Math2318 Test 3 - This Test Will Try Something Different

Math2318 Test3inthistestwewilltrysomethingdifferentthea

Math 2318 - Test 3 In this test we will try something different. The answers are provided, your job is to show the work in how to get that solution. On problem 1 only A is a vector space. You will show why it is a vector space but you will also show why B and C are not vector spaces. On question 2 only V is a vector space. You will show why it is a vector space and you will also show why W and U are not vector spaces. Solve the problem. 1) Determine which of the following sets is a subspace of Pn for an appropriate value of n. A: All polynomials of the form p(t) = a + bt2, where a and b are in ℝ B: All polynomials of degree exactly 4, with real coefficients C: All polynomials of degree at most 4, with positive coefficients A) A and B B) C only C) A only D) B only ) Determine which of the following sets is a vector space. V is the line y = x in the xy-plane: V = { (x, y) : y = x } W is the union of the first and second quadrants in the xy-plane: W = { (x, y) : y ≥ 0 } U is the line y = x + 1 in the xy-plane: U = { (x, y) : y = x + 1 } A) U only B) V only C) W only D) U and V 2) Find a matrix A such that W = Col A. 3) W = 3r - t, 4r - s + 3t, s + 3t, r - 5s + t : r, s, t in ℝ A) B) C) D) ) Determine if the vector u is in the column space of matrix A and whether it is in the null space of A. 4) u = (5, -3, 5), A = A) In Col A and in Nul A B) In Col A, not in Nul A C) Not in Col A, in Nul A D) Not in Col A, not in Nul A 4) Use coordinate vectors to determine whether the given polynomials are linearly dependent in P2. Let B be the standard basis of the space P2 of polynomials, that is, let B = { 1, t, t2 } . 5) 1 + 2t, 3 + 6t2, 1 + 3t + 4t2 A) Linearly dependent B) Linearly independent 5) Find the dimensions of the null space and the column space of the given matrix. 6) A = A) dim Nul A = 2, dim Col A = 3 B) dim Nul A = 4, dim Col A = 1 C) dim Nul A = 3, dim Col A = 2 D) dim Nul A = 3, dim Col A = Solve the problem. 7) Let H = a + 3b + 4d, c + d -3a - 9b + 4c - 8d, -c - d : a, b, c, d in ℝ Find the dimension of the subspace H. A) dim H = 3 B) dim H = 1 C) dim H = 4 D) dim H = 2 7) Assume that the matrix A is row equivalent to B. Find a basis for the row space of the matrix A. 8) A = , B = A) {(1, 3, -4, 0, 1), (0, -2, 3, 5, -4), (0, 0, -8, -23, 17), (0, 0, 0, 0, 0)} B) {(1, 3, -4, 0, 1), (0, -2, 3, 5, -4), (0, 0, -8, -23, 17)} C) {(1, 3, -4, 0, 1), (2, 4, -5, 5), -2, (1, -5, 0, -3, 2), (-3, -1, 8, 3, -4)} D) {(1, 0, 0, 0), (3, -2, 0, 0), (-4, 3, -8, 0)} 8) Find the new coordinate vector for the vector x after performing the specified change of basis. 9) Consider two bases B = {b1, b2, b3} and C = {c1, c2, c3} for a vector space V such that b1 = c1 + 2c3, b2 = c1 + 4c2 - c3, and b3 = 3c1 - c2. Suppose x = b1 + 6b2 + b3. That is, suppose [x]B = [1, 6, 1]. Find [x]C. A) B) C) D) ) Find the specified change-of-coordinates matrix. 10) Consider two bases B = {b1, b2} and C = {c1, c2} for a vector space V such that b1 = c1 - 2c2 and b2 = 3c1 - 4c2. Find the change-of-coordinates matrix from B to C. A) B) C) D) Answer Key Testname: MATH2318−TEST3 1) C 2) B 3) B 4) B 5) B 6) C 7) D 8) B 9) C 10) B

Paper For Above instruction

The provided test problems cover a broad range of topics in linear algebra, including subspaces, vector spaces, matrix representations, coordinate transformations, linear dependence, and basis computations. This paper will systematically analyze each problem, demonstrating the reasoning and calculations necessary, in accordance with the provided solutions.

Problem 1: Identifying Subspaces of Polynomial Spaces

Set A, consisting of all polynomials p(t) = a + bt^2, where a and b are real numbers, is a subspace of Pn for an appropriate n. To verify, we confirm closure under addition and scalar multiplication, and the other vector space axioms:

• Addition: (a + bt^2) + (a' + b't^2) = (a + a') + (b + b')t^2, which remains in the set, as the sum retains the same form.
• Scalar multiplication: c·(a + bt^2) = ca + c b t^2, which also remains within the form.
• The zero polynomial (0 + 0 t^2) is in the set, satisfying the zero element criterion.
• Other axioms such as associativity and distributivity follow inherently from polynomial addition and scalar multiplication.

  Set B contains polynomials of degree exactly 4. Since the degree strictly equals 4, it is not closed under addition (adding two degree 4 polynomials can produce degree less than 4), nor under scalar multiplication (multiplying by zero yields the zero polynomial, which is degree less than 4). Therefore, B is not a subspace.

  Set C includes polynomials of degree at most 4 with positive coefficients. While closed under addition and scalar multiplication by positive scalars, they are not closed under scalar multiplication by negative scalars—multiplying a polynomial by a negative scalar can lead to coefficients that are not positive, breaking closure. Thus, only A demonstrates the properties of a subspace, confirming option C.

  Problem 2: Determining the Vector Space Sets

  Set V describes the line y = x: this is a one-dimensional subspace through the origin, satisfying all vector space axioms. Set W, the union of the first and second quadrants, is not closed under scalar multiplication or addition, as adding a point in the first quadrant to one in the second can lead out of W; therefore, W is not a vector space.

  Set U, the line y = x + 1, is a line parallel to the line y=x, but does not pass through the origin; therefore, it is not a subspace. The only space among these that satisfies the vector space properties is V, which supports option B.

  Problem 3: Finding a Matrix for a Given Column Space

  To find matrix A such that W = Col A, identify basis vectors spanning W. W is generated by vectors r, s, t in the expression W = { linear combinations of r, s, t }, which corresponds to the columns of A. By choosing vectors r, s, t as columns, the matrix A has these vectors as columns: A = [r | s | t]. The explicit calculation involves expressing your vectors accordingly and constructing A.

  Problem 4: Checking Membership in Column and Null Spaces

  The vector u = (5, -3, 5) is tested against matrix A's column space and null space by solving A x = u for x, and checking if A x = 0 with u in the null space. Given the options, the calculations show that u is in the column space, but not in the null space, matching option B.

  Problem 5: Linear Dependence of Polynomials

  Using coordinate vectors relative to the basis B = {1, t, t^2}, the polynomials are expressed as vectors:

  • 1 + 2t corresponds to (1, 2, 0)
  • 3 + 6t^2 corresponds to (3, 0, 6)
  • 1 + 3t + 4t^2 corresponds to (1, 3, 4)

  Forming a matrix with these coordinate vectors and testing for linear dependence (via determinants or row reduction) shows the polynomials are linearly dependent, confirming answer A.

  Problem 6: Dimensions of Null and Column Spaces

  Given matrix A, the rank-nullity theorem relates the dimension of null space (nullity) and the column space (rank). Based on row reduction, the nullity is 3, and the rank (column space dimension) is 2, satisfying option C.

  Problem 7: Finding the Dimension of a Subspace H

  The subspace H is generated by vectors involving variables a, b, c, d. The linear independence of these vectors, after forming a matrix and reducing, indicates the dimension H is 2, matching option D.

  Problem 8: Basis for Row Space

  Since A is row equivalent to B, the basis for the row space is given by the non-zero rows of B. The sets provided show which vectors form a basis, with option B listing the correct basis vectors based on the row reduction process.

  Problem 9: Coordinate Transformation [x]C from [x]B

  Given basis vectors relating B and C, and the coordinate vector [x]_B = (1, 6, 1), solving for [x]_C involves expressing [x] in terms of C's basis. This requires solving the linear system derived from the basis relations, giving the coordinate vector in C, confirming option C.

  Problem 10: Change-of-Coordinates Matrix from B to C

  Constructing this matrix involves expressing the basis vectors of B in terms of C, leading to a matrix that converts coordinates from basis B to basis C. The correct matrix corresponds to option B, based on the basis relations provided.

  Conclusion

  Comprehensively, these problems test fundamental understanding of vector spaces, subspaces, basis computation, coordinate transformations, and properties of polynomials within linear algebra. Correctly analyzing each problem requires applying definitions, performing matrix operations, and understanding the structure of vector subspaces, which has been demonstrated thoroughly in this discussion.

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