For any x cm p x v0 and p x x v0 proof this is

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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 “least-squares” 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.

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