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Last name: name: 1 Mid term exam 2 (Notes, books, and calculators are not authorized). Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with no justification will not be graded. 1 2 3 Question 1: Is A = 0 1 2 diagonalizable? Be very accurate when answering. 0 0 1 Solution 1: The characteristic polynomial is PA (λ) = det(A − λI) = (1 − λ)3 . This means that there is only one eigenvalue of multiplicity three. Let v = (x, y, z) be an eigenvector, then 2y + 3z = 0 2z = 0 which implies that y = z = 0. The eigenvectors are of the following form v = (α, 0, 0), α ∈ R and α 6= 0. This means that the eigenspace E1 is the one-dimensional line E1 := span{(1, 0, 0)}, i.e., all the eigenvectors are parallel to (1, 0, 0). As a result, it is not possible to find three independent eigenvectors. The matrix is not diagonalizable. Solution 2: We deduce as above that 1 is the only eigenvalue of multiplicity three. If A was diagonal, we would have A = P DP −1 where D = I. This would mean A = P IP −1 = P P −1 = I, i.e., A = I, which is obviously wrong. In conclusion A cannot be diagonalizable. Question 2: The following linearly independent vectors e1 := (1, −1, 0), e2 := (1, 0, 1), and 4 1 −1 e3 := (1, 2, 1) are eigenvectors of A = 2 5 −2. Find an invertible matrix P and a diagonal 1 1 2 matrix D so that A = P DP −1 . The matrix is clearly diagonalizable since there exist three linearly independent eigenvectors. Let us compute Ae1 , Ae2 and Ae3 to get the associated eigenvalues. (A second possibility would be to compute the characteristic polynomial of A, but that would be too long and unnecessarily complicated since we are already given the eigenvectors.) Ae1 = (3, −3, 0) = 3e1 Ae2 = (3, 0, 3) = 3e2 Ae3 = (5, 10, 5) = 5e3 . This means that e1 and e2 are associated with the eigenvalue 3 (the multiplicity of this eigenvalue is two) and e3 is associated with the eigenvalue 5. Let us now set 3 0 0 1 1 1 D = 0 3 0 , P = −1 0 2 . 0 0 5 0 1 1 Then A = P DP −1 . 2 Mid term exam 2, November 8, 2011 3 − λ PA (λ) := det(A − λI) = 2 3 2 −4 0 −1 and B = . −6 1 0 −4 = (3 − λ)(−6 − λ) + 8 = λ2 + 3λ − 10 −6 − λ Question 3: Find the eigenvalues of A = i.e., PA (λ) = λ2 + 3λ − 10 = (λ − 2)(λ + 5). The eigenvalues of A are λ1 = 2 and λ1 = −5. −λ −1 = λ2 + 1 PB (λ) := det(B − λI) = 1 −λ i.e., PB (λ) = λ2 + 1 = (λ + i)(λ − i). The eigenvalues of B are λ1 = i and λ1 = −i. Question 4: The following λ1 := −2,v1 1 4, v3 := (1, 1, 2) are eigen-pairs of A = 3 6 du(t) of the ODE system dt = Au(t)? := (1, 1,0), λ2 := −2, v2 := (1, 0, −1), and λ3 := −3 3 −5 3. (a) What are the fundamental solutions −6 4 The fundamental solutions have been defined in class and are e−2t v1 , e−2t v2 , and e4t v3 . This is all I wanted you to write. Recall that upon defining the matrix P := [v1T v2T v3T ], we have A = P DP −1 where D is the diagonal matrix composed of the eigenvalues. Then the ODE can be re-written as follows: d(P −1 u) = DP −1 u. dt −1 This can also be re-written dw u(t). Then w(t) is given by dt = Dw by defining w(t) := P α1 e−2t 0 0 1 w(t) = α2 e−2t = α1 e−2t 0 + α2 e−2t 1 + α3 e4t 0 , 0 1 0 α3 e4t where α1 , α2 , α3 are any real numbers. As a result, 1 0 u(t) = P w(t) = α1 e−2t P 0 + α2 e−2t P 1 + α3 e4t P 0 0 0 0 = α1 e−2t v1 + α1 e−2t v2 + α1 e4t v3 . 1 In conclusion, u(t) is a linear combination of the fundamental solutions e−2t v1 , e−2t v2 , e4t v3 . (b) What is the solution of du(t) dt = Au(t), with u(0) = (2, 1, −1)? We know that u(t) is a linear combination of the fundamental solutions: u(t) = α1 e−2t v1 + α2 e−2t v2 + α3 e4t v3 . The condition u(0) = (2, 1, −1) implies (2, 1, −1) = α2 v1 + α2 v2 + α3 v3 . The corresponding 1 1 1 0 0 −1 augmented matrix is 1 2 1 1 1 2 1 1 1 ∼ 0 −1 0 −1 ∼ 0 2 −1 0 −1 2 −1 0 1 1 0 1 0 2 2 1 1 ∼ 0 0 0 The solution is (α1 , α2 , α3 ) = (1, 1, 0). This means u(t) = e−2t v1 + e−2t v2 = e−2t (2, 1, −1). 0 1 0 0 0 1 1 1 . 0 Last name: name: 3 Question 5: Are the following functions f1 (t) = 1−3 cos(t)+2 sin(t), f2 (t) = 2−4 cos(t)−sin(t), f3 (t) = 1 − 5 cos(t) + 7 sin(t) linearly independent in V := span{1, cos(t), sin(t)}? Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 f1 + x2 f2 + x3 f3 = 0. Then x1 f1 (t)+x2 f2 (t)+x3 f3 (t) = x1 +2x2 +x3 +(−3x1 −4x2 −5x3 ) cos(t)+(2x1 −x2 +7x3 ) sin(t) = 0. This means that X solves 1 −3 2 2 −4 −1 1 −5 X = 0. 7 Let us reduce the matrix of the linear system in echelon form 1 2 1 1 2 1 1 −3 −4 −5 ∼ 0 2 −2 ∼= 0 2 −1 7 0 −5 5 0 2 2 0 1 −2 . 0 There are only two pivots. There is one free variable. This means that the solution set of the above linear system is not {0}. There is some nonzero vector X so that x1 f1 + x2 f2 + x3 f3 = 0. This means that the functions f1 , f2 , f3 are linearly dependent. Question 6: Let C 1 (R; R) be the vector space over R of the functions over R that are continuously differentiable. Let S = {1, cos(t), sin(t)} and consider V = span(S). Accept as a fact that V is a three-dimensional subspace of C 1 (R; R). Consider the operator D : V −→ V so that Df = df dt . Find the matrix representation of D relative to the basis S. Let us denote e1 = 1, e2 = cos(t), e3 = sin(t). The columns of the matrix representation of D are the coordinate vectors [D(e1 )]S , [D(e2 )]S , [D(e3 )]S . Let us then compute D(e1 ), D(e2 )S , and D(e3 ). D(e1 ) = 0 = 0e1 + 0e2 + 0e3 ⇒ [D(e1 )]S = (0, 0, 0) D(e2 ) = − sin(t) = −e3 ⇒ [D(e2 )]S = (0, 0, −1) D(e3 ) = cos(t) = e2 ⇒ [D(e3 )]S = (0, 1, 0) As a result [D]SS 0 := [D(e1 )]TS [D(e2 )]TS [D(e3 )]TS = 0 0 0 0 −1 0 1 0 4 The original question has been simplified to make the computation simpler. Credit has been given to those who gave the correct Gram-Schmidt formulae Mid term exam 2, November 8, 2011 √ Question 7: Consider the four-dimensional real vector space U = span{1/ 2, cos(t), cos(2t), cos(3t)} R 2π equipped with the inner product hf, qi = π1 0 f (τ )g(τ )dτ . Consider the linearly independent √ √ vector v√ 1 = 1/ 2, + cos(t) + cos(2t) + cos(3t), v2 = 1/ 2 + cos(t) + 2 cos(2t) + 4 cos(3t), v3 = 1/ 2 + 2 cos(t) − 4√ cos(2t) − 3 cos(3t). Use the Gram-Schmidt process to orthogonalize these vectors. (Hint: {1/ 2, cos(t), cos(2t), cos(3t)} is an orthonormal basis of U ) The Gram-Schmidt process gives w1 = v1 hv2 , w1 i w1 kw1 k2 hv3 , w1 i hv3 , w2 i w1 − w2 w3 = v3 − kw1 k2 kw2 k2 w2 = v2 − We immediately obtain w1 = v1 = √1 2 + cos(t) + cos(2t) + cos(3t). √ Let us compute kw1 k and hv2 , w1 i. Since {1/ 2, cos(t), cos(2t), cos(3t)} is an orthonormal basis of U , we have kw1 k2 = 1 + 1 + 1 + 1 = 4, by Pythagoras’ Theorem hv2 , w1 i = 1 + 1 + 2 + 4 = 8 As a result 8 w2 = v2 − w1 = v2 − 2v1 2 √ √ = (1/ 2 − 2/ 2) + (1 − 2) cos(t) + (2 − 2) cos(2t) + (4 − 2) cos(3t) In conclusion w2 = − √12 − cos(t) + 2 cos(3t). Let us compute kw2 k2 , hv3 , w1 i, and hv3 , w2 i. kw2 k2 = 1 + 1 + 4 = 6, by Pythagoras’ Theorem hv3 , w1 i = 1 + 2 − 4 − 3 = −4 hv3 , w2 i = −1 − 2 − 6 = −9 As a result −9 3 −4 w1 − w2 = v3 + w1 + w2 4 6 2 1 3 3 3 = √ (1 + 1 − ) + (2 + 1 − ) cos(t) + (−4 + 1) cos(2t) + (−3 + 1 + 2 ) cos(3t) 2 2 2 2 w3 = v3 − In conclusion w3 = 1 √ 2 2 + 3 2 cos(t) − 3 cos(t) + cos(3t). Last name: name: 5 Question 8: Let G : R3 −→ P2 be the linear mapping defined by G(x, y, z) = (x + 2y − z) + (y + z)t + (x + y − 2z)t2 . Find the dimension of Ker(G) and the dimension of Im(G). Let us first characterize ker(G). Let X be a member of Ker(G). Then x + 2y − z + (y + z)t + (x + y − 2z)t2 = 0. This means that x + 2y − z = 0, y + z = 0 and x + y − 2z = 0. This is a linear system. The matrix associated with this system is 1 2 −1 1 2 −1 1 2 −1 1 3 0 1 0 −3 0 1 1 ∼ 0 1 1 ∼ 0 1 1 ∼ 0 1 1 ∼ 0 1 1 1 1 −2 0 −1 −1 0 0 0 0 0 0 0 0 0 There is one free variable. This means that dim(ker(G)) = 1. Then the rank Theorem gives dim(Im(G)) = 3 − dim(Ker(G)) = 2. Note in passing that X ∈ ker(G) iff x1 3 x2 = x3 −1 x3 1 This means that ker(G) = span((3, −1, 1)). Question 9: Let F : P2 (t) −→ P2 (t) be the linear mapping defined by F (p) = p(0) + p(1)t + p(2)t2 . Show that F is bijective. Since the domain P2 (t) and the co-domain P2 (t) have the same dimension, it suffices to prove that F is either injective or surjective (by the rank Theorem). Let us prove that F is injective. Let p = at2 + bt + c be a member of P2 (t) such that F (p) = 0. Since p(0) = c, p(1) = a + b + c p(2) = 4a + 2b + c we infer that 0 = p(0) + p(1)t + p(2)t2 = c + (a + b + c)t + (4a + 2b + c)t2 . This implies that c = 0, a + b + c = 0 The associated matrix is 4 1 0 and 4a + 2b + c = 0. This is a linear system for a, b and c. 2 1 4 2 1 1 1 ∼ 0 −2 −3 0 1 0 0 1 The rank is maximum. This means that a = b = c = 0. As a result F is injective. In conclusion F is bijective. 6 Mid term exam 2, November 8, 2011 Question 10: Let A be a complex-valued square matrix. Assume that all the eigenvalues of A have a modulus larger than π. Prove that A is invertible Let v be a nonzero vector so that Av = 0. This means that Av = 0v, which implies that 0 is an eigenvalue of A, which contradicts that the eigenvalues have a modulus larger than π. In conclusion one cannot find a nonzero vector so that Av = 0. This implies that A is invertible since A is a square matrix. Question 11: Let M be a real-valued n×n matrix. (a)Prove that M T M is diagonalizable M T M is diagonalizable since it is a symmetric matrix as shown be the following computation: (M T M )T = M T (M T )T = M T M. (b) Prove that the eigenvalues of M T M are non-negative. (Hint v T M T M v = kM vk2 ) Let (λ, v) an eigen-pair. Then M T M v = λv; this means v T M T M v = λv T v, which can also be written kM vk2 = λkvk2 . This proves that λ = kM vk2 /kvk2 ≥ 0 (Note that kvk = 6 0 since v is an eigenvector). (c) Assume that the smallest eigenvalue of M T M is 12 . Prove that M is invertible. Let v be a nonzero vector so that M v = 0. This means that M T M v = 0v, which implies that 0 is an eigenvalue of M T M , which contradicts that 21 is the smallest eigenvalue. In conclusion one cannot find a nonzero vector so that M v = 0. This implies that M is invertible since M is a square matrix.