aj for all 1 j n we recall that zj cj aj with

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Unformatted text preview: sis {zj }n=1 j j −1 ￿ < aj , zl > zl ) with zj ∈ span{a1 , a2 , . . . , aj } for all 1 ≤ j ≤ n. We recall that zj = cj (aj − with some cj chosen such that cj ￿= 0 and ||zj || = 1. We let cj > 0, then j −1 l=1 j −1 ￿1 ￿ 1 aj = z j + rl,j zl < aj , zl > zl = rj,j zj + cj cj l=1 l=1 where rj,j = c1j and rl,j = c1j < aj , zl >. Since zj is in the span, it implies that aj ∈ span{z1 , . . . , zj }. In matrix notation, we get in upper triangular matrix R with entries rl,j and 1 c1 0 A=U 0 0 ￿ 1 ￿ c2 .. .. . . 00 ￿ ...
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This note was uploaded on 01/16/2014 for the course MATH 6304 taught by Professor Bernhardbodmann during the Fall '12 term at University of Houston.

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