ + ... + aₖuₖ for some u₁,...,uₖ ∈ S₁ and scalars a₁,...,aₖ. Since S₁ ⊆ S₂, each uᵢ ∈ S₂. So v is a linear combination of vectors in S₂, hence v ∈ span(S₂).
Q: Prove that span(S) is a subspace (sketch the three conditions of Theorem 1.3).
A: (a) 0 ∈ span(S) since 0 = 0z for any z ∈ S. (b) If x = Σaᵢuᵢ and y = Σbⱼvⱼ are in span(S), then x + y = Σaᵢuᵢ + Σbⱼvⱼ is a linear combination of vectors in S. (c) cx = Σ(caᵢ)uᵢ is also a linear combination. By Theorem 1.3, span(S) is a subspace.
Linear combinations and span are the bridge to linear dependence and independence (Section 1.5), which asks when span(S) can be generated by a smaller subset. The concept of a basis (Section 1.6) unifies spanning and independence: a basis is a linearly independent spanning set. The solving technique here is formalised as Gaussian elimination in Chapter 3, where the augmented matrix and row echelon form provide a systematic framework.
linear combination, span, spanning set, generates, system of linear equations, Gaussian elimination, row reduction, free variable, leading variable, inconsistent system, subspace generated by a set, Theorem 1.5, Friedberg chapter 1 section 4, abstract linear algebra