# solhw6 - Solutions for Homework 6 Foundations of...

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Solutions for Homework 6 Foundations of Computational Math 1 Fall 2010 Problem 6.1 Suppose you are attempting to solve Ax = b using a linear stationary iterative method deﬁned by x k = Gx k - 1 + f that is consistent with Ax = b . Suppose the eigenvalues of G are real and such that | λ 1 | > 1 and | λ i | < 1 for 2 i n . Also suppose that G has n linearly independent eigenvectors, z i , 1 i n . 6.1.a . Show that there exists an initial condition x 0 such that x k converges to x = A - 1 b . 6.1.b . Does your answer give a characterization of selecting x 0 that could be used in practice to create an algorithm that would ensure convergence? Solution: Let e ( k ) = x - x k be the error on step k . We know that e (0) = α 1 z 1 + n X i =2 α i z i , uniquely e ( k ) = λ k 1 α 1 z 1 + n X i =2 λ k i α i z i So if x 0 is such that α 1 = 0 then e (0) = n X i =2 α i z i , uniquely e ( k ) = n X i =2 λ k i α i z i Since | λ i | < 1 for 2 i n it follows that lim k →∞ e ( k ) = 0 This is not a practical characterization for two main reasons. The ﬁrst is obvious: e (0) and z 1 are not known therefore we cannot be sure α 1 = 0. However, even if x 0 was such that α 1 = 0 on each iteration there would be roundoﬀ error that would most likely have a piece in the direction of z 1 . This would then give an iteration x j such that its α 1 6 = 0 and divergence in practice would follow. 1

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Problem 6.2 Suppose you are attempting to solve Ax = b using a linear stationary iterative method deﬁned by x k = M - 1 Nx k - 1 + M - 1 b where A = M - N . Suppose further that M = D + F where D = diag ( α 11 ,...,α nn ) and F is made up of any subset of the oﬀ-diagonal elements of A . The matrix N is therefore the remaining oﬀ-diagonal elements of A after removing those in F . Show that if A is a strictly diagonally dominant M -matrix then the iteration is convergent to x = A - 1 b . Solutions: Given the method of choosing F it is clear that any sum (row or column) of the mag- nitudes of oﬀ-diagonal elements in M must be smaller than the corresponding sum of the magnitudes of oﬀ-diagonal elements in A (since we are simply removing terms). So if A is strictly diagonally dominant then so is M . Since A is an M -matrix we know that α ij 0. It follows that N 0 and for the oﬀ-diagonal elements of M , either μ ij = α ij 0 or μ ij = 0. Since M has nonpositive oﬀ-diagonal elements and is strictly diagonally dominant, M is an M -matrix. Therefore A = M - N is a regular splitting and the iteration is convergent. Problem 6.3 6.3.a . Textbook page 241, Problem 2 6.3.b . Textbook page 241, Problem 4 6.3.c . Textbook page 241, Problem 5 Material in textbook Sections 1.7 and 5.1 is useful for these problems. Solution:
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## This note was uploaded on 07/25/2011 for the course MAD 5403 taught by Professor Gallivan during the Spring '11 term at University of Florida.

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solhw6 - Solutions for Homework 6 Foundations of...

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