Addition/Subtraction flops. ... k k + 1, n (n k)(n k) (n k)(n + 1 k) n 1 n, n (1)(1) (1)(2)

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1 1 CHAPTER The flop counts for LU decomposition can be determined in a similar fashion as was done for Gauss elimination The major difference is that the elimination is only implemented for the left-hand side coefficients Thus, for every iteration of the inner loop, there are n multiplications/divisions and n 1 addition/subtractions The computations can be summarized as Outer Loop k Inner Loop i Addition/Subtraction flops Multiplication/Division flops 1, n (n 1)(n 1) (n 1)n, n (n )(n ) (n )(n 1) k k + 1, n (n k)(n k) (n k)(n + 1 k) n 1 n, n (1)(1) (1)() Therefore, the total addition/subtraction flops for elimination can be computed as n1 n1 ( nk)( nk) n nk k k1 k1 Applying some of the relationships from Eq (814) yields n1 k 1 n n n n nk k 6 A similar analysis for the multiplication/division flops yields n1 k 1 n n ( nk)( n1 k) 1 n n O( n ) n O( n) n O( n ) O( n ) Summing these results gives n n n 6 For forward substitution, the numbers of multiplications and subtractions are the same and equal to n1 ( n1) n n n i i1 Back substitution is the same as for Gauss elimination: n / n/ subtractions and n / + n/ multiplications/divisions The entire number of flops can be summarized as PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

2 Forward elimination Forward substitution Back substitution Total Mult/Div Add/Subtr Total n n n n n n n n 6 6 n n n n n n n n n n n n n n n 5n n n 7 n n 6 6 The total number of flops is identical to that obtained with standard Gauss elimination 10 Equation (106) is [ L][ U]{} x {} d [ A]{} x {} b (106) Matrix multiplication is distributive, so the left-hand side can be rewritten as [ L][ U]{} x [ L]{} d [ A]{} x {} b Equating the terms that are multiplied by {x} yields, [ L][ U]{} x [ A]{} x and, therefore, Eq (107) follows [ L][ U] [ A] (107) Equating the constant terms yields Eq (108) [ L]{ d} { b} (108) 10 (a) The coefficient a 1 is eliminated by multiplying row 1 by f 1 = /7 and subtracting the result from row a 1 is eliminated by multiplying row 1 by f 1 = 1/7 and subtracting the result from row The factors f 1 and f 1 can be stored in a 1 and a a is eliminated by multiplying row by f = 18571/ = 090 and subtracting the result from row The factor f can be stored in a Therefore, the LU decomposition is PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

3 [ L] [ U ] These two matrices can be multiplied to yield the original system For example, using MATLAB to perform the multiplication gives >> L=[1 0 0; ; ]; >> U=[7 -; ; ]; >> L*U (a) Forward substitution: [L]{D} = {B} d d d 6 Solving yields d 1 = 1, d = , and d = Back substitution: 7 x x x x ( 1486)( ) x ( ) (146875) x (b) Forward substitution: [L]{D} = {B} d d d 6 Solving yields d 1 = 1, d = , and d = 4887 Back substitution: 7 x x x 4887 x PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

4 ( 1486)(0565) x ( )(0565) (565) x The system can be written in matrix form as [ A] 1 6 { b} 4 [ P] Partial pivot: [ A] 1 7 [ P] Compute factors: f 1 = /(8) = 075 f 1 = /(8) = 05 Forward eliminate and store factors in zeros: 8 1 [ LU ] Pivot again [ LU ] [ P] Compute factors: f = 175/(575) = 091 Forward eliminate and store factor in zero: 8 1 [ LU ] Therefore, the LU decomposition is PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

5 [ L]{ U] Forward substitution First multiply right-hand side vector {b} by [P] to give [ P]{ b} Therefore, { d} d1 40 d 8 ( 05)( 40) 48 d 4 075( 40) 091( 48) Back substitution: 8 1 x x x x ( 15)( 10581) x (8685) ( )( 10581) x Here is an M-file to generate the LU decomposition without pivoting function [L, U] = LUNaive(A) % LUNaive(A): % LU decomposition without pivoting % input: % A = coefficient matrix % output: % L = lower triangular matrix % upper triangular matrix [m,n] = size(a); if m~=n, error('matrix A must be square'); end L = eye(n); A; % forward elimination for k = 1:n-1 PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

6 6 for i = k+1:n L(i,k) = U(i,k)/U(k,k); U(i,k) = 0; U(i,k+1:n) = U(i,k+1:n)-L(i,k)*U(k,k+1:n); end end Test with Prob 10 >> A = [10-1;- -6 ;1 1 5]; >> [L,U] = LUnaive(A) L = Verification that [L][U] = [A] >> L*U Check using the lu function, >> [L,U]=lu(A) L = The result of Example 105 can be substituted into Eq (1014) to give T [ A] [ U] [ U] The multiplication can be implemented as in a a a a a a a PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

7 7 a a (a) For the first row (i = 1), Eq (1015) is employed to compute u11 a Then, Eq (1016) can be used to determine u a u u a u For the second row (i = ), u a u u 1 80 ( ) a u u ( ) 1 1 u For the third row (i = ), u a u u 1 60 ( ) (18574) Thus, the Cholesky decomposition yields [ U ] The validity of this decomposition can be verified by substituting it and its transpose into Eq (1014) to see if their product yields the original matrix [A] >> [ ; ; ]; >> U *U (b) >> A = [8 0 16; ; ]; >> chol(a) (c) The solution can be obtained by hand or by MATLAB Using MATLAB: PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

8 8 >> b = [100;50;100]; >> d=u'\b d = >> x=u\d x = Here is an M-file to generate the Cholesky decomposition without pivoting function cholesky(a) % cholesky(a): % cholesky decomposition without pivoting % input: % A = coefficient matrix % output: % upper triangular matrix [m,n] = size(a); if m~=n, error('matrix A must be square'); end for i = 1:n s = 0; for k = 1:i-1 s = s + U(k, i) ^ ; end U(i, i) = sqrt(a(i, i) - s); for j = i + 1:n s = 0; for k = 1:i-1 s = s + U(k, i) * U(k, j); end U(i, j) = (A(i, j) - s) / U(i, i); end end Test with Prob 108 >> A = [8 0 16; ; ]; >> cholesky(a) Check with the chol function >> chol(a) The system can be written in matrix form as PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

9 [ A] 6 4 { b} 44 [ P] Partial pivot: [ A] 6 4 [ P] Compute factors: f 1 = /(8) = 05 f 1 = /(8) = 075 Forward eliminate and store factors in zeros: 8 5 [ LU ] Pivot again [ LU ] [ ] P Compute factors: f = 55/(75) = Forward eliminate and store factor in zero: 8 5 [ LU ] Therefore, the LU decomposition is [ L]{ U] Forward substitution First multiply right-hand side vector {b} by [P] to give [ P]{ b} PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

10 10 Therefore, { d} d1 6 d 10 ( 075)( 6) 1975 d 44 ( 05)( 6) ( )( 1975) Back substitution: 8 5 x x x x ( ) x ( ) ( ) x (a) Multiply first row by f 1 = /8 = 075 and subtract the result from the second row to give Multiply first row by f 1 = /8 = 05 and subtract the result from the third row to give Multiply second row by f = 175/515 = and subtract the result from the third row to give [ U ] As indicated, this is the U matrix The L matrix is simply constructed from the f s as PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

11 [ L] Merely multiply [L][U] to yield the original matrix (b) The determinant is equal to the product of the diagonal elements of [U]: D (c) Solution with MATLAB: >> A=[8 5 1; 7 4; 9]; >> [L,U]=lu(A) L = >> L*U >> det(a) (a) The determinant is equal to the product of the diagonal elements of [U]: D (b) Forward substitution: >> L=[1 0 0; ; ]; >> U=[ - 1; ; ]; >> b=[ ]'; >> d=l\b d = Back substitution: >> x=u\d x = PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission

12 Using MATLAB: >> A=[ -1 0;-1-1;0-1 ]; >> U=chol(A) The result can be validated by >> U'*U Using MATLAB: >> A=[9 0 0;0 5 0;0 0 16]; >> U=chol(A) Thus, the factorization of this diagonal matrix consists of another diagonal matrix where the elements are the square root of the original This is consistent with Eqs (1015) and (1016), which for a diagonal matrix reduce to u ii a ii u 0 for i j ij PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual individual course preparation If you are a student using this Manual, you are using it without permission