Mathematics_1_oneside.pdf

For k 1 the statement is obvious as span v 1 span u 1

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For k = 1 the statement is obvious as span( v 1 ) = span( u 1 ) and k v 1 k = 1. Now suppose the result holds for k 1. By the induction hypothesis, { v 1 ,..., v k } forms an orthonormal basis for span( u 1 ,..., u k ). In particular we have v 0 j v i = δ ji . Let w k + 1 = u k + 1 - k X j = 1 p v j ( u k + 1 ) = u k + 1 - k X j = 1 ( v 0 j u k + 1 ) v j . First we show that w k + 1 and v i are orthogonal for all i = 1,..., k . By construction we have w 0 k + 1 v i = ˆ u k + 1 - k X j = 1 ( v 0 j u k + 1 ) v j ! 0 v i = u 0 k + 1 v i - k X j = 1 ( v 0 j u k + 1 ) v 0 j v i = u 0 k + 1 v i - k X j = 1 ( v 0 j u k + 1 ) δ ji = u 0 k + 1 v i - v 0 i u k + 1 = 0. Now w k + 1 cannot be 0 since otherwise u k + 1 - k j = 1 p v j ( u k + 1 ) = 0 and consequently u k + 1 span( v 1 ,..., v k ) = span( u 1 ,..., u k ), a contradiction to our assumption that { u 1 ,..., u k , u k + 1 } is a subset of a basis of U . Thus we may take v k + 1 = w k + 1 k w k + 1 k . Then by Lemma 8.19 the vectors { v 1 ,..., v k + 1 } are linearly independent and consequently form a basis for span( u 1 ,..., u k + 1 ) by Theorem 5.21 . Thus the result holds for k + 1, and by the principle of induction, for all k = 1,..., n and in particular for k = n . 9.3 Orthogonal Complement We want to generalize Theorem 9.3 and Lemma 9.1 . Thus we need the concepts of the direct sum of two vector spaces and of the orthogonal complement . Direct sum. Let U , V R n be two subspaces with U V = { 0 } . Then Definition 9.7 U V = { u + v : u U , v V } is called the direct sum of U and V . Let U , V R n be two subspaces with U V = { 0 } and dim( U ) = k 1 Lemma 9.8 and dim( V ) = l 1. Let { u 1 ,..., u k } and { v 1 ,..., v l } be bases of U and V , respectively. Then { u 1 ,..., u k } { v 1 ,..., v l } is a basis of U V .
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9.3 O RTHOGONAL C OMPLEMENT 67 P ROOF . Obviously { u 1 ,..., u k } { v 1 ,..., v l } is a generating set of U V . We have to show that this set is linearly independent. Suppose it is lin- early dependent. Then we find α 1 ,..., α k R not all zero and β 1 ,..., β l R not all zero such that k i = 1 α i u i + l i = 1 β i v i = 0. Then u = k i = 1 α i u i 6= 0 and v = - l i = 1 β i v i 6= 0 where u U and v V and u - v = 0. But then u = v , a contradiction to the assumption that U V = { 0 } . Decomposition of a vector. Let U , V R n be two subspaces with U Lemma 9.9 V = { 0 } and U V = R n . Then every x R n can be uniquely decomposed into x = u + v where u U and v V . P ROOF . See Problem 9.6 . Orthogonal complement. Let U be a subspace of R n . Then the or- Definition 9.10 thogonal complement of U in R n is the set of vectors v that are or- thogonal to all vectors in R n , that is, U = { v R n : u 0 v = 0 for all u U } . Let U be a subspace of R n . Then the orthogonal complement U is also Lemma 9.11 a subspace of R n . Furthermore, U U = { 0 } . P ROOF . See Problem 9.7 . Let U be a subspace of R n . Then Lemma 9.12 R n = U U . P ROOF . See Problem 9.8 . Orthogonal decomposition. Let U be a subspace of R n . Then every Theorem 9.13 y R n can be uniquely decomposed into y = u + u where u U and u U . u is called the orthogonal projection of y into U . We denote this projection by p U ( y ).
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