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Unformatted text preview: 6.4 Continued, 6.5 Least Squares Method Consider n n matrix A = [ vectorx 1   vectorx n ] with lin early independent columns Col A = Span { vectorx 1 , , vectorx n } . Applying GramSchmidt process to A , get orthonormal { vectoru 1 , , vectoru n } such that Span { vectorx 1 , , vectorx l } = Span { vectoru 1 , , vectoru l } for l = 1 , 2 , ,n . Question: What is the matrix relationship between A = [ vectorx 1   vectorx n ] and U = [ vectoru 1   vectoru n ]? Answer: vectorx l = r 1 l vectoru 1 + + r ll vectoru l + 0 vectoru l +1 + + 0 vectoru n = ( vectorx l vectoru 1 ) vectoru 1 + + ( vectorx l vectoru l ) vectoru l + 0 vectoru l +1 + + 0 vectoru n or vectorx l = [ vectoru 1   vectoru n ] r 1 l . . . r ll . . . or A = [ vectorx 1   vectorx n ] = [ vectoru 1   vectoru n ] r 11 r 12 r 1 n r 22 r 2 n ....
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This note was uploaded on 04/09/2011 for the course MATH 232 taught by Professor Russel during the Spring '10 term at Simon Fraser.
 Spring '10
 Russel
 Linear Algebra, Algebra, Least Squares

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