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Unformatted text preview: 0 = span{z1 , z2 , . . . , zn } by
P : x → P x = n
j =1 x, zj zj . The name is motivated by a fundamental property of P .
1.2.5 Lemma. For any x ∈ Cm , P x ∈ V0 and P x − x ⊥ V0 .
Proof. This is because
1. P x is by deﬁnition in V0 and
2 2. P x − x is orthogonal to each vector zj , thus also to V0 .
The orthogonality to each zj results from the linearity of the dot product and from the orthogonality relation among the zj ’s,
P x − x, zj =
= n
l=1
l=1 x, zl zl − x, zj x, zl zl , zj − x, zj = x, zj − x, zj = 0 . According to the deﬁnition, P seems to depend on the choice of the orthonormal system.
However, a diﬀerent choice of orthonormal system with the same span results in the same P , as
we see below. This justiﬁes calling it an orthogonal projection onto V0 .
To verify this, we prove that the orthogonality property P x − x ⊥ V0 is equivalent to the
“leastsquares” property.
1.2.6 Proposition. Let V0 be a subspace of Cm ,...
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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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