Week_10

# Week_10 - Week 10 Section 5.3: Orthogonality We looked at...

This preview shows pages 1–3. Sign up to view the full content.

Week 10 Section 5.3: Orthogonality We looked at dot product and length of vectors in R 2 and R 3 in Chapter 4. We can extend these deﬁnitions to R n . 1. If ~u = [ u 1 u 2 . . . u n ] T and ~v = [ ~v 1 ~v 2 . . . ~v n ] T are vectors in R n , then ~u · ~v = u 1 v 2 + u 2 v 2 + . . . + u n v n . 2. Two vectors ~u and ~v R n are orthogonal if ~u · ~v = 0. 3. The length (or norm) of a vector ~u = [ u 1 u 2 ··· u n ] T is k ~u k = p u 2 1 + u 2 2 + . . . + u 2 n and is always positive. 4. k ~u k 2 = ~u · ~u Cauchy Inequality If ~u,~v R n , then | ~u · ~v | ≤ k ~u kk ~v k . Proof: The general case is proven in the textbook, we will consider the case where n = 2. In R 2 , we can use the deﬁnition of dot product involving cos θ to prove this. | ~u · ~v | = |k ~u kk ~v k cos θ | = k ~u kk ~v k| cos θ | ≤ k ~u kk ~v k (1), since | cos θ | ≤ 1 Triangle Inequality k ~u + ~v k ≤ k ~u k + k ~v k Proof: We will prove that k ~u + ~v k 2 ( k ~u k + k ~v k ) 2 , and since k ~u + ~v k ≥ 0 and k ~u k + k ~v k ≥ 0 we can take square roots of both sides and the desired inequality will hold. k ~u + ~v k 2 = ( ~u + ~v ) · ( ~u + ~v ) = ~u · ~u + ~u · ~v + ~v · ~u + ~v · ~v = k ~u k 2 + 2 ~u · ~v + k ~v k 2 ≤ k ~u k 2 + 2 | ~u · ~v | + k ~v k 2 since any real number is less than equal to its absolute value ≤ k ~u k 2 + 2 k ~u kk ~v k + k ~v k 2 by the Triangle Inequality = ( k ~u k + k ~v k ) 2 as desired. 1

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

View Full Document
Deﬁnition: A set { ~u 1 ,~u 2 , . . . ,~u n } of non-zero vectors in R n is called an orthogonal set if ~u i · ~u j = 0 for all i 6 = j . This set { ~u 1 ,~u 2 , . . . ,~u n } is called orthonormal if it is orthogonal and each ~u i is a unit vector. (ie. if k ~u i k = 1 for all i .) Examples 1. The standard basis { ~e 1 ,~e 2 , . . . ,~e n } is an orthonormal set in R n . 2. If { ~u 1 ,~u 2 , . . . ,~u n } is orthogonal, then so is { a 1 ~u 1 , a 2 ~u 2 , . . . , a n ~u n } . We can create an orthonormal set from any orthogonal set simply by dividing each vector by its length making it a unit vector. That is, if { ~u 1 ,~u 2 , . . . ,~u k } is an orthogonal set, then ± 1 k ~u 1 k ~u 1 , 1 k ~u 2 k ~u 2 , . . . , 1 k ~u k k ~ u k ² is an orthonormal set. Example: If ~ f 1 = 1 1 1 - 1 , ~ f 2 = 1 0 1 2 , ~ v 3 = - 1 0 1 0 , and ~ f 4 = - 1 3 - 1 1 , then show that { ~ f 1 , ~ f 2 , ~ f 3 , ~ f 4 } is orthogonal then normalize this set. Pythagorean Theorem in
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 08/05/2008 for the course MATH 115 taught by Professor Dunbar during the Fall '07 term at Waterloo.

### Page1 / 8

Week_10 - Week 10 Section 5.3: Orthogonality We looked at...

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

View Full Document
Ask a homework question - tutors are online