Vector Spaces: Definition & Properties

🎯 Learning Objectives

• State the eight vector space axioms
• Determine whether a given set with operations is a vector space
• Prove simple properties from the axioms
• Recognise $\mathbb{R}^n$, $\mathbb{R}^{m \times n}$, and polynomial spaces as vector spaces
• Identify when a set fails to be a vector space

📋 Prerequisites

• Vectors Introduction (Week 1) — vector operations in $\mathbb{R}^n$
• Matrices Introduction (Week 1) — matrix algebra
• Basic set notation — set membership $\in$, subset $\subseteq$, set builder $\{x \mid \text{condition}\}$
• Vector spaces are the universe in which all of linear algebra takes place

1. Intuition: What Makes a "Space"?

1.1 The Big Idea

Can we add two things of this type? Can we multiply one by a scalar? If the answer is yes and the operations behave nicely (commute, associate, distribute), then the set is a vector space. The eight axioms below capture this "nice behaviour" precisely. They are the rules of the game for linear algebra. (Diagram)

1.2 Why Axioms?

2. The Eight Axioms

Definition (Vector Space). A vector space $V$ over the real numbers $\mathbb{R}$ is a set equipped with two operations:

• Addition: $+ : V \times V \to V$ (maps two vectors to a vector)
• Scalar multiplication: $\cdot : \mathbb{R} \times V \to V$ (maps a scalar and a vector to a vector)

satisfying the following eight axioms for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and all $a, b \in \mathbb{R}$:

Note: Some textbooks list 10 axioms (counting closure separately). Either way, all 10 rules must hold.

3. Examples of Vector Spaces

3.1 $\mathbb{R}^n$ — Euclidean Space

3.2 $\mathbb{R}^{m \times n}$ — Matrices

3.3 $P_n$ — Polynomials of Degree $\leq n$

Example 1: Polynomial vector space operations

Let $p(x) = 2 + 3x - x^2$, $q(x) = 1 - 2x + 4x^2$ in $P_2$.

$p + q = (2+1) + (3-2)x + (-1+4)x^2 = 3 + x + 3x^2$

$-2p = -4 - 6x + 2x^2$

3.4 $C[a,b]$ — Continuous Functions

3.5 $\mathbb{R}^\infty$ — Sequences

4. Non-Examples (Failure Cases)

4.1 $\mathbb{Z}^n$ — Integer Vectors

4.2 First Quadrant in $\mathbb{R}^2$

4.3 Unit Circle

5. Simple Consequences of the Axioms

Proof: $0 \cdot \mathbf{v} = \mathbf{0}$

Now subtract $0 \cdot \mathbf{v}$ from both sides (axiom 5 says we can):

Thus $0 \cdot \mathbf{v} = \mathbf{0}$. ∎

6. Edge Cases & Gotchas

7. Common Pitfalls

❌ Pitfall 1: Checking only closure

❌ Pitfall 2: Confusing the zero vector with the number 0

❌ Pitfall 3: Thinking any subset of $\mathbb{R}^n$ is a vector space

8. Formula Summary Table

9. 📝 Practice Questions

Q1: Verifying axioms for ℝ²

Verify axiom 9 ($a(\mathbf{u}+\mathbf{v}) = a\mathbf{u} + a\mathbf{v}$) for $\mathbf{u} = (1,2)$, $\mathbf{v} = (3,4)$, $a = 5$ in $\mathbb{R}^2$.

Solution: LHS: $a(\mathbf{u}+\mathbf{v}) = 5((1,2)+(3,4)) = 5(4,6) = (20,30)$ RHS: $a\mathbf{u} + a\mathbf{v} = 5(1,2) + 5(3,4) = (5,10) + (15,20) = (20,30)$ LHS = RHS ✓ Q2: Is this a vector space?

Is $V = \{(x,y) \in \mathbb{R}^2 \mid x = y\}$ a vector space with standard operations?

Solution: Yes. Check:

• Closure under addition: $(a,a)+(b,b) = (a+b, a+b)$ ✓ (first = second)
• Closure under scalar multiplication: $c(a,a) = (ca, ca)$ ✓
• Zero: $(0,0)$ is in $V$ ✓
• Negatives: $-(a,a) = (-a,-a)$ is in $V$ ✓
• Other axioms inherited from $\mathbb{R}^2$

So $V$ is a vector space (it's the line $y = x$). Q3: Is this a vector space?

Is $V = \{(x,y) \in \mathbb{R}^2 \mid xy = 0\}$ a vector space?

Solution: No. Check closure under addition: $(1,0) \in V$, $(0,1) \in V$, but $(1,0)+(0,1) = (1,1)$ is not in $V$ because $1\cdot1 = 1 \neq 0$. Closure fails, so $V$ is not a vector space. Q4: Polynomial vector space

In $P_2$, compute $2p - 3q$ where $p(x) = x^2 - 2x + 1$, $q(x) = 3x + 2$.

Solution: $2p = 2x^2 - 4x + 2$ $3q = 9x + 6$ $2p - 3q = 2x^2 + (-4-9)x + (2-6) = 2x^2 - 13x - 4$ Q5: Zero vector in matrix space

What is the zero vector in $\mathbb{R}^{2 \times 3}$?

Solution: The $2 \times 3$ matrix of all zeros:

. Q6: Proof from axioms

Prove that $c \cdot \mathbf{0} = \mathbf{0}$ for any scalar $c$.

Solution: $c \cdot \mathbf{0} = c \cdot (\mathbf{0} + \mathbf{0})$ (axiom 4) $= c \cdot \mathbf{0} + c \cdot \mathbf{0}$ (axiom 9) Subtract $c \cdot \mathbf{0}$ from both sides: $c \cdot \mathbf{0} - c \cdot \mathbf{0} = (c \cdot \mathbf{0} + c \cdot \mathbf{0}) - c \cdot \mathbf{0}$ $\mathbf{0} = c \cdot \mathbf{0} + (c \cdot \mathbf{0} - c \cdot \mathbf{0}) = c \cdot \mathbf{0} + \mathbf{0} = c \cdot \mathbf{0}$ ✓ Q7: Identifying axiom violations

Consider $V = \mathbb{R}^2$ with modified addition: $(a,b) \oplus (c,d) = (a+c+1, b+d)$. Is $V$ a vector space? If not, which axiom fails?

Solution: Let's check the zero vector. We need $\mathbf{0}$ such that $(a,b) \oplus \mathbf{0} = (a,b)$. If $\mathbf{0} = (z_1, z_2)$, then $(a+z_1+1, b+z_2) = (a,b)$. So $a+z_1+1 = a \Rightarrow z_1 = -1$, $b+z_2 = b \Rightarrow z_2 = 0$. So $\mathbf{0} = (-1, 0)$.

But now axiom 1 (closure) and others might work... Let's check commutativity: $(a,b) \oplus (c,d) = (a+c+1, b+d) = (c+a+1, d+b) = (c,d) \oplus (a,b)$ ✓.

What about axiom 4 with $\mathbf{0} = (-1,0)$? $(a,b) \oplus (-1,0) = (a-1+1, b+0) = (a,b)$ ✓.

But axiom 2 (usual numbering) corresponds to existence of negatives: for $(a,b)$, we need $(-a-2, -b)$ because $(a,b) \oplus (-a-2, -b) = (a+(-a-2)+1, b+(-b)) = (-1, 0) = \mathbf{0}$ ✓.

So this actually works! But wait — is scalar multiplication standard? If scalar multiplication is also non-standard, we need to check.

If scalar multiplication is standard $c(a,b) = (ca, cb)$, then check axiom 9: $c((a,b) \oplus (d,e)) = c(a+d+1, b+e) = (c(a+d+1), c(b+e)) = (ca+cd+c, cb+ce)$ $c(a,b) \oplus c(d,e) = (ca, cb) \oplus (cd, ce) = (ca+cd+1, cb+ce)$

These are equal only if $c = 1$! So the distributive axiom fails for $c \neq 1$.

Thus $V$ is not a vector space. Q8: Trivial vector space

Show that $V = \{\mathbf{0}\}$ with $\mathbf{0} + \mathbf{0} = \mathbf{0}$ and $c\mathbf{0} = \mathbf{0}$ is a vector space.

Solution: All axioms hold trivially because every operation returns the only element $\mathbf{0}$. Q9: Function space

Is $V = \{f: \mathbb{R} \to \mathbb{R} \mid f(0) = 0\}$ a vector space?

Solution: Yes.

• Closure under addition: $(f+g)(0) = f(0) + g(0) = 0 + 0 = 0$ ✓
• Closure under scalar multiplication: $(cf)(0) = c \cdot f(0) = c \cdot 0 = 0$ ✓
• Zero function $z(x) = 0$ satisfies $z(0) = 0$ ✓
• Negative: $(-f)(0) = -f(0) = 0$ ✓ Q10: Non-example with wrong scalar field

Explain why $V = \{(x,y) \mid x,y \in \mathbb{Q}\}$ (vectors with rational entries) is not a vector space over $\mathbb{R}$.

Solution: Scalar multiplication by an irrational $c$ (e.g., $\sqrt{2}$) gives $\sqrt{2}(1,0) = (\sqrt{2}, 0)$, and $\sqrt{2}$ is not rational. So closure under scalar multiplication fails when the scalars are $\mathbb{R}$.

(Note: $V$ is a vector space over $\mathbb{Q}$, but the course uses $\mathbb{R}$ as scalars.)

🔗 Cross-References

• Next topic: Subspaces — smaller vector spaces inside larger ones
• Week 4 (Basis & Dimension): Every vector space has a basis
• Week 5 (Linear Transformations): Maps between vector spaces
• BSMA1001 (Maths 1): Functions, sets, and basic algebra
• BSCS2004 (ML Foundations): Feature spaces are vector spaces Join Discord Previous2.2 Cramer's Rule & InversesNext3.2 Subspaces