MTH2021 · Linear Algebra with Applications
Vector Spaces & Subspaces
Weeks 3–4 make the leap from computation to structure: a vector space is a set with addition and scalar multiplication satisfying ten axioms, and a subspace is a nonempty subset closed under both operations. The examinable skill is the subspace test — proving a set is (or, by counterexample, is not) a subspace — across ℝⁿ, matrix spaces, polynomial spaces Pₙ and function spaces. These proofs sit in Test 2 (Weeks 3–7) and are prime handwritten-section material in the final.
What this chapter covers
- 01The ten vector-space axioms (closure, commutativity, associativity, zero, negatives, scalar-multiplication laws)
- 02Consequences: 0 is unique, each v has a unique −v, 0·u = 0, k·0 = 0, (−1)u = −u, ku = 0 ⇒ k = 0 or u = 0
- 03Standard examples: {0}, ℝⁿ, ℝ^∞, matrix space ℝ^{m×n}, polynomials Pₙ and P∞, function space C[0,1]
- 04The subspace test: nonempty W ⊆ V is a subspace ⇔ closed under addition and under scalar multiplication
- 05A subspace must contain the zero vector (a fast disqualifier)
- 06Subspace examples: lines/planes through the origin, symmetric matrices in ℝ^{2×2}, Pₙ ⊂ P∞
- 07Non-examples: a half-plane {x ≥ 0}, an affine plane 3x + y + z = 5 (fails to contain 0)
- 08Proving closure directly from an arbitrary element and an arbitrary scalar
Subspace test: a plane through the origin vs an affine plane
- +1(a) Nonempty / contains 0: the zero vector (0,0,0) satisfies 0 + 2·0 − 0 = 0, so 0 ∈ W and W is nonempty.
- +1(a) Closed under addition: take u = (x₁,y₁,z₁), v = (x₂,y₂,z₂) ∈ W, so x₁+2y₁−z₁ = 0 and x₂+2y₂−z₂ = 0. Then u + v = (x₁+x₂, y₁+y₂, z₁+z₂) and (x₁+x₂) + 2(y₁+y₂) − (z₁+z₂) = 0 + 0 = 0, so u + v ∈ W.
- +1(a) Closed under scalar multiplication: for any scalar k, ku = (kx₁, ky₁, kz₁) gives kx₁ + 2ky₁ − kz₁ = k(x₁+2y₁−z₁) = k·0 = 0, so ku ∈ W. By the subspace test W is a subspace (it is a plane through the origin, dim 2).
- +1(b) Contains 0? Test the origin in U: 0 + 2·0 − 0 = 0 ≠ 1, so (0,0,0) ∉ U. A subspace must contain the zero vector, so U fails immediately.
- +1(b) Confirm closure fails: (1,0,0) ∈ U and (0,0,−1) ∈ U (both satisfy x+2y−z = 1), but their sum (1,0,−1) gives 1 + 0 − (−1) = 2 ≠ 1, so U is not closed under addition. U is an affine plane, not a subspace.
Key terms
- Vector space
- A set V with vector addition and scalar multiplication satisfying ten axioms (closure, commutativity, associativity, an additive identity 0 and inverses, and four scalar-multiplication laws).
- Subspace
- A nonempty subset W of a vector space V that is itself a vector space under the inherited operations; equivalently, W is closed under addition and scalar multiplication.
- Subspace test
- The shortcut: a nonempty W ⊆ V is a subspace iff for all u, v ∈ W and scalars k, both u + v ∈ W and ku ∈ W. Checking that 0 ∈ W confirms nonemptiness.
- Zero-vector criterion
- Every subspace contains the zero vector; if 0 ∉ W then W is not a subspace — a one-line disqualifier for affine sets like 3x + y + z = 5.
- Polynomial space Pₙ
- The vector space of polynomials of degree at most n; it has dimension n+1 with standard basis {1, x, …, xⁿ}.
- Function space
- A vector space whose vectors are functions, e.g. C[0,1] (continuous functions on [0,1]) with pointwise addition and scalar multiplication.
Vector Spaces & Subspaces FAQ
What exactly do I check in the subspace test?
Three things: the set is nonempty (show 0 is in it), it is closed under addition (u, v in W ⇒ u + v in W), and it is closed under scalar multiplication (u in W, scalar k ⇒ ku in W). If all three hold, W is a subspace. If any fails, it is not — and failing the 0 check is the quickest disqualifier.
Why does a plane through the origin count but an off-origin plane does not?
A subspace must be closed under scalar multiplication, so it must contain 0 = 0·v. A plane like x + 2y − z = 0 passes through the origin and is closed under both operations; a parallel plane x + 2y − z = 1 misses the origin and is not closed under addition, so it is only an affine set, not a subspace.
Are polynomials and matrices really 'vectors'?
Yes. A vector space is any set satisfying the ten axioms, so Pₙ (polynomials of degree ≤ n) and ℝ^{m×n} (matrices) are genuine vector spaces, with the symmetric matrices and the even/odd polynomials as subspaces. This abstraction is exactly what lets the same rank, basis and dimension machinery apply everywhere.
How do I disprove that a set is a subspace?
One explicit counterexample suffices. Either show 0 is not in the set, or produce two elements whose sum leaves the set, or an element and a scalar whose product leaves it. A single concrete failure beats any amount of hand-waving, and it is what the marking scheme wants.
Where is this assessed?
Vector-space and subspace proofs fall in Test 2 (Weeks 3–7) and are a favourite of the final's handwritten section (for example, proving U ∩ W is a subspace but U ∪ W generally is not). Practise writing the three-part test cleanly, because presentation earns method marks.
Exam move
Learn the subspace test as a fixed three-line ritual — 0 ∈ W, closure under +, closure under scalar · — and always test the zero vector first. Build a stock of examples and non-examples across ℝⁿ, matrix spaces and Pₙ so you can produce a counterexample instantly, and rehearse the classic U ∩ W (subspace) vs U ∪ W (not a subspace) argument since it is exam-favoured. This material sits in Test 2 (Weeks 3–7) and the proof-heavy final. When a proof feels shaky, ask Sia to review your closure argument step by step.
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