Question

The subset {(1,0,0), (0,1,0), (0,0,1)} of vector space R3 is

(A) Linearly dependent

(B) Linearly independent

(C) both

(D) none​

Answer 1

Answer:

B :- linearly independent

Answer 2

Answer :

(B) Linearly independent

Solution :

• Given. : {(1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1)} is a subset of R³
• To find : Whether the given subset is linearly dependent or independent .

Here ,

The given vectors are (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1) .

Now ,

Let a , b , c be any scalars such that ,

a(1 , 0 , 0) + b(0 , 1 , 0) + c(0 , 0 , 1) = (0 , 0 , 0)

→ (a , 0 , 0) + (0 , b , 0) + (0 , 0 , c) = (0 , 0 , 0)

→ (a , b , c) = (0 , 0 , 0)

→ a = b = c = 0

→ The vectors (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1) are linearly independent .

Hence ,

The subset {(1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1)} of R³ is linearly independent .

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) .

Linear combination :

A vector v in a vector space V is called a linear combination of the vectors v₁ , v₂ , v₃ , . . . , vₖ if v can be expressed in the form :

v = c₁v₁ + c₂v₂ + c₃v₃ + . . . + cₖvₖ

where c₁ , c₂ , c₃ , . . . , cₖ are scalars and are called weights of linear combination .

Linear dependence :

Let v₁ , v₂ , . . . , vₙ be the n non-zero vectors of a vector space V(F) . If for c₁v₁ + c₂v₂ + . . . + cₙvₙ = 0 (cᵢ ∈ F are scalars) , there exists atleast one cᵢ ≠ 0 , then v₁ , v₂ , . . . , vₙ are called linearly dependent .

♦ If the vectors v₁ , v₂ , . . . , vₙ are linearly dependent , then atleast one of these vectors can be expressed as a linear combination of the remaining vectors .

♦ Examples :

1. (1 , 2 , 3) and (2 , 4 , 6) are linearly dependent vectors since (2 , 4 , 6) = 2(1 , 2 , 3)
2. (1 , 3 , 4) , (1 , 2 , 3) and (0 , 1 , 1) are linearly dependent vectors since (1 , 3 , 4) = (1 , 2 , 3) + (0 , 1 , 1)
3. (3 , 2 , 5) , (2 , 1 , 2) and (-1 , 0 , 1) are linearly dependent vectors since (3 , 2 , 5) = 2(2 , 1 , 2) + (-1 , 0 , 1) .

Linearly independence :

Let v₁ , v₂ , . . . , vₙ be the n non-zero vectors of a vector space V(F) . If for c₁v₁ + c₂v₂ + . . . + cₙvₙ = 0 (cᵢ ∈ F are scalars) , all cᵢ = 0 , then v₁ , v₂ , . . . , vₙ are called linearly independent .

♦ If the vectors v₁ , v₂ , . . . , vₙ are linearly dependent , then none of these vectors can be expressed as a linear combination of the remaining vectors .

♦ Examples :

1. (1 , 0) and (0 , 1) are linearly independent vectors .
2. (1 , 0 , 0) , (0 , 1 , 0) and (0 , 0 , 1) are linearly independent vectors .
3. (1 , 2 , 3) and (0 , 3 , 4) are linearly independent vectors .