Let V and W be vector spaces, and let T: VW be a linear transformation. Given...

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Let V and W be vector spaces, and let T: V→W be a linear transformation. Given a subspace 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 U, 0, is in V. Since T is linear, T(0)=0w, where 0w is the zero vector of W. So 0 is in T(U). OB. Since U is a subspace of W, the zero vector of W, Ow, is in U. Since T is linear, T(0w)=0, where Oy is the zero vector of V. So Ow is in T(U). O C. Since V is a subspace of U, the zero vector of V, 0v, 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 D. Since U is a subspace of V, the zero vector of V, 0, is in U. Since T is linear, T(0)=0w, where 0w is the zero vector of W. So 0w is in T(U). Let v and w be in T(U). Relate v and w to vectors in U. Since 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. there exist x and y in U such that and ▼ O A. Since x and y are in 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 U is a subspace of V, x+y is also in U. O C. Since x and y are in T(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 T(U) is a subspace of W, x+y is also in W. Use these results to explain why T(U) is closed under vector addition in W. Choose the correct answer below. Use these results to 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 T(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 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 V. Thus, T(cx) is in T(U). O B. Since x is in U and U is a subspace of V, cx is in U. Thus, T(cx) is in T(U). O C. Since x is in U and U is a subspace of V, cx is in W. Thus, T(cx) is in T(V). O D. Since x is in U and U is a subspace of W, cx is in U. Thus, T(cx) is in T(W). The preceding result helps to show why T(U) is closed under multiplication by scalars. Recall that every element of T(U) can be written as T(x) for some x in U. Choose the correct answer below. O A. Since T is linear, T(cx)=cT(x) and cT(x) is in U. Thus, T(U) is closed under multiplication by scalars. O B. Since T is linear, T(cx)=cT(x) and cT(x) is in V. Thus, T(U) is closed under multiplication by scalars. O C. Since Tis linear, T(cx) = cT(x) and cT(x) is in T(U). Thus, T(U) is closed under multiplication by scalars. O D. Since T is linear, T(cx)=cT(x) and cT(x) is in W. Thus, T(U) is closed under multiplication by scalars. Use these results to explain why T(U) is a subspace of W. Choose the correct answer below. OA. The image of the transformation T(U) is a subspace of W because T(U) contains the zero vector of W and is closed under vector addition and multiplication by a scalar. OB. The image of the transformation T(U) is a subspace of W because T(U) contains the zero vector of V and is closed under vector addition and multiplication by a scalar. O C. The image of the transformation T(U) is a subspace of W because T(U) is closed under vector addition and multiplication by a scalar. O D. The image of the transformation T(U) is a subspace of W because U contains the zero vector of W and T(U) is closed under vector addition and multiplication by a scalar. Let V and W be vector spaces, and let T: V→W be a linear transformation. Given a subspace 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 U, 0, is in V. Since T is linear, T(0)=0w, where 0w is the zero vector of W. So 0 is in T(U). OB. Since U is a subspace of W, the zero vector of W, Ow, is in U. Since T is linear, T(0w)=0, where Oy is the zero vector of V. So Ow is in T(U). O C. Since V is a subspace of U, the zero vector of V, 0v, 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 D. Since U is a subspace of V, the zero vector of V, 0, is in U. Since T is linear, T(0)=0w, where 0w is the zero vector of W. So 0w is in T(U). Let v and w be in T(U). Relate v and w to vectors in U. Since 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. there exist x and y in U such that and ▼ O A. Since x and y are in 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 U is a subspace of V, x+y is also in U. O C. Since x and y are in T(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 T(U) is a subspace of W, x+y is also in W. Use these results to explain why T(U) is closed under vector addition in W. Choose the correct answer below. Use these results to 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 T(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 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 V. Thus, T(cx) is in T(U). O B. Since x is in U and U is a subspace of V, cx is in U. Thus, T(cx) is in T(U). O C. Since x is in U and U is a subspace of V, cx is in W. Thus, T(cx) is in T(V). O D. Since x is in U and U is a subspace of W, cx is in U. Thus, T(cx) is in T(W). The preceding result helps to show why T(U) is closed under multiplication by scalars. Recall that every element of T(U) can be written as T(x) for some x in U. Choose the correct answer below. O A. Since T is linear, T(cx)=cT(x) and cT(x) is in U. Thus, T(U) is closed under multiplication by scalars. O B. Since T is linear, T(cx)=cT(x) and cT(x) is in V. Thus, T(U) is closed under multiplication by scalars. O C. Since Tis linear, T(cx) = cT(x) and cT(x) is in T(U). Thus, T(U) is closed under multiplication by scalars. O D. Since T is linear, T(cx)=cT(x) and cT(x) is in W. Thus, T(U) is closed under multiplication by scalars. Use these results to explain why T(U) is a subspace of W. Choose the correct answer below. OA. The image of the transformation T(U) is a subspace of W because T(U) contains the zero vector of W and is closed under vector addition and multiplication by a scalar. OB. The image of the transformation T(U) is a subspace of W because T(U) contains the zero vector of V and is closed under vector addition and multiplication by a scalar. O C. The image of the transformation T(U) is a subspace of W because T(U) is closed under vector addition and multiplication by a scalar. O D. The image of the transformation T(U) is a subspace of W because U contains the zero vector of W and T(U) is closed under vector addition and multiplication by a scalar.

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1 option D because given U subspace of V 2since v and w in TU there exist ... View the full answer