exam4 - Version 055 EXAM04 gilbert(56780 This print-out should have 14 questions Multiple-choice questions may continue on the next column or page nd

# exam4 - Version 055 EXAM04 gilbert(56780 This print-out...

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This preview shows page 1 out of 8 pages. Unformatted text preview: Version 055 – EXAM04 – gilbert – (56780) This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – ﬁnd all choices before answering. 001 10.0 points Compute the component of the spectral decomposition determined by λ1 when A is the symmetric 2 × 2 matrix with eigenvalues λ1 = −4, λ2 = 11 and corresponding eigenvectors 2 , −4 x1 = 1. 11 1 5 −2 x2 = 2 . 1 4 1 2. − 5 −2 Then 1 1 1 1 −2 [ 1 −2 ] = , 5 −2 5 −2 4 T v1 v1 = while Thus λ1 5 002 5. − 4 1 5 2 6. − 4 4 5 2 λ2 −2 + 4 5 4 2 2 . 1 4 1 5 −2 −2 4 . 10.0 points If A is symmetric, then the change of variable x = P y transforms −2 1 2 1 Q(x) = xT A x into a quadratic form with no cross-product term for any orthogonal matrix P . Explanation: When v1 , v2 are orthonormal eigenvectors of A corresponding to respective eigenvalues λ1 and λ2 , then the spectral decomposition of A is given by T A = λ1 v1 v1 + λ2 v2 vT , and the component determined by λ1 is Now the given eigenvectors x1 , x2 are orthogonal, but not orthonormal, so set 1 √ 2 = 2 5 −4 True or False? 1. TRUE 2. FALSE correct Explanation: When P is orthogonal and x = P y, then Q(x) = xT A x = (P y)T A (P y) T λ1 v1 v1 . v1 = 1 −2 − 11 4 2 5 2 1 x1 x1 1 2 1 4 2 [2 1] = . 5 1 5 2 1 T v2 v2 = 2 4 4. 1 2 = √ . 5 1 x2 x2 v2 = Consequently, the component of the spectral decomposition of A determined by λ1 is −2 correct 4 11 4 3. 5 −2 and A = −2 4 1 1 1 = √ 5 −2 = yT P T A P y = yT (P T AP )y . But this will contain cross-product terms unless P T AP is a diagonal matrix, i.e., unless A is orthogonally diagonalized by P . Version 055 – EXAM04 – gilbert – (56780) is an orthogonal matrix such that Consequently, the statement is FALSE . 003 2 1 0 P −1 = P DP T 0 −19 A = P 10.0 points is an orthogonal diagonalization of A. Make an orthogonal change of variables that reduces Now set xT A x = −3x2 + 16x1 x2 − 15x2 = −5 1 2 x = to a quadratic equation in y1 , y2 with no crossproduct term given that λ1 ≥ λ2 . 1. 2 2 y1 + 19y2 = 5 2. 2 2 y1 − 19y2 = 5 3. 2 2 y1 − 19y2 = −5 correct 4. 2 2 y1 + 19y2 = −5 x Ax = −3x2 1 = [x1 x2 ] −3 − λ 8 Then = yT 1 0 2 2 y = y1 − 19y2 = −5 . 0 −19 + 16x1 x2 − −3 8 15x2 2 8 −15 2 2 −3x2 + 16x1 x2 − 15x2 = y1 − 19y2 = −5 . 1 2 x1 . x2 = λ2 + 18λ − 19 004 10.0 points Find the x-intercept of the Least Squares Regression line y = mx + b that best ﬁts the data points (−2, −2), (−1, 1), = (λ − 1)(λ + 19) = 0 , i.e., λ1 = 1 and λ2 = −19. Associated eigenvectors are u1 = u2 = −1 , 2 and these are orthogonal since λ1 = λ2 . The normalized eigenvectors 1 2 , v1 = √ 5 1 x = P y. Consequently, when x = P y, 8 −15 − λ 2 , 1 y1 , y2 = yT (P T P )D(P T P )y = yT Dy The eigenvalues λ1 , λ2 of A are the solutions of det y = xT A x = (P y)T (P DP T )P y Explanation: In matrix terms, T x1 , x2 1 −1 v2 = √ , 5 2 are thus orthonormal, and 1 2 −1 P = [v1 v2 ] = √ 5 1 2 1. x-intercept = − 14 5 2. x-intercept = − (0, 2), 12 5 3. x-intercept = −2 correct 4. x-intercept = − 11 5 5. x-intercept = − 13 5 Explanation: (1, −4) . Version 055 – EXAM04 – gilbert – (56780) The design matrix and list of observed values for the data (−2, −2), (−1, 1), are given by (0, 2), and so its x-intercept = −2 . (1, −4) . 005 1 1 A = 1 1 −2 1 b = . 2 −4 −2 −1 , 0 1 ˆ AT A x = AT b , b m ˆ x = 10.0 points A vector space V is a subspace of itself. True or False? 1. FALSE The least squares regression line for this data ˆ is y = mx + b where x is the solution of the normal equation 2. TRUE correct Explanation: A set H is a subspace of a vector space V when . (i) H contains the zero vector 0, Now (ii) the sum u + v of any u, v in H is in H, 1 −2 AT b = 1 −1 1 0 1 1 while −2 1 2 = −4 −3 −1 , (iii) the scalar multiple cu of any scalar c and any u in H is in H. Since V has these three properties, V is a subspace of itself. Consequently, the statement is 1 −2 AT A = 1 −1 4 −2 = 1 1 0 1 −2 . 6 1 −2 1 −1 1 0 1 1 TRUE . 006 −2 6 ˆ x = 4 −2 −2 6 10.0 points When Thus the normal equation is 4 −2 3 u1 = b m = −3 . −1 1 , −2 u2 = 2 , 1 are eigenvectors of a symmetric 2 × 2 matrix A corresponding to eigenvalues So b m = = 4 −2 1 6 20 2 −2 6 −1 2 4 −3 −1 λ1 = 2 , −3 −1 = −1 −1 2 λ2 = −3 , ﬁnd matrices D and P in an orthogonal diagonalization of A. . Consequently, the Least Squares Regression line is 1 y = − x−1 2 1. D = 2. D = correct −2 2 0 , P = 1 0 −3 2 0 , 0 −3 1 2 1 1 P = √ −2 5 2 1 Version 055 – EXAM04 – gilbert – (56780) Explanation: Note that for B to be symmetric, B T = B. Also note that all diagonal matrices are by deﬁnition symmetric. Consider B T . 1 2 2 0 , P = 3. D = −2 1 0 −3 1 2 0 , P = −2 1 2 4. D = −3 0 5. D = 1 −2 2 0 , P =√ 0 −3 5 1 1 2 6. D = −3 0 1 0 1 , P =√ 2 5 −2 2 1 B T = (P DP T )T = (P T )T DT P T = P DT P T = P DP T = B. Thus B T = B and B is by deﬁnition symmetric. Consequently, the statement is TRUE . Explanation: When D= 008 0 , Q = [u1 u2 ], λ λ1 0 then Q has orthogonal columns and A = QDQ−1 is a diagonalization of A, but it is not an orthogonal diagonalization because Q is not an orthogonal matrix. We have to normalize u1 and u2 : set 1 1 = √ 5 −2 u1 u1 v1 = u2 u2 v2 = 1 2 = √ 5 1 , is an orthogonal diagonalization of A when 2 0 1 1 0 , P =√ −3 5 −2 007 10.0 points Determine the singular value σ1 for the matrix 3 0 A = 0 −3 . 1 1 √ 1. σ1 = 7 √ 2. σ1 = 10 √ 3. σ1 = 11 correct 4. σ1 = 3 . Then P = [v1 v2 ] is an orthogonal matrix and A = P DP −1 D= 4 2 1 . √ 5. σ1 = 2 2 Explanation: By deﬁnition, σ is a singular value of A √ when σ = λ and λ is an eigenvalue of AT A; σ1 is the largest of these singular values. Now AT A = 10.0 points If B = P DP with P an orthogonal matrix and D a diagonal matrix, then B is a symmetric matrix. 3 0 0 −3 −1 True or False? 1. FALSE 2. TRUE correct = 3 1 0 1 1 10 1 . 1 10 0 −3 1 But then det(AT A − λI) = 10 − λ 1 1 10 − λ = λ2 − 20λ + 99 = (λ − 11)(λ − 9) . Version 055 – EXAM04 – gilbert – (56780) Consequently, σ1 = √ 11 . 5 Explanation: The normal equations for a least-squares solution of Ax = b are by deﬁnition AT Ax = AT b. 009 10.0 points If A is an m × n matrix and b is in Rm , the general least-squares problem is to ﬁnd an x that makes Ax as close as possible to b. Now, AT A = True or False? = 1. TRUE correct 2. FALSE Consequently, the statement is TRUE . 010 10.0 points Find the least-squares solution of Ax = b when 0 −1 −4 A = −1 −1 , b = −7 . 1 0 5 1 −21 1. 3 −8 12 2. 11 −4 −7 4. 6 7 5. 0 1 −1 0 1 −1 −1 2 1 1 2 −1 −1 0 and Explanation: ˆ A least squares solution of Ax = b is an x n in R such that b − Aˆ ≤ b − Ax for all x x in Rn . Note that b − Ax is the distance from Ax to b, thus the goal is to minimize that distance. 3. 0 −1 1 13 correct 3 10 AT b = 0 −1 −1 −1 −4 12 1 . −7 = 11 0 5 Hence the least squares solution of Ax = b is the solution x to the equation 12 2 1 . x= 11 1 2 This can be solved with row reduction or inverse matrices to determine that the solution is 1 2 −1 3 −1 2 1 13 . = 3 10 (AT A)−1 (AT b) = 12 11 Consequently, the least squares solution to Ax = b is 1 13 3 10 011 . 10.0 points Which one of the following is the graph of 5x2 − 4x1 x2 + 2x2 = 16 ? 1 2 (All axes drawn to the same scale.) Version 055 – EXAM04 – gilbert – (56780) 6 x2 x2 1. 5. x1 x1 x2 x2 6. 2. correct x1 x1 x2 Explanation: In matrix terms, 3. 5x2 − 4x1 x2 + 2x2 = xT Ax 1 2 with x1 x = x1 , x2 A = 5 −2 −2 2 . But if λ1 , λ2 are the eigenvalues of A and v1 , v2 are respective corresponding normalized eigenvectors, then by the Principal Axes theorem, x2 xT Ax = yT 4. λ1 0 0 2 2 y = λ1 y1 + λ2 y2 λ2 where x1 P = [v1 v2 ], y = y1 , y2 x = Py. Now det[A − λI] = 5−λ −2 −2 2−λ = λ2 − 7λ + 6 = (λ − 6)(λ − 1) , Version 055 – EXAM04 – gilbert – (56780) so λ1 = 6, λ2 = 1 and 1 2 v1 = √ , −1 5 1 1 v2 = √ . 5 2 are orthogonal. Thus u and v must be orthogonal, that is, u · v = 0. Consequently, the statement is TRUE . Thus the graph of 5x2 − 4x1 x2 + 2x2 = 16 1 2 013 is that of the ellipse 2 6y1 10.0 points When + 2 y2 A = = 16 with respect to the y1 y2 -axes. Since x = P y = [y1 y2 ] y1 y2 x2 y2 1 2 −1 1. V = √ correct 5 1 2 2 1 −1 2 1 2 −2 1 1 1 −2 4. V = √ 5 2 1 Explanation: Since AT A = 10.0 points If AT = A and if vectors u and v satisfy Au = 3u and Av = 4v, then u · v = 0. True or False? 1. FALSE 2. TRUE correct Explanation: The given vectors u and v are eigenvectors of A corresponding to eigenvalues λ1 = 3 and λ2 = 4. But when A is symmetric, then any two eigenvectors from diﬀerent eigenspaces , of A? 3. V = y1 3 2 A = U ΣV T 2. V = x1 2 0 which of the following could be a choice for the matrix V in a Singular Value Decomposition = y1 v1 + y2 v2 , the y1 -axis is the line tv1 , while the y2 -axis is the line tv2 . Consequently, the graph of the ellipse is given by 012 7 2 3 0 2 13 6 2 0 , = 6 4 3 2 the eigenvalues of AT A are the solutions of (13 − λ)(4 − λ) − 36 = λ2 − 17λ + 16 = (λ − 16)(λ − 1) = 0 , i.e., λ1 = 16, λ2 = 1. Eigenvectors x1 and x2 associated with λ1 and λ2 are x1 = 2 , 1 x2 = −1 ; 2 these are orthogonal. Associated orthonormal eigenvectors are thus 1 2 v1 = √ , 5 1 1 −1 v2 = √ . 5 2 Version 055 – EXAM04 – gilbert – (56780) and Consequently, one choice for V is 1 2 V = [v1 v2 ] = √ 5 1 −1 2 . Since multiples of x1 and x2 are again eigenvectors of AT A corresponding respectively to λ1 = 16 and λ2 = 1, there are other choices for V . But these are not among the choices listed. 014 10.0 points Use an orthogonal matrix P to identify 3x2 − 4xy + 6y 2 = 6 as a conic section in x1 , y1 without cross-term when x1 x . = P y1 y 2 1. ellipse 2x2 + 7y1 = 6 correct 1 2 2. straight lines 2x2 + 7y1 = 0 1 2 3. ellipse 2x2 − 7y1 = 5 1 2 4. hyperbola 2x2 + 7y1 = 6 1 5. point (0, 0) 2 6. hyperbola 2x2 − 7y1 = 5 1 Explanation: The quadratic relation 3x2 − 4xy + 6y 2 = 6 can be written in matrix terms as xT Ax = xT 3 −2 −2 x = 6 6 x . y To eliminate the cross-term we orthogonally diagonalize A by ﬁnding the eigenvalues λ1 , λ2 and corresponding normalized eigenvectors v1 , v2 of A. For then A = P DP T with λ1 0 , P = [v1 v2 ], D = 0 λ2 where x = 8 2 xT Ax = yT Dy = λ1 x2 + λ2 y1 , 1 setting P y = x, y = x1 , y1 x = x . y Since A is symmetric, P will be orthogonal if λ1 = λ2 . But det[A − λI] = (3 − λ)(6 − λ) − 4 = λ2 − 9λ + 14 = (λ − 2)(λ − 7) = 0. Thus λ1 = 2 and λ2 = 7, so that in x1 , y1 coordinates the quadratic relation becomes 2 2x2 + 7y1 = 6, 1 which as a conic section is the familiar form of an ellipse. ...
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