Vector Spaces: Theorems Part 4

Vector Spaces

Theorems Part 4

Theorem 9

The span of any subset "S" of a vector space "V" is a subspace of V. Moreover any subspace of V that contains must also contain the span of S.

Proof:

Let V be a vector space over F and S contained in V then L (S) is a subspace of V.

Let v, w ∈ L (S) then

 v = α₁v₁ + α₂v₂ + ... + α_nv_n

 w= β₁w₁ + β₂w₂+ ... + β_mw_m

where

α¡ ∈ F ; i = 1, 2, ... n and

β_i ∈ F ; i = 1, 2, ... m

and v₁, v₂, ... v_n, w₁, w₂, w₃ …. w_m ∈ S also m and n are finite.

Let α ∈ F.

 αv+w = α (α₁v₁ + α₂v₂ + ... + α_nv_n) + (β₁w₁ + β₂w₂ + …. + β_mw_m)

 = αα₁v₁ + αα₂v₂ + ... + αα_nv_n + β₁w₁+ β₂w₂ + …. + β_mw_m

 = γ₁v₁ + γ₂v₂ + ... + γ_nv_n + γ_n+1v_n+1 + γ_n+2v_n+2 + … + γ_n+mv_n+m

where w_i= v_n+i; αα₁ = γ_i; β₁ = γ_n+i;

  αv + w ∈ L(S).

  L(S) is a subspace of V.

Theorem 10

Let V be a vector space over F and S is contained in V and W is a subspace of V then L (S)  W if and only if SW.

Proof:

Given: V is a vector space over F and W is a subspace.

Assume that L (S) is contained in W.

To prove: SW

 v∈S

1.v  L(S)  W

  ν ∈ W

 SW

Conversely: Assume SW

To prove L(S)  W

Let x ∈ L(S)

x= α₁v₁ +α₂v₂ + ... + α_nv_n

where

 α_i ∈ F; v_i ∈ SW; v_i ∈ W

W is a subspace of V.

⇒  α₁v₁ + α₂v₂ + . ... + α_nv_n ∈  W

 X  W

 L(S)  W

Theorem 11

The span of any subsets S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S.

Proof:

Part (i) L (S) is a subspace of V.

Case (i): If S=ϕ, It is trivial.

Because span (ϕ) = {0} which is a subspace of V that is also contained in any subspace of V.

Case (ii): If S≠ϕ, then S contains a vector z.

 0z=0 ∈ span (S);

 0 ∈ F, z ∈ S.

Let x, y ∈ span (S). Then there exists vectors u₁, u₂, ..... u_m, v₁, v₂, ... v₂ in S and scalars a₁, a₂, ... a_m, b₁, b₂, ......., b₁ ∈ F,

 such that x = a₁u₁ + α₂u₂ + ... + a_mu_m

 and y=b₁v₁+ b₂v₂ + ... + b_nv_n

  x + y = α₁u₁ + а₂u₂ + ... + a_mu_m + b₁v₁ + b₂v₂ + ... + b_nv_n and for any scalar c,

 we have cx=(ca₁)u₁ + (ca₂)u₂ + ... + (ca_m)u_m

are clearly linear combinations of the vectors in S.

  x + y ∈ L (S)

and cx ∈ L (S)

  L(S) is a subspace of V.

Part (ii) Let W be any subspace of V containing S.

To prove W  L (S).

If w ∈ L(S) then w=c₁w₁ + c₂w₂ + ... + c_kw_k for some vectors w₁, w₂, ... w_k in S and some scalars c₁, c₂, ... c_k in F.

Since

S W

 w₁, w₂, ... w_k ∈ W

⇒ c₁w₁ + c₂w₂ + ... c_kw_k ∈ W

('.' W is a subspace)

⇒ W ∈ W

 { Since w is an arbitrary element of L (S) and w ∈ W}

  L(S)  W

Theorem 12

Direct sum theorem: Let W₁ and W₂ be subspaces of a vector space V. Prove that V is the direct sum of W₁ and W₂ iff each vector in V can be uniquely written as x₁+x₂ where x₁ ∈ W₁ and x₂ ∈ W₂.

Proof:

Assume that V is the direct sum of W₁ and W₂ then V=W₁+W₂ and W₁∩W₂ = { 0 }.

Since V=W₁+ W₂; every v ∈ V can be written as v=x₁+x₂ where x₁ ∈ W₁ and x₂ ∈ W₂:

This expression is unique.

Suppose if v = x₁′+x₂′ where x₁' ∈ W₁, x₂' ∈ W₂

 v = x₁+x₂ = x₁′ + x₂′

x₁−x₁′ = x₂′−x₂ ∈ W₁ ∩ W₂

But W₁ ∩ W₂ = {0}.

⇒ x=x₁' and x₂=x₂'

 x₁, x₁'∈ W₁

 x₂, x₂' ∈ W₂

x₁−x₁' ∈ W₁

x₂−x₂' ∈ W₂

Hence the expression is unique.

Conversely,

Assume that v ∈ V can be expressed uniquely as v = x₁ + x₂ ; x₁ ∈ W₁ and x₂ ∈ W₂.

Then clearly V=W₁  W₂⋅

 Let w ≠ 0 ∈ W₁ ∩ W₂

⇒ w ∈ V

then w=w+0; w ∈ W₁, 0 ∈ W₂

w=0+w; 0 ∈ W₁, w ∈ W₂

Since the expression for w is unique we have w=0.

Hence W₁ ∩ W₂ = {0}.

Hence V is the direct sum of W₁ and W₂.

Theorem 13

Let V be a vector space over F and S₁ S₂  V then L(S₁)  L(S₂)

Proof:

Let v_i ∈ S₁,

then α_iv_i ∈ L(S₁)

 S₁  S₂

⇒ v_i ∈ S₂ and α_iv_i ∈ L(S₂)

  S₁S₂

L(S₁)  L(S₂).