Differential Equations Solutions 27

# Differential Equations Solutions 27 - O kn 2 instead of O n...

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Chapter 7 Solutions: Case Study: Updating and Downdating Matrix Factorizations: A Change in Plans CHALLENGE 7.1. (a) Set the columns of Z to be the diFerences between the old columns and the new columns, and set the columns of V to be the 6th and 7th columns of the identity matrix. (b) The ±rst column of Z can be the diFerence between the old column 6 and the new one; the second can be the 4th column of the identity matrix. The ±rst column of V is then the 6th column of the identity matrix and the second is the diFerence between the old row 4 and the new one but set the 6th element to zero. CHALLENGE 7.2. (a) This is veri±ed by direct computation. (b) We use several facts to get an algorithm that is
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Unformatted text preview: O ( kn 2 ) instead of O ( n 3 ) for dense matrices: • x = ( A − ZV T ) − 1 b = ( A − 1 + A − 1 Z ( I − V T A − 1 Z ) − 1 V T A − 1 ) b . • ²orming A − 1 from the LU decomposition takes O ( n 3 ) operations, but forming A − 1 b as U \ ( L \ b ) uses forward- and back-substitution and just takes O ( n 2 ). • ( I − V T A − 1 Z ) is only k × k , so factoring it is cheap: O ( k 3 ). (²orming it is more expensive, with a cost that is O ( kn 2 ).) • Matrix multiplication is associative. Using MATLAB notation, once we have formed [L,U]=lu(A) , the resulting algo-rithm is y = U \ (L \ b); Zh = U \ (L \ Z); 37...
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## This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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