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Unformatted text preview: sis {zj }n=1
j
j âˆ’1
ï¿¿
< aj , zl > zl )
with zj âˆˆ span{a1 , a2 , . . . , aj } for all 1 â‰¤ j â‰¤ n. We recall that zj = cj (aj âˆ’ with some cj chosen such that cj ï¿¿= 0 and zj  = 1. We let cj > 0, then
j âˆ’1 l=1 j âˆ’1 ï¿¿1
ï¿¿
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 âˆˆ
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 ï¿¿
1
ï¿¿
c2
.. ..
.
.
00 ï¿¿
ï...
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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.
 Fall '12
 BernhardBodmann
 Math, Matrices

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