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APSC 174J, Solutions #5 Posted: March 30, 2016 Section 8, Problem 1(b). Let ~v = (0, 0, 1), then easy to see ~v cannot be written as a linear combination of ~v2 = (1, 1, 0). So {~v2 } is not a generating set for R3 . Section 8, Problem 1(g). Let ~v = (1, 1, 1), then ~v cannot be written as a linear combination of ~v2 = (1, 1, 0) and ~v3 = (1, 2, 1). To see this, we solve the linear system: x1~v2 + x2~v3 = ~v : 1 1 1 1 1 1 1 1 1 1 2 1 → 0 1 0 → 0 1 0 0 1 1 0 1 1 0 0 1 From the last row of the row-echelon from, we see there are no solutions for x1 , x2 , which means ~v is not in the linear span of {~v2 , ~v3 }. Hence {~v2 , ~v3 } is not a generating set for R3 . Note: Since the dimension of R3 is 3, we know any generating set for R3 should have at least three vectors. So it is easy to see that {~v2 } and {~v2 , ~v3 } are not generating sets for R3 . Section 8, Problem 1(k). To show {~v1 , ~v3 , ~v4 } is a generating set for R3 , we need to show that any vector ~v = (b1 , b2 , b3 ) in R3 can be written as a linear combination of ~v1 , ~v3 , ~v4 . That is, the linear system: x1~v1 + x2~v3 + x3~v4 = ~v always have a solution. 1 1 0 b1 1 1 0 b1 0 2 0 b2 → 0 2 0 b 2 0 1 3 b3 0 0 3 b3 − b2 /2 From the row-echelon form, we know the linear system always has a solution no matter what the values for b1 , b2 , b3 are. So {~v1 , ~v3 , ~v4 } is a generating set for R3 . 2 Section 8, Problem 3(a). Let ~v = (1, −3, 0, 0), then ~v is in W , since the entries of ~v satisfy the conditions of W : 3(1) + (−3) − 0 = 0 and 0 − 2 · 0 = 0 Since ~v = (1, −3, 0, 0) cannot be written as a linear combination of ~v1 = (1, 0, 6, 3), we know {~v1 } does not generate W , i.e. it is not a generating set for W . Section 8, Problem 3(f ). To show {~v2 , ~v3 } is a generating set for W , we need to show that the vector equation: α1 (0, 1, 2, 1) + α2 (1, 1, 8, 4) = (x, y, z, w) has a solution for any (x, y, z, w) in W . Claim: α1 = y − x, α2 = x is a solution for the equation. In fact, if (x, y, z, w) is in W , then 3x + y − w = 0 and z − 2w = 0. Hence if we express z, w using x, y, we get z = 2w = 6x + 2y, w = 3x + y. Thus α1 (0, 1, 2, 1) + α2 (1, 1, 8, 4) = (y − x)(0, 1, 2, 1) + x(1, 1, 8, 4) = (x, y, 6x + 2y, 3x + y) i.e. α1 = y − x, α2 = x is a solution for the vector equation, so any vector in W can be written as a linear combination of ~v2 , ~v3 . That is, {~v2 , ~v3 } is a generating set for W . Section 8, Problem 4(e). Now we know the dimension of R2 is 2, so any basis of R2 must have exactly two vectors. Since B1 = (~v1 , ~v2 ) contains two vectors, to show it is a basis, we only need to show it is linearly independent. Let α1 (1, 1) + α2 (−1, 1) = (0, 0): 1 −1 0 1 1 0 → 1 −1 0 0 2 0 the only solution is α1 = α2 = 0, so {~v1 , ~v2 } is linearly independent. Hence it is a basis for R2 . Section 8, Problem 4(l). Since any basis of R2 must have exactly two vectors, we know immediately that (~v1 , ~v2 , ~v4 ) is not a basis for R2 . 3 Section 8, Problem 5(b). Let ~v = (1, 0, 0, 1), then ~v is in W , since the entries of ~v satisfy the conditions of W : 1 + 0 + 0 − 1 = 0 and 0 − 0 = 0 Since ~v = (1, 0, 0, 1) cannot be written as a linear combination of ~v2 = (0, 1, 1, 2), we know {~v2 } does not generate W , i.e. it is not a generating set for W . Section 8, Problem 5(g). To show B3 = (~v3 , ~v1 ) is a basis of W , we need to show {~v1 , ~v3 } is a generating set for W and also {~v1 , ~v3 } is linearly independent. 1. To show {~v1 , ~v3 } is a generating set for W , we need to show that the vector equation: α1 (1, 0, 0, 1) + α2 (1, 1, 1, 3) = (x, y, z, w) has a solution for any (x, y, z, w) in W . Claim: α1 = x − y, α2 = y is a solution for the equation. In fact, if (x, y, z, w) is in W , then x + y + z − w = 0 and y − z = 0. Hence if we express z, w using x, y, we get z = y, w = x + 2y. Thus α1 (1, 0, 0, 1, ) + α2 (1, 1, 1, 3) = (x − y)(1, 0, 0, 1, ) + y(1, 1, 1, 3) = (x, y, y, x + 2y) i.e. α1 = x − y, α2 = y is a solution for the vector equation, so any vector in W can be written as a linear combination of ~v1 , ~v3 . That is, {~v1 , ~v3 } is a generating set for W . 2. To show {~v1 , ~v3 } is linearly independent, we solve the vector equation: α1 (1, 0, 0, 1) + α2 (1, 1, 1, 3) = (0, 0, 0, 0) Easy to see the solution is α1 = α2 = 0, which implies that {~v1 , ~v3 } is linearly independent. Combining 1. and 2. we conclude that B3 = (~v3 , ~v1 ) is a basis of W . Section 8, Problem 5(o). 1. The component vector of ~v2 with respect to the basis B1 = (~v1 , ~v2 ) is the pair of numbers (α1 , α2 ) so that α1~v1 + α2~v2 = ~v2 . α1 (1, 0, 0, 1) + α2 (0, 1, 1, 2) = (0, 1, 1, 2) The solution is α1 = 0, α2 = 1. So the component vector of ~v2 with respect to the basis B1 is (0, 1). 4 2. The component vector of ~v2 with respect to the basis B2 = (~v1 , ~v3 ) is the pair of numbers (α1 , α2 ) so that α1~v1 + α2~v3 = ~v2 . α1 (1, 0, 0, 1) + α2 (1, 1, 1, 3) = (0, 1, 1, 2) The solution is α1 = −1, α2 = 1. So the component vector of ~v2 with respect to the basis B1 is (−1, 1). 3. The component vector of ~v2 with respect to the basis B3 = (~v3 , ~v1 ) is the pair of numbers (α1 , α2 ) so that α1~v3 + α2~v1 = ~v2 . α1 (1, 1, 1, 3) + α2 (1, 0, 0, 1) = (0, 1, 1, 2) The solution is α1 = 1, α2 = −1. So the component vector of ~v2 with respect to the basis B1 is (1, −1). 4. The component vector of ~v2 with respect to the basis B4 = (~v1 , ~v4 ) is the pair of numbers (α1 , α2 ) so that α1~v1 + α2~v4 = ~v2 . α1 (1, 0, 0, 1) + α2 (1, −1, −1, −1) = (0, 1, 1, 2) The solution is α1 = 1, α2 = −1. So the component vector of ~v2 with respect to the basis B1 is (1, −1).