lecture_33 - MA 265 LECTURE NOTES: WEDNESDAY, APRIL 2...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 Gram-Schmidt 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 ) #...
View Full Document

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.

Page1 / 3

lecture_33 - MA 265 LECTURE NOTES: WEDNESDAY, APRIL 2...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online