final-s09-soln

# final-s09-soln - 18.06 Final Solution Hold on Tuesday 19...

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18.06 Final Solution Hold on Tuesday, 19 May 2009 at 9am in Walker Gym. Total: 100 points. Problem 1: A sequence of numbers f 0 ,f 1 ,f 2 ,... is deﬁned by the recurrence f k +2 = 3 f k +1 - f k , with starting values f 0 = 1, f 1 = 1. (Thus, the ﬁrst few terms in the sequence are 1 , 1 , 2 , 5 , 13 , 34 , 89 ,... .) (a) Deﬁning u k = ± f k +1 f k ² , re-express the above recurrence as u k +1 = A u k , and give the matrix A . (b) Find the eigenvalues of A , and use these to predict what the ratio f k +1 /f k of successive terms in the sequence will approach for large k . (c) The sequence above starts with f 0 = f 1 = 1, and | f k | grows rapidly with k . Keep f 0 = 1, but give a diﬀerent value of f 1 that will make the sequence (with the same recurrence f k +2 = 3 f k +1 - f k ) approach zero ( f k 0) for large k . Solution (18 points = 6+6+6) (a) We have ± f k +2 f k +1 ² = ± 3 - 1 1 0 ²± f k +1 f k ² A = ± 3 - 1 1 0 ² . (b) Eigenvalues of A are roots of det( A - λI ) = λ 2 - 3 λ + 1 = 0. They are λ 1 = 3 + 5 2 and λ 2 = 3 - 5 2 . Note that λ 1 > λ 2 , so the ratio f k +1 /f k will approach λ 1 = 3 + 5 2 for large k . (c) Let v 1 , v 2 be the eigenvectors with eigenvalues λ 1 and λ 2 , respectively. So, we can write u 0 = c 1 v 1 + c 2 v 2 and then u k = c 1 λ k 1 v 1 + c 2 λ k 2 v 2 . If we need f k 0, 1

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we have to make c 1 = 0. In other words, u 0 must be proportional to the eigenvector v 2 . A - λ 2 I = 3 + 5 2 - 1 1 - 3 - 5 2 v 2 = 3 - 5 2 1 . Hence, we need to take f 1 = 3 - 5 2 so that f k will approach zero for large k . 2
Problem 2: For the matrix A = 1 0 - 1 1 1 1 2 1 0 with rank 2, consider the system of equations A x = b . (i) A x = b has a solution whenever b is orthogonal to some nonzero vector c . Explicitly compute such a vector c . Your answer can be multiplied by any overall constant, because c is any basis for the space of A . (ii) Find the orthogonal projection p of the vector b = 9 9 9 onto C ( A ). ( Note : The matrix A T A is singular, so you cannot use your formula P = A ( A T A ) - 1 A T to obtain the projection matrix P onto the column space of A . But I have repeatedly discouraged you from computing P explicitly, so you don’t need to be reminded anyway, right?) (iii) If p is your answer from (ii), then a solution y of A y = p minimizes what? [You need not answer (ii) or compute y for this part.] Solution (18 points = 7+7+4) (i) The system of equations A x = b has a solution if and only if b lies in the column space of A , which is orthogonal to the left nullspace of A . We solve for a (nonzero) vector c in the left nullspace using Gaussian elimination, as follows. A

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## This note was uploaded on 01/14/2011 for the course EECS 18.06 taught by Professor Strang during the Spring '05 term at University of Michigan.

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final-s09-soln - 18.06 Final Solution Hold on Tuesday 19...

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