Let V and W be vector spaces, and let T:V-W be a linear transformation. Given a subspace U of V, let T(U) denote the set of all images of the form T(x), where x is in U. Show that T(U) is a subspace of W. To show that T(U) is a subspace of W, first show that the zero vector of W is in T(U). Choose the correct answer below. O A. Since V is a subspace of U, the zero vector of V, 0y, is in U. Since T is linear, T(0,) = 0w. where Ow is the zero vector of W. So Ow is in T(U). O B. Since U is a subspace of W, the zero vector of W, Ow, is in U. Since T is linear, T(0w) = 0y, where 0, is the zero vector of V. So Ow is in T(U). Oc. Since U is a subspace of V, the zero vector of V, Oy, is in U. Since T is linear, T(0,) = 0w, where 0w is the zero vector of W. So Ow is in T(U). O D. Since V is a subspace of U, the zero vector of U, 0, is in V. Since T is linear, T(0) = 0w, where Ow is the zero vector of W. So Ow is in T(U). Let v and w be in T(U). Relate v and w to vectors in U. Since V there exist x and y in U such that and Next, show that T(U) is closed under vector addition in W. Let T(x) and T(y) be in T(U), for some x and y in U. Choose the correct answer below. O A. Since x and y are in T(U) and T(U) is a subspace of W, x + y is also in W. O B. Since x and y are in U and T(U) is a subspace of W, x + y is also in W. O c. Since x and y are in U and U is a subspace of V, x +y is also in U. O D. Since x and y are in T(U) and U is a subspace of V, x + y is also in U. Use these results explain why T(U) is closed under vector addition in W. Choose the correct answer below. O A. Since T is linear, T(x) + T(y) = T(x + y). So T(x) + T(y) is in U, and T(U) is closed under vector addition in W. O B. Since T is linear, T(x) + T(y) = T(x + y). So T(x) + T(y) is in V, and T(U) is closed under vector addition in W. O c. Since T is linear, T(x) + T(y) = T(x + y). So T(x) + T(y) is in W, and T(U) is closed under vector addition in W. O D. Since T is linear, T(x) + T(y) = T(x +y). So T(x) + T(y) is in T(U), and T(U) is closed under vector addition in W. Next, show that T(U) is closed under multiplication by scalars. Let c be any scalar and x be in U. Choose the correct answer below. O A. Since x is in U and U is a subspace of V, cx is in W. Thus, T(cx) is in T(V). O B. Since x is in U and U is a subspace of V, cx is in V. Thus, T(cx) is in T(U). OC. Since x is in U and U is a subspace of W, cx is in U. Thus, T(cx) is in T(W). O D. Since x is in U and U is a subspace of V, cx is in U. Thus, T(cx) is in T(U).