Question

10 (a) Define linear dependence and linear independence of vectors. Prove that if
two vectors are linearly dependent one of them, then it is a scalar multiple of
the other.

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Answer 1

Answer:

Two vectors are linearly dependent if and only if they are collinear, i.e., one is a scalar multiple of the other. Any set containing the zero vector is linearly dependent. If a subset of { v 1 , v 2 ,..., v k } is linearly dependent, then { v 1 , v 2 ,..., v k } is linearly dependent as well.

Answer 2

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 .

To prove :

If two vectors are linearly dependent , then one of them is a scalar multiple of the other .

Proof :

Let x , y be any two linearly dependent vectors of a vector space V(F) .

ie. If ax + by = 0 where a , b are scalars , then atleast one of them is non-zero scalar .

Case 1 :

Let a ≠ 0 , then the multiplicative inverse of a , i.e. a⁻¹ exists .

Now ,

→ ax + by = 0

→ ax = -by

→ x = a⁻¹(-by)

→ x = (-a⁻¹b)y

→ x = ky , where k = -a⁻¹b is scalar

→ x is a scalar multiple of y .

Case 2 :

Let b ≠ 0 , then the multiplicative inverse of b , i.e. b⁻¹ exists .

Now ,

→ ax + by = 0

→ by = -ax

→ y = b⁻¹(-ax)

→ y = (-b⁻¹a)x

→ y = kx , where k = -b⁻¹a is scalar

→ y is a scalar multiple of x .

Hence ,

If two vectors are linearly dependent , then one of them is a scalar multiple of the other .