Answer:
1) W₁ is a subspace of Pₙ (R)
2) W₂ is not a subspace of Pₙ (R)
4) W₃ is a subspace of Pₙ (R)
Step-by-step explanation:
Given that;
1.Let W₁ be the set of all polynomials of the form p(t) = at², where a is in R
2.Let W₂ be the set of all polynomials of the form p(t) = t² + a, where a is in R
3.Let W₃ be the set of all polynomials of the form p(t) = at² + at, where a is in R
so
1)
let W₁ = { at² ║ a∈ R }
let ∝ = a₁t² and β = a₂t² ∈W₁
let c₁, c₂ be two scalars
c₁∝ + c₂β = c₁(a₁t²) + c₂(a₂t²)
= c₁a₁t² + c²a₂t²
= (c₁a₁ + c²a₂)t² ∈ W₁
Therefore c₁∝ + c₂β ∈ W₁ for all ∝, β ∈ W₁ and scalars c₁, c₂
Thus, W₁ is a subspace of Pₙ (R)
2)
let W₂ = { t² + a ║ a∈ R }
the zero polynomial 0t² + 0 ∉ W₂
because the coefficient of t² is 0 but not 1
Thus W₂ is not a subspace of Pₙ (R)
3)
let W₃ = { at² + a ║ a∈ R }
let ∝ = a₁t² +a₁t and β = a₂t² + a₂t ∈ W₃
let c₁, c₂ be two scalars
c₁∝ + c₂β = c₁(a₁t² +a₁t) + c₂(a₂t² + a₂t)
= c₁a₁t² +c₁a₁t + c₂a₂t² + c₂a₂t
= (c₁a₁ +c₂a₂)t² + (c₁a₁t + c₂a₂)t ∈ W₃
Therefore c₁∝ + c₂β ∈ W₃ for all ∝, β ∈ W₃ and scalars c₁, c₂
Thus, W₃ is a subspace of Pₙ (R)
.