辅导案例-MATH2099

  • August 14, 2020

Term 2, 2019 MATH2099 Page 2 Part A – Linear Algebra 1. Answer in a separate book marked Question 1 a) [4 marks] i) Suppose that M1 = ( 1 3 2 −1 ) , M2 = ( 3 −1 4 −5 ) , M3 = ( 2 −3 7 −11 ) , and M2,2 is the vector space of all 2× 2 matrices. Is the set S = {M1,M2,M3} a spanning set for M2,2? Give reasons. ii) Let V = C[0, 1] be the vector space of continuous real-valued func- tions on [0, 1], where addition and scalar multiplication are defined by (f + g)(t) = f(t) + g(t) and (λf)(t) = λf(t) for all f, g ∈ V and all λ ∈ R. Is the set W = { f ∈ V : ∫ 1 0 f(t) dt = 1 } a vector subspace of V ? Give reasons. b) [6 marks] Let B = {1− t, 2− t+ t2, 1 + 3t+ 2t2} and S = {1, t, t2, t3}. Suppose that T : P2 → P3 is a linear map. i) Giving full reasons, prove that B is a basis for P2. ii) Find the coordinate vector of 3 + 4t+ 3t2 with respect to B. iii) Suppose that the matrix of T with respect to the bases B for the domain and S for codomain is M =  1 0 0 −1 2 1 −1 0 1 0 −1 2  . Find T (3 + 4t+ 3t2). c) [4 marks] Find the line y = a+ bx which best fits, in the least squares sense, the points (0,−1), (1, 0), (2,−1), (4, 3). Please see over . . . Term 2, 2019 MATH2099 Page 3 d) [6 marks] Let x1 =  1 2 2 0  , x2 =  −1 −4 0 1  and x3 =  3 −1 7 2  , and write W = span{x1,x2,x3 }. i) Find an orthonormal basis for W . ii) Find a QR factorisation of B =  1 −1 3 2 −4 −1 2 0 7 0 1 2  . Please see over . . . Term 2, 2019 MATH2099 Page 4 2. Answer in a separate book marked Question 2 a) [4 marks] Consider the surface in R3 with equation x2 + 2y2 + z2 + 8xy + 6xz − 8yz = 12. This equation can be written in the form ( x y z ) A xy z , where A = 1 4 34 2 −4 3 −4 1  . i) Show that 4 is an eigenvalue of A. ii) If −6 is another eigenvalue of A, find the third eigenvalue of A. iii) Identify the surface and find the shortest distance from the surface to the origin. b) [3 marks] Suppose that A is a 10 × 10 matrix. You are given that A has only one eigenvalue 3, and nullity(A− 3I) = 4, nullity(A− 3I)2 = 6, nullity(A− 3I)3 = 8. Find all possible Jordan forms of A. c) [10 marks] Let C = ( 7 −4 1 3 ) . i) Find the eigenvalues and eigenvectors of C. ii) Is C diagonalizable? Give reasons for your answer. iii) Find an invertible matrix P and a Jordan form J such that C = PJP−1. iv) Calculate etC . v) Solve the system of differential equations.{ y′1 = 7y1 − 4y2 y′2 = y1 + 3y2 where y1(0) = 1 and y2(0) = 3. d) [3 marks] Let M be an n× n matrix. Prove that nullity(M) ≤ nullity(M2). Please see over . . . MATH2099 Term 2, 2019 Examination Question 1 Solutions/Marking Scheme a) i) The dimension of M2,2 is 4. A spanning set for M2,2 must contain at least 4 elements. The S of three elements cannot be a spanning set of M2,2. ii) The zero function, f(t) = 0 for all t ∈ [0, 1], is not in W because∫ 1 0 f(t) dt = ∫ 1 0 0 dt = 0 6= 1. Hence W is not a subspace of V . [For each part, 2 marks for the truth value and the explanation; deduct one mark for poorly presented explanation.] b) The sets B = {1 − t, 2 − t + t2, 1 + 3t + 2t2} and S = {1, t, t2, t3}, and T : P2 → P3 is a linear map. Let p1(t) = 1− t, p2(t) = 2− t+ t2, p3(t) = 1 + 3t+ 2t2 and q(t) = 3 + 4t+ 3t2. i) Since dim(P2) = 3 and the set B has three elements, to prove that B is a basis for P2 we only need to prove that B is linearly independent. For both (i) and (ii), we reduce the following augmented matrix for λ1p1 + λ2p2 + λ3p3 = q . 1 2 1−1 −1 3 0 1 2 ∣∣∣∣∣∣ 3 4 3  −→ 1 2 10 1 4 0 1 2 ∣∣∣∣∣∣ 3 7 3  −→ 1 2 10 1 4 0 0 −2 ∣∣∣∣∣∣ 3 7 −4  Since all columns on the left are leading, B is a linearly independent set and hence it is a basis for P2. ii) Solve the system in (i), we obtain λ3 = 2, λ2 = −1, λ1 = 3. Hence the coordinate vector of q with respect to B is [3 + 4t+ 3t2]B =  3−1 2  . iii) As [T (q)]S = M  3−1 2  =  1 0 0 −1 2 1 −1 0 1 0 −1 2   3−1 2  =  3 −3 −1 5  , Hence T (3 + 4t+ 3t2) = 3− 3t− t2 + t3. [1 mark for the augmented matrix; 1 mark for row reduction, 1 mark for explana- tion/answer of each of (i) and (ii). For (iii), 1 mark for M multiply with the coordinate vector in (ii); 1 mark for the answer (in vector form).] c) If the four given points lie exactly on the line y = a+bx, we have in matrix form Ax = y, where A =  1 0 1 1 1 2 1 4  , x = (ab ) , y =  −1 0 −1 3  . We thsn solve the normal equations ATAx = ATy. 1 2( 4 7 7 21 ) x = ( 1 10 ) . Hence, x = ( 4 7 7 21 )−1( 1 10 ) = 1 35 ( 21 −7 −7 4 )( 1 10 ) = (−7 5 33 35 ) . Therefore, the line of best fit is y = −7 5 + 33 35 x. [1 mark for Ax = y with A and y calculated. 1 mark for the products ATA and ATy. 1 mark for the inverse of the 2 × 2 matrix. 1 mark for the equation of the line of best fit (or the values of a and b.] d) i) Apply Gram-Schmidt process. v1 = x1 =  1 2 2 0  v2 = x2 − projv1x2 = x2 − x2 · v1 v1 · v1v1 = x2 + v1 =  0 −2 2 1  v3 = x3 − projv1x3 − projv2x3 = x3 − x3 · v1 v1 · v1v1 − x3 · v2 v2 · v2v2 = x3 − 5 3 v1 − 2v2 = 1 3  4 −1 −1 0  Normalise v1, v2, v3 to obtain an orthonormal basis {u1, u2, u3} for W , where u1 = 1 3 v1, u2 = 1 3 v2, u3 = 1√ 2 v3 ii) By writing x1, x2, x3 as subjects x1 = 3u1 x2 = −3u1 + 3u2 x3 = 5u1 + 6u2 + √ 2u3 Hence, B = (x1|x2|x3) = (u1|u2|u3) 3 −3 50 3 6 0 0 √ 2  = QR, where Q =  1 3 0 2 √ 2 3 2 3 −2 3 − 1 3 √ 2 2 3 2 3 − 1 3 √ 2 0 1 3 0  , R = 3 −3 50 3 6 0 0 √ 2  . [1 mark for an attempt to use Gram-Schmidt, one mark for each of v2,v3. 1 mark for normalisation. 1 mark for Q. 1 mark for R.] MATH2099 Term 2, 2019 Examination Question 2 Solutions/Marking Scheme a) A = 1 4 34 2 −4 3 −4 1 . i) As the first row and third row of A− 4I = −3 4 34 −2 −4 3 −4 −3  are scalar multiple of each other, det(A− 4I) = 0. Hence 4 is an eigenvalue. ii) Let λ be the last eigenvalue. Hence 4− 6 + λ = tr(A) = 1 + 2 + 1 = 4. Hence λ = 6. iii) The equation of the surface with respective to the principal axes is 6X2 − 6Y 2 + 4Z2 = 12 ⇔ X 2 2 − Y 2 2 + Z2 3 = 1. The surface is a hyperboloid of one sheet. The shortest distance to the original is√ 2. [1 mark for each of part (i) and part (ii). For part (iii), one mark for the equations with respect to the principal axes or the type of surface and one mark for the shortest distance or the closest point w.r.t. principal axes.] b) Let dk = nullity(A− 3I)k. The sequence {dk} can only be d1 = 4, d2 = 6, d3 = 8, d4 = 9, d5 = 10, . . . or d1 = 4, d2 = 6, d3 = 8, d4 = 10, . . . Hence the possible Jordan forms of A are J5(3)⊕ J3(3)⊕ J1(3)⊕ J1(3) or J4(3)⊕ J4(3)⊕ J1(3)⊕ J1(3) [One mark for drawing diagrams or writing at least one sequence of dk. One mark for each possible Jordan form. Do not deduct marks for using the wrong notation Jλ(k). One mark if a student read the question wrong and gave an answer J3(3)⊕ J3(3)⊕ J1(3)⊕ J1(3).] c) C = ( 7 −4 1 3 ) . i) The eigenvalues of C are roots of det(C − λI) = 0. det ( 7− λ −4 1 3− λ ) = (7− λ)(3− λ) + 4 = λ2 − 10λ+ 25 = (λ− 5)2 The eigenvectors are non-zero solutions of (C − 5I)v = 0. C − 5I = ( 2 −4 1 −2 ) −→ ( 1 −2 0 0 ) Hence the eigenvectors are non-zero element in span {( 2 1 )} . 1 2ii) The matrix C is not diagonalisable because the algebraic multiplicity of 5 is 2 while the geometric multiplicity is 1. iii) Take a generalised eigenvector which is not in E5, v1 = ( 1 0 ) . We obtain v2 = (C − 5I)v1 = ( 2 1 ) . Hence take P = ( 2 1 1 0 ) and J = ( 5 1 0 5 ) , then C = PJP−1. iv) From (iii), etC = PetJP−1 = Pe5t ( 1 t 0 1 ) P−1 = e5t ( 2 1‘ 1 0 )( 1 t 0 1 )( 0 1 1 −2 ) = e5t ( 1 + 2t −4t t 1− 2t ) v) The system of differential equations can be written as y = Cy with initial condition y(0) = ( 1 3 ) . Hence, the solution is y = etCy(0) = e5t ( 1 + 2t −4t t 1− 2t )( 1 3 ) = e5t ( 1− 10t 3− 5t ) Hence, y1 = e 5t(1− 10t) and y2 = e5t(3− 5t). [(i) 1 mark for the eigenvalue and 1 mark for the eigenvectors. (ii) 1 mark for an answer with explanation. (iii) 1 mark for the chain of generalised eigenvectors. 1 mark each for the matrices P and J . (iv) 1 mark for etJ . 1 mark for using etC = PetJP−1. 1 mark for etC . (v) 1 mark for the solution.] d) For any vector v ∈ ker(M), we have Mv = 0. Then M2v = M(Mv) = M0 = 0. Hence v ∈ ker(M2). Therefore nullity(M) ≤ nullity(M2). [2 marks for a correct proof for ker(M) ⊂ ker(M2). 1 mark for the conclusion about nullity.]

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