LU Decomposition Notes

# LU Decomposition Notes - Chapter 04.07 LU Decomposition...

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Chapter 04.07 LU Decomposition After reading this chapter, you should be able to: 1. identify when LU decomposition is numerically more efficient than Gaussian elimination, 2. decompose a nonsingular matrix into LU, and 3. show how LU decomposition is used to find the inverse of a matrix. I hear about LU decomposition used as a method to solve a set of simultaneous linear equations. What is it? We already studied two numerical methods of finding the solution to simultaneous linear equations – Naïve Gauss elimination and Gaussian elimination with partial pivoting. Then, why do we need to learn another method? To appreciate why LU decomposition could be a better choice than the Gauss elimination techniques in some cases, let us discuss first what LU decomposition is about. For a nonsingular matrix [ ] A on which one can successfully conduct the Naïve Gauss elimination forward elimination steps, one can always write it as [ ] [ ] [ ] U L A = where [ ] L = Lower triangular matrix [ ] U = Upper triangular matrix Then if one is solving a set of equations ] [ ] ][ [ C X A = , then [ ] [ ] [ ] [ ] C X U L = as [ ] [ ] ( 29 U L A ] [ = Multiplying both sides by [ ] 1 - L , [ ] [ ] [ ] [ ] [ ] [ ] C L X U L L 1 1 - - = [ ] [ ] [ ] X U I = [ ] [ ] C L 1 - as [ ] [ ] ( 29 ] [ 1 I L L = - [ ] [ ] [ ] [ ] C L X U 1 - = as [ ] [ ] ( 29 ] [ U U I = Let [ ] [ ] [ ] Z C L = - 1 04.07.1

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04.07.2 Chapter 04.07 then [ ] [ ] [ ] C Z L = (1) and [ ] [ ] [ ] Z X U = (2) So we can solve Equation (1) first for ] [ Z by using forward substitution and then use Equation (2) to calculate the solution vector [ ] X by back substitution. This is all exciting but LU decomposition looks more complicated than Gaussian elimination. Do we use LU decomposition because it is computationally more efficient than Gaussian elimination to solve a set of n equations given by [ A ] [ X ]=[ C ]? For a square matrix ] [ A of n n × size, the computational time 1 DE CT | to decompose the ] [ A matrix to ] ][ [ U L form is given by DE CT | = - + 3 20 4 3 8 2 3 n n n T , where T = clock cycle time 2 . The computational time FS CT | to solve by forward substitution [ ] [ ] [ ] C Z L = is given by FS CT | = ( 29 n n T 4 4 2 - The computational time BS CT | to solve by back substitution [ ] [ ] [ ] Z X U = is given by BS CT | = ( 29 n n T 12 4 2 + So, the total computational time to solve a set of equations by LU decomposition is LU CT | = DE CT | + FS CT | + BS CT | = - + 3 20 4 3 8 2 3 n n n T + ( 29 n n T 4 4 2 - + ( 29 n n T 12 4 2 + = + + 3 4 12 3 8 2 3 n n n T Now let us look at the computational time taken by Gaussian elimination. The computational time FE CT | for the forward elimination part, FE CT | = - + 3 32 8 3 8 2 3 n n n T , and the computational time BS CT | for the back substitution part is 1 The time is calculated by first separately calculating the number of additions, subtractions,
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## This note was uploaded on 05/01/2011 for the course CHBE 2120 taught by Professor Gallivan during the Spring '07 term at Georgia Tech.

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LU Decomposition Notes - Chapter 04.07 LU Decomposition...

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