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Unformatted text preview: NOTES ON SECTION 5 . 4 MA265, SECTIONS 41, 52 Some key words Orthogonal Orthonormal basis QR factorization 1. Gram Schmidt Process Theorem 1. Let W = span { u 1 ,...,u k } . There is an orthogonal basis { v 1 ,...,v l } for the subspace W . The vector v i is obtained as follows (assuming v 1 ,...,v i 1 have been computed before): v i = u i ( u i ,v 1 ) ( v 1 ,v 1 ) v 1 ( u i ,v 2 ) ( v 2 ,v 2 ) v 2 ···   ( u i ,v i 1 ) ( v i 1 ,v i 1 ) v i 1 We’ll do an example of how to implement this in lecture. One reason why orthogonal bases are extremely useful: Theorem 2. Suppose we have an orthogonal basis { v 1 ,...,v n } for an inner product space V . Then if w ∈ V is a vector, we have w = n X i =1 a i · v i , where a i = ( w,v i ) ( v i ,v i ) for each i If the v i s were not orthogonal, we would have to solve Ax = w, where A is the matrix whose columns are the vectors v i , for x . This is clearly more difficult than computing n inner products to get the coefficients. It is ABSOLUTELY ESSENTIAL that the basis { v i } is orthogonal for this theorem to apply. Example 3. Let W be the subspace of R 4 defined as the span of the vectors u 1 = 1 ,u 2 = 1 1 ,u 3 = 1 1 1 ,u 4 = 1 , Use GramSchmidt to find an orthonormal basis. Following GramSchmidt: v 1 = u 1 = 1 , Stage 2: v 2 = u 2 ( u 2 ,v 1 ) ( v 1 ,v 1 ) = 1 1  1 1 1 = 1 1 Stage 3: v 3 = u 3 ( u 3 ,v 1 ) ( v 1 ,v 1 ) ( u 3 ,v 2 ) ( v 2 ,v 2 ) = 1 1 1  1 1 1  1 1 1 = 1 Thus, the orthogonal set obtained from the original set is v 1 = 1 ,v 2 = 1 ,v 3 = 1 Remark 4. The GramSchmidt process is very sensitive to the order of the vectors. If we performed GramThe GramSchmidt process is very sensitive to the order of the vectors....
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This note was uploaded on 03/31/2012 for the course MA 265 taught by Professor Bens during the Spring '08 term at Purdue.
 Spring '08
 Bens
 Linear Algebra, Algebra

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