Math 110 Homework Assignment 9 due date: Nov. 18, 2016 1. (a) Prove or disprove: the vectors ~v1 = (2, 3) and ~v2 = (1, 5) form a basis for R2 . (b) Prove or disprove: the vectors w ~ 1 = (1, 9) and w ~ 2 = (2, 3) form a basis for R2 . (c) How many bases can a subspace have? (d) Let V be the set of vectors (x, y, z, w) in R4 which are the solutions to the equations: x + 0y + 3z 2w = 0 0x + y 4z 9w = 0 The subset V is a subspace of R4 . Find a basis for V . (Suggestion: V is given as the set of solutions to a system of linear equations. You know how to parameterize all the solutions. . . ) 2. Let ~v1 ,. . . , ~vk be vectors in Rn , and let A = [~v1 | ~v2 | ~vk ], i.e., the matrix whose columns are ~v1 ,. . . , ~vk . Prove that ~v1 ,. . . , ~vk are linearly independent if and only if Rank(A) = k. Reminder: Proving a statement with an \"if and only if\" requires proving both directions. You assume that Rank(A) = k and then deduce that ~v1 ,. . . , ~vk are linearly independent. Then assume that ~v1 , . . . , ~vk are linearly independent and prove that Rank(A) = k. (If you can do both steps at the same time that is fine too.) One other reminder: looking for c1 ,. . . , ck so that c1~v1 + + ck~vk = ~0 is the same as solving a system of linear equations. 3. Linear transformation puzzlers (a) Consider a linear transformation T : Rn Rm . If ~v1 ,. . . , ~vk are linearly dependent vectors in Rn , are the vectors T (~v1 ),. . . , T (~vk ) necessarily linearly dependent in Rm ? If so, why? (b) If A is an n p matrix, and B is a p m matrix, with Im(B) Ker(A), what can you say about the product AB? 1 (c) if A is a p m matrix, and B a q m matrix, and we make a (p + q) m matrix C by \"stacking\" A on top of B: \u0014 \u0015 A C= , B what is the relation between Ker(A), Ker(B), and Ker(C)? Note: So far, we have only used the symbols Ker and Im when talking about a linear transformation T . In this homework we're going to extend this notation and also use Ker and Im when talking about a matrix. If A is an m n matrix, then n o Ker(A) = ~v Rn | A~v = ~0 . While n o Im(A) = w ~ Rm | there is a v Rn so that A~v = w ~ . The connection between this notation and our usual notation about linear transformations is that if T : Rn Rm is a linear transformation and A the standard matrix of T , then Ker(T ) = Ker(A) and Im(T ) = Im(A). 4. (a) Suppose that we have a system of linear equations in n variables. For instance, we might have m equations: a11 x1 + a12 x2 + + a1n xn = 0 a21 x1 + a22 x2 + + a2n xn = 0 .. .. .. . . . am1 x1 + ak2 x2 + + amn xn = 0 where the aij are any numbers in R. Show that the set of solutions to this system of equations forms a subspace of Rn . (b) The vectors ~v1 = (1, 3, 1, 2), ~v2 = (2, 3, 2, 7), and ~v3 = (2, 1, 1, 6) span a 3dimensional subspace of R4 . Find a single equation of the form ax+by+cz+dw = 0 whose solutions are this subspace. 2 Math 110 Homework Assignment 8 due date: Nov. 11, 2016 1. Determine if the linear transformations described by the following matrices are invertible. If not, explain why, and if so, find the matrix of the inverse transformation. (a) \u0014 4 0 0 3 \u0015 (b) 3 1 5 (e) 6 3 1 0 0 0 \u0014 2 0 6 0 3 1 1 0 3 1 (f) 5 6 7 10 \u0015 0 0 1 4 (c) \u0014 7 3 9 4 \u0015 (d) \u0014 3 6 2 4 \u0015 0 0 0 1 2. Suppose that A is the matrix 5 2 4 A = 2 3 1 . 5 6 3 (a) Find the inverse of A. (b) Explain why, for any values of a, b, and c, the equations 5x + 2y + 4z = a 2x + 3y + z = b 5x + 6y + 3z = c always have a unique solution. (c) Find this unique solution (in terms of a, b, and c). 3. Suppose that T1 : Rn Rm and T2 : Rm Rp are linear transformations. (a) If T1 and T2 are injective, prove that T2 T1 is injective. (b) If T1 and T2 are surjective, prove that T2 T1 is surjective. (c) If T1 and T2 are invertible, prove that T2 T1 is invertible. 1 4. For each of the following subsets W of R3 , either show that they are subspaces, or show why they aren't subspaces by explaining which of the three conditions don't hold. (a) W = {(x, y, z) R3 | x2 + y 2 = z 2 }. (b) W = {(x, y, z) R3 | (x, y, z) is orthogonal to (3, 1, 2)}. (c) W = {(x, y, z) R3 | x + y + z > 0}. 2