Postulates of vector space

 

Question:

If P(x) denotes the set of all polynomials in one indeterminate x over a field F then to show that P(x) is a vector space over F with addition defind as addition of polynomials and scalar multiplication defined as the product of polynomial by an element of F. 

Solution:

Given P(x) = {p(x): p(x) = a₀ + a₁x + ·····+ aₙxⁿ +... = ∑ aₙxⁿ for aₙ, s ∈ F}

Addition of polynomials is defined as follows:

If P(x) =∑ aₙxⁿ ∈ P(x).q(x) =∑bₙxⁿ ∈ P(x). 

then p(x) + q(x) = ∑ (aₙ + bₙ)xⁿ .......(1)

Scalar multiplication is defined as follows : 

If p(x) =∑ aₙxⁿ ∈ P(x), α ∈ F

Then α p(x) =∑ (α aₙ) xⁿ .............(2) 

Now to show that P(x) is a vector space, we have to verify all the postulates of vector space. Let us verify them one by one. 

Closure axiom for addition:

 For each p(x) . q(x) ∈ p(x) ⟹ p(x) + q(x) =∑ (aₙ + bₙ)xⁿ ∈ P(x) 

(as if aₙ .bₙ ∈ F ⟹ aₙ + bₙ ∈ F as field is closed for +). 

V₁: (i) For any p(x) = ∑ aₙxⁿ . q(x) = ∑bₙxⁿ.r(x) =∑ cₙxⁿ ∈ P(x). 

We have 

[p(x) + q(x)] +r(x) ={ ∑ aₙxⁿ + ∑bₙxⁿ} + ∑ cₙxⁿ

= ∑ (aₙ + bₙ)xⁿ + ∑ cₙxⁿ

= ∑ (aₙ + bₙ + cₙ)xⁿ

= ∑ (aₙ + (bₙ + cₙ))xⁿ

= ∑ aₙxⁿ + ∑(bₙ + cₙ)xⁿ

= p(x) + [ q(x) + r(x) ]

[as aₙ , bₙ , cₙ ∈ F ⟹ aₙ + (bₙ + cₙ) =aₙ + (bₙ + cₙ)]

Hence elements of P(x) arc associative for addition. 

(ii) We know that 

o̅(x) = ∑o̅xⁿ ∈ F(x) 

∴ o̅ (x) + p(x) = ∑oxⁿ + ∑ aₙxⁿ ∀ p(x) ∈ P(x) 

= ∑ (0 + aₙ)xⁿ 

= ∑ aₙxⁿ (0 ∈ F is additive identity in F)

= p(x)

 Similarly, p(x) + o̅(x) = p(x). 

Hence o̅(x) is additive identity in P(x).

(iii) For each p(x) =∑ aₙxⁿ ∈ P(x), ∃ - p(x) ∑ (-aₙ)xⁿ ∈ P(x) such that

p(x) + (-p(x)) =∑ aₙxⁿ + ∑ (-aₙ)xⁿ

= ∑ [aₙx + (-aₙ)]xⁿ

(as -z is additive inverse of aₙ in F)

=∑ 0xⁿ = o̅(x)

Similarly, -p(x) + p(x) = o̅(x)

Hence additive inverse of each element in P(x) exists.

(iii) For any p(x). q(x) ∈ P(x), we have

p(x) + q(x) =∑ aₙxⁿ + ∑ bₙxⁿ

= ∑ (aₙ + bₙ)xⁿ

= ∑ (bₙ + aₙ)xⁿ 

 as addition is commutative in F

= ∑ bₙxⁿ + ∑ aₙxⁿ 

= q(x) + p(x).

Hence addition is commutative in P(x). Thus (P(x), +) is an abelian group. 

Closure axiom for scalar multiplication: 

For a ∈ F . p(x) ∈ P(x) ⟹ a ap(x) = a∑ aₙxⁿ = ∑ (aaₙ)xⁿ ∈ P(x) 

(as a ∈ F. aₙ ∈ F =aaₙ ∈ F, because F is closed for multiplication) 

V₂ : For any a ∈ F , p(x).(x) ∈ P(x). we have 

a (p(x) + q(x)) = a(∑ aₙxⁿ + ∑ bₙxⁿ) 

= a (∑ (aₙ + bₙ)xⁿ)

= ∑ a(aₙ + bₙ)xⁿ

= ∑ (aₙ + bₙ)xⁿ

= ∑ (aaₙ + abₙ)xⁿ

(as elements of F satisfy the distributive law) 

= ∑ (aaₙ)xⁿ + ∑(abₙ)xⁿ

= a∑aₙxⁿ + a∑bₙxⁿ

= ap(x) + aq(x)

V₃: For any α, β ∈ F, p(x) ∈ P(x), we have 

(α + β) p(x) = (α + β)∑aₙxⁿ

= ∑((α + β)aₙ)xⁿ

= ∑(αaₙ + βaₙ)xⁿ

= ∑αaₙxⁿ + ∑βaₙxⁿ

= α∑aₙxⁿ + β∑aₙxⁿ

= α p(x) + βp(x)

V₄: For any α, β ∈ F, p(x) ∈ P(x), we have

(αβ) p(x) = (αβ)∑aₙxⁿ

= ∑((αβ)aₙ)xⁿ

= ∑(α(βaₙ))xⁿ

(as elements of F are associative for multiplication)

= α(∑(βaₙ)xⁿ)

= α(β∑aₙxⁿ)

= α (βp(x))

V₅: As 1 ∈ F, so for any p(x) ∈ P(x), we have

1p(x) = 1 ∑aₙxⁿ = ∑(1aₙ)xⁿ = ∑aₙxⁿ = p(x)

(as 1 is multiplication identity of F)

Hence showing that P(x) is a vector space.