mat 2310 0607_2_note8_2

mat 2310 0607_2_note8_2 - Lecture Note 8-2: March 12...

Info iconThis preview shows pages 1–8. 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

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight 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: Lecture Note 8-2: March 12 & 14, 2007 Dr. Jeff Chak-Fu WONG Department of Mathematics Chinese University of Hong Kong jwong@math.cuhk.edu.hk MAT 2310C Linear Algebra and Its Applications Spring, 2007 Produced by Jeff Chak-Fu WONG 1 O RTHOGONAL C OMPLEMENTS ORTHOGONAL COMPLEMENTS 2 Example 1 Suppose W 1 and W 2 are subsets of a vector space V . Define W 1 + W 2 . Solution W 1 + W 2 consists of all sums of w 1 a + w 2 a , where w 1 a ∈ W 1 and w 2 a ∈ W 2 : W 1 + W 2 = { w 1 a + w 2 a | w 1 a ∈ W 1 , w 2 a ∈ W 2 } . ORTHOGONAL COMPLEMENTS 3 Example 2 Suppose W 1 and W 2 are subspaces of a vector space V . Show that W 1 + W 2 is a subspace of V . Solution Since W 1 and W 2 are subspaces, ∈ W 1 and ∈ W 2 . Hence = + ∈ W 1 + W 2 . Suppose v a , v b ∈ W 1 + W 2 . Then there exist w 1 a , w 1 b ∈ W 1 and w 2 a , w 2 b ∈ W 2 such that v a = w 1 a + w 2 a and v b = w 1 b + w 2 b . Since W 1 and W 2 are subspaces, w 1 a + w 1 b ∈ W 1 and w 2 a + w 2 b ∈ W 2 , and for any scalar c , c w 1 a ∈ W 1 and c w 2 a ∈ W 2 . Accordingly, v a + v b = ( w 1 a + w 2 a )+( w 1 b + w 2 b ) = ( w 1 a + w 1 b )+( w 2 a + w 2 b ) ∈ W 1 + W 2 and for any scalar c , c v a = c ( w 1 a + w 1 b ) = c w 1 a + c w 1 b ∈ W 1 + W 2 . Thus W 1 + W 2 is a subspace of V ORTHOGONAL COMPLEMENTS 4 Example 3 Define the direct sum V = W 1 ⊕ W 2 . Solution The vector space V is said to be the direct sum of its subspaces W 1 and W 2 , denoted by V = W 1 + W 2 if every vector v ∈ V can be written in one and only one way as v = w 1 + w 2 , where w 1 ∈ W 1 and w 2 ∈ W 2 ORTHOGONAL COMPLEMENTS 5 Example 4 The vector space V is the direct sum of its subspaces W 1 and W 2 , i.e., V = W 1 ⊕ W 2 if and only if 1. V = W 1 + W 2 2. W 1 ∩ W 2 = { } Solution ( ⇒ ) Suppose V = W 1 ⊕ W 2 . Then any v ∈ V can uniquely written in the form v = w 1 + w 2 , where w 1 ∈ W 1 and w 2 ∈ W 2 . Thus, in particular, V = W 1 + W 2 . Now suppose v = W 1 ∩ W 2 . Then v = v + , where v ∈ W 1 , ∈ W 2 and v = + v , where ∈ W 1 , v ∈ W 2 Since such a sum for v must be unique, v = . Accordingly, W 1 ∩ W 2 = { } . ORTHOGONAL COMPLEMENTS 6 ( ⇐ ) On the other hand, suppose V = W 1 + W 2 and W 1 ∩ W 2 = { } . Let v ∈ V . Since V = W 1 + W 2 , there also exist w 1 ∈ W 1 and w 2 ∈ W 2 such that v = w 1 + w 2 . We need to show that a sum is unique. Suppose also that v = w * 1 + w * 2 , where w * 1 ∈ W 1 and w * 2 ∈ W 2 . Then w 1 + w 2 = w * 1 + w * 2 and so w 1- w * 1 = w 2- w * 2 . But w 1- w * 1 ∈ W 1 = w 2- w * 2 ∈ W 2 ; hence by W 1 ∩ W 2 = { } , w 1- w * 1 = , w 2- w * 2 = and so w 1 = w * 1 and w 2 = w * 2 . Thus such a sum for v ∈ V is unique and V = W 1 ⊕ W 2 ....
View Full Document

Page1 / 59

mat 2310 0607_2_note8_2 - Lecture Note 8-2: March 12...

This preview shows document pages 1 - 8. Sign up to view the full document.

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