Question

0. The basis set for the vector space C(R) is

(A) {(0,1), (1,0)}

(B) {1, i}

(C) {(1,2)}​

Answer 1

Answer:

Answer B

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Answer 2

Answer :

(B) {1 , i}

Explanation :

Here ,

The given vector space is C(R) which represents the vector space of complex numbers over the field of real numbers .

Now ,

Let z € C(R) be an arbitrary complex number , then

→ z = x + iy , where x , y € R

→ z = 1•x + i•y where x , y € R

→ Any z € C(R) is a linear combination of 1 and i

→ The vector space C(R) is spanned by {1 , i}

Also ,

{1 , i} are linearly independent set .

Hence ,

{1 , i} is a basis of the vector space C(R) .

Moreover ,

Dimension of the vector space C(R) is the number of elements in its basis .

•°• Dim(C(R)) = 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 .

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. (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 .

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 .