AMS526: Numerical Analysis I (Numerical Linear Algebra)
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1 AMS526: Numerical Analysis I (Numerical Linear Algebra) Lecture 12: Gaussian Elimination and LU Factorization Xiangmin Jiao SUNY Stony Brook Xiangmin Jiao Numerical Analysis I 1 / 10
2 Gaussian Elimination and LU Factorization Gaussian elimination can be viewed as triangular triangularization of nonsingular A C m m Lm 1 L2L1 A = U }{{} L 1 analogous to Householder QR factorization of matrix A C m n Q n Q 2 Q }{{} 1 A = R Q Example of LU factorization of 4 4 matrix A L 1 0 x x x 0 x x x L 2 0 x x 0 x x x 0 x x } {{ } L 1 A } {{ } L 2 L 1 A L 3 0 x } {{ } L 3 L 2 L 1 A Xiangmin Jiao Numerical Analysis I 2 / 10
3 What is Matrices L k At step k, eliminate entries below a kk : let x k be kth column of L k 1 L 1 A, x k = [x 1,k, x 2,k, x k,k, x k+1,k, x m,k ] T L k x k = [x 1,k, x 2,k, x k,k, 0, 0] T The multipliers l jk = x jk /x kk appear in L k 1... L k = 1 l k+1,k l m,k 1 Let l k = [0,, 0, l k+1,k,, l m,k ] T and e k = [0,, 0, 1,, 0] }{{} T, k 1 then L k = I l k e k Xiangmin Jiao Numerical Analysis I 3 / 10
4 Forming L Luckily, the L matrix contains the multipliers l jk = x jk /x kk 1 l 21 1 L = L 1 1 L 1 2 L 1 m 1 = l 31 l l m1 l m2... l m,m 1 1 and is said to be a unit lower triangular matrix First, L 1 k = I + l k e k, because e k l k = 0 and (I l k e k )(I + l ke k ) = I l ke k l ke k = I Second, L 1 1 L 1 2 L 1 k+1 = I + k+1 j=1 l je j, since (prove by induction) (I + k j=1 l je j )(I +l k+1e k+1 ) = I + k+1 j=1 l je j + k j=1 l j(e j l k+1)e k+1 where e j l k+1 = 0 for j < k + 1 In other words, L is union of L 1 1, L 1 2,, L 1 m 1 Xiangmin Jiao Numerical Analysis I 4 / 10
5 Gaussian Elimination without Pivoting Factorize A C m m into A = LU Gaussian elimination without pivoting U = A, L = I for k = 1 to m 1 for j = k + 1 to m l jk = u jk /u kk u j,k:m = u j,k:m l jk u k,k:m Flop count m k=1 2(m k)(m k) 2 m k=1 k2 2m 3 /3 In actually, L often overwrites lower-triangular part of A and U overwrites upper-triangular part of A Question: What if u kk is 0? Answer: The algorithm would break. Xiangmin Jiao Numerical Analysis I 5 / 10
6 Partial Pivoting At step k, we divide by u kk, which would break if u kk is 0 (or close to 0), which can happen even if A is nonsingular x kk x x x x kk 0 x x x 0 x x x 0 x x x However, any nonzero entry in kth column below diagonal can also be used as pivot x ik x x x 0 x x x 0 x x x x ik 0 x x x and we permute (interchange) row i with row k In general, we take nonzero entry with largest absolute value Xiangmin Jiao Numerical Analysis I 6 / 10
7 More on Partial Pivoting kth step of Gaussian elimination of partial pivoting x ik x x P k x kk x x x x x L k Pivot selection Row interchange x kk 0 x x 0 x x Elimination and we interchange row i with row k In terms of matrices, it becomes L m 1 P m 1 L 2 P 2 L 1 P }{{} 1 A = U L 1 P P = P m 1 P 2 P 1 and L = ( L m 1 L 2L 1, 1) where L k = P m 1 P k+1 L k P 1 k+1 P 1 m 1 It is easy to verify that L m 1 P m 1 L 2 P 2 L 1 P 1 = ( L m 1 L 2L 1) (Pm 1 P 2 P 1 ) L k = I P m 1 P k+1 l k e k and L is union of ( ) L 1 k I + Pm 1 P k+1 l k e k Xiangmin Jiao Numerical Analysis I 7 / 10
8 Algorithm of Gaussian Elimination with Partial Pivoting Factorize A C m m into PA = LU Gaussian elimination with partial pivoting U = A, L = I, P = I for k = 1 to m 1 i arg max i k u ik u k,k:m u i,k:m l k,1:k 1 l i,1:k 1 p k,: p i,: for j = k + 1 to m l jk = u jk /u kk u j,k:m = u j,k:m l jk u k,k:m Question: What if u kk is 0? Flop count m k=1 2(m k)(m k) 2 m k=1 k2 2m 3 /3, same as without pivoting Xiangmin Jiao Numerical Analysis I 8 / 10
9 An Alternative Implementation In practice, L and U overwrite A and P is represented by a vector Gaussian elimination with partial pivoting (alternative) p = [1, 2,, m]; for k = 1 to m 1 i arg max i k a ik a k,1:m a i,1:m p k p i a k+1:m,k a k+1:m,k /a k,k A k+1:m,k+1:m A k+1:m,k+1:m a k+1:m,k a k,k+1:m Using LU factorization to solve Ax = b : 1 PA = LU; (LU factorization with partial pivoting) 2 Ly = Pb; (Forward substitution) 3 Ux = y; (Back substitution) Xiangmin Jiao Numerical Analysis I 9 / 10
10 Complete Pivoting More generally, we can use any nonzero entry In theory, any nonzero entry (i, j), i k, j k x x ij x x 0 x x ij x 0 x and we then permute row i with row k, column j with column k In matrix operations, it can be expressed as L m 1 P m 1 L 2 P 2 L 1 P }{{} 1 A Q 1 Q 2 Q m 1 = U }{{} L 1 P Q Therefore, PAQ = LU where P = P m 1 P 2 P 1 and L = ( L m 1 L 2L ) 1 1 However, complete pivoting is typically not used in practice because it increases cost in search of pivot and complexity of implementation Xiangmin Jiao Numerical Analysis I 10 / 10
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