1/16 The Direct Methods xiaozhouli@uestc.edu.cn http://xiaozhouli.com School of Mathematical Sciences University of Electronic Science and Technology of China Chengdu, China
Does the LU factorization always work? That A = LU, where 1. m.. L = 21.... 1 m n1 m n,n 1 1 and a 11 a 12 a 1n U = A (n 1) a (1) = 22 a (1) 2n.... a nn (n 1) 2/16
3/16 Theorem a (k 1) kk 0(k = 1, 2,..., n) if and only if the matrix A satisfies that its leading principle minors a 11 a 12 a 1k a 21 a 22 a D k = 2k... 0 k = 1,..., n a k1 a k2 a kk Theorem If the LU factorization of the matrix A exists, then it is unique.
4/16 where L = A = a 11 a 12 a 1n a 21 a 22 a 2n... a n1 a n2 a nn 1. l.. 21. U =... 1 l n1 l n,n 1 1 = LU u 11 u 12 u 1n u 22 u 2n.... u nn
5/16 1st row 1st column u 1j = a 1j, j = 1,..., n; l i1 = a i1 u 11, i = 2,..., n.
rth row n r 1 a rj = l rk u kj = l rk u kj + u rj k=1 k=1 r 1 = u rj = a rj l rk u kj, k=1 j = r,..., n rth column n r 1 a ir = l ik u kr = l ik u kr + l ir u rr k=1 k=1 ( ) r 1 = l ir = a ir l rk u kj /u rr, i = r + 1,..., n k=1 6/16
Example: 7/16
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Crout 9/16 where L = A = a 11 a 12 a 1n a 21 a 22 a 2n... a n1 a n2 a nn l 11. l.. 21. Ū =... 1 l n1 l n,n 1 l nn = LŪ 1 ū 12 ū 1n 1 ū 2n.... 1 Cholesky : A is symmetric positive definite (SPD) matrix.
Another Example 10/16 The solution of the system { 0.0003x1 + 1.566x 2 = 1.569 0.3454x 1 2.436x 2 = 1.018 is x 1 = 10, x 2 = 1. Solving this by elimination with four-decimal floating arithmetic. Pivoting Strategy 1 [ 0.0003 1.566 1.569 0.3454 2.436 1.018 ] [ 0.0003 1.566 1.569 1.804 1.805 ] So we have x 2 = 1.001 and x 1 = 3.333
11/16 Pivoting Strategy 2 [ 0.0003 1.566 1.569 0.3454 2.436 1.018 ] [ 0.3454 2.436 1.018 1.568 1.568 So we have x 2 = 1 and x 1 = 10 [ 0.3454 2.436 1.018 0.0003 1.566 1.569 It is not possible at present to give a best pivoting strategy for a general linear system, nor is it even clear what such a term might mean. ] ]
12/16 For the sake of economy, the pivotal equation for each step must be selected on the basis of the current state of the system under consideration at the beginning of the step, i.e., without foreknowledge of the effect of the selection on later steps. A currently accepted strategy is partial pivoting.
13/16 Comparing numbers before carrying out each elimination step; Then, at the beginning of the kth step of the elimination, one picks as pivotal equation that one from the available n k candidates which has the absolutely largest coefficient of x k, assume it to be p (k + 1 p n); Exchanges the kth row and the pth row if necessary. Using partial pivoting ensures that all multipliers, or entires of L, will be no greater than 1.
Permutation Matrices 14/16 Definition A permutation matrix is an n n matrix consisting of all zeros, except for a single 1 in every row and column. Equivalently, a permutation matrix P is created by applying arbitrary row exchanges to the n n identity matrix (or arbitrary column exchanges). Theorem Let P be the n n permutation matrix formed by a particular set of row exchanges applied to the identity matrix. Then, for any n n matrix A, PA is the matrix obtained by applying exactly the same set of row exchanges to A.
Example 15/16 Find the PA = LU factorization of the matrix 2 1 5 A = 4 4 4 1 3 1 The solution 0 1 0 2 1 5 1 1 0 0 1 4 4 4 = 4 1 1 1 0 0 1 3 1 2 1 2 1 If we rewrite A = PLU, then what is P? 4 4 4 2 2 8 P = P 1 = P T.
Comments on 16/16 It will increase the computational cost of elimination. Considering a variation of a previous example { 3x 1 + 1566x 2 = 1569 0.3454x 1 2.436x 2 = 1.018 Is partial pivoting stable here? The scaled partial pivoting (code). Does pivoting always better than without pivoting? Considering the tri-diagonal matrix, d-diagonal matrix. Complete pivoting elimination.