Find whether the vector (347] [203] [823] [556] are linearly Independent ar. Linearly dependent
Answers
Find whether the vector (347] [203] [823] [556] are linearly Independent
yes they r linearly independent
Answer :
Linearly dependent
Explanation :
When the number of vectors is greater than the dimension of the vector space to which those vectors belongs , then those vectors are linearly dependent .
Here ,
(3 , 4 , 7) , (2 , 0 , 3) , (8 , 2 , 3) and (5 , 5 , 6) are the vectors of the vector space R³ .
Also ,
{(1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1)} is a basic of R³
→ dim(R³) = 3
Clearly ,
The no. of given vectors , i.e. 4 > dim(R³) = 3
Hence ,
The given vectors are linearly dependent .
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 :
- (V , +) is an abelian group .
- ku ∈ V ∀ u ∈ V and k ∈ F
- k(u + v) = ku + kv ∀ u , v ∈ V and k ∈ F .
- (a + b)u = au + bu ∀ u ∈ V and a , b ∈ F .
- (ab)u = a(bu) ∀ u ∈ V and a , b ∈ F .
- 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 .
Span / spanning set / generating set :
Let v₁ , v₂ , . . . , vₙ be the n vectors of a vector space V(F) , then the set of all linear combinations of v₁ , v₂ , . . . , vₙ , i.e. span{v₁ , v₂ , . . . , vₙ} = {c₁v₁ + c₂v₂ + . . . + cₙvₙ : cᵢ ∈ F}
♦ The spanning set is also called the subset of V spanned (or generated) by v₁ , v₂ , . . . , vₙ .
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 , 2 , 3) and (2 , 4 , 6) are linearly dependent vectors since (2 , 4 , 6) = 2(1 , 2 , 3)
- (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 , 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 , 0) and (0 , 1) are linearly independent vectors .
- (1 , 0 , 0) , (0 , 1 , 0) and (0 , 0 , 1) are linearly independent vectors .
- (1 , 2 , 3) and (0 , 3 , 4) are linearly independent vectors .
Basis of a vector space :
A set B of vectors in a vector space V is called a basis if all the elements of B are linearly independent and every element of V can be written as a linear combination of elements of B (i.e. B must spans V) .
Dimension of a vector space :
Dimension of a vector space is defined as the number of elements in its basis . The dimension of a vector space V is denoted by dim(V) .
♦ If B = {v₁ , v₂ , v₃ , . . . , vₖ} is a basis of vector space V , then Dimension of V = Cardinality of V , i.e. dim(V) = n(B) = k .
♦ A vector space can have more than one basis .
♦ Every basis of a vector space has the same number of vectors .