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Unformatted text preview: Lecture Note 81: March 12, 2007 Dr. Jeff ChakFu WONG Department of Mathematics Chinese University of Hong Kong jwong@math.cuhk.edu.hk MAT 2310C Linear Algebra and Its Applications Spring, 2007 Produced by Jeff ChakFu WONG 1 O RTHONORMAL B ASES IN R n ORTHONORMAL BASES IN R n 2 From our work with the standard (or natural) bases for R 2 , R 3 , and, in general, for R n , we know that when these bases are present, the computations are kept to a minimum. A subspace W of R n need not contain any of the natural basis, but in this lecture, we want to show that it has a basis with the same properties. That is, we want to show that W contains a basis S such that • every vector in S is of unit length and • every two vectors in S are orthogonal. The method used to obtain such a basis is the GramSchmidt process , which is presented in this lecture. ORTHONORMAL BASES IN R n 3 DEFINITIONA set S = { u 1 , u 2 ,..., u k } in R n is called orthogonal if every pair of distinct vectors in S are orthogonal, that is, if u i · u j = 0 for i 6 = j . An orthonormal set of vectors is an orthogonal set of unit vectors . That is, S = { u 1 , u 2 ,..., u k } is orthonormal if u i · u j = 0 for i 6 = j and u i · u i = 1 for i = 1 , 2 ,...,k. Normalizing refers to the process of dividing each vector in an orthogonal set S by its length so S is transformed into an orthonormal set . ORTHONORMAL BASES IN R n 4 Example 1 If x 1 = (1 , , 2) , x 2 = ( 2 , , 1) and x 3 = (0 , 1 , 0) , then { x 1 , x 2 , x 3 } is an orthogonal set in R 3 . The vectors u 1 = ( 1 √ 5 , , 2 √ 5 ) and u 2 = ( 2 √ 5 , , 1 √ 5 ) are unit vectors in the directions of x 1 and x 2 , respectively. Since x 3 is also of unit length, it follows that { u 1 , u 2 , x 3 } is an orthonormal set. Also, span { x 1 , x 2 , x 3 } is the same as span { u 1 , u 2 , x 3 } . ORTHONORMAL BASES IN R n 5 Example 2 The natural basis { (1 , , 0) , (0 , 1 , 0) , (0 , , 1) } is an orthonormal set in R 3 . More generally, the natural basis in R n is an orthonormal set. ORTHONORMAL BASES IN R n 6 Theorem 0.1 Let S = { u 1 , u 2 ,..., u k } be an orthogonal set of nonzero vectors in R n . Then S is linearly independent. Proof Consider the equation c 1 u 1 + c 2 u 2 + ··· c k u k = . (1) Taking the dot product of both sides of (1) with u i , 1 ≤ i ≤ k , we have ( c 1 u 1 + c 2 u 2 + ··· c k u k ) · u i = · u i . (2) By properties (c) and (d) of Theorem 0.3, Properties of dot Product, the left side of (2) is c 1 ( u 1 · u i ) + c 2 ( u 2 · u i ) + ··· + c k ( u k · u i ) , and the right side is . Since u j · u i = 0 if i 6 = j , (2) becomes 0 = c i ( u i · u i ) = c i k u i k 2 . (3) By (a) of Theorem 0.3, Properties of dot Product, k u i k 6 = 0 , since u i 6 = ....
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This note was uploaded on 06/05/2011 for the course MATH 3333 taught by Professor Jeffwong during the Spring '11 term at CUHK.
 Spring '11
 JeffWong
 Linear Algebra, Algebra

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