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Unformatted text preview: Numerical Linear Algebra Midterm Exam Takehome Exam Open Notes, Textbook, Homework Solutions Only Calculators Allowed Tuesday 19 October, 2010 Question Points Points Possible Awarded 1. LU 25 2. Structured 30 Factorizations 3. Symmetric Tridiagonal 25 Eigenvalue Problem 4. SVD 30 Total 110 Points Name: Alias: to be used when posting anonymous grade list. 1 Problem 1 Let A ∈ R n × n be a diagonally dominant matrix with A = LU . Suppose you have computed the first i 1 rows of L and the first i 1 rows of U by reading and processing the first i 1 rows of A , i.e., you have not touched rows i to n of A so the algorithm is a delayed update version. 1.a . (10 points) Derive the ith step of the algorithm where you read row i of A and compute row i of L and row i of U . 1.b . (5 points) Identify the computational primitives used and the level of BLAS in which they appear. 1.c . (10 points) Suppose you want to add pivoting to guarantee stability and exis tence of the factorization for any nonsingular matrix A . What form of pivoting can be introduced into this algorithm? Solution: Let A 11 ∈ R i 1 × i 1 , L 11 ∈ R i 1 × i 1 , U 11 ∈ R i 1 × i 1 and conformally partition A = LU as follows A 11 A 12 A 21 A 22 = L 11 L 21 L 22 U 11 U 12 U 22 This yields the standard identities L 11 U 11 = A 11 L 11 U 12 = A 12 L 21 U 11 = A 21 L 22 U 22 + L 21 U 12 = A 12 The first i 1 rows of A are known and given by ( A 11 A 12 ) . The first i 1 rows of L are known and given by ( L 11 ) . The first i 1 rows of L are known and given by ( U 11 U 12 ) . We are given e T i A and are to determine e T i L and e T i U . Therefore, we first must find an expression for the latter two rows and relate it to the former. We have e T i A = ( e T 1 A 21 e T 1 A 22 ) e T i L = ( e T 1 L 21 e T 1 L 22 ) = ( e T 1 L 21 e T 1 ) e T i U = ( T e T 1 U 22 ) The row e T i L is the solution to the triangular system ( e T 1 L 21 ) U 11 = e T 1 A 21 where U 11 and e T 1 A 21 are known. Given e T 1 L 21 , the row e T i U is the result of a matrixvector product ( e T 1 L 22 ) U 22 = e T 1 A 22 ( e T 1 L 21 ) U 12 e T 1 U 22 = e T 1 A 22 ( e T 1 L 21 ) U 12 2 These two BLAS2 primitives are essentially a large structure triangular solve. Complete pivoting requires the entire Schur complement which we do not have, rook pivoting requires potentially several elements in arbitrary positions in the Schur comple ment, standard partial pivoting (row interchanges) requires the first column of the Schur complement which we do not have. Since the algorithm is a delayed update version that produces the entire first row of the appropriate Schur complement, we can use partial pivoting on the row, i.e., interchange elements in a row which translates into a column permutation matrix on the right of A ....
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This note was uploaded on 07/21/2011 for the course MAD 5932 taught by Professor Gallivan during the Fall '06 term at FSU.
 Fall '06
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