# solhw3 - Homework 3 Foundations of Computational Math 1...

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Homework 3 Foundations of Computational Math 1 Fall 2010 The solutions will be posted on Wednesday, 9/22/09 Problem 3.1 Suppose A R n × n is a nonsymmetric nonsingular diagonally dominant matrix with the following nonzero pattern (shown for n = 6) * * * * * * * * 0 0 0 0 * 0 * 0 0 0 * 0 0 * 0 0 * 0 0 0 * 0 * 0 0 0 0 * It is known that a diagonally dominant (row or column dominant) matrix has an LU factor- ization and it can be computed stably without pivoting. 3.1.a . Describe an algorithm that solves Ax = b in a stable manner as eﬃciently as possible. 3.1.b . Given that the number of operations in the algorithm is of the form Cn k + O ( n k - 1 ), where C is a constant independent of n and k > 0, what are C and k ? Solution: Since the matrix is assumed to be diagonally dominant, no pivoting is required for stabil- ity and existence as was discussed in class and the text. Suppose we simply apply Gaussian elimination without pivoting. After one step of LU factorization the n - 1 × n - 1 matrix to be factored on steps 2 to n is dense and 2 3 n 3 + O ( n 2 ) computations are needed to complete the factorization. That is, the linear combination of the ﬁrst row with each row 2 i 6 replaces all of the 0 elements with nonzero elements: M - 1 1 A = * * * * * * 0 * * * * * 0 * * * * * 0 * * * * * 0 * * * * * 0 * * * * * These new nonzeros must be eliminated on subsequent steps. Since the matrix is diagonally dominant any permuted PAP T , i.e., using the same gen- eral permutation on the left and right, keeps diagonal elements on the diagonal but changes 1

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the ordering. Oﬀ-diagonal elements stay in their original rows and columns but occupy dif- ferent positions. Therefore, PAP T is also diagonally dominant and can be factored without pivoting. Now consider step one of the LU factorization where we interchange rows and columns to choose the pivot element. We do not need to do so for stability or existence as long as we choose a pivot element from the diagonal elements of the matrix, i.e., P i = Q i . Choose P 1 = ( e n e 2 ... e n - 1 e 1 ) = Q 1 i.e., interchange rows n and 1 and columns n and 1 thereby choosing α nn as the pivot for step one. For example, with n = 6 the matrix structure for the permuted matrix is P 1 AP T 1 = * 0 0 0 0 * 0 * 0 0 0 * 0 0 * 0 0 * 0 0 0 * 0 * 0 0 0 0 * * * * * * * * So M - 1 1 only needs to combine row 1 with row 6 (in general with row n ) to produce a ﬁrst column of 0 below the diagonal, i.e., M - 1 1 P 1 AP T 1 = * 0 0 0 0 * 0 * 0 0 0 * 0 0 * 0 0 * 0 0 0 * 0 * 0 0 0 0 * * 0 * * * * * No further pivoting is needed and each step combines row i with row
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solhw3 - Homework 3 Foundations of Computational Math 1...

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