Question

A non empty subset U of a vector space v(F) is a subspace of V iff​

Answer 1

Answer:

A subset W of a vector space V is a subspace if (1) W is non-empty (2) For every ¯v, ¯w ∈ W and a, b ∈ F, a¯v + b ¯w ∈ W. ... So a non-empty subset of V is a subspace if it is closed under linear combinations.

Step-by-step explanation:

Hope it will help you!!

Answer 2

Answer :

A non empty subset U of a vector space V(F) is a subspace of V iff ax + by ∈ W ∀ a , b ∈ F and x , y ∈ W .

Explanation :

Let U be the subspace of V(F) , then

→ ax ∈ U and by ∈ U ∀ x , y ∈ U and a , b ∈ F

(°•° Scalar multiplication is closed)

→ ax + by ∈ U ∀ x , y ∈ U and a , b ∈ F

(°•° Addition is closed)

Conversely ,

Let ax + by ∈ U ∀ x , y ∈ U and a , b ∈ F be given .

Now ,

Let a = b = 0 ∈ F , then

→ 0•x + 0•y ∈ U ∀ x , y ∈ U

→ 0 + 0 ∈ U

→ 0 ∈ U

→ U ≠ ∅

Now ,

Let a = b = 1 ∈ F , then

→ 1•x + 1•y ∈ U ∀ x , y ∈ U

→ x + y ∈ U ∀ x , y ∈ U

→ Vector addition is closed in U

Also ,

Let b = 0 ∈ F , then

→ ax + 0•y ∈ U ∀ x , y ∈ U , a ∈ F

→ ax + 0 ∈ U ∀ x ∈ U , a ∈ F

→ ax ∈ U ∀ x ∈ U , a ∈ F

→ Scalar multiplication is closed in U

Since ,

The vector addition and the scalar multiplication are closed in U , thus U is a subspace of V .

Some important information :

Vector space :

(V , +) be an algebraic structure and (F , + , •) be a field , then V is called a vector space over the field F if the following conditions hold :

1. (V , +) is an abelian group .
2. ku ∈ V ∀ u ∈ V and k ∈ F
3. k(u + v) = ku + kv ∀ u , v ∈ V and k ∈ F .
4. (a + b)u = au + bu ∀ u ∈ V and a , b ∈ F .
5. (ab)u = a(bu) ∀ u ∈ V and a , b ∈ F .
6. 1u = u ∀ u ∈ V where 1 ∈ F is the unity .

♦ Elements of V are called vectors and the lements of F are called scalars .

♦ If V is a vector space over the field F then it is denoted by V(F) .

Subspace :

A non empty subset W of the vector space V(F) is said to be a subspace of V if it itself forms a vector space over the same field F .