Final answer:
The given map D: P³ → P² is a linear map that preserves addition and scalar multiplication. The null space of D is a plane in P³ and the range of D is all polynomials in P². It is not stated in the question if the given map is an inverse.
Step-by-step explanation:
To show that the given map D:P³ → P² is a linear map, we need to prove that it satisfies the properties of linearity: preservation of addition and scalar multiplication. Let's start by checking the preservation of addition:
Let P and Q be two polynomials in P³, such that P(x) = a₀ + a₁x + a₂x² + a₃x³ and Q(x) = b₀ + b₁x + b₂x² + b₃x³. The map D preserves addition if D(P + Q) = D(P) + D(Q).
Using the definitions of P and Q, we have D(P + Q) = D((a₀ + b₀) + (a₁ + b₁)x + (a₂ + b₂)x² + (a₃ + b₃)x³) = (a₁ + b₁) + (a₂ + b₂)x + (a₃ + b₃)x² = (a₁ + a₂x + a₃x²) + (b₁ + b₂x + b₃x²) = D(P) + D(Q).
Therefore, the given map D preserves addition and hence is a linear map.
To check if the given map is an inverse, we need to verify if applying D and its inverse D⁻¹ to a polynomial results in the identity function. Since the inverse map is not provided in the question, we cannot determine if D is an inverse map or not.
The null space of a linear map is the set of vectors that map to the zero vector. In this case, the null space of D is the set of polynomials P(x) in P³ such that D(P) = 0. From the definition of D, we obtain D(P) = a₁ + a₂x + a₃x² = 0. This equation represents a plane in P³ and the null space of D is the set of all polynomials that lie on this plane.
The range of a linear map is the set of vectors that can be obtained by applying the map to any vector in the domain. In this case, the range of D is the set of polynomials Q(x) in P² such that there exists a polynomial P(x) in P³ for which D(P) = Q. From the definition of D, we can see that the range of D is the set of all polynomials Q(x) in P².