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Linear Algebra — Homework 4 Solutions 1. Write the following vectors as linear combinations of (2, 1, −1), (3, 1, 1), (−4, 0, 6). (a) (1, 2, 6) (1, 2, 6) = 1(2, 1, −1) + 1(3, 1, 1) + 1(−4, 0, 6) (b) (1, 23 , 1) (1, 23 , 1) = 13 (2, 1, −1) + 31 (3, 1, 1) + 61 (−4, 0, 6) 2. Determine whether each of the following vectors is a linear combination of (1, 1, 1, 0), (1, 0, 0, 1), (0, 1, 0, 1). (a) (1, 5, 3, 0) (1, 5, 3, 0) = 3(1, 1, 1, 0) − 2(1, 0, 0, 1) + 2(0, 1, 0, 1). (b) (1, 1, 1, −3) (1, 1, 1, −3) is not a linear combination of these vectors. 3. Write the polynomial 4x3 − 7x2 − 3x + 8 as a linear combination of x + 1, x2 + x, x3 + x2 . 4x3 − 7x2 − 3x + 8 = 8(x + 1) − 11(x2 + x) + 4(x3 + x2 ). 4. (a) Show that {(2, 4, −2), (3, 2, 0), (1, −2, −2)} spans R3 . To show that {(2, 4, −2), (3, 2, 0), (1, −2, −2)} spans R3 , we will choose a representative element of R3 and show that it can be written as a linear combination of the three vectors. Let (x, y, z) ∈ R3 . We attempt to solve the equation a(2, 4, −2) + b(3, 2, 0) + c(1, −2, −2) = (x, y, z) (2a + 3b + c, 4a + 2b − 2c, −2a − 2c) = (x, y, z) So we get the system of equations 2a + 3b + c = x 4a + 2b − 2c = y −2a − 2c = z. Adding the first and third equations, we get 3b − c = x + z. (1) Now take 2 times the third equation and add it to the second to get 2b − 6c = y + 2z. (2) . Plugging Take 6 times (1) and subtract (2) to get 16b = 6x − y + 4z. So b = 6x−y+4z 16 18x−3y+12z − x − z. Then from the first equation in the system we into (1) we get c = 16 18x−3y+12z 18x−3y+12z 1 get a = 2 x − −( − x − z) . Thus for any x, y and z the system has 16 16 3 a solution. Hence the vectors span R . (b) Show that {(2, 4, −2), (3, 2, 0), (1, −2, 2)} does not span R3 . We will show that (1, 0, 0) is not in the span of {(2, 4, −2), (3, 2, 0), (1, −2, 2)} and hence this set does not span R3 . We attempt to solve the equation a(2, 4, −2) + b(3, 2, 0) + c(1, −2, 2) = (1, 0, 0) (2a + 3b + c, 4a + 2b − 2c, −2a + 2c) = (1, 0, 0) So we get the system of equations 2a + 3b + c = 1 4a + 2b − 2c = 0 −2a + 2c = 0. The third equation tells us c = a. Plugging into the second equation we get b = −a. Plugging these into the first equation gives us 0 = 1. Thus the system has no solution. 5. (a) If possible, give a set with five elements that spans P3 . {1, x, x2 , x3 , 2x2 − 5x + 9}. (b) If possible, give a set with four elements that spans P3 . {1, x, x2 , x3 }. (c) If possible, give a set with three elements that spans P3 . Impossible. 6. Let W = {ax3 + 2bx2 − 3(a + b)x + (a − 2b) ∈ P3 | a, b ∈ R} (a) Write down two different elements of W . Letting a = 0 and b = 0 we get 0x3 + 0x2 + 0x + 0, which is the 0-polynomial. Letting a = 1 and b = 0 we get x3 − 3x + 1. (b) Find an element of P3 that is not an element of W . We claim that the constant polynomial 1 is not in W . If it were, then we could find a and b such that ax3 + 2bx2 − 3(a + b)x + (a − 2b) = 1. Setting coefficients on the left equal to those on the right, we get a = 0 and b = 0, but a − 2b = 1. Thus there are no such a and b. (c) Find a finite set of elements of W that spans W . Note: {1, x, x2 , x3 } is wrong. Rewrite W as follows: W = {ax3 + 2bx2 − 3(a + b)x + (a − 2b) ∈ P3 | a, b ∈ R} = {ax3 + 2bx2 − 3ax − 3bx + a − 2b ∈ P3 | a, b ∈ R} = {a(x3 − 3x + 1) + b(2x2 − 3x − 2) ∈ P3 | a, b ∈ R} Thus every element of W is a linear combination of x3 − 3x + 1 and 2x2 − 3x − 2. That is, {x3 − 3x + 1, 2x2 − 3x − 2} spans W . Letting a = 1 and b = 0 demonstrates that x3 − 3x + 1 ∈ W . Similarly, letting a = 0 and b = 1 demonstrates that 2x2 − 3x − 2 ∈ W . (d) Show that W is a subspace of P3 . You could use the subspace theorem, but it’s much more efficient to cite Homework 3, problem 8. Since W is the span of two vectors in P3 , W is a subspace of P3 . 7. Suppose {~v1 , ~v2 , . . . , ~vk } spans the vector space V , and let w ~ be another vector in V . Prove that {~v1 , ~v2 , . . . , ~vk , w} ~ also spans V . Proof: We wish to show that {~v1 , ~v2 , . . . , ~vk , w} ~ spans V . So we must choose a representative element of V and show that it can be written as a linear combination of those vectors. Let ~x ∈ V . Since {~v1 , ~v2 , . . . , ~vk } spans V , we know we can write ~x = a1~v1 + a2~v2 + · · · + ak~vk . But then ~x = a1~v1 + a2~v2 + · · · + ak~vk + 0w, ~ since 0w ~ = ~0. Thus ~x is a linear combination of ~v1 , ~v2 , . . . , ~vk , w. ~ Since ~x is representative, this means {~v1 , ~v2 , . . . , ~vk , w} ~ spans V.