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Unformatted text preview: mal system {z1 , z2 , . . . , zj } with the preceding ones and
zj = uj,1 x1 + uj,2 x2 + · · · + uj,j xj ,
with appropriate coeﬃcients uj,k , so zj ∈ span{xl : l ≤ j }. As a consequence, zk ∈ span{xl :
l ≤ j } for k ≤ j and thus
span{zl : l ≤ j } ⊂ span{xl : l ≤ j }
but since the dimension on both sides is equal, the two spans must be the same.
1 The inductive choice of zj is as follows:
z1 = x1
.
x 1 Given an orthonormal system {z1 , z2 , . . . , zk−1 } with the same span as {x1 , x2 , . . . , xk−1 }, then
we let
y k = x k − x k , z k − 1 z k −1 − x k , z k − 2 z k −2 − · · · − x k , z 1 z 1 .
We have by the assumed orthogonality of {z1 , z2 , . . . , zk−1 } that yk , zj = xk , zj −xk , zj = 0
for j < k . On the other hand yk = 0 because xk is not in the span of the preceding xj ’s, which
is by induction assumption...
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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
 Multiplication, Matrices, Counting

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