Let V be a vector space over F≡R or C and let S be a subset of V.

1. Linear combination:

Let x₁, x₂, …, x_k ∈ V and α₁, α₂, ..., α_k ∈ F. Then α₁x₁+ α₂x₂+…+ α_kx_k is called a linear combination of x₁, x₂, …, x_k.

2. Examples:

Consider the vector spaces R² over R

Then some linear combinations of (3, 0), (1, 4)∈R² are given below:
Case (i) α₁=2∈R, α₂=1∈R
α₁x₁+α₂x₂= 2(3, 0) +1(1, 4) =(6, 0)+(1, 4)=(7, 4)

Case (ii) α₁=1∈R, α₂=0∈R
α₁x₁+α₂x₂= 1(3, 0) + 0(1, 4) = (3, 0)+(0, 0) = (3, 0)

Case (iii) α₁=0∈R, α₂=1∈R
α₁x₁+α₂x₂= 0(3, 0)+1(1, 4)=(0, 0)+(1, 4)=(1, 4)

Case (iv) α₁=-2∈R, α₂=√2∈R
α₁x₁+α₂x₂= -2(3, 0)+ √2(1, 4)=(-6, 0)+(√2, 4√2)=(√2-6, 4√2)

3. Example:

Consider the vector space (P₂ (x), +, .) over R denoted by P₂ (x) where P₂ (x)={ a₂x²+a₁x¹+a₀, where a₁, a₂, a₀ ∈ R }.

The operation addition is defined as (a₂x²+a₁x¹+a₀)+(b₂x²+b₁x¹+b₀)={(a₂+b₂)x²+(a₁+b₁)x¹+(a₀+b₀)};
where a₀, a₁, a₂, b₀, b₁, b₂∈R

And the operation scalar multiplication is defined as α.(a₂x²+a₁x¹+a₀)=α.a₂x²+ α.a₁x¹+ α.a₀∈R

Then x²+3x+3 ∈ P₂ (x) is a linear combination of p₁=x+1, p₂=x²+x+1, for α₁=2∈R, α₂=1∈R, s.t. α₁p₁+α₂p₂=2(x+1)+1(x²+x+1)=(2x+2)+ (x²+x+1)=x²+3x+3

4. Linear span:

Case (i) Let φ ≠ S ⊆ V. Then the linear span of S consists of all possible linear combinations of elements of S. It is denoted by L(S). Thus L(S)= { α₁x₁+ α₂x₂+ …+α_kx_k : x₁, x₂, ..., x_k ∈ V, α₁, α₂, …, α_k ∈ F}.

Case (ii) If S= φ, L(S) is defined to be {0}.

5. Spanning set:

If L(S)=V, that is if V is the linear span of S then, S is called a spanning set of V and it is said that S spans V.

6. Examples-I:

Let R² be the vector space over R where S ⊆ R². Then

(i) S={(1, 0)}does not span R², i.e. L(S) ≠ R² Justification: By definition, L(S)={x(1, 0)|x ∈ R}={(x, 0)|x ∈ R}. We show that (1, 1) ∉ L(S) because if not, then (1, 1) ∈ L(S) which implies (1, 1)=α(x, 0)=(αx, 0) i.e. 1=0, a contradiction. Thus (1, 1) ∉ L(S). Hence L(S) ≠ R².

(ii) S={(1, 0),(1, 1)} spans R², i.e. S is a spanning set of R², i.e. L(S)=R². Justification: By definition, L(S)={x(1, 0)+y(1, 1)|x, y ∈ R}={(x, 0)+(y, y)|x, y ∈ R}={(x+y, y)|x, y ∈ R}. Clearly L(S) ⊆ R². We show that R² ⊆ L(S). Let (a, b) ∈ R2. Then (a, b)=α(1, 0)+β(1, 1) for some α, β ∈ R. Therefore α=a-b and β=b. Clearly α(1, 0)+β(1, 1)=(a, b). Thus (a, b) ∈ L(S). So that R² ⊆ L(S). Hence L(S)=R².

(iii) S={(1, 0), (2, 0)} does not span R² i.e. L(S) ≠ R². Justification: By definition, L(S)={x(1, 0)+y(2, 0)|x, y ∈ R}={(x, 0)+(2y, 0)|x, y ∈ R}={(x+2y, 0)|x, y ∈ R}. We show that (0, 1) ∉ L(S). For if (0, 1) ∈ L(S), then (0, 1)=α(1, 0)+β(2, 0) for some α, β ∈ R. Therefore 1=0, a contradiction. Thus (0, 1) ∉ L(S). Hence L(S) ≠ R²

7. Examples-II:

i. Consider the vector space P₂x over R and let S={1, x+1, x²+x+1} ⊆ P₂x . Then S spans P₂x , i.e. L(S)=P₂x .

Justification: By definition, L(S)={α.1+β(x+1)+γ(x²+x+1): α, β, γ ∈ R}. Clearly L(S) ⊆ P₂x . To prove that P₂x ⊆ L(S). Let a2x²+a₁x¹+a₀ ∈ P₂x , where a₀, a₁, a₂ ∈ R. Then a₂x²+a₁x¹+a₀= α.1+β(x+1)+γ(x²+x+1) for some α, β, γ ∈ R. Therefore α=a₀-a₁, β=a₁-a₂ and γ=a₂. Clearly, α.1+β(x+1)+γ(x²+x+1)= a₂x²+a₁x¹+a₀. Thus a₂x²+a₁x¹+a₀ ∈ L(S), so that P₂x ⊆ L(S). Hence L(S)= P₂x .

ii. Consider the vector space P₂x over R and let S={ x+1, x²+1, x-x²} ⊆ P₂x . Then S does not span P₂x , i.e. L(S) ≠ P₂x . Justification: By definition, L(S)={α(x+1)+β(x²+1)+γ(x-x²): α, β, γ ∈ R}.We show that 1 ∉ L(S)because if not, then 1 ∈ L(S). Thus 1=α(x+1)+β(x²+1)+γ(x-x²) for some α, β, γ ∈ R. Therefore α+β=1, α+γ=0, β-γ=0. By solving these equations, we get α+β=1and α+β=0, a contradiction. Thus 1 ∉ L(S). Hence L(S) ≠ P₂x .

8. Properties:

(i) S is a subset of L(S).

(ii) L(S) is the smallest subspace of V containing S. Thus the smallest subspace of V containing φ is {0}.

(iii) Let A ⊆ B ⊆ V. Then L(A) ⊆ L(B).

9. Summary:

For S ⊆ V,

L(S) is the collection of all possible linear combinations of S, if S ≠ φ and L(S)={0}, if S= φ. In other words, S ⊆ L(S), L(S) is a subspace and if W is a subspace of V such that S ⊆ W, then L(S) ⊆ W. If L(S)=V then S is called a spanning set of V. A vector space V can have more than one spanning sets. For example both {(1, 0), (0, 1), (2, 2) and {(1, 2), (2, 3)} span the vector space R² over R.