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Amath/Math 516 First Homework Set Solutions

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Amath/Math 516 First Homework Set Solutions AMATH/MATH 516 FIRST HOMEWORK SET SOLUTIONS The purpose of this problem set is to have you brush up and further develop your multi-variable calculus and linear algebra skills. The problem set will be very difficult for some and straightforward for others. If you are having any difficulty, please feel free to discuss the problems with me at any time. Don't delay in starting work on these problems! 1. Let Q be an n × n symmetric positive definite matrix. The following fact for symmetric matrices can be used to answer the questions in this problem. Fact: If M is a real symmetric n×n matrix, then there is a real orthogonal n×n matrix U (U T U = I) T and a real diagonal matrix Λ = diag(λ1; λ2; : : : ; λn) such that M = UΛU . (a) Show that the eigenvalues of Q2 are the square of the eigenvalues of Q. Note that λ is an eigenvalue of Q if and only if there is some vector v such that Qv = λv. Then Q2v = Qλv = λ2v, so λ2 is an eigenvalue of Q2. (b) If λ1 ≥ λ2 ≥ · · · ≥ λn are the eigen values of Q, show that 2 T 2 n λnkuk2 ≤ u Qu ≤ λ1kuk2 8 u 2 IR : T T Pn 2 We have u Qu = u UΛUu = i=1 λikuk2. The result follows immediately from bounds on the λi. (c) If 0 < λ < λ¯ are such that 2 T ¯ 2 n λkuk2 ≤ u Qu ≤ λkuk2 8 u 2 IR ; then all of the eigenvalues of Q must lie in the interval [λ; λ¯]. T 2 ¯ Take u1 to be the eigenvector for λ1. Then u1 Qu1 = λ1ku1k2, so λ1 ≤ λ. The analogous argument applied to un, the eigenvector associated to λn, gives λ ≤ λn. (d) Let λ and λ¯ be as in Part (c) above. Show that ¯ n λkuk2 ≤ kQuk2 ≤ λkuk2 8 u 2 IR : 2 T 2 Hint: kQuk2 = u Q u. 2 2 From part (a), we have that the eigenvalues of Q are λi , where λi are the eigenvalues of Q. From part b 2 2 2 2 2 n we have that λnkuk1 ≤ kQuk2 ≤ λ1kuk2 8u 2 R , and the result follows immediately. 2. Consider the quadratic function f : IRn 7! IR given by 1 f(x) := xT Qx − aT x + α ; 2 where Q 2 IRn×n, a 2 IRn, and α 2 IR: (a) Write expressions for both rf(x) and r2f(x). Since it is not assumed that f is symmetric, be careful in how you express r2f(x). 1 T 2 1 T rf(x) = 2 (Qx + Q x) − a, and r f(x) = 2 (Q + Q ). 1 (b) If it is further assumed that Q is symmetric, what is r2f? It is Q. (c) State first– and second{order necessary conditions for optimality in the problem minff(x): x 2 IRng. We want rf(x) = 0 and Q to be positive semidefinite. (d) State a sufficient condition on the matrix Q under which the problem min f has a unique global solution and then display this solution in terms of the data Q and a. 1 T −1 Q should be positive definite; in which case the solution is given by 2 (Q + Q ) a. We might as well assume Q is symmetric. 3. Consider the linear equation Ax = b; where A 2 IRm×n and b 2 IRm. When n < m it is often the case that this equation is over-determined in the sense that no solution x exists. In such cases one often attempts to locate a `best' solution in a least squares sense. That is one solves the linear least squares problem 1 (lls) : minimize kAx − bk2 2 2 for x. Define f : IRn 7! IR by 1 f(x) := kAx − bk2: 2 2 (a) Show that f can be written as a quadratic function, that is, it can be written in the same form as the function of the preceding exercise. 1 f(x) = 2 hAx − b; Ax − bi 1 T T T 1 T = 2 x A Ax − b Ax + 2 b b (b) What are rf(x) and r2f(x)? rf(x) = AT Ax − AT b; ; r2f(x) = AT A: (c) Show that r2f(x) is positive semi-definite. T T 2 u A Au = kAuk2 ≥ 0: (d) Show that Nul(AT A) = Nul(A) and Ran(AAT ) = Ran(A). It is clear that Nul(A) ⊂ Nul(AT A). For the other direction, if w 2 Nul(AT A), then AT Ax = 0 so that 0 = xT AT Ax = kAxk2, or equivalently, Ax = 0. By the Fundamental Theorem of the Alternative for matrices and the fact that Nul(AAT ) = Nul(AT ) (obtained by replacing A with AT in what was just proved), we have Ran(AAT ) = Nul(AAT )? = Nul(AT )? = Ran(A): 2 (e) Show that a solution to (lls) must always exist. The necessary and sufficient conditions for optimality are that rf = 0 and r2f is positive semidefinite (which we showed in part c) since this implies that f is convex. Now note that the equation AT Ax = AT b always has a solution by part (d), since the ranges of the matrices AT A and AT agree. (f) Provide a necessary and sufficient condition on the matrix A under which (lls) has a unique solution and then display this solution in terms of the data A and b. By 2d), we know we need r2f to be positive definite. Since we know r2f = AT A is already positive semidefinite by 3c), we just need it to have a trivial nullspace. By 2d), we know AT A will have a trivial nullspace if A does. From this it immediately follows that Nul(A) = f0g, and so m ≥ n. In this case, the solution is given by x = (AT A)−1AT b. 4. A mapping h·; ·i : IRn 7! IRn is said to be an inner product on IRn is for all x; y; z 2 IRn (i) hx; xi ≥ 0 Non-Negative (ii) hx; xi = 0 , x = 0 Positive (iii) hx + y; zi = hx; zi + hy; zi Additive (iv) hαx; yi = α hx; yi 8 α 2 IR Homogeneous (v) hx; yi = hy; xi Symmetric Two vectors x; y 2 IRn are said to be orthogonal in the inner product h·; ·i if hx; yi = 0 Unless otherwise specified, we use the notation hx; yi to designate the usual Euclidean inner product: n X hx; yi = xiyi : i=1 (a) Let hx; yi be the Euclidean inner product on IRn. Given A 2 IRn×n, show that A = 0 if and only if hx; Ayi = 0 8 x; y 2 IRn : If A = 0, then hx; Ayi = 0. Conversely, note that Aij = hei; Aeji = 0, where ei is the vector with all zeros and a 1 in the ith position, so all the entries of A must be zero. (b) Let H 2 IRn×n be symmetric and positive definite (i.e. H = HT and xT Hx > 0 8 x 2 IRn n f0g). Show that the bi-linear form given by T n hx; yiH = x Hy 8 x; y 2 IR defines an inner product on IRn. Properties (i) and (ii) are immediate from the definition of positive definite matrices. Properties (iii) and (iv) follow from the linearity of matrix multiplication. Property (v) follows from the symmetry of H. (c) Every inner product defines a transformation on the space of linear operators called the adjoint. For the Euclidean inner product on IRn, this is just the usual transpose. Given a linear transformation M : IRn 7! IRn, the adjoint is defined by the relation hy; Mxi = hM ∗y; xi ; for all x; y; 2 IRn: TH The inner product given above, h·; ·iH , also defines an adjoint mapping which we can denote by M . Show that M TH = H−1M T H: T T −1 −1 T hy; MxiH = hy; HMxi = M Hy; x = M Hy; H x H = H M Hy; x H Comparing the leftmost and rightmost expressions gives the result. 3 (d) The matrix P 2 IRn×n is said to be a projection if P 2 = P . Clearly, if P is a projection, then so is I − P . The subspace P IRn = Ran (P ) is called the subspace that P projects onto. A projection is said to be orthogonal with respect to a given inner product h·; ·i on IRn if and only if h(I − P )x; P yi = 0 8 x; y 2 IRn ; that is, the subspaces Ran (P ) and Ran (I − P ) are orthogonal in the inner product h·; ·i. Show that the n×n projection P is orthogonal with respect to the inner product h·; ·iH (defined above), where H 2 IR is symmetric and positive definite, if and only if P = H−1P T H: Note that T h(I − P )x; P yiH = h(I − P )x; HP yi = P H(I − P )x; y If P = H−1P T H, then P T H(I − P ) = HPH−1H(I − P ) = HP (I − P ) = H(P − P ) = 0, so the result follows. For the converse, we have P T H(I − P )x; y = 0, so P T H(I − P ) = 0 by (a). Then P T H = P T HP , so P T = P T HPH−1. Taking the transpose, we have P = H−1(P T HP ) = H−1P T H since P T H = P T HP . 5. Consider the minimization problem P : minimize f(x) subject to Ax = b ; where f : IRn 7! IR is assumed to be twice continuously differentiable, A 2 IRm×n has full rank with m ≤ n, and b 2 IRm.
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