MGU Mathematics Series — Linear Algebra

Part III: Vector Spaces, Subspaces, and Dimension

Span · Linear Independence · Basis · Dimension · Coordinates

Chapter 9 of 14

MGU Mathematics Series · Linear Algebra · Library Textbook Edition

Chapter 9:
Span, Linear Independence,
and Basis

Linear combinations, the reach of a vector set, redundancy and independence, efficient coordinate systems, and dimension — the structural heart of linear algebra

■ Math Level: Linear Algebra ■

■ 14 Sections • Part III of 5 ■

Follows Chapter 8: Vector Spaces and Subspaces · Prepares students for Chapter 10: Rank, Null Space, and the Fundamental Subspaces

Chapter 9 — Front Matter

Purpose and Overview

This chapter develops the structural heart of linear algebra. In Chapter 8, you learned what a vector space is — a collection of objects that can be added and scaled according to consistent rules. In this chapter, you will learn how to describe the inside of a vector space: how vectors build subsets of a space, which vectors carry genuinely new information, and how to choose an efficient coordinate system for any space.

The three central ideas of this chapter are span, linear independence, and basis. These ideas are connected. Span tells you what a collection of vectors can reach. Linear independence tells you whether every vector in the collection is contributing something that the others cannot. Basis brings these ideas together: a basis is a minimal, nonredundant collection of vectors that can describe every element of a space.

These ideas are not just abstract. They determine whether a system of equations has a unique solution, whether data contains redundancy, whether a set of forces can produce any desired motion, and whether a transformation can be fully described. Every major topic in the second half of linear algebra — rank, null space, orthogonality, eigenvalues, diagonalization, and least squares — depends on understanding span, independence, and basis.

Opening Question

Suppose you are describing locations on a map using two directions: east and north. Any location can be described as some combination of miles east and miles north. Now suppose someone proposes a third direction: northeast. Does adding "northeast" give you any new descriptive power that east and north alone did not already have? Or is northeast just a combination of east and north that you could already build?

This is the question behind linear independence. When a collection of vectors can already describe everything in a space, adding more vectors does not increase the space — it only increases the redundancy. Linear algebra gives you precise tools for detecting and eliminating that redundancy.

14Sections

13Objectives

14Key Terms

10+Worked Examples

Learning Objectives

1Compute linear combinations of two or more vectors.
2Describe the span of a set of vectors geometrically and algebraically.
3Determine whether a vector belongs to the span of a given set.
4Define linear dependence and linear independence precisely.
5Apply the definition of independence to test whether a set of vectors is independent.
6Use row reduction to test linear independence of a set of vectors.
7Define a basis for a vector space or subspace.
8Identify the standard basis for ℝⁿ.
9State and use the definition of dimension.
10Find coordinates of a vector relative to a given basis.
11Explain why different bases for the same space have the same number of vectors.
12Recognize common errors in reasoning about span, independence, and basis.
13Apply these ideas to systems, data, geometry, and modeling contexts.

Chapter 9 — Contents

Sections at a Glance

9.1Linear CombinationsDefinition of linear combination; coefficients; the trivial combination; linear combinations and systems; row reduction connection
9.2SpanDefinition of span; spanning sets; span as a subspace; testing span by row reduction
9.3Span in ℝ² and ℝ³Geometric cases in two and three dimensions; point, line, plane, full space; adding vectors does not always expand span
9.4Span in Higher DimensionsExtending span to ℝⁿ; algebraic test for spanning; minimum vectors needed to span
9.5Linear DependenceDefinition of dependence; nontrivial combinations; dependence relation; redundancy; geometric view
9.6Linear IndependenceDefinition of independence; trivial solution only; independence as lack of redundancy; single nonzero vector
9.7Testing Independence with Row ReductionRow reduction method; pivot columns; why the test works; more vectors than components are always dependent
9.8BasisDefinition of basis; spanning and independence together; uniqueness of representation; finding a basis by row reduction
9.9The Standard BasisStandard basis vectors e₁, …, eₙ; natural coordinate system; standard bases for other spaces
9.10DimensionDimension theorem; all bases have the same size; dimension as structure; key dimension facts
9.11Coordinates Relative to a BasisCoordinate vectors; finding coordinates by solving a system; why non-standard coordinates matter
9.12Changing Descriptions of a SpaceMany bases for one space; purposes of different bases; parametric and implicit descriptions; row reduction between forms
9.13Common Span and Independence MistakesEight recurring errors; explanations and corrections; prevention strategies
9.14Preparing for Rank and NullityColumn space, null space, rank, nullity; Rank-Nullity Theorem preview; connections to Chapters 3–9 and forward

Section 9.1

Linear Combinations

What Is a Linear Combination?

A linear combination of a collection of vectors is any expression formed by scaling each vector by a number (called a scalar or coefficient) and adding the results.

If v₁, v₂, …, vₖ are vectors and c₁, c₂, …, cₖ are scalars, then the vector

c₁v₁ + c₂v₂ + … + cₖvₖ

is a linear combination of v₁, v₂, …, vₖ. The scalars c₁, c₂, …, cₖ are called the coefficients of the linear combination.

You have seen this idea before. In Chapter 2, when you added two vectors and scaled the result, you were building a linear combination. In Chapter 3, when you solved systems of equations by elimination, you were forming linear combinations of rows. The concept is elementary, but its consequences run through all of linear algebra.

Examples of Linear Combinations

Let v₁ = (1, 0) and v₂ = (0, 1).

The vector (3, −2) is the linear combination 3v₁ + (−2)v₂. The coefficients are 3 and −2.

The vector (0, 0) is the linear combination 0·v₁ + 0·v₂. This is called the trivial combination — every collection of vectors produces the zero vector when all coefficients are zero. This is always possible, and it is never surprising.

In fact, any vector in ℝ² can be written as a linear combination of v₁ and v₂. This observation leads directly to the idea of span.

Linear Combinations and Systems

Asking whether a vector b is a linear combination of vectors v₁, …, vₖ is equivalent to asking whether the system c₁v₁ + c₂v₂ + … + cₖvₖ = b has a solution. This connects linear combinations directly to the row reduction methods of Chapter 3. To determine whether b is reachable, form the augmented matrix [v₁ v₂ … vₖ | b] and reduce.

Example 9.1.1

Writing a Vector as a Linear Combination

ProblemDetermine whether b = (1, 4, 2) is a linear combination of v₁ = (1, 0, 1) and v₂ = (0, 2, −1).
ThinkFind scalars c₁ and c₂ (if they exist) such that c₁v₁ + c₂v₂ = b. This gives a system of three equations in two unknowns.
RepresentForm the augmented matrix [v₁ v₂ | b]: [ 1 0 | 1 ] [ 0 2 | 4 ] [ 1 -1 | 2 ]
CalculateSubtract row 1 from row 3, then divide row 2 by 2, then add row 2 to row 3: [ 1 0 | 1 ] [ 1 0 | 1 ] [ 1 0 | 1 ] [ 0 2 | 4 ] → [ 0 1 | 2 ] → [ 0 1 | 2 ] [ 0 -1 | 1 ] [ 0 -1 | 1 ] [ 0 0 | 3 ]
CheckThe last row says 0 = 3, a contradiction. The system is inconsistent.
InterpretThe vector b is not a linear combination of v₁ and v₂. Geometrically, v₁ and v₂ together describe a plane through the origin in ℝ³, and b does not lie in that plane.

Answer: b is not a linear combination of v₁ and v₂. No scalars c₁, c₂ satisfy the equation.

The Zero Vector Is Always a Linear Combination

The zero vector 0 is always a linear combination of any collection of vectors: simply set all coefficients to zero. This trivial observation is important because it becomes the starting point for the definition of linear independence.

Practice 9.1Linear Combination Practice

Let u = (2, −1, 0) and v = (0, 3, 1). Compute: (a) 3u + v, (b) −2u + 4v, (c) u − v.
Write (5, 1) as a linear combination of (1, 0) and (0, 1). Then write it as a linear combination of (1, 1) and (1, −1).
Determine whether b = (3, 1, 2) is a linear combination of v₁ = (1, 0, 1) and v₂ = (1, 1, 0). Show all row reduction steps.
Find all vectors in ℝ² that can be written as linear combinations of (2, 4) and (1, 2). Describe the result geometrically.

Section 9.2

Span

Definition

The span of a collection of vectors {v₁, v₂, …, vₖ} is the set of all linear combinations that can be formed from those vectors:

span{v₁, v₂, …, vₖ} = { c₁v₁ + c₂v₂ + … + cₖvₖ : c₁, c₂, …, cₖ ∈ ℝ }

The span is a set of vectors — specifically, every vector that can be reached by combining v₁, …, vₖ with real-number scalars. A key result from Chapter 8: the span of any collection of vectors is always a subspace of the ambient vector space. It contains the zero vector, is closed under addition, and is closed under scalar multiplication. Span is one of the most natural ways to construct subspaces.

Spanning Sets

If the span of a collection equals an entire space V, then the collection is called a spanning set for V. This means every vector in V can be written as a linear combination of vectors in the set. A spanning set covers everything.

For example, the vectors e₁ = (1, 0) and e₂ = (0, 1) span all of ℝ², because any vector (a, b) = ae₁ + be₂.

Example 9.2.1

Determining Whether a Set Spans ℝ³

ProblemDo the vectors v₁ = (1, 0, 0), v₂ = (0, 1, 0), v₃ = (1, 1, 0) span all of ℝ³?
ThinkForm the matrix A = [v₁ v₂ v₃] and row-reduce. If there is a pivot in every row, the columns span ℝ³.
Calculate A = [ 1 0 1 ] Row echelon form: [ 1 0 1 ] [ 0 1 1 ] [ 0 1 1 ] [ 0 0 0 ] [ 0 0 0 ]
CheckThe third row is all zeros — no pivot in row 3. Choose b = (0, 0, 1); the augmented matrix gives 0 = 1, a contradiction.
InterpretThese three vectors do not span ℝ³. All three lie in the xy-plane (their third components are zero), so their span is only that plane. Any vector with a nonzero third component is unreachable.

Answer: No. The span is the xy-plane, not all of ℝ³.

Section 9.3

Span in ℝ² and ℝ³

Setting	Collection	Geometric Span
ℝ²	Only the zero vector	A single point: {0}
ℝ²	One nonzero vector v	A line through the origin in the direction of v
ℝ²	Two vectors pointing in different directions	All of ℝ²
ℝ²	Two vectors on the same line (v₂ = cv₁)	A line through the origin — adding v₂ adds nothing
ℝ³	One nonzero vector	A line through the origin
ℝ³	Two independent vectors	A plane through the origin
ℝ³	Three independent vectors	All of ℝ³

This table reveals a crucial point: adding more vectors does not automatically increase the span. If the new vectors are already reachable from the existing ones, the span does not grow. The question of whether the span grows when you add a vector is precisely the question of linear independence.

Diagram description: In ℝ³, draw the origin. A single nonzero vector v₁ produces a line — a one-dimensional thread through space. Adding a second independent vector v₂ stretches that thread into a plane — a flat two-dimensional sheet through the origin. Adding a third vector v₃ that does not lie in the plane lifts the structure into all three dimensions. The span grows only when a genuinely new direction is introduced.

Section 9.4

Span in Higher Dimensions

The same ideas extend beyond ℝ² and ℝ³, even when direct geometric visualization is no longer available. In ℝⁿ, the span of a collection of vectors is still the set of all linear combinations, and we still ask whether that span equals all of ℝⁿ or something smaller.

The algebraic test remains the same: form the matrix A whose columns are the vectors in the collection, row-reduce, and check whether every row has a pivot. If yes, the columns span ℝⁿ. If any row lacks a pivot, the columns fail to span.

An Important General Principle

To span ℝⁿ, you need at least n vectors. More than n vectors may span ℝⁿ, but only if at least n of them are independent. Fewer than n vectors cannot span ℝⁿ no matter what.

Example 9.4.1

Span Test in ℝ⁴

ProblemDo the vectors v₁ = (1,0,0,0), v₂ = (0,1,0,0), v₃ = (0,0,1,1) span ℝ⁴?
ThinkThere are three vectors in ℝ⁴. The matrix A has only three columns, so it can have at most three pivots — fewer than the four required to span ℝ⁴.
InterpretThree vectors cannot span ℝ⁴. The maximum number of independent directions three vectors can cover is three, which corresponds to a three-dimensional subspace (a hyperplane) inside ℝ⁴. To span all of ℝ⁴, you need at least four independent vectors.

Answer: No. Three vectors cannot span ℝ⁴. At least four independent vectors are required.

Practice 9.2 – 9.4Span Practice

Describe the span of each set geometrically (point, line, plane, or all of ℝ³): (a) {(1,0,0)}, (b) {(1,2,3),(2,4,6)}, (c) {(1,0,0),(0,1,0)}, (d) {(1,0,0),(0,1,0),(0,0,1)}.
Do the vectors (1,0,2), (0,1,−1), (3,2,4) span ℝ³? Use row reduction to decide.
Find a vector in ℝ³ that is not in the span of v₁ = (1,0,1) and v₂ = (0,1,1). Justify your answer.
True or false: If three vectors span ℝ³, they must be linearly independent. Explain.

Section 9.5

Linear Dependence

The Question of Redundancy

Given a set of vectors, we can ask: is any vector in the set already a linear combination of the others? If so, that vector is redundant — it does not contribute anything new to the span. We say the set is linearly dependent.

Definition — Linear Dependence

A set of vectors {v₁, v₂, …, vₖ} is linearly dependent if there exist scalars c₁, c₂, …, cₖ, not all zero, such that:

c₁v₁ + c₂v₂ + … + cₖvₖ = 0

This equation is called a dependence relation. The key phrase is not all zero — a nontrivial combination that gives the zero vector reveals redundancy.

Why This Definition Detects Redundancy

Suppose c₁v₁ + c₂v₂ + c₃v₃ = 0 with c₁ ≠ 0. Solving: v₁ = −(c₂/c₁)v₂ − (c₃/c₁)v₃. This says v₁ is already a linear combination of v₂ and v₃. Removing v₁ from the set does not shrink the span. The set is redundant.

Also: any set containing the zero vector is automatically linearly dependent, because you can set the coefficient of 0 to 1 and all others to 0, obtaining a nontrivial combination that gives 0.

Example 9.5.1

Identifying a Dependent Set

ProblemDetermine whether the set {(1,2), (3,6)} is linearly dependent.
ThinkLook for scalars c₁, c₂, not both zero, such that c₁(1,2) + c₂(3,6) = (0,0).
Calculate c₁ + 3c₂ = 0 2c₁ + 6c₂ = 0 The second equation is twice the first. Setting c₂ = 1 gives c₁ = −3 — a nontrivial solution.
Check−3(1,2) + 1(3,6) = (−3+3, −6+6) = (0,0) ✓
InterpretThe set is linearly dependent. The second vector is exactly three times the first. Both vectors lie on the same line through the origin. Adding the second vector to the set did not expand the span.

Answer: Linearly dependent. Dependence relation: −3(1,2) + 1(3,6) = 0.

Section 9.6

Linear Independence

Definition — Linear Independence

A set of vectors {v₁, v₂, …, vₖ} is linearly independent if the only solution to:

c₁v₁ + c₂v₂ + … + cₖvₖ = 0

is the trivial solution c₁ = c₂ = … = cₖ = 0.

No nontrivial combination works. Every vector in the set contributes something new — none is a linear combination of the others.

The definition is phrased in terms of the zero vector because it provides a clean, unified test. Instead of asking "is v₁ a combination of the others?" for each vector, we ask a single question: can the zero vector be produced nontrivially? If yes, there is redundancy. If no, the set is independent.

A single nonzero vector is always independent. If v ≠ 0, the only solution to cv = 0 is c = 0. Also, independence is a property of the collection, not of any individual vector. Two vectors are independent if neither is a scalar multiple of the other. Three vectors are independent if none lies in the plane spanned by the other two.

Example 9.6.1

Testing Linear Independence in ℝ³

ProblemDetermine whether v₁ = (1,0,1), v₂ = (0,1,1), v₃ = (1,1,3) are linearly independent.
ThinkSet c₁v₁ + c₂v₂ + c₃v₃ = 0. If the only solution is c₁ = c₂ = c₃ = 0, the set is independent.
CalculateRow reduce the coefficient matrix: [ 1 0 1 ] [ 1 0 1 ] [ 1 0 1 ] [ 0 1 1 ] R₃−R₁→ [ 0 1 1 ] R₃−R₂→ [ 0 1 1 ] [ 1 1 3 ] [ 0 1 2 ] [ 0 0 1 ]
CheckAll three columns have pivots. The system has only the trivial solution.
InterpretThe vectors are linearly independent. No one of them is a linear combination of the other two. They span all of ℝ³ and do not lie in any common plane through the origin.

Answer: Linearly independent. All three pivot columns confirm only the trivial solution.

Example 9.6.2

Detecting Dependence by Inspection

ProblemAre v₁ = (2,1,−1) and v₂ = (4,2,−2) linearly independent?
CalculateNotice v₂ = 2v₁. Therefore 2v₁ − v₂ = 0, a nontrivial dependence relation (coefficients 2 and −1 are not both zero).
InterpretThe set is linearly dependent. The second vector adds nothing to the span. The span of {v₁, v₂} is the same as the span of {v₁} alone: a single line through the origin.

Answer: Linearly dependent. v₂ = 2v₁ is a nontrivial dependence relation.

Section 9.7

Testing Independence with Row Reduction

The Row Reduction Test for Linear Independence

Step 1: Form the matrix A whose columns are v₁, …, vₖ.
Step 2: Row reduce A to echelon form.
Step 3: Count the pivots.

The vectors are linearly independent if and only if every column has a pivot (exactly k pivots). If any column lacks a pivot, the corresponding vector is a linear combination of the others and the set is dependent.

The equation c₁v₁ + … + cₖvₖ = 0 is the same as Ac = 0. The system has only the trivial solution if and only if there are no free variables — which happens if and only if every column of A has a pivot.

Important Size Constraint

If you have more vectors than components — that is, if k > n for vectors in ℝⁿ — then there are more columns than rows, and the system Ac = 0 must have free variables. Therefore: more than n vectors in ℝⁿ are always linearly dependent. You cannot have more independent vectors than the dimension of the space allows.

Example 9.7.1

Row Reduction Test for Independence in ℝ⁴

ProblemTest the linear independence of v₁ = (1,2,0,1), v₂ = (0,1,1,0), v₃ = (1,4,2,1) in ℝ⁴.
Calculate A = [ 1 0 1 ] R₂−2R₁, R₄−R₁ [ 1 0 1 ] [ 2 1 4 ] ──────────────→ [ 0 1 2 ] [ 0 1 2 ] [ 0 1 2 ] [ 1 0 1 ] [ 0 0 0 ] R₃−R₂ [ 1 0 1 ] ─────────→ [ 0 1 2 ] [ 0 0 0 ] [ 0 0 0 ]
CheckOnly two columns have pivots. The third column has no pivot — it is a free variable column.
InterpretThe vectors are linearly dependent. Reading the echelon form: v₃ = v₁ + 2v₂. The three vectors span the same subspace as v₁ and v₂ alone.

Answer: Linearly dependent. v₃ = v₁ + 2v₂. The set has only two independent directions.

Practice 9.5 – 9.7Independence Practice

Determine whether each set is linearly independent or dependent. If dependent, find a dependence relation. (a) {(1,2),(−2,−4)}, (b) {(1,0,1),(0,1,1),(1,1,2)}, (c) {(1,0,0),(0,1,0),(0,0,1)}.
Can four vectors in ℝ³ be linearly independent? Explain your reasoning.
If {v₁, v₂, v₃} is linearly independent, is {v₁, v₂} also independent? Justify.
If {v₁, v₂} is linearly dependent, can {v₁, v₂, v₃} be independent for some v₃? Explain.

Section 9.8

Basis

Definition — Basis

A basis for a vector space V is a set of vectors {b₁, b₂, …, bₙ} that satisfies two conditions simultaneously:

The set spans V: every vector in V is a linear combination of the basis vectors.
The set is linearly independent: the basis vectors contain no redundancy.

A basis is the most efficient possible description of a space — it covers everything, but wastes nothing.

Why Both Conditions Are Required

Spanning alone is not enough. If you have a spanning set that contains redundant vectors, you can remove one and still have a spanning set. A basis has no redundancy to remove.

Independence alone is not enough. A set of independent vectors might not cover the whole space. A basis strikes the perfect balance: large enough to span, small enough to contain no redundancy.

Uniqueness of Representation

One of the most important properties of a basis: every vector in V can be written as a linear combination of basis vectors in exactly one way. This is why a basis functions as a coordinate system. To prove uniqueness: suppose v = c₁b₁ + … + cₙbₙ and also v = d₁b₁ + … + dₙbₙ. Subtracting gives 0 = (c₁−d₁)b₁ + … + (cₙ−dₙ)bₙ. Since the basis vectors are independent, all coefficients must be zero: cᵢ = dᵢ for every i.

Example 9.8.1

Verifying a Basis for ℝ²

ProblemDetermine whether ℬ = {(1,1), (1,−1)} is a basis for ℝ².
Check IndependenceSet c₁(1,1) + c₂(1,−1) = (0,0). System: c₁ + c₂ = 0 and c₁ − c₂ = 0. Adding: 2c₁ = 0 → c₁ = 0, then c₂ = 0. Only the trivial solution. ✓
Check SpanningCan any (a,b) be written as c₁(1,1) + c₂(1,−1)? System: c₁ + c₂ = a, c₁ − c₂ = b. Solution: c₁ = (a+b)/2, c₂ = (a−b)/2. Always defined for any real a, b. ✓
Interpretℬ is a valid basis for ℝ². Notice it is different from the standard basis {e₁, e₂} but equally valid. There are infinitely many bases for any space.

Answer: Yes, {(1,1),(1,−1)} is a basis for ℝ². It is independent and spans ℝ².

Example 9.8.2

Finding a Basis for a Subspace

ProblemFind a basis for the subspace W = span{v₁, v₂, v₃} where v₁ = (1,2,3), v₂ = (2,4,6), v₃ = (0,1,1).
ThinkRow reduce the matrix A = [v₁ v₂ v₃]. The original vectors corresponding to pivot columns form a basis for W.
Calculate [ 1 2 0 ] R₂−2R₁, R₃−3R₁ [ 1 2 0 ] R₃−R₂ [ 1 2 0 ] [ 2 4 1 ] ─────────────────→ [ 0 0 1 ] ────────→ [ 0 0 1 ] [ 3 6 1 ] [ 0 0 1 ] [ 0 0 0 ] Pivots are in columns 1 and 3.
InterpretA basis for W is {v₁, v₃} = {(1,2,3), (0,1,1)}. The vector v₂ = 2v₁ is redundant and is dropped. The subspace W is two-dimensional: a plane through the origin in ℝ³.

Answer: Basis for W = {(1,2,3), (0,1,1)}. W is a plane through the origin, dimension 2.

Section 9.9

The Standard Basis

The standard basis for ℝⁿ is the collection {e₁, e₂, …, eₙ}, where eᵢ has a 1 in position i and 0 in every other position.

Space	Standard Basis	Why It's a Basis
ℝ²	`e₁ = (1,0), e₂ = (0,1)`	Any (a,b) = ae₁ + be₂. The two vectors are independent.
ℝ³	`e₁ = (1,0,0), e₂ = (0,1,0), e₃ = (0,0,1)`	Any (a,b,c) = ae₁ + be₂ + ce₃. Three independent vectors.
ℝⁿ	`e₁, e₂, …, eₙ` (unit coordinate vectors)	n independent vectors spanning ℝⁿ.
Polynomials ≤ degree 2	`{1, x, x²}`	Any a + bx + cx² = a(1) + b(x) + c(x²). Independent.

When you write a vector as (3, −1, 2), you are implicitly using the standard basis. The numbers 3, −1, 2 are coordinates relative to e₁, e₂, e₃. This feels natural because it is the basis we always use by default — but it is still a choice, not a universal truth. Every valid basis is equally legitimate.

Section 9.10

Dimension

The Dimension Theorem

Different bases for the same vector space always have the same number of vectors. The dimension of a vector space V, written dim(V), is the number of vectors in any basis for V. This number does not depend on which basis you choose.

Dimension Is Structure, Not Appearance

Dimension measures how many genuinely independent directions exist in a space — not simply how many coordinates we use to describe it.

Space	Dimension	Reason
`ℝⁿ`	n	Standard basis has n vectors.
{0} (zero space)	0	No nonzero vectors; empty set is its basis.
A line through the origin in ℝ³	1	One independent direction.
A plane through the origin in ℝ³	2	Two independent directions.
Polynomials of degree ≤ n	n+1	Basis: {1, x, x², …, xⁿ}.

Key Dimension Facts

A spanning set with exactly dim(V) vectors is a basis (it must be independent).
An independent set with exactly dim(V) vectors is a basis (it must span).
Any independent set in V can be extended to a basis.
Any spanning set can be reduced to a basis by removing redundant vectors.

Example 9.10.1

Computing the Dimension of a Subspace

ProblemFind the dimension of the solution set of the system: x₁ + 2x₂ − x₃ = 0 and 2x₁ + 4x₂ − 2x₃ = 0.
Calculate [ 1 2 -1 ] R₂−2R₁ [ 1 2 -1 ] [ 2 4 -2 ] ─────────→ [ 0 0 0 ] One pivot. Two free variables: x₂ = s, x₃ = t. Then x₁ = −2s + t.
Result Solution = s(-2, 1, 0) + t(1, 0, 1)
InterpretThe solution set is the span of two independent vectors. The dimension is 2. It is a plane through the origin in ℝ³. This dimension equals the number of free variables, which Chapter 10 will call the nullity of the coefficient matrix.

Answer: Dimension 2. Basis: {(−2,1,0), (1,0,1)}. The solution set is a plane through the origin.

Section 9.11

Coordinates Relative to a Basis

When we use the standard basis, the components of a vector are its coordinates. But every basis produces its own coordinate system. If ℬ = {b₁, b₂, …, bₙ} is a basis for V, and v = c₁b₁ + c₂b₂ + … + cₙbₙ, then the coordinate vector of v relative to ℬ is:

[v]_ℬ = (c₁, c₂, …, cₙ)

The numbers c₁, c₂, …, cₙ are the coordinates of v in the basis ℬ. Different bases are useful for different purposes: in computer graphics, a basis aligned with a character's body frame is more useful than the global world frame; in data science, a basis aligned with directions of greatest variance reveals more information. Understanding that a vector has different coordinates in different bases — but remains the same vector — is a key conceptual step.

Example 9.11.1

Finding Coordinates in a Non-Standard Basis

ProblemLet ℬ = {b₁, b₂} = {(1,1), (1,−1)}. Find the coordinates of v = (4,2) relative to ℬ.
ThinkSolve c₁b₁ + c₂b₂ = v for c₁ and c₂.
Calculate c₁ + c₂ = 4 c₁ - c₂ = 2 Adding: 2c₁ = 6 → c₁ = 3 Subtract: 2c₂ = 2 → c₂ = 1
Check3(1,1) + 1(1,−1) = (3+1, 3−1) = (4, 2) ✓
InterpretIn standard coordinates, the vector is (4, 2). In the ℬ coordinate system, it is [v]_ℬ = (3, 1). The vector itself has not changed — only its description relative to a different reference frame.

Answer: [v]_ℬ = (3, 1). In standard form (4,2); in basis ℬ coordinates (3,1).

Section 9.12

Changing Descriptions of a Space

A given subspace has infinitely many bases. All of them are equally valid; they simply offer different coordinate systems for the same underlying space. For example, the xy-plane in ℝ³ can be described with {e₁, e₂}, or with {(1,1,0), (1,−1,0)}, or with countless other pairs of independent vectors that lie in the plane.

There are also two natural forms for describing a subspace. In parametric (spanning) form, you list vectors whose span equals the subspace — for example, the xz-plane is span{e₁, e₃}. In implicit (equation) form, you write the condition every vector must satisfy — the xz-plane satisfies x₂ = 0. Row reduction lets you move between these descriptions: pivot columns give a spanning set, free variables give a parametric form, and the system structure gives the implicit equations.

Practice 9.8 – 9.12Basis and Dimension Practice

Which of the following sets are bases for ℝ²? (a) {(1,0),(0,1)}, (b) {(2,3),(4,6)}, (c) {(1,1),(2,1)}, (d) {(1,0),(0,1),(1,1)}. Justify each answer.
Find a basis for the subspace of ℝ³ satisfying x₁ − x₂ + 2x₃ = 0. What is the dimension?
Find the coordinates of v = (3,5) relative to the basis ℬ = {(1,2),(−1,1)}.
A subspace W of ℝ⁴ has dimension 2. How many vectors are in any basis for W? How many in any spanning set?
Find a basis for the column space of A = [[1,2,3],[0,1,1],[1,3,4]]. What is the dimension?

Section 9.13

Common Span and Independence Mistakes

Mistake 1 — Confusing span with independence

"Span and independence are two names for the same idea."

Error: Span is a set of vectors (all combinations reachable from a collection). Independence is a property of a set (does it contain redundancy?). A spanning set may be dependent; an independent set may not span the full space. These are separate conditions that together define a basis.

Mistake 2 — Thinking a larger set is always better

"Adding more vectors to a spanning set always improves it."

Error: If the new vectors are already in the span, adding them introduces redundancy without expanding the span. A larger, dependent spanning set is less useful than a smaller basis.

Mistake 3 — Assuming span equals the whole space

"The span of {v₁, v₂} in ℝ³ is all of ℝ³."

Error: The span of {v₁, v₂} is only a subspace — whether it equals all of ℝ³ depends on independence and the number of vectors. Two vectors can span at most a plane in ℝ³, not all of ℝ³.

Mistake 4 — Using reduced pivot columns instead of original ones as a basis

"The pivot columns of the row-reduced matrix form a basis for the column space."

Correct: Use the pivot columns of the ORIGINAL matrix. Row reduction changes the column vectors, so the pivot column positions identify which original columns form the basis — but you must take those vectors from the original matrix, not the row-reduced one.

Mistake 5 — Thinking the standard basis is the only basis

"Coordinates must always be given as standard (x, y, z) components."

Error: Every independent spanning set is a valid basis. The standard basis is simply the most familiar. Countless other bases exist for any vector space, and many applications use non-standard bases deliberately.

Mistake 6 — Confusing independence with orthogonality

"Linearly independent vectors must be perpendicular to each other."

Error: Independent vectors do not need to be perpendicular. For example, (1,0) and (1,1) are independent (neither is a scalar multiple of the other) but not perpendicular. Orthogonality is a stronger, separate condition studied in Chapter 11.

Mistake 7 — Checking dependence only pairwise

"I checked every pair of vectors — no two are multiples of each other, so the set is independent."

Error: Linear dependence is a property of the entire collection, not of pairs. Three vectors can be dependent even if no two of them are scalar multiples of each other. The correct test is the homogeneous system Ac = 0, not pairwise comparison.

Mistake 8 — Concluding that any n vectors in ℝⁿ form a basis

"I have exactly 3 vectors in ℝ³, so they form a basis."

Correct: Exactly n vectors in ℝⁿ form a basis IF AND ONLY IF they are linearly independent. If they are dependent, they do not span ℝⁿ and are not a basis. Count alone is not sufficient — independence must be verified.

Section 9.14

Preparing for Rank and Nullity

What Comes Next

Chapter 10 builds directly on the ideas of this chapter. The column space of a matrix A is the span of its columns. The null space of A is the solution set of Ax = 0 — which, as you have seen, is a subspace. The rank of A is the dimension of its column space (equivalently, the number of pivot columns). The nullity of A is the dimension of its null space (equivalently, the number of free variables).

The Rank-Nullity Theorem states:

rank(A) + nullity(A) = n

where n is the number of columns of A. This equation connects the two fundamental subspaces associated with A and shows that a matrix's columns either contribute to the span or to the null space — and the two dimensions sum to the total column count. It ties together virtually everything from Chapters 3 through 9.

Connections to Earlier Chapters

The pivot columns of A (found in Chapters 3 and 5) are now understood as a basis for the column space. The free variables in the solution of Ax = 0 parametrize the null space; the solution basis vectors form a basis for the null space. The number of pivot positions (the rank) determines whether Ax = b can be solved and whether the solution is unique.

Practice 9.13 – 9.14Error Analysis

Each problem shows a student's error. Identify it and write the correct reasoning.

Student: "The vectors (1,2,3) and (2,4,6) are linearly independent because they have different entries." What is wrong?
Student: "I have three vectors that span ℝ³. Therefore they form a basis for ℝ³." Identify the error. Give a counterexample.
Student: "I row-reduced and found pivot columns 1 and 3 in the reduced form. So the pivot columns of the reduced matrix are my basis." What should the student have done instead?
Student: "v₁, v₂, v₃ are independent in pairs (no two are multiples), so the set of all three is independent." Why is this wrong? Construct a counterexample.
Student: "The span of {(1,0),(0,1),(1,1)} in ℝ² is better than just {(1,0),(0,1)} because it has more vectors." Explain why this reasoning is flawed.

Chapter 9 — Back Matter

Chapter Summary

Sec 9.1A linear combination of vectors v₁, …, vₖ is any expression c₁v₁ + … + cₖvₖ with scalar coefficients. Testing whether b is a linear combination of given vectors is equivalent to solving an augmented linear system by row reduction.
Sec 9.2The span of a set is the collection of all linear combinations — the reachable part of the space. The span of any finite set is always a subspace. If the span equals all of V, the set is a spanning set for V.
Sec 9.3In ℝ² and ℝ³, span takes geometric form: a point, a line, a plane, or the full space, depending on how many independent directions the vectors cover. Adding more vectors does not always expand the span.
Sec 9.4In ℝⁿ, the algebraic test for spanning is row reduction — check for a pivot in every row. To span ℝⁿ, you need at least n vectors, and they must be independent.
Sec 9.5A set is linearly dependent if a nontrivial combination of its vectors gives the zero vector. Dependence means at least one vector is redundant — already reachable from the others. Any set containing the zero vector is dependent.
Sec 9.6A set is linearly independent if the only way to form the zero vector is the trivial combination (all-zero coefficients). Independence means every vector contributes genuinely new directional information.
Sec 9.7The row reduction test: form matrix A with vectors as columns, reduce, check for pivot in every column. More than n vectors in ℝⁿ are always dependent. Independence requires k = n and k pivots.
Sec 9.8A basis is an independent spanning set — covers everything, contains no redundancy. Every vector in the space has a unique representation as a linear combination of basis vectors. Bases are found by row reduction and extracting pivot columns of the original matrix.
Sec 9.9The standard basis {e₁, …, eₙ} for ℝⁿ is the most familiar basis. Using standard components is an implicit choice of basis. Many other valid bases exist, including non-standard coordinate systems used in graphics, robotics, and data science.
Sec 9.10The dimension of V is the number of vectors in any of its bases — a well-defined number independent of which basis is chosen. Dimension measures genuinely independent directions. Key facts: spanning set with dim(V) vectors is a basis; independent set with dim(V) vectors is a basis.
Sec 9.11Coordinates of a vector relative to a basis ℬ are the unique coefficients in the basis expansion. The coordinate vector [v]_ℬ describes the same vector in a different reference frame. Non-standard coordinates are essential in graphics, robotics, and machine learning.
Sec 9.12Any subspace has infinitely many bases. Subspaces can be described parametrically (spanning form) or implicitly (equation form). Row reduction allows movement between these descriptions. Change of basis is an algebraic operation handled by an invertible matrix.
Sec 9.13Eight recurring errors: span ≠ independence; more vectors is not always better; span ≠ whole space; using reduced rather than original pivot columns for a basis; standard basis is not the only basis; independence ≠ orthogonality; pairwise comparison is insufficient; n vectors in ℝⁿ do not automatically form a basis.
Sec 9.14Chapter 10 develops rank (dimension of column space), nullity (dimension of null space), and the Rank-Nullity Theorem rank(A) + nullity(A) = n, which unifies everything from Chapters 3–9. Pivot columns of the original matrix form a basis for the column space; free variables parametrize the null space.

Key Terms Review

linear combination

A sum c₁v₁ + c₂v₂ + … + cₖvₖ formed by scaling vectors and adding the results. The scalars cᵢ are the coefficients.

span

The set of all linear combinations of a collection of vectors. The span is always a subspace of the ambient space.

spanning set

A collection of vectors whose span equals an entire given space V. Every vector in V is reachable from the spanning set.

linearly dependent

A set of vectors for which a nontrivial combination (not all coefficients zero) gives the zero vector. At least one vector is redundant.

linearly independent

A set for which only the trivial combination (all-zero coefficients) gives the zero vector. No vector is a combination of the others.

trivial solution

The all-zero coefficient vector. Every set of vectors trivially produces the zero vector this way.

nontrivial solution

A nonzero coefficient vector satisfying Ac = 0. Its existence means the columns of A are linearly dependent.

dependence relation

An equation c₁v₁ + … + cₖvₖ = 0 with at least one nonzero cᵢ, showing that the set is dependent.

basis

An independent spanning set for a vector space. Every vector in the space has a unique representation as a linear combination of basis vectors.

standard basis

The collection {e₁, e₂, …, eₙ} of unit coordinate vectors for ℝⁿ. Each eᵢ has a 1 in position i and 0 elsewhere.

dimension

The number of vectors in any basis for a space. A well-defined number independent of which basis is chosen.

coordinates

The unique coefficients c₁, …, cₙ when a vector is expressed as a linear combination of basis vectors.

coordinate vector

The ordered list [v]_ℬ = (c₁, …, cₙ) of coordinates of v relative to a basis ℬ.

ordered basis

A basis with a specified ordering of its vectors, used to assign unique coordinate vectors to each element of the space.

Skill Practice

Part ALinear Combinations

Compute 2u − v for u = (3,−1,2), v = (1,4,0).
Is (5,3) a combination of (1,1) and (2,1)?
Is (1,2,3) a combination of (1,0,1) and (0,1,1)?
Write (a,b) as a combination of (2,1) and (1,2).
Find all combinations of (1,2) and (2,4). Describe geometrically.
Is (4,0,4) in span{(1,0,1),(0,1,1)}? Show work.

Part BSpan Tests

Do (1,0,1),(0,1,0),(1,1,1) span ℝ³?
Do (1,2),(2,4) span ℝ²? Why not?
Do (1,0,0),(0,1,0),(1,1,0) span ℝ³?
What is the span of {(0,0)} in ℝ²?
Minimum vectors needed to span ℝ⁵?
Can two vectors span ℝ³? Explain.

Part CIndependence Tests

{(1,3),(2,6)}: independent?
{(1,0),(0,1),(1,1)}: independent?
{(1,0,0),(0,1,0),(0,0,1)}: independent?
{(1,2,3),(4,5,6),(7,8,9)}: independent?
Can 5 vectors in ℝ³ be independent?
If {v₁,v₂} is independent, is {v₁,v₂,v₃} for any v₃?

Part DBasis and Dimension

Is {(1,2),(3,1)} a basis for ℝ²?
Is {(1,0,0),(0,2,0),(0,0,3)} a basis for ℝ³?
Find a basis for span{(1,1,0),(2,2,0),(0,1,1)}.
Dimension of the plane x + y + z = 0 in ℝ³?
Dimension of ℝ⁷?
Find [v]_ℬ for v = (5,1), ℬ = {(1,0),(1,1)}.

Applications

Navigation. A ship records displacements d₁ = (3,1) km (east-north) and d₂ = (−1,2) km on two legs. Write the total displacement as a linear combination. Can all 2D directions be reached by combinations of these two displacements? Explain using independence.
Data redundancy. A dataset records three features for each sample: height in cm, weight in kg, and height in inches. The third feature is always 0.3937 times the first. Explain why these three feature vectors form a linearly dependent set. What is the effective dimension of the data?
Computer graphics. A 2D system uses basis {b₁, b₂}. A character's position is 3b₁ − 2b₂. If the system switches to basis ℬ' = {b₁ + b₂, b₁ − b₂}, find the new coordinates of the character's position.
Resource allocation. A factory uses resource vectors r₁ = (2,1) and r₂ = (1,3) (units of two raw materials per product type). Does span{r₁, r₂} cover all resource combinations in ℝ²? Justify using independence.
Network signals. Three signal vectors in ℝ⁴ represent different transmission patterns. Explain what it would mean for these signals to be linearly dependent, and what consequence this would have for the information content of the transmission system.

Geometric Interpretation

Sketch (or describe) the span of (2,−1) in ℝ². Is this a line, plane, or all of ℝ²?
Explain geometrically why two vectors in ℝ² that both point in the same direction cannot be a basis for ℝ².
Explain why three vectors in ℝ² are always linearly dependent using a geometric argument.
You have two independent vectors in ℝ³. Describe the subspace they span. What is its dimension? What does it look like geometrically?
You have three vectors in ℝ³ that all lie in the same plane through the origin. Can they be independent? Can they span ℝ³? Justify both answers.

Chapter 9 Checkpoint

Complete without assistance — assesses all key skills and concepts from Chapter 9

Part A — Computation

Determine whether b = (2,3,1) is a linear combination of v₁ = (1,1,0) and v₂ = (0,1,1). If yes, find the coefficients. Show all row reduction steps.
Test the linear independence of v₁ = (1,−1,2), v₂ = (2,0,1), v₃ = (0,−2,3) using row reduction. State your conclusion and explain it.
Find a basis for the subspace spanned by the columns of: A = [[1,2,3],[0,1,1],[1,3,4]] What is the dimension of this column space?
Find the coordinates of w = (5,−1) relative to the basis ℬ = {(2,1),(1,−1)}.

Part B — Concepts

A student claims the vectors (1,0,0), (0,1,0), (1,1,0) form a basis for ℝ³. Are they correct? Justify your answer completely — test both required conditions.
Let W be a subspace of ℝ⁵ with dimension 3. What is the minimum number of vectors needed to span W? What is the maximum number of independent vectors you can find in W?
Explain in your own words why a basis for ℝⁿ must contain exactly n vectors — not fewer, not more. What goes wrong if you have n − 1 vectors? What goes wrong if you have n + 1 vectors?

Part C — Application

A data table has three columns: temperature in Celsius, temperature in Fahrenheit, and humidity percentage. The first two columns are related by F = 1.8C + 32. Discuss whether the three columns form a linearly independent set as vectors in ℝᵐ (where m is the number of data rows). What is the effective dimension of this data, and what are the implications for a statistical analysis that treats these three columns as separate features?

Looking Ahead

Chapter 10: Rank, Null Space, and the Fundamental Subspaces

Chapter 9 gave you the structural language of vector spaces: span, independence, basis, dimension, and coordinates. Chapter 10 applies this language to the study of matrices and their associated subspaces — column space, row space, null space, and left null space.

The central result of Chapter 10 is the Rank-Nullity Theorem: rank(A) + nullity(A) = n. This equation ties together every computational skill you have developed — row reduction, pivot columns, free variables, parametric solutions — into a single structural statement about what a matrix preserves and what it collapses. The ideas of span and independence, which felt abstract in Chapter 9, will become concrete tools for understanding systems, data, and transformations.

Chapters 11 through 13 then add orthogonality, eigenvalues, and diagonalization to the picture — all built on the foundation of bases and dimension developed here.

MGU Library Connection

Appendix K: Span, Independence, Basis, and Dimension Reference — this textbook
Appendix E: Row Reduction Reference — practice for the independence test
Appendix R: Calculus Readiness Bridge Guide — how dimension and bases appear in multivariable calculus
Chapter 8: Vector Spaces and Subspaces — prerequisite material on vector space rules
Chapter 10: Rank, Null Space, and the Fundamental Subspaces — direct continuation
Chapter 11: Orthogonality and Projections — basis refinement with perpendicularity
Chapter 12: Eigenvalues and Eigenvectors — special bases (eigenbases) for linear transformations
Data Science Foundations: Principal Components Analysis — non-standard basis aligned with data directions
MGU Dictionary entries: basis, dimension, span, linear independence, coordinate vector

Chapter 9:Span, Linear Independence,and Basis