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HARVARD MATH 21B - FIRST PRACTICE EXAM SECOND HOURLY

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FIRST PRACTICE EXAM SECOND HOURLY Math 21b, Fall 2003Name:MWF10 Izzet CoskunMWF11 Oliver Knill• Start by writing your name in the above box andcheck your section in the box to the left.• Try to answer each question on the same page asthe question is asked. If needed, use the back or thenext empty page for work. If you need additionalpaper, write your name on it.• Do not detach pages from this exam packet or un-staple the packet.• Please write neatly. Answers which are illegible forthe grader can not be given credit.• No notes, books, calculators, computers, or otherelectronic aids can be allowed.• You have 90 minutes time to complete your work.1 202 103 104 105 106 107 108 109 10Total: 100Problem 1) (20 points) True or False? No justifications are needed.T FSimilar matrices have the same determinant.T FThe matrix1 100 1 1100 1 1 11 1 1 1001 1 100 1is invertible.T FIf A is a 3 × 3 matrix for which every entry is 1, then det(A) = 1.T FIf ~v is an eigenvector of A and of B and A is invertible, then ~v is aneigenvector of 3A−1+ 2B.T Fdet(−A) = det(A) for every 5 × 5 matrix A.T FIf ~v is an eigenvector of A and an eigenvector of B and A is invertible, then~v is an eigenvector of A−3B2.T FIf a 11 × 11 matrix has the eigenvalues 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, then A isdiagonalizable.T FFor any n × n matrix, the matrix A has the same eigenvectors as AT.T FIf ~v =abcis a vector of length 1, then ~v~vTis a diagonalizable 3 × 3matrix.T FThe span of m orthonormal vectors is m-dimensional.T FA square matrix A can always be expressed as the sum of a symmetricmatrix and a skew-symmetric matrix as follows A =12(A+AT)+12(A−AT).T FThere exists an invertible n × n matrix A which satisfies ATA = A2, but Ais not symmetric.T FIf two n × n matrices A and B commute, then (AT)2commutes with (B3)T.T F−AATis skew-symmetric for every n × n matrix A.T FA maps a least squares solution ~x∗of A~x =~b to the projection of~b ontoim(A).T FA matrix which is obtained from the identity matrix by an arbitrary numberof switching of rows or columns is an orthogonal matrix.T FThe trace of a real skew-symmetric n × n matrix is always equal to 0.T FThere exists a real 3 × 3 matrix A which satisfies A4= −I3.T FIf A has four different integer eigenvalues, then there exists a vector ~v suchthat ||An~v|| → ∞ for n → ∞.T FGiven 5 data points (x1, y1), ..., (x5, y5), then a best fit with a polynomiala + bt + ct2+ dt3+ et4+ ft5is possible in a unique way.TotalProblem 3) (10 points)Consider the matrix A =0 1 1 1 11 0 1 1 11 1 0 1 11 1 1 0 11 1 1 1 0.a) Find all eigenvalues of A with their multiplicities.b) Write down all eigenvectors of A. What are the geometric multiplicities of the eigenvalues?c) What is det(A).Hint. You might want to look at A + I5first.Problem 4) (10 points)a) Let A be a n × n matrix such that A2= 2A − I. What are the possible eigenvalues of A?b) Let A be a real n × n matrix such that A4= −In. Show that n must be even.Problem 5) (10 points)Let An="1 10 1 + 1/n#, where n is a positive integer.a) Find all eigenvalues for An.b) Find an eigenbasis Anfor each n.c) We hope to get an eigenbasis for A ="1 10 1#by taking the limit in b), when n goes toinfinity and possibly rescaling the eigenvectors so that they converge. Does this work? Whathappens?Problem 6) (10 points)Given n + 1 numbers a0, a1, ..., an, define the matrix A =1 1 ... 1 1a0a1... an−1ana20a21... a2n−1a2n... ... ... ... ...an0an1... ann−1ann. It is calleda van der Monde matrix.a) If we call an= x, then the determinant of A becomes a function of x. Show that is apolynomial of degree n in x.b) Verify that f(a0) = f(a1) = ... = f(an−1) = 0.c) Conclude that f(x) = k(x − a1) . . . , (x − an−1) for some constant k.d) Verify that the constant k is a determinant of a n × n van der Monde matrix.e) Use this to conclude that det(A) =Qi>j(ai− aj).Problem 7) (10 points)A discrete dynamical system is given byT ("xy#) ="2x − 2y−x + 3y#= A"xy#.Find a closed formula for T100("12#).Problem 8) (10 points)a) Show that for an arbitrary matrix A for which ATA is invertible, the least squares solutionof A~x =~b simplifies to~x = R−1QT~b ,if A = QR is the QR decomposition of A.b) Find the least square solution in the case A =1 10 10 1and~b =121using this formula.Problem 9) (10 points)Find the least square solution for the system A~x =~b given by the equationsx + y = 4y = 2x = −1


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