7 - Gram Schmit and Orthogonal Projection

7 - Gram Schmit and Orthogonal Projection - Math 1b Prac...

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Math 1b Prac — The Gram-Schmidt process; orthogonal projection January 24, 2011 — extended/revised January 28, 2011 To warm up, we start with Theorem 1. Nonzero pairwise orthogonal vectors u 1 , u 2 ,..., u k are linearly independent. Proof: Suppose u 1 , u 2 u k are nonzero orthogonal vectors and that c 1 u 1 + c 2 u 2 + ... + c k u k = 0 . Given i , take the inner product of both sides with u i to get c 1 h u 1 , u i i + + c i h u i , u i i + + c k h u k , u i i = h 0 , u i i =0 . Most of the inner products on the left are zero, and the only surviving term is c i h u i , u i i = c i || u i || 2 , and because || u i || 2 6 = 0, it must be that c i = 0. This holds for all i =1 , 2 ,...,k ,andth is means that u 1 , u 2 u k are linearly independent. ut To save typing, we borrow the description of of the Gram-Schmidt process from the Wikipedia article.
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The defnition oF proj u ( v ) should be modifed to include the case when u = 0 ,where we understand proj 0 ( v )= 0 .E x e r c i s e :I F x c u is orthogonal to u 6 =0 ,th en c = h x , u i / h u , u i , i.e. c u =proj u ( x ). Example. Let v 1 =(1 , 1 , 1 , 1 , 1) , v 2 =(0 , 1 , 2 , 3 , 4) , v 3 , 1 , 4 , 9 , 16) . Then u 1 , 1 , 1 , 1 , 1) , u 2 =( 2 , 1 , 0 , 1 , 2) , u 3 =(2 , 1 , 2 , 1 , 2) . Theorem 2. The vectors u 1 , u 2 , u 3 ,... produced by the Gram-Schmidt process are pair- wise orthogonal. Proof: We need to check that h u i , u j i =0Fo r i 6 = j . We proceed by induction on k . Suppose we know h u i , u j i =0For1 i<j k 1. Then For i<k , h u i , u k i = h u i , v k ( proj u 1 ( v k )+proj u 2 ( v k )+ ... +proj u k 1 ( v k ) ) i = h u i , v k i− ( h u i , u 1 ( v k ) i + + h u i , u k 1 ( v k ) i ) . Since proj u j ( x ) is a scalar multiple oF u j , all the terms h u i , u j ( v k ) i are 0 except possibly when i = j , and we have h u i , u k i = h u i , v k i−h u i , u i ( v k ) i = h u i , v k u i , h u i , v k i h u i , u i i u i i = h u i , v k h u i , v k i h u i , u i i h u i , u i i . ut Note that iF u 1 , u 2 , u 3 are nonzero orthogonal vectors, and we let e i = u i / || u i || , then e 1 , e 2 , e 3 are orthonormal vectors. Orthonormal means that the vectors are or- thogonal and that they each have length 1. The standard basis in R n is only one example oF an orthonormal basis For R n . Theorem 3. Let u 1 , u 2 ,..., u k be the vectors produced from v 1 , v 2 v k by the Gram- Schmidt orthogonalization process; let U and V be the matrices whose rows are, respec- tively, u 1 , u 2 u k and v 1 , v 2 v k . Then there are (square) lower triangular matrices
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7 - Gram Schmit and Orthogonal Projection - Math 1b Prac...

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