This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 5.2 ORTHOGONAL SUBSPACES KIAM HEONG KWA pp. 214215 Two subspaces X,Y ⊆ R n are said to be orthogonal , denoted X ⊥ Orthogonal sub spaces Y , if h x , y i = x T y = 0 for all x ∈ X and all y ∈ Y . If X ⊥ Y , then X ∩ Y = { } . To see this, let z ∈ X ∩ Y , so that k z k 2 = h z , z i = 0 and thus z = . p. 215 Let Y be a subspace of R n . The orthogonal complement of Y is Orthogonal com plement the set Y ⊥ = { x ∈ R n h x , y i = x T y = 0 for all y ∈ Y } . Clearly, ∈ Y ⊥ , so that Y ⊥ 6 = ∅ . Let x 1 , x 2 ∈ Y ⊥ and let α,β ∈ R . Then for any y ∈ Y , h α x 1 + β x 2 , y i = α h x 1 , y i + β h x 2 , y i = 0 , so that α x 1 + β x 2 ∈ Y ⊥ . This shows that Y ⊥ is a subspace of R n . Exercise 1. Prove that if Y ⊆ R n is a subspace, then Y ⊆ ( Y ⊥ ) ⊥ . Example 1 (Exercise 5.2.8 in the text) . If S = Span( x 1 , x 2 , ··· , x k ) ⊆ R n , then for any y ∈ R n , y ∈ S ⊥ if and only if y ⊥ x i for all i ∈ { 1 , 2 , ··· ,k } . Proof. If y ∈ S ⊥ , then y ⊥ x i in view of the definition of S ⊥ and the fact that x i ∈ S for all i ∈ { 1 , 2 , ··· ,k } . Conversely, suppose that y ⊥ x i for all i ∈ { 1 , 2 , ··· ,k } . Let x ∈ S but otherwise arbitrary. Since S = Span( x 1 , x 2 , ··· , x k ) , there are scalars c i ’s such that x = ∑ k i =1 c i x i . Hence h y , x i = * y , k X i =1 c i x i + = k X i =1 c i h y , x i i = k X i =1 c i · 0 = 0 , indicating that y ∈ S ⊥ . pp. 215216 Let A ∈ R m × n . Recall that a vector w ∈ R m is in the column space of A if and only if A x = w for a vector x ∈ R n . Considering A as a map from R n into R m , it follows that such a vector b is in the Date : July 12, 2011. 1 2 KIAM HEONG KWA range R ( A ) of A . In other words, the column space of A is in fact the range of A as a map: R ( A ) = { w ∈ R m  A x = w for some x ∈ R n } = the column space of A. Likewise, the column space of A T is the range of A T as a map: R ( A T ) = { y ∈ R n  A T z = y for some z ∈ R m } = the column space of A T . Note that y ∈ R ( A T ) if and only if y T is in the row space of A . More important is the fact that Theorem 5.2.1: Fundamental Sub spaces Theorem N ( A ) = R ( A T ) ⊥ and N ( A T ) = R ( A ) ⊥ ....
View
Full
Document
This note was uploaded on 11/11/2011 for the course MATH 571 taught by Professor Kim during the Summer '08 term at Ohio State.
 Summer '08
 KIM
 Linear Algebra, Algebra

Click to edit the document details