Examination paper for TMA4145 Linear Methods

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1 Department of Mathematical Sciences Examination paper for TMA4145 Linear Methods Academic contact during examination: Franz Luef Phone: Examination date: Examination time (from to): 09:00-13:00 Permitted examination support material: D:No written or handwritten material. Calculator Casio fx-8es PLUS, Citizen SR-70X, Hewlett Packard HP30S Other information: The exam consists of 1 problems, but the grade is based on your 10 best solutions. Although the problems are only formulated in English you can answer either in English or your favorite Norwegian language. Language: English Number of pages: 9 Number of pages enclosed: 0 Checked by: Informasjon om trykking av eksamensoppgave Originalen er: 1-sidig -sidig sort/hvit farger skal ha flervalgskjema Date Signature

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3 TMA4145 Linear Methods Page 1 of 9 Problem 1 Let A be a non-empty subset of the real line R. a) Define the following notions: (a) the infimum of A; (b) the supremum of A; (3) the closure of A; (4) the interior of A; (5) the boundary of A. 1. If m is a lower bound of A such that m m for every lower bound m, then m is called the infimum of A, denoted by m = inf A.. If M is an upper bound of A such that M M for every upper bound M, then M is called the supremum of A, denoted by M = sup A. 3. The closure of A, denoted by A, is the intersection of all closed sets containg A. 4. The interior of a subset of A of R, denoted by inta, is the union of all open subsets of R contained in A. 5. The boundary of a subset A of R, denoted by bda, is the set A\intA. b) Assume that A is bounded from above. Show that the supremum of A lies in the closure of A. Let A R be bounded from above. By the axiom of completeness, the supremum of A exists (as a real number), that is, sup A R. Let ε > 0. Since sup A is the least upper bound of A, we have that sup A ε cannot be an upper bound for A, so there is some element a ε A such that a ε > sup A ε. Furthermore, since sup A is an upper bound of A, and since a ε A, we must have that a ε sup A. Thus sup A ε < a ε sup A. For every n 1 we may choose ε = 1. n element a n A such that Using the above, there is some sup A 1 n < a n sup A. We have obtained a sequence (a n ) A such that sup A 1 n < a n sup A for all n 1. Subtracting sup A from all sides of this inequality, we get that 1 n < a n sup A 0,

4 Page of 9 TMA4145 Linear Methods which implies a n sup A 1 n 0 as n, so a n sup A. Problem Consider the initial value problem: dx dt = f(t, x), and x(t 0) = x 0, where f is a function f : U V R defined on U V of R such that t 0 lies in the interior of the interval U and x 0 in the interior of the interval V, respectively. a) Formulate the theorem of Picard-Lindelöf. Assume that f is continuous in t and uniformly Lipschitz in x: f(t, x) f(t, x ) L x x for all t U, x, x V. Then the IVP has a unique local solution, i.e. there exists a δ > 0 such that the IVP has a solution x on (x 0 δ, x 0 + δ). b) Solve the initial value problem dx dt = t(1 + x), and x(0) = 0, by applying the theorem of Picard-Lindelöf. Compute the first three Picard iterations x 1 (t), x (t) and x 3 (t) starting from x 0 (t) = 0. In this problem f(x, t) = t(1 + x), which is continuous in t and uniformly continuous on the closed interval [ B, B] for any B > 0. Hence there exists a unique local solution. The formula for the Picard iteration is x n+1 (t) = Since x 0 (t) = 0 we have t 0 s(1 + x n (s))ds = t + t 0 sx n (s)ds. x 1 (t) = t, x (t) = t + t4!, x 3(t) = t + t4! + t6 3!. Note that x n (t) is the Taylor expansion of e t 1. Hence x n (t) e t 1 and thus the solution actually exists for all t R.

5 TMA4145 Linear Methods Page 3 of 9 Problem 3 Given the matrix 1 A =. 1 a) Compute the singular value decomposition of A. ( ) A A = has as characteristic polynomial x x + 17 = (x 17)(x 1). Hence the eigenvalues of A A are λ 1 = 17 and ) λ = 1. ( The corresponding normalized eigenvectors are v 1 = ) 1 ( 1 1. Consequently, we have ( ). and v =. The singular values of A are σ 1 = 17 and σ = 1. Thus 17 0 Σ = The first two columns of U are given by and by u 1 = 1 1 Av 1 = 1 1 ( ) 1 = u = 1 1 Av = 1 1 ( ) 1 = Consequently, U has the form 1 34 x x 1 34 x 3

6 Page 4 of 9 TMA4145 Linear Methods The last column is determined by the assumption that it has to be orthogonal to the first two columns. The choice u 3 = satisfies these conditions, but there are many other ways to complete the first two columns to become an orthonormal basis for C 3, The SVD of A is ( b) Use the result of a) to find: (1) Bases for the following vector spaces: ker(a), ker(a ), ran(a), ran(a ). ker(a ) = {u 3 }, ker(a) = {0}, ran(a ) = {v 1, v }, ran(a) = {u 1, u }. (3) The pseudo-inverse of A. ). A = ( ) ( ) Problem 4 Let. a and. b be two norms on a vector space X. a) Show that x := ( x a + x b) 1/ is a norm on X. Furthermore, show if a sequence (x n ) converges in (X,. ), then it converges in (X,. a ) and in (X,. b ). x := ( x a + x b) 1/ is a norm, because

7 TMA4145 Linear Methods Page 5 of 9 1. x = 0 if and only if x a = 0 and x b = 0, which is the case only if x = 0.. λx = ( λx a + ax b) 1/ = (λ x a + λ x b) 1/ = λ ( x a + x b) 1/ = λ x. 3. There is a direct way to demonstrate the triangle inequality via brute force computation or one can use the Minkowski inequality. Here is the first approach: Let us compute x + y and ( x + y ). ( x + y ) = ( x a + x b ) + ( y a + y b ) = x a + x b + y a + y b+ x a y a + y a x b + x a y b + x b y b x + y = x + y a + x + y b ( x a + y a ) + ( x b + y b ) = x a + x b + y a + y b+ which gives us x a x b + y a y b, ( x + y ) x + y = y a x b + x a y b 0 and consequently x + y x + y. Second approach: Let us use the Minkowski inequality on R. Take x = ( x a, x b ) and y = ( y a, y b ). Then we have (( x a + y a ) +( x b + y b ) ) 1/ (( x a+ x b) 1/ +( y a+ y b)) 1/ by Minkowski s inequality for p =. The argument is valid for a general p [1, ) and thus x = ( x p a + x p b )1/p defines a norm on X, too. For the convergence statement, we just observe that x a x and x b x, which yields the desired assertions. b) Suppose there exist constants C 1, C > 0 such that C 1 x b x a C x b holds for all x X, i.e.. a and. b are equivalent norms on X.

8 Page 6 of 9 TMA4145 Linear Methods Show that there exist constants C 1, C > 0 such that holds for all x X. C 1 x a x b C x a implies that C 1 x b x a, hence and from x a C x b we deduce. Consequently, we have so C 1 = C 1, C = C 1 1. C 1 x b x a C x b x b C 1 1 x a, C 1 x a x b C 1 x a x b C 1 1 x a, Determine the constants C 1 and C for the sup-norm. and. p -norm, 1 p <. on R n : C 1 x x p C x. We have C 1 = 1 and C = n 1/p, because max{ x i : i = 1,..., n} = max{ x i p : i = 1,..., n} 1/p ( n i=1 x i p ) 1/p and ( n i=1 x i p ) 1/p ( n i=1 max{ x i : i = 1,..., n} p ) 1/p n 1/p max{ x i : i = 1,..., n}. Problem 5 Let M be the subspace of l defined by M = {x = (x k ) k N l : x k = 0 for k = 1,,...}. a) Show that M is a closed subspace of l and determine its orthogonal complement M.

9 TMA4145 Linear Methods Page 7 of 9 Suppose (x n ) n N, where x n = (x n k) k N, is a sequence in M converging to x = (x k ) k N in l. Since x n j = 0 for j = 1,,... we have x j = x j x n j = ( x j x n j ) 1/ ( x j x n j ) 1/ = x x k j N for all j, j N. Hence in the limit as k we get that x j = 0 for all j N. Thus x M and M is a closed subspace of l. The natural candidate for the orthogonal complement of M is the subspace N = {(x k ) k N x j 1 = 0, j = 1,,...}. Now, for x M and y N we have x, y = 0. Hence N M. Suppose y M. Then x, y = 0 for all x M. Let us take the standard basis {e k : k N}. Then we e j 1 M for all j and we have for x = e j 1 that x, e j 1 = x j 1 = 0. Consequently, y N and thus M = N. b) Determine the orthogonal projection P from l onto M without using the projection theorem and show that P = P and its operator norm P = 1. The orthogonal projection P x of x onto M is the best approximation of x in M such that its error is in the orthogonal complement. Hence x P x M, i.e x P x, y = 0 for all y M. Hence x P x = (x j 1 ), so P x = (x j ) = (x, x 4, x 6,...). We also have that M M = {0} since (x 1, x, x 3,...) = (x, x 4,...) + (x 1, x 3,...) for all x l. Hence P = P and P x, y = x ky k = x, P y, k=1 i.e. P = P. By Pythagoras we have x = P x + y where y M. Thus P x x. On the other hand there exists an x l such that P x 0, but P (P x) = P x. Thus P (P x) = P x, so we have P 1. Hence P = 1. Now for x l to M. Problem 6 Let X be a separable Hilbert space and {e k : k = 0, 1,,...} an orthonormal basis for X. We define the linear operator S by S(e k ) = e k+1 for k = 0, 1,,....

10 Page 8 of 9 TMA4145 Linear Methods a) Suppose a = (a 0, a 1,...) l is the coefficient sequence of x X: x = a k e k. Describe the operator S in terms of the coefficient sequence (a 0, a 1,...), i.e. as an operator on l. Determine S on l and find S (e k ) for k = 0, 1,... Compute the operator norm of S. By definition we have S(x) = a k S(e k ) = a k e k+1. In order to get the action of S on l we have to change the summation index:. a k S(e k ) = a k 1 e k = 0e 0 + a 0 e 1 + a 1 e + k=1 Hence for a = (a 0, a 1,...) we have S(a 0, a 1,...) = (0, a 0, a 1, a,...) is a shift operator. Then we have that the adjoint of S is defined by Sa, b = a k 1 b k = a k b k+1 = a, S b. Hence S is the other shift operator on l S (a 0, a 1,...) = (a 1,...), and in terms of the basis elements S is defined by S e 0 = 0 and S e k = e k 1 for k = 1,,....The operator norm of S equals S = sup{ Sa : a l with a = 1}. Since Sa = a 0 + a 1 + a + = a we have that S = 1. b) Determine if S and S are injective and/or surjective, respectively. Determine S S and SS, their kernels and ranges, respectively.

11 TMA4145 Linear Methods Page 9 of 9 We consider S and S as linear operators on l. S is not injective, (1, 0, 0,...) gets mapped to (0, 0,...), and it is surjective: any a l lies in the range of S : S (0, a 0, a 1,...) = (a 0, a 1,...). The map S is injective and not surjective: Sa = (0, a 0, a 1,...) = 0 if and only if a = 0; and (1, 0, 0,..) does not lie in the range of S. S Sa = (a 0, a 1, a,...) and SS = (0, a 1, a,...). Hence the kernel of SS is {(α, 0, 0,...) : α C} and the range of SS is l. Since S S = I its kernel is just {0} and its range is l.