📐 Concept diagram

08-01 — Vector Spaces

Phase: 8 — Linear Algebra (Rigorous) Subject: 08-01 Prerequisites: Phase 7 complete; familiarity with vectors in R^n and basic matrix algebra Next subject: 08-02 — Linear Independence and Basis

Learning Objectives

By the end of this subject, you will be able to:

1. State the eight axioms of a vector space over a field and verify whether a given set with operations satisfies them
2. Determine whether a subset is a subspace using the subspace test (closed under addition and scalar multiplication)
3. Express vectors as linear combinations of a given set, and compute the span of a set of vectors
4. Define and compute the null space and column space of a matrix, and relate them to the solutions of Ax = 0 and Ax = b
5. Distinguish between concrete vector spaces (R^n, P_n, M_{m×n}) and abstract vector spaces, recognizing the unifying structure

Core Content

1. Definition: Vector Space

A vector space V over a field F (for us, usually F = R or C) is a set equipped with two operations — addition and scalar multiplication — satisfying eight axioms.

Vector addition: For every u, v ∈ V, there is a unique u + v ∈ V.

Scalar multiplication: For every c ∈ F and v ∈ V, there is a unique cv ∈ V.

CRITICAL — Foundational: The eight vector space axioms are the bedrock of linear algebra. Every theorem — subspaces, linear transformations, dimension — traces back to these. Master these before proceeding.

The eight axioms: For all u, v, w ∈ V and all scalars a, b ∈ F:

1. Associativity of addition: u + (v + w) = (u + v) + w
2. Commutativity of addition: u + v = v + u
3. Existence of additive identity: There exists 0 ∈ V such that v + 0 = v for all v
4. Existence of additive inverses: For each v ∈ V, there exists −v ∈ V such that v + (−v) = 0
5. Compatibility of scalar multiplication with field multiplication: a(bv) = (ab)v
6. Identity element of scalar multiplication: 1·v = v (where 1 is the multiplicative identity in F)
7. Distributivity of scalar multiplication over vector addition: a(u + v) = au + av
8. Distributivity of scalar multiplication over field addition: (a + b)v = av + bv

Examples of vector spaces:

• R^n: n-tuples of real numbers with componentwise addition and scalar multiplication
• P_n: Polynomials of degree ≤ n with usual polynomial addition and scalar multiplication
• M_{m×n}: All m × n matrices with matrix addition and scalar multiplication
• C[a, b]: Continuous functions on [a, b] with pointwise addition and scalar multiplication
• {0}: The zero vector space containing only the zero vector

Non-examples (common pitfalls):

• The set of integers Z over R is NOT a vector space because scalar multiplication by 1/2 does not stay in Z (violates closure)
• The set of polynomials of degree exactly 3 is NOT a vector space because adding two degree-3 polynomials can cancel the leading term, yielding degree ≤ 2 (not closed under addition)
• R^2 with addition defined as (x₁, y₁) + (x₂, y₂) = (x₁ + x₂, 0) fails the existence of additive inverses: what is the inverse of (1, 1)?

Theorem: Basic properties of vector spaces. For any v ∈ V and scalar c: - 0·v = 0 (the zero vector) - c·0 = 0 - (−1)·v = −v - If cv = 0, then c = 0 or v = 0

Proof of 0·v = 0:

0·v = (0+0)·v = 0·v + 0·v     (axiom 8)
Subtract 0·v from both sides: 0 = 0·v.

2. Subspaces

A subspace W of a vector space V is a subset of V that is itself a vector space under the same operations. Rather than checking all eight axioms, we use:

Subspace Test: W ⊆ V is a subspace iff: 1. Contains zero: 0 ∈ W (or equivalently, W is nonempty) 2. Closed under addition: u, v ∈ W ⟹ u + v ∈ W 3. Closed under scalar multiplication: c ∈ F, v ∈ W ⟹ cv ∈ W

Common Pitfall: Always check 0 ∈ W FIRST — if it fails, you're done. The most common mistake is forgetting that subspaces must contain the origin.

Examples of subspaces:

• In R^3: any line or plane through the origin
• In P_n: the set of all even polynomials (only even-degree terms)
• In M_{n×n}: the set of all symmetric matrices
• {0} (the trivial subspace) and V itself are always subspaces

Non-examples: A line or plane NOT through the origin is NOT a subspace — it fails the zero vector test. A quadrant in R^2 (e.g., all (x, y) with x ≥ 0, y ≥ 0) is NOT a subspace — not closed under scalar multiplication by negative scalars.

Intersection of subspaces: If W₁ and W₂ are subspaces, then W₁ ∩ W₂ is a subspace. Proof: 0 ∈ W₁ and 0 ∈ W₂, so 0 ∈ W₁ ∩ W₂. If u, v ∈ W₁ ∩ W₂, then u+v ∈ W₁ and u+v ∈ W₂, so u+v ∈ W₁ ∩ W₂. Closure under scalar multiplication is similar.

But the UNION of subspaces is generally NOT a subspace (unless one contains the other). Consider the x-axis and y-axis in R^2: (1,0) + (0,1) = (1,1) is in neither axis.

3. Linear Combinations and Span

Definition: A linear combination of vectors v₁, v₂, ..., v_n is an expression of the form:

$c₁v₁ + c₂v₂ + ... + c_n v_n
$

where c₁, ..., c_n ∈ F are scalars.

Definition: The span of a set S = {v₁, ..., v_n} is:

span(S) = {c₁v₁ + ... + c_n v_n : c₁, ..., c_n ∈ F}

— the set of ALL linear combinations of vectors in S.

Key facts about span:

• span(S) is always a subspace (the smallest subspace containing S)
• If W is a subspace containing S, then span(S) ⊆ W
• span(∅) = {0} (by convention)
• A vector w ∈ span(S) iff the system c₁v₁ + ... + c_n v_n = w has a solution

Example: In R^3, let v₁ = (1, 0, 0) and v₂ = (0, 1, 0). Then span{v₁, v₂} = {(x, y, 0) : x, y ∈ R} — the xy-plane through the origin.

Example: Does (2, 6, −4) ∈ span{(1, 2, −1), (0, 1, 1)}? Solve:

$c₁(1,2,-1) + c₂(0,1,1) = (2,6,-4)
$

This gives the system:

$c₁ = 2
2c₁ + c₂ = 6  →  4 + c₂ = 6  →  c₂ = 2
-c₁ + c₂ = -4  →  -2 + 2 = 0 ≠ -4
$

Inconsistent! So (2, 6, −4) ∉ span{(1,2,−1), (0,1,1)}.

4. Null Space and Column Space

For an m × n matrix A over R:

Null space (kernel): N(A) = {x ∈ R^n : Ax = 0}

The null space is the set of all solutions to the homogeneous equation. It is a subspace of R^n.

Column space (range/image): C(A) = {Ax : x ∈ R^n} = span{columns of A}

The column space is a subspace of R^m. It tells us which vectors b make Ax = b consistent.

Key relationship: The system Ax = b has a solution iff b ∈ C(A).

Finding N(A) by Gaussian elimination:

A = [1  2  1]
    [2  4  2]

Reduce: [1  2  1] → x₁ + 2x₂ + x₃ = 0
        [0  0  0]

Free variables: x₂ = s, x₃ = t
Then x₁ = -2s - t

N(A) = {(-2s-t, s, t) : s, t ∈ R}
      = span{(-2, 1, 0), (-1, 0, 1)}

Dimension relationships (preview of rank-nullity): dim(N(A)) = number of free variables = n − rank(A) dim(C(A)) = rank(A) = number of pivot columns

Common misconception: N(A) is a subspace of the DOMAIN R^n, while C(A) is a subspace of the CODOMAIN R^m. They live in different spaces!

Key Terms

• 08 01 Vector Spaces
• Correct: A) R^3
• Correct: B) 1
• Correct: B) R^n
• Correct: B) The xy-plane
• Correct: B) {p ∈ P₂ : p′(0) = 0}
• Correct: C) 4
• Correct: C) The set of all integers Z
• Correct: D) Both B and C
• Definition: Vector Space
• End-of-Subject Quiz
• Example 1: Verifying a Vector Space

Worked Examples

Example 1: Verifying a Vector Space

Problem: Prove that V = {(x, y, z) ∈ R^3 : x + y + z = 0} is a vector space under the usual operations on R^3.

Solution: Since V is a subset of the known vector space R^3, we just need the subspace test.

Step 1 — Contains zero: 0 + 0 + 0 = 0, so (0, 0, 0) ∈ V. ✓

Step 2 — Closed under addition: Let u = (x₁, y₁, z₁), v = (x₂, y₂, z₂) ∈ V. Then x₁ + y₁ + z₁ = 0 and x₂ + y₂ + z₂ = 0. u + v = (x₁+x₂, y₁+y₂, z₁+z₂). Sum of components: (x₁+x₂) + (y₁+y₂) + (z₁+z₂) = (x₁+y₁+z₁) + (x₂+y₂+z₂) = 0 + 0 = 0. ✓

Step 3 — Closed under scalar multiplication: Let c ∈ R, v ∈ V. cv = (cx, cy, cz). Sum: cx + cy + cz = c(x + y + z) = c·0 = 0. ✓

All three conditions satisfied, so V is a subspace (and hence a vector space). Geometrically, this is the plane through the origin with normal vector (1, 1, 1).

Example 2: Span and Linear Combinations

Problem: Determine whether the polynomial p(x) = 2x² − 3x + 1 belongs to span{1, x, x² − x} in P₂.

Solution: We need scalars a, b, c such that:

$a(1) + b(x) + c(x² − x) = 2x² − 3x + 1
$

Equating coefficients of like powers:

$x²: c = 2
x¹: b − c = −3  →  b − 2 = −3  →  b = −1
x⁰: a = 1
$

We found a = 1, b = −1, c = 2. So:

$1·(1) + (−1)·(x) + 2·(x² − x) = 1 − x + 2x² − 2x = 2x² − 3x + 1 ✓
$

Yes, p(x) ∈ span{1, x, x² − x}.

Example 3: Null Space Computation

Problem: Find the null space of A = [1 2 3; 2 4 6; 1 1 1].

Solution (step-by-step):

Step 1 — Row reduce A:

$[1  2  3]    R₂-2R₁    [1  2  3]
[2  4  6]  ---------->  [0  0  0]
[1  1  1]    R₃-R₁     [0 -1 -2]
$

Step 2 — Swap R₂↔R₃ and eliminate:

$[1  2  3]    R₁+2R₂    [1  0 -1]
[0 -1 -2]  ---------->  [0  1  2]
[0  0  0]               [0  0  0]
$

Step 3 — Read from RREF: x₁ − x₃ = 0 → x₁ = x₃ x₂ + 2x₃ = 0 → x₂ = −2x₃ x₃ = free = t

So x = (t, −2t, t) = t(1, −2, 1).

N(A) = span{(1, −2, 1)}. It's a 1-dimensional subspace of R^3 — a line through the origin.

Step 4 — Verify: A(1, −2, 1)^T = (1−4+3, 2−8+6, 1−2+1)^T = (0, 0, 0)^T. ✓

Example 4: Column Space and Consistency

Problem: Does Ax = b where A = [1 2; 3 6] and b = (3, 9)^T have a solution?

Solution: We check whether b ∈ C(A).

The columns of A are v₁ = (1, 3)^T and v₂ = (2, 6)^T. Note v₂ = 2v₁, so C(A) = span{(1, 3)^T}.

We need b = c·(1, 3)^T for some c: (3, 9)^T = 3·(1, 3)^T. Yes! So b ∈ C(A), and the system is consistent.

Indeed: x₁ + 2x₂ = 3, 3x₁ + 6x₂ = 9. The second is 3× the first, so one equation. Solutions: x₁ = 3 − 2t, x₂ = t (infinitely many).

Edge case: If b = (3, 10)^T, then b is not a scalar multiple of (1, 3)^T, so b ∉ C(A), and the system is inconsistent.

Quiz

Q1: Which of the following is NOT one of the eight vector space axioms?

A) u + v = v + u (commutativity of addition) B) a(u + v) = au + av (distributivity) C) u · v = v · u (commutativity of dot product) D) There exists 0 ∈ V such that v + 0 = v

Correct: C)

• If you chose C: The dot product is not part of the vector space axioms — it's an additional structure (inner product) not required for a vector space. Correct!
• If you chose A: Commutativity of addition IS one of the eight axioms.
• If you chose B: Distributivity of scalar multiplication over vector addition IS an axiom.
• If you chose D: Existence of additive identity IS an axiom (axiom 3).

Q2: To verify that a subset W ⊆ V is a subspace, you must check:

A) W is nonempty, and W is closed under addition B) W contains the zero vector, is closed under addition, and is closed under scalar multiplication C) W satisfies all eight vector space axioms D) W is closed under scalar multiplication only

Correct: B)

• If you chose B: The subspace test requires: 0 ∈ W, and for any u, v ∈ W and scalar c, u + v ∈ W and cv ∈ W. Correct!
• If you chose A: Missing closure under scalar multiplication.
• If you chose C: If W is a subset and passes the subspace test, the eight axioms are inherited from V.
• If you chose D: Both closure properties are required.

Q3: The span of vectors {v₁, ..., v_k} is:

A) The set of all scalar multiples of v₁ B) The set of all linear combinations c₁v₁ + ... + c_k v_k C) The set of all vectors perpendicular to each v_i D) The dot product of all v_i

Correct: B)

• If you chose B: The span is exactly the set of all linear combinations of the given vectors. Correct!
• If you chose A: That's only the span of {v₁}, not the full set.
• If you chose C: That describes the orthogonal complement, not the span.
• If you chose D: The dot product of multiple vectors is not defined this way.

Q5: The null space of a matrix A is:

A) The set of all vectors b such that Ax = b has a solution B) The set of all vectors x such that Ax = 0 C) The set of all columns of A D) The set of all rows of A

Correct: B)

• If you chose B: Null(A) = {x : Ax = 0} — it's the solution set of the homogeneous equation. Correct!
• If you chose A: That describes the column space (range) — vectors b for which Ax = b is consistent.
• If you chose C: The set of columns spans the column space, not the null space.
• If you chose D: The row space is the span of the rows, distinct from the null space.

Q6: Which set is NOT a vector space (over R)?

A) All polynomials of degree exactly 3 B) All 2 × 2 matrices with real entries C) All continuous functions on [0, 1] D) R^n with componentwise operations

Correct: A)

• If you chose A: The sum of two degree-3 polynomials could have degree < 3 (cancellation of leading terms), violating closure. Polynomials of degree ≤ 3 form a vector space. Correct!
• If you chose B: M_{2×2} is a vector space.
• If you chose C: C[0, 1] is a vector space under pointwise operations.
• If you chose D: R^n is the canonical vector space.

Q4: If vectors v₁, ..., v_k span a vector space V, then:

A) Every vector in V can be written as a linear combination of v₁, ..., v_k B) v₁, ..., v_k are linearly independent C) k is the dimension of V D) The vectors are orthogonal

Correct: A)

• If you chose A: Spanning means every vector is a linear combination of the spanning set. Correct!
• If you chose B: Spanning sets can be linearly dependent (e.g., {v, 2v} still spans the same line).
• If you chose C: k could be larger than the dimension if the set is dependent.
• If you chose D: Spanning sets aren't required to be orthogonal.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Determine whether W = {(x, y) ∈ R^2 : x ≥ 0} is a subspace of R^2.

Problem 2: Prove that the set of all 2×2 symmetric matrices is a subspace of M_{2×2}.

Problem 3: In R^3, does v = (4, −1, 8) belong to span{(1, 2, −1), (2, −1, 3)}?

Problem 4: Find N(A) and C(A) for A = [1 1 0; 2 1 1; 3 2 1].

Problem 5: Show that span{u, v} = span{u+v, u−v} when u, v are vectors in a vector space over F where 2 ≠ 0.

Problem 6: Determine whether the set of all polynomials p in P₃ satisfying p(1) = 0 is a subspace of P₃.

Problem 7: Express (7, 7, 7) as a linear combination of (1, 2, 3), (3, 2, 1), and (1, 1, 1).

$c₁ + 2c₂ = 4
2c₁ − c₂ = −1
−c₁ + 3c₂ = 8
$

$[1  1  0]    R₂-2R₁    [1  1  0]    R₂/(-1)    [1  1  0]
[2  1  1]  ---------->  [0 -1  1]  ---------->  [0  1 -1]  R₁-R₂  [1  0  1]
[3  2  1]    R₃-3R₁    [0 -1  1]                [0  0  0]          [0  0  0]
$

$a + 3b + c = 7
2a + 2b + c = 7
3a + b + c = 7
$

Summary

1. A vector space is defined by eight axioms; verifying a set is a vector space means checking each one (unless the set is a subset of a known vector space, where the 3-condition subspace test suffices)
2. The subspace test (contains zero, closed under addition, closed under scalar multiplication) is the primary tool for identifying subspaces — a line through the origin is a subspace, a line not through the origin is not
3. The span of a set of vectors is the vector space of ALL linear combinations; determining membership in span reduces to solving a linear system
4. The null space N(A) = {x : Ax = 0} is a subspace of the domain R^n; the column space C(A) = {Ax : x ∈ R^n} is a subspace of the codomain R^m — they exist in different universes
5. Ax = b is consistent iff b ∈ C(A); the general solution is x_particular + N(A)

Pitfalls

1. Forgetting to check for closure before verifying the other axioms. A set that doesn't contain the zero vector or isn't closed under addition/scalar multiplication is immediately not a vector space. Always do the subspace test (zero, closed under addition, closed under scalar multiplication) before checking the eight axioms individually.

2. Confusing the domain and codomain when identifying N(A) and C(A). The null space N(A) is a subspace of R^n (the domain, where x lives). The column space C(A) is a subspace of R^m (the codomain, where Ax lives). They exist in different spaces with possibly different dimensions.

3. Assuming a set is a vector space just because it's a subset of one. Subsets of vector spaces are not automatically subspaces. A line in R^2 is a subspace only if it passes through the origin. The first quadrant of R^2 is not a subspace because it fails closure under scalar multiplication by negative numbers.

4. Misapplying the span test. Determining whether a vector w belongs to span{v₁, …, v_n} requires solving a linear system. A common error is to check only one or two equations instead of all of them. The system must be consistent for EVERY component, not just some.

5. Confusing "span" with "basis." The span of a set of vectors is the subspace of all linear combinations. A basis is a linearly independent spanning set. An arbitrary spanning set may be redundant (dependent); converting a spanning set to a basis requires removing dependent vectors.

Next Steps

Move on to 08-02 — Linear Independence and Basis to learn how to identify when vectors are independent, how to construct bases — minimal spanning sets — and how to define the dimension of a vector space, including coordinate representations and change-of-basis transformations.