orthogonalcomplements

orthogonalcomplements - Orthogonal Complements (Revised...

Info iconThis preview shows pages 1–3. 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: Orthogonal Complements (Revised Version) Math 108A: May 19, 2010 John Douglas Moore 1 The dot product You will recall that the dot product was discussed in earlier calculus courses. If x = ( x 1 ....,x n ) and y = ( y 1 ....,y n ) are elements of R n , we define their dot product by x · y = x 1 y 1 + ··· + x n y n . The dot product satisfies several key axioms: 1. it is symmetric: x · y = y · x ; 2. it is bilinear: ( a x + x ) · y = a ( x · y ) + x · y ; 3. and it is positive-definite: x · x ≥ 0 and x · x = 0 if and only if x = . The dot product is an example of an inner product on the vector space V = R n over R ; inner products will be treated thoroughly in Chapter 6 of [1]. Recall that the length of an element x ∈ R n is defined by | x | = √ x · x . Note that the length of an element x ∈ R n is always nonnegative. Cauchy-Schwarz Theorem. If x 6 = and y 6 = , then- 1 ≤ x · y | x || y | ≤ 1 . (1) Sketch of proof: If v is any element of R n , then v · v ≥ 0. Hence ( x ( y · y )- y ( x · y )) · ( x ( y · y )- y ( x · y )) ≥ . Expanding using the axioms for dot product yields ( x · x )( y · y ) 2- 2( x · y ) 2 ( y · y ) + ( x · y ) 2 ( y · y ) ≥ or ( x · x )( y · y ) 2 ≥ ( x · y ) 2 ( y · y ) . 1 Dividing by y · y , we obtain | x | 2 | y | 2 ≥ ( x · y ) 2 or ( x · y ) 2 | x | 2 | y | 2 ≤ 1 , and (1) follows by taking the square root. The key point of the Cauchy-Schwarz Inequality (1) is that it allows us to define angles between vectors x and y in R n . (It is a first step towards extending geometry from R 2 and R 3 to R n .) It follows from properties of the cosine function that given a number t ∈ [- 1 , 1], there is a unique angle θ such that θ ∈ [0 ,π ] and cos θ = t. Thus we can define the angle between two nonzero vectors x and y in R n by requiring that θ ∈ [0 ,π ] and cos θ = x · y | x || y | . Then the dot product satisfies the formula x · y = | x || y | cos θ. In particular, we can say that two vectors vectors x and y in R n are perpen- dicular or orthogonal if x · y = 0. This provides much intuition for dealing with vectors in R n . Thus if a = ( a 1 ,...a n ) is a nonzero element of R n , the homogeneous linear equation a 1 x 1 + ··· + a n x n = 0 describes the set of all vectors x = ( x 1 ,...,x n ) ∈ R n that are perpendicular to a . The set of solutions W = { x ∈ R n : a · x = 0 } to this homogeneous linear equation is a linear subspace of R n . We remind you that to see that W is a linear subspace, we need to check three facts: 1. a · = 0, so ∈ W . 2. If x ∈ W and y ∈ W , then a · x = 0 and a · y = 0, and it follows from the axioms for dot product that a · ( x + y ) = 0 so x + y ∈ W ....
View Full Document

This note was uploaded on 06/05/2010 for the course MATH Math 108a taught by Professor Moore during the Spring '10 term at UCSB.

Page1 / 8

orthogonalcomplements - Orthogonal Complements (Revised...

This preview shows document pages 1 - 3. Sign up to view the full document.

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