Summary of "Vector Space | Definition Of Vector Space | Examples Of Vector Space | Linear Algebra"

Topic and context

Introduction to vector spaces as a foundational concept in linear algebra.
Lecturer: Dr. Gajendra Purohit (YouTube channel on Engineering Mathematics & BSc; also posts CSIR NET material).
Prerequisite knowledge: basic group theory (group, ring, field).

Two types of composition involved:
- Internal composition: vector addition — an operation on two vectors producing another vector in the same set.
- External composition: scalar multiplication — an element of the field F times an element of V produces another vector in V.
Formal scalar-multiplication mapping:

V × F → V
A set V together with a field F is a vector space over F only if certain properties are satisfied (both for addition and scalar multiplication).

(V, +) must be an abelian group, i.e., satisfy:

For all a, b ∈ F and u, v ∈ V:

Closure under scalar multiplication: a·v ∈ V.
Distributivity over vector addition: a·(u + v) = a·u + a·v.
Distributivity over scalar addition: (a + b)·v = a·v + b·v.
Compatibility with field multiplication: (ab)·v = a·(b·v).
Identity scalar: 1·v = v, where 1 is the multiplicative identity in the field F.

Examples:
- The n‑tuples over a field F, denoted Vn(F) or F^n, are vector spaces (elementwise addition, zero vector, additive inverses, scalar multiplication all satisfy the axioms).
- Function spaces such as C(R) (continuous real-valued functions) and similar spaces over appropriate fields are vector spaces (note: R is a subfield of C in relevant contexts when needed).
Non-example:
- Q(Z) as described in the lecture is not a vector space when the scalars are taken from Z (the integers). Z is not a field (nonzero integers do not all have multiplicative inverses), so the scalar set must be a field.

To prove V with operations is a vector space over a field F:

Show (V, +) is an abelian group:
- Prove closure of addition.
- Prove associativity.
- Prove commutativity.
- Identify the additive identity (zero vector).
- Show every vector has an additive inverse.
Show closure under scalar multiplication: ∀a ∈ F, v ∈ V, a·v ∈ V.
Verify scalar multiplication axioms:
- a·(u + v) = a·u + a·v (distributivity over vector addition).
- (a + b)·v = a·v + b·v (distributivity over scalar addition).
- (ab)·v = a·(b·v) (compatibility).
- 1·v = v (multiplicative identity acts as scalar 1).
Conclude V is a vector space if all axioms hold.

To show a set is not a vector space:

Find a single axiom that fails. Common tactics:
- Show addition is not closed: pick two elements of V whose sum is not in V.
- Show scalar multiplication is not closed: pick a ∈ F and v ∈ V with a·v not in V.
- Exhibit failure of distributivity or identity property with a concrete counterexample.
The lecturer’s illustrative counterexample: pick two vectors satisfying a defining equation; their sum produced a value (−36 in his example) that did not satisfy the defining equation, so closure under addition fails → not a vector space.

Follow-up videos are planned on subspaces and their properties.
The presentation emphasizes standard exam-style proofs: use elementwise checking for F^n, and use concrete counterexamples to disprove vector-space claims.