To prove that the range of a linear map T: V → V is a subset of its null space (range(T) ⊆ null(T)) if and only if T^2 = 0 (where 0 denotes the zero map V → V), we need to show that the two statements are logically equivalent. Proof: Part 1: Prove that if range(T) ⊆ null(T), then T^2 = 0. Assume that range(T) ⊆ null(T). For any vector v in V, T(v) belongs to the range of T. Since range(T) ⊆ null(T), we have T(T(v)) = 0 for all v in V. In other words, T^2(v) = 0 for all v in V. Therefore, T^2 = 0 (the zero map V → V). Part 2: Prove that if T^2 = 0, then range(T) ⊆ null(T). Assume that T^2 = 0. Let w be a vector in the range of T. Then, there exists a vector v in V such that T(v) = w. Since T^2 = 0, we have T(T(v)) = T(w) = 0. This means that w belongs to the null space of T. Therefore, every vector in the range of T is also in the null space of T, which implies range(T) ⊆ null(T). By proving both directions, we have shown that range(T) ⊆ null(T) if and only if T^2 = 0 (the zero map V → V). It's important to note that this proof holds for any finite-dimensional vector space V over a field F and any linear map T: V → V.