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Unformatted text preview: MA 265 LECTURE NOTES: WEDNESDAY, APRIL 2 Orthogonal Complements Properties. Let ( V, + , · , ( , ) ) be an inner product space. Let W is a subspace of V , and W ⊥ be its orthogonal complement: W ⊥ = v ∈ V ( v , w ) = 0 for all w ∈ W . The following properties hold. (1) W ⊥ is also a subspace of V . (2) W ∩ W ⊥ = { } . (3) V = W + W ⊥ , i.e., each v ∈ V can be expressed in the form v = w + u where w ∈ W and u ∈ W ⊥ . We explain why these are true. Last time we showed (1). For (2), let u ∈ W ∩ W ⊥ . Write w = u ∈ W and v = u ∈ W ⊥ , so that ( v , w ) = 0. But then  u  = p ( u , u ) = p ( v , w ) = √ 0 = 0 = ⇒ u = . For (3), let v ∈ V . We wish to show v = w + u for some w ∈ W and u ∈ W ⊥ . To this end, choose a basis S = { u 1 , u 2 , ..., u m } for W . Using the GramSchmidt process, we can construct an orthonormal basis T = { w 1 , w 2 , ..., w m } for W . Define the linear combination w = ( v , w 1 ) w 1 + ( v , w 2 ) w 2 + ··· + ( v , w m ) w m ∈ W. Let u = v w . It suffices to show that u ∈ W ⊥ . Each element in W can be expressed as a linear combination of the elements from T , so we have the inner product u , m X j =1 a j w j = m X j =1 a j ( u , w j ) = m X j =1 a j " ( v , w j ) m X i =1 ( v , w i )( w i , w j ) #...
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.
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