Summary of "Vector Subspace | Subspace Theorems & Examples | Linear Algebra"

Summary — main ideas and lessons

Definition

A subset W of a vector space V over a field F is a subspace iff W itself is a vector space under the same operations (vector addition and scalar multiplication). Equivalently, W must satisfy the vector-space axioms when using the operations inherited from V.

Key theorem (necessary and sufficient condition)

Single combined condition:
- A nonempty subset W ⊆ V is a subspace of V iff for all α, β ∈ W and all scalars a, b ∈ F, the linear combination aα + bβ ∈ W.
Equivalent two-condition form:
1. α + β ∈ W for all α, β ∈ W (closed under addition).
2. aα ∈ W for all a ∈ F and α ∈ W (closed under scalar multiplication).
Note: In proofs, show both directions — necessity (if W is a subspace then closure of combinations holds) and sufficiency (if the closure condition holds then W satisfies all vector-space axioms).

Proof ideas (high level)

Necessity:
- If W is already a subspace, closure under addition and scalar multiplication is immediate, so aα + bβ ∈ W for any scalars a, b and vectors α, β ∈ W.
Sufficiency (starting from the closure property):
- Show W is nonempty and contains the zero vector (choose scalars appropriately).
- Show additive inverses exist (use scalar −1).
- Show closure under addition (take a = b = 1).
- Conclude W is an abelian group under addition; associativity and commutativity are inherited from V.
- Use closure under scalar multiplication and distributive properties inherited from V to verify the remaining vector-space axioms.
- Therefore W is a vector space (subspace of V).

Methodology — step-by-step checklist to prove a subset W is a subspace

Preferred compact test (single-step):
- Verify that for all α, β ∈ W and all scalars a, b ∈ F, aα + bβ ∈ W. If true and W is nonempty, this suffices.
Equivalent two-step test:
1. Show W is closed under vector addition: for all α, β ∈ W, α + β ∈ W.
2. Show W is closed under scalar multiplication: for all a ∈ F and α ∈ W, aα ∈ W. - If both hold and W is nonempty, W is a subspace.
Practical proof steps when using either test:
- Check nonemptiness (or show 0 ∈ W).
- Use a = b = 1 to get closure under addition.
- Use a = 0 or a = −1 to demonstrate zero in W and additive inverses.
- Use a general scalar a to get closure under scalar multiplication.
- Remember: associativity, commutativity and distributive laws are inherited from V, so it suffices to show the group and closure properties plus scalar-closure to conclude W is a vector space.

Examples and short justifications

Symmetric n×n matrices (subset of all n×n matrices over F):
- If A and B are symmetric (A^T = A, B^T = B) then (A + B)^T = A^T + B^T = A + B, so A + B is symmetric.
- For scalar c, (cA)^T = c A^T = cA, so scalar multiples are symmetric.
- Therefore symmetric n×n matrices form a subspace.
Solution set of a homogeneous linear differential equation (subset of continuous real functions C(R)):
- If y1 and y2 solve L[y] = 0, linearity gives L[a y1 + b y2] = a L[y1] + b L[y2] = 0.
- Hence any linear combination of solutions is a solution, so the solution set is a subspace.
W = {(a, b, 0) : a, b ∈ F} as a subset of F^3:
- Sum: (a1, b1, 0) + (a2, b2, 0) = (a1+a2, b1+b2, 0) ∈ W.
- Scalar multiple: c(a, b, 0) = (ca, cb, 0) ∈ W.
- Hence W is a subspace (or verify using the combined linear-combination test).

Additional remarks / next topics

Both the single linear-combination condition and the two-condition version are interchangeable; use whichever is more convenient.
Upcoming material: algebra of subspaces (intersection and union) and more examples.
Background in group theory (groups, rings, fields) aids understanding of vector-space properties.

Speakers / sources