This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture Note 82: Oct 31  Nov 3, 2006 Dr. Jeff ChakFu WONG Department of Mathematics Chinese University of Hong Kong [email protected] MAT 2310 Linear Algebra and Its Applications Fall, 2006 Produced by Jeff ChakFu WONG 1 R EAL V ECTOR S PACES 1. Vector Spaces 2. Subspaces 3. Linear Independence 4. Basis and Dimension 5. Homogeneous Systems 6. The Rank of a Matrix and Applications 7. Coordinates and Change of Basis 8. Orthonormal Bases in R n 9. Orthogonal Complements REAL VECTOR SPACES 2 O RTHOGONAL C OMPLEMENTS ORTHOGONAL COMPLEMENTS 3 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 4 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 5 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 6 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 7 ( ⇐ ) 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
 Spring '09
 JeffWong
 Linear Algebra, Algebra, RTHOGONAL C OMPLEMENTS

Click to edit the document details