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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
1)
2) 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) 0 3 -1 4 -1 3 0 1 3 1 -5 1
B) 3 0 -1 4 -1 3 0 1 3 1 -5 1
C) 3 -1 4 3 1 3 1 -5
D) 3 4 0 1 0 -1 1 -5 -1 3 3 1
3)
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 =
1 -3 4 -1 0 -5 3 -3 6
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 = 1 -5 -4 3 0 -2 3 -1 -4 1
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 = 3
6)
1
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 =
1 3 -4 0 1 2 4 -5 5 -2 1 -5 0 -3 2 -3 -1 8 3 -4
, B =
1 3 -4 0 1 0 -2 3 5 -4 0 0 -8 -23 17 0 0 0 0 0
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) 10 23 8
B) 3 24 -3
C) 10 23 -4
D) 10 25 -6
9)
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)
1 -2 3 -4
B) 1 3
-2 -4
C) 0 3
-2 -4
D) 1 3 2 4
10)
2
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
3
