201y05mt1 - B U Department of Mathematics Math 201 Matrix...

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Unformatted text preview: B U Department of Mathematics Math 201 Matrix Theory Summer 2005 First Midterm This archive is a property of Bo˘azi¸i University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. gc This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 3 1.) a)[4] Find c such that the following set of columns is a basis for R : 1 2 1 , 1 , 1 . 1 −1 0 c Solution: 1 21 12 1 12 1 −r1 +r2 →r 2r +r →r 0 − 2− 3 −3 0 −1 0 . Hence A = 1 1 1 − − − −2 0 −1 − − −→ −− → − r1 +r3 →r3 −1 0 c 0 2 c+1 0 0 c+1 c = −1, i.e., ∀c ∈ R \ {−1} the given set of columns is a basis for R3 . b)[4] Is the set of polynomials S = {1 − x, 1 + x, 1 − x2 } linearly independent? Solution: Consider a(1 − x) + b(1 + x) + c(1 − x2 ) = 0 Then −cx2 = 0 implies c = 0. So a + b = 0 and −a + b = 0 give that a = 0, b = 0. Thus S is linearly independent. c)[2] If a matrix A is n × (n − 1) and its rank is (n − 2) what is the dimension of its null space? Solution: 2.) Since the dimension of the null space is the difference of the number of unknowns and the rank, we get dim(Null(A)) = (n − 1) − (n − 2) = 1 1 21 Let A = 2 −1 1 −1 3 0 a) [10] Find the LU decomposition of A. Solution: 1 21 12 1 12 1 −2r1 +r2 →r r2 +r →r3 − − −→ −−− A = 2 −1 1 − − − −2 0 −5 −1 − − 3− → 0 −5 −1 = U , where r1 +r3 →r3 −1 3 0 05 1 00 0 −1 −1 −1 E3 E2 E1 A = U , i.e., A = E1 E2 E3 U . Writing explicitly 100 1 00 1 0 0 1 2 1 A = 2 1 0 0 1 0 0 1 0 0 −5 −1 0 0 1 −1 0 1 0 −1 1 0 0 0 1 1 0012 1 0 0 −5 −1 = 2 −1 −1 1 0 0 0 1 00 2 1 0 where L = −1 −1 1 b) [6] Find a basis for the column space and the null space of A. What is the rank of A? Solution: From U we see pivots 1 and -5 appear in the first and second columns. Therefore that 2 1 2 , −1 is a basis for the column space of A. To find a basis for the null space −1 3 recall that Ax = 0 ⇐⇒ U x = 0. Then x1 + 2x2 + x3 = 0 −5x2 − x3 = 0 3 3 1 . Thus 1 is a basis for the implies x3 = −5x2 and x1 = 3x2 , hence, x = x2 −5 −5 null space of A. Since rank equals to the dimension of the column space, rank (A) = 2. c) [4] Using the LU decomposition of A find the complete solution to 4 3 Ax = 1 Solution: 4 4 3 implies Ly = 3 , since Ax = LU x. Then using L Setting y = U x, Ax = 1 1 from part (a), y1 =4 2y1 +y2 =3 −y1 −y2 +y3 = 1 4 entails y1 = 4, y2 = −5 and y3 = 0. Now, U x = −5 gives that 0 x1 +2x2 x3 =4 −5x2 −x3 = −5 −1 3 0 + x2 1 . hence x1 = 3x2 − 1 and x3 = −5x1 + 5, i.e., x = 5 −5 3.) a) [6] Let A be an m × n and B be an n × m matrix, and m > n. What can you say about the invertibility of AB ? Solution: We claim that AB is singular. Given m > n there exists a nonzero solution to Bx = 0, i.e., ∃x0 = 0 such that Bx0 = 0. Then (AB )x0 = A(Bx0 ) = 0. But AB being an m × m matrix and (AB )x0 being zero with x0 = 0 implies dim(Null(AB )) = 0, hence rank (AB ) = m. Thus AB is not invertible. b) [6] Let A and B be n × n matrices. Show that if A is singular then AB is also singular. Solution: Assume that A is singular. Then AT is also singular, i.e., AT x has a non-trivial solution, say, AT x0 = 0 for some x0 = 0. But then we get that (AB )T x0 = B T (AT x0 ) = 0, so that (AB )T is singular. Thus AB is singular. c) [3] If A is an n × n matrix with A2 = A and rank(A)=n, find A. Solution: rank (A) = n implies that A is invertible, i.e., A−1 exists. Then multiplying both sides of A2 = A by A−1 we get A2 A−1 = AA−1 = I and so A = I. 4.) Let T : R3 → R3 be defined by T (x, y, z ) = (x + 2y + z, x + y, 2y + z ). a) [2] Write down what we must show to prove that T is a linear transformation. (Do not carry out the computations). b) [5] What is the matrix representing this transformation in the standard basis for R3 . c) [8] Show that T is non-singular and find its inverse transformation. Solution: a) We have to show that given two points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) in R3 T ((x1 , y1 , z1 ) + (x2 , y2 , z2 )) = T (x1 , y1 , z1 ) + T (x2 , y2 , z2 ), T (c(x1 , y1 , z1 )) = cT (x1 , y1 , z1 ), ∀c ∈ R. b) Since T (1, 0, 0) = (1, 1, 0), T (0, 1, 0) = (2, 1, 2) and T (0, 0, 1) = (1, 0, 1), we get that the matrix representing T is 121 A = 1 1 0 . 021 c) To show that T is non-singular, it suffices to show that A is row equivalent to I3×3 . Using Gauss-Jordan method we get that 12110 [A|I ] = 1 1 0 0 1 02100 1 2 0 −1 r2 −r →r2 − − 3− → 0 1 0 −1 −−− r1 −r3 →r1 001 2 0 12 r −r →r2 0 −1− 2− → 0 1 −−− 2r2 −r3 →r3 1 00 2 1 r1 −2r →r 1 1 − − 2 −1 −− → − −2 −1 11 0 0 1 1 −1 0 1 2 −2 −1 100 1 0 −1 1 = [I |A−1 ]. 0 1 0 −1 1 0 0 1 2 −2 −1 . Knowing A−1 1 0 −1 1 , we get that = −1 1 2 −2 −1 T −1 (x, y, z ) = (x − z, −x + y + z, 2x − 2y − z ). ...
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