Let V and W be vector spaces, and let T: VW be a linear transformation. Given...
Fantastic news! We've Found the answer you've been seeking!
Question:
Transcribed Image Text:
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.
Expert Answer:
Answer rating: 100% (QA)
1 option D because given U subspace of V 2since v and w in TU there exist ... View the full answer
Related Book For
Posted Date:
Students also viewed these mathematics questions
-
Let V and W be vector spaces of dimensions n and m. Show that the space L(V,W) of linear maps from V to W is isomorphic to Mmn.
-
Let V, W be vector spaces with ordered bases F and F, respectively. If L : V W is a linear transformation and A is the matrix representing L relative to F and F, show that (a)v ker(L) if and only...
-
Let V and W be vector spaces. If X is a , subset of V, define X = [T in L(V, W) | T(v) = 0 for all v in X) (a) Show that X is a subspace of L(F, IV). (b) If Xcxu show diat X. (c) If U and U\ are...
-
Consider an investor's choice of a farm unit in the Corn Belt, one in the California Central Valley, or the one in the Great Plains region. An investor added these three assets in one portfolio. The...
-
What types of variables should the researcher consider when screening for each of the following: export markets, foreign direct investment, and global sourcing?
-
The following accounts and corresponding balances were drawn from Jogger Companys Year 2 and Year 1 year-end balance sheets: The Year 2 income statement is shown next: Income Statement Sales...
-
A confidence interval for a mean response and a prediction interval for an individual response are to be constructed from the same data. True or false: The number of degrees of freedom for the...
-
This year Andrews achieved an ROE of 5.6%. Suppose the Board of Directors of Andrews mandates that management take measures to increase financial Leverage (=Assets/Equity) next year. Assuming Sales,...
-
Kraft Heinz was compelled to restate nearly three years of financial reporting following an examination. As a financial analyst, do you believe that restatements of this nature should raise warning...
-
Your shoe factory has a production capacity of 10,000 units per month. Your fixed costs are $200,000, your variable cost of production is $30, and you sell each pair for $35. The problem is that this...
-
What is the difference between a conceptual, a logical, and a physical data model? Where do relationship diagrams (RDs) and entity/relationship diagrams (ERDs) fit in?
-
Based on the information for J's Lemonade Stand and C's Orange Juicery Answer the following two questions in your discussion post. Which company do you think performed better during the year? Give...
-
What are entity relationship diagrams (ERDs) important in database design, even in smaller databases?
-
Budgeting is not just about numbers; it can significantly influence how an organization operates and how employees perceive their roles within it. Budgeting is a critical aspect of financial...
-
What industrial target (industry, business, or location) do you believe is the most inviting to a cyber attack and why ?
-
public class ParaT 2 0 2 3 extends RecursiveTask { int lo , hi , val; int [ ] arr; static final int CUT = 5 ; ParaT 2 0 2 3 ( int [ ] a , int l , int h , int v ) { lo = l; hi = h; arr = a; val = v; }...
-
Refer to the table to answer the following questions. Year Nominal GDP (in billions) Total Federal Spending (in billions) Real GDP (in billions) Real Federal Spending (in billions) 2000 9,817 578...
-
In each case, describe the graph of the equation (where z denotes a complex number). (a) |z - 1| = 2 (b) z = - (c) im z = m re z, m a real number
-
Let U and W be subspaces of a vector space V. (a) If dim V = 3, dim U = dim W = 2, and U W, show that dim(U W) = 1. (b) Interpret (a) geometrically if V= R3.
-
If {u, v, w} is a basis of V, determine which of the following are bases. (a) {2u + v + 3w, 3u + v - w, u - 4w) (b) {u, u + w, u - w, v + w}
-
How do you allocate requirements?
-
How do you flow down requirements?
-
What is requirements flow down?
Study smarter with the SolutionInn App