Lecture_8_2 - Lecture Note 8-2: Oct 31 - Nov 3, 2006 Dr....

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Unformatted text preview: Lecture Note 8-2: Oct 31 - Nov 3, 2006 Dr. Jeff Chak-Fu WONG Department of Mathematics Chinese University of Hong Kong jwong@math.cuhk.edu.hk MAT 2310 Linear Algebra and Its Applications Fall, 2006 Produced by Jeff Chak-Fu 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 ....
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Lecture_8_2 - Lecture Note 8-2: Oct 31 - Nov 3, 2006 Dr....

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