Vector Subspaces: Theorems Part 2

Vector Subspaces: Theorems Part 2

Theorem 6

Any intersection of subspace of a vector space is a subspace of V.

Proof:

W= {∩W_i/W_i is subspace of V)

  Every subspace contains the zero vector

0 ∈ W_i

⇒ 0 ∈ W_i

 0 ∈ W.

Let α ∈ F and x, y ∈ W

x, y ∈ W_i

x+y ∈ W_i      (. W_i is a subspace)

⇒ x+y ∈ ∩ W_i

 x + y ∈ W;

Let x ∈ W_i, α ∈ F

α x ∈ Wi­     (. W_i is a subspace)

 αx ∈ ∩ W­i

 W is a subspace of V.

 (OR)

Let C be a collection of subspaces of V, W be the intersection of subspaces in C. Since every subspace contains the zero vector, 0 ∈ W. Let a ∈ F and x, y ∈ W. Then x and y are contained in each subspace in C. Because each subspace in C is closed under addition and scalar multiplication, it follows that x+y and ax are contained in each subspace in C. Hence x + y and ax are also contained in W, so that W is a subspace of V.

Note: The union of two subspaces of V need not be a subspace of V.

For example:

Let V=R³ is a vector space over R.

W₁ = {(0, x, y)/x, y ∈ R); W₂ = {(x, 0, y)/x, y∈R) are subspaces of V.

W₁∪W₂ = {(x, y, z)/ either x = 0 or y = 0}

 (0, 2, 3) + (5, 0, 4) = (5, 2, 7) ∉ W₁ ∪ W₂

  W₁∪W₂ is not a subspace of V.

Theorem 7

W₁ and W₂ be subspaces of V. Prove that W₁ ∪ W₂ is a subspaces of V if and only if W₁ W₂ (or) W₂  W₁.

Proof:

Given W₁ and W₂ are subspaces. Assume that W₁ ∪ W₂ is a subspace of V.

To prove

 W₁W₂ such that w₁ ∈ W₁  ⇒ w₁ ∉ W₂

Let w₂ ∈ W₂     ⇒   w₁ + w₂ ∈ W₁ ∪ W₂

 w₁+w₂ ∈ W₁ or w₁+w₂ ∈ W₂

Suppose w₁+w₂ ∈ W₂.

Given w₂ ∈ W₂ then there exists − w₂ ∈ W₂

         (W₂ is a subspace)

then w₁+w₂‒w₂ ∈ W₂

    ⇒ w₁ ∈ W₂.

This is a contradiction to our initial assumption.

 W₁ ∉ W₂ (ie) w₁+w₂ ∈ W₂ is not true.

.. w₁ + W₂ ∈ W₁.

Hence W₁W₂.

Conversely,

Assume W₁W₂ or W₂  W₁

To prove: W₁ ∪ W₂ is a subspace

Since W₁  W₂ or W₂  W₁

  W₁ ∪ W₂ = W₂ and W₁ ∪ W₂ = W₁

Here W₁ and W₂ are subspaces of V.

 W₁ ∪ W₂ is a subspace of V.

Example 6

Let V be a vector space over F and let V₁, V₂ be the subspaces of V, then prove that W=V₁+V₂ = { v = v₁ + v₂ / v₁ ∈ V₁ ; v₂ ∈ V₂} is a subspace of V.

Solution:

Let v, v' ∈ W and α ∈ F then

 v=v₁+v₂;      v₁, v₁' ∈ v₁

v=v₁' + v₂';     v₂, v₂' ∈ V₂

 αv + v′ = α (v₁ + v₂) + (v₁′ + v₂′)

= αv₁ + v₁' + αv₂ + v₂′

        (αv₁ + v₁′ ∈ V₁, αv₂ + v₂′ ∈ V₂)

V₁+V₂ = W

  W is a subspace.

Definition:

Direct sum: A vector space V is called the direct sum of W₁ and W₂ if W₁ and W₂ are subspaces of V such that W₁∩W₂ = {0} and W₁ + W₂ = V. We denote that V is the direct sum of W₁ and W₂ by writing V=W₁  W₂.

Definition

It S₁ and S₂ are non−empty sub sets of a vector space V, then the sum of S₁ and S₂ denoted by S₁ + S₂ is the set (x+y; x ∈ S₁ and y ∈ S₂}.