91影视

To prove that T(V鈧�) is a subspace of W, we will show that it satisfies the three properties of a subspace. 1. The zero vector of W is in T(V鈧�): Let 0_V be the zero vector in V, then T(0_V) = 0_W (since T is linear). Thus, the zero vector of W is in T(V鈧�). 2. T(V鈧�) is closed under addition: Let u, v be elements in T(V鈧�). Then there exist u鈧�, v鈧� in V鈧� such that T(u鈧�) = u and T(v鈧�) = v. Since V鈧� is a subspace of V, u鈧� + v鈧� is in V鈧�. Therefore, T(u鈧� + v鈧�) is in T(V鈧�). And, as T is linear, T(u鈧� + v鈧�) = T(u鈧�) + T(v鈧�) = u + v. This shows that the sum of any two elements in T(V鈧�) is also in T(V鈧�). 3. T(V鈧�) is closed under scalar multiplication: Let u be an element in T(V鈧�) and c be a scalar. There exists u鈧� in V鈧� such that T(u鈧�) = u. Since V鈧� is a subspace of V, c * u鈧� is in V鈧�. Therefore, T(c * u鈧�) is in T(V鈧�). And, as T is linear, T(c * u鈧�) = c * T(u鈧�) = c * u. This shows that the scalar multiple of any element in T(V鈧�) is also in T(V鈧�). Since T(V鈧�) satisfies all three properties of a subspace, it is a subspace of W.

Let S be the set of all x in V such that T(x) belongs to W鈧�. To prove that S is a subspace of V, we will again show that it satisfies the three properties of a subspace. 1. The zero vector of V is in S: As T is linear, T(0_V) = 0_W (the zero vector of W). Since W鈧� is a subspace of W, the zero vector of W is in W鈧�. Hence, the zero vector of V is in S. 2. S is closed under addition: Let x鈧�, x鈧� be elements in S. Then T(x鈧�) and T(x鈧�) are in W鈧�. Since W鈧� is a subspace of W, T(x鈧�) + T(x鈧�) is in W鈧�. Moreover, due to the linearity of T, T(x鈧�) + T(x鈧�) = T(x鈧� + x鈧�). Thus, T(x鈧� + x鈧�) is in W鈧�, which means x鈧� + x鈧� is in S. 3. S is closed under scalar multiplication: Let x be an element in S and c be a scalar. Then T(x) is in W鈧�. Since W鈧� is a subspace of W, c * T(x) is in W鈧�. And, as T is linear, c * T(x) = T(c * x). Thus, T(c * x) is in W鈧�, which means c * x is in S. Since S satisfies all three properties of a subspace, it is a subspace of V.