lecture5

# lecture5 - CSE/MATH 6643 Numerical Linear Algebra Haesun...

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CSE/MATH 6643: Numerical Linear Algebra Haesun Park { hpark } @cc.gatech.edu School of Computational Science and Engineering College of Computing Georgia Institute of Technology Atlanta, GA 30332, USA Lecture 5

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LU Factorization 3.9 Applications of PA = LU 3.9.1 Linear System with Multiple RHS Ax (1) = b (1) , Ax (2) = b (2) ,..., Ax ( p ) = b ( p ) ⇐⇒ AX = B where X = h x (1) · · · x ( p ) i is n × p , B = h b (1) · · · b ( p ) i is n × p . 1. Compute G.E. with p.p on A : PA = LU . PAx ( i ) = Pb ( i ) ⇐⇒ LUx ( i ) = Pb ( i ) 2 3 n 3 flops 2. Solve Ly ( i ) = Pb ( i ) for y ( i ) . n 2 × p flops 3. Solve Ux ( i ) = y ( i ) for x ( i ) n 2 × p flops CSE/MATH 6643: Numerical Linear Algebra – p.1/13
LU Factorization Special case, when B = I n = h e 1 · · · e n i , X is A 1 . 2 3 n 3 + 2 n 3 flops? No, less. 1. PA = LU 2 3 n 3 flops 2. Ly ( i ) = P e i e.g. 2 6 6 6 6 6 4 x x x x x x x x x x x x x x x 3 7 7 7 7 7 5 2 6 6 6 6 6 4 0 0 0 x x 3 7 7 7 7 7 5 = 2 6 6 6 6 6 4 0 0 0 1 0 3 7 7 7 7 7 5 . No computation required for rows above ’ 1 ’. It takes n P j =1 ( n j )( n j + 1) , i.e. n 3 3 flops 3. Solve Ux ( i ) = y ( i ) . Same usual case, n 3 flops. Totally takes 2 3 n 3 + 1 3 n 3 + n 3 = 2 n 3 flops. CSE/MATH 6643: Numerical Linear Algebra – p.2/13

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LU Factorization 3.9.2 Want α = c T A 1 b method 1: 1. Compute A 1 : 2 n 3 flops 2. Compute d = A 1 b : 2 n 2 flops 3. Compute α = c T d : 2 n flops Totally 2 n 3 flops. method 2: 1. Subsystem: d = A 1 b . We solve Ad = b for d . 2 3 n 3 flops. 2. Compute α = c T d , 2 n flops Totally 2 3 n 3 flops. CSE/MATH 6643: Numerical Linear Algebra – p.3/13
Condition Number and Error Analysis 4 Condition Number and Error Analysis Some useful expressions : Ae i = a i , a ij = e T i Ae j 4.1 Error Analysis How good is the computed solution x ?

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