To solve this problem, we need to understand the concept of the dual map and its properties. The dual map T' : W' → V' is a linear transformation between the dual spaces W' and V', defined as follows: For any linear functional φ in W', T'(φ) is a linear functional in V' given by: (T'(φ))(v) = φ(T(v)) for all v in V. In other words, T' maps a linear functional φ in W' to a linear functional T'(φ) in V' by composing φ with the linear map T. With this definition in mind, we can prove the following properties of the dual map: 1. The dual map T' is a linear transformation. Proof: Let φ₁, φ₂ be linear functionals in W', and let α, β be scalars in F. We need to show that T'(αφ₁ + βφ₂) = αT'(φ₁) + βT'(φ₂). For any v in V, we have: (T'(αφ₁ + βφ₂))(v) = (αφ₁ + βφ₂)(T(v)) = α(φ₁(T(v))) + β(φ₂(T(v))) = α((T'(φ₁))(v)) + β((T'(φ₂))(v)) Therefore, T'(αφ₁ + βφ₂) = αT'(φ₁) + βT'(φ₂), which proves the linearity of T'. 2. The kernel of T' is the annihilator of the range of T. Proof: Let φ be a linear functional in W'. φ is in the kernel of T' (i.e., T'(φ) = 0) if and only if (T'(φ))(v) = 0 for all v in V. By the definition of T', this is equivalent to φ(T(v)) = 0 for all v in V. In other words, φ is in the kernel of T' if and only if φ annihilates the range of T. Therefore, ker(T') = {φ in W' | φ(T(v)) = 0 for all v in V} = ann(range(T)). 3. The range of T' is the annihilator of the null space of T. Proof: Let ψ be a linear functional in V'. ψ is in the range of T' if and only if there exists a linear functional φ in W' such that ψ = T'(φ). By the definition of T', this is equivalent to ψ(v) = φ(T(v)) for all v in V. In other words, ψ is in the range of T' if and only if ψ annihilates the null space of T. Therefore, range(T') = {ψ in V' | ψ(v) = 0 for all v in null(T)} = ann(null(T)). These properties are fundamental in understanding the relationship between a linear map T and its dual map T', and they have important applications in linear algebra and functional analysis.