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# handout7 - Math 171A Linear Programming So far we have...

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Math 171A: Linear Programming Lecture 7 Properties of Incompatible Systems Philip E. Gill c 2011 http://ccom.ucsd.edu/~peg/math171a Wednesday, January 19, 2011 So far, we have focused on compatible equations Ax = b . i.e., equations Ax = b with b range( A ). Question Given A R m × n , how do we characterize vectors b R m such that b 6∈ range( A )? UCSD Center for Computational Mathematics Slide 2/24, Wednesday, January 19, 2011 y y = Ax x Set of all rhs vectors R n R m unreachable part of R m y y = Ax x Set of all rhs vectors R n R m unreachable part of R m

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Example: A = 1 5 3 - 1 - 5 - 3 ! , rank( A ) = 1 range( A ) = y : y = α a 1 = α 1 - 1 for some α Observe that z = 1 1 ! 6∈ range( A ) because there is no α such that 1 1 ! = α 1 - 1 ! UCSD Center for Computational Mathematics Slide 5/24, Wednesday, January 19, 2011 The vectors z = 1 1 ! and a 1 = 1 - 1 ! are independent. Moreover, they are orthogonal , i.e., z T a 1 = ( 1 1 ) 1 - 1 ! = 0 But a 1 spans all of range( A ) z is orthogonal to every vector in range( A ) z satisfies ( 1 1 ) 1 5 3 - 1 - 5 - 3 ! = z T A = 0 . UCSD Center for Computational Mathematics Slide 6/24, Wednesday, January 19, 2011 Result Let z be a nonzero vector that is orthogonal to every column of A = a 1 a 2 · · · a n i.e., z T a j = 0 for j = 1, 2, . . . , n . Then z 6∈ range( A ). Proof: Let z range( A ) with z T a j = 0 for j = 1, 2, . . . , n . As z range( A ), there must be some y such that z = Ay , i.e., z = Ay = a 1 a 2 · · · a n y = n X j =1 a j y j If we form z T z , we get z T z = z T Ay = n X j =1 z T a j y j = n X j =1 ( z T a j ) y j = 0 UCSD Center for Computational Mathematics Slide 7/24, Wednesday, January 19, 2011 z T z = 0.
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handout7 - Math 171A Linear Programming So far we have...

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