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handout8 - Math 171A Linear Programming Class Announcements...

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Math 171A: Linear Programming Lecture 8 Linear Programming with Equality Constraints Philip E. Gill c ± 2011 http://ccom.ucsd.edu/~peg/math171a Friday, January 21, 2011 Class Announcements 1 The midterm will be held in class next Wednesday, January 26. 2 The midterm is based on material covered in Homework Assignments #1, #2 and #3. Bring a ruler! 3 The egregious typo in Lecture 6 has been fixed. Did you notice it? UCSD Center for Computational Mathematics Slide 2/29, Friday, January 21, 2011 Recap: Range- and null-space portions of a vector If A R m × n , then every b R m has the unique expansion b = b R |{z} range( A ) + b N |{z} null( A T ) range( A ) = { y : y = Ax for some x R n } null( A T ) = { z : A T z = 0 } In other words: R m = range( A ) null( A T ) UCSD Center for Computational Mathematics Slide 3/29, Friday, January 21, 2011 Given A R m × n : range( A ) is a subspace of dimension rank( A ) null( A T ) is a subspace of dimension m - rank( A ) basis for R m = n a β 1 , a β 2 , ..., a β r , | {z } column basis z 1 , z 2 , ..., z m - r | {z } basis for null( A T ) o UCSD Center for Computational Mathematics Slide 4/29, Friday, January 21, 2011
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Matlab can be used to find a basis for null( A T ). In Matlab , type: Z = null(A’) The m × ( m - r ) matrix has columns Z = ± z 1 z 2 ··· z m - r ² Matlab gives an orthonormal basis , i.e., z T i z j = ³ 0 , i 6 = j 1 , i = j UCSD Center for Computational Mathematics Slide 5/29, Friday, January 21, 2011 Computing b R and b N Example: A = 1 - 2 5 2 - 4 10 ! , and b = 4 3 ! Then rank( A ) = 1 with range( A ) = ³ y : y = α ´ 1 2 µ for all α null( A T ) = ³ z : z = γ ´ - 2 1 µ for all γ Matlab gives z = - 2 5 / 5 5 / 5 ! = 5 5 - 2 1 ! UCSD Center for Computational Mathematics Slide 6/29, Friday, January 21, 2011 b = 4 3 ! = α
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handout8 - Math 171A Linear Programming Class Announcements...

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