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Lecture

Lecture15-16f.pdf

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Department
Mathematics
Course
MATB24H3
Professor
Sophie Chrysostomou
Semester
Winter

Description
University of Toronto at Scarborough Department of Computer & Mathematical Sciences MAT B24S Fall 2011 MAT B24 Lectures 15. 1 The Gram-Schmidt Process In this section we will develop a process that will give us a basis of a subspace W of R with the property that any two vectors in this basis are orthogonal and unit vectors. DEFINITION: Let {v ,v ,...v }1be 2 set kf nonzero vectors in R . We n say that this set is orthogonal if v ▯ i = 0jfor all i,j ∈ {1,2,...k} such that i ▯= j THEOREM: Let {v ,v ,...v } be an orthogonal set of vectors in R Then n 1 2 k this set is independent. proof. : Suppose that 0 = r v + r v + ... + r v 1 1 2 2 k k then v ▯i0 = v ▯ ir v 1 r1v + 2..2+ r v ) Thksk 0 = r 1v i v )1+ r (2 ▯ i ) 2 ... + r (vi▯ vi) +i... + r (v k vi) k 2 or 0 = r i|v i| Since v ▯=i0 then ||v || =i ▯ 0 therefore r =i0. This is true for all i, therefore r = r = ... = r = 0 and the vectors v ,v ,...v are linearly 1 2 k 1 2 k independent. n If we have an orthogonal basis for a subspace W of R , then it is easy to n ﬁnd b W the projection of b on W for any vector b ∈ R using the following theorem: 1 THEOREM: Projection Using an Orthogonal Basis. If {v1,v2,...k } be an orthogonal basis for W in R , then for any b ∈ R the projection of b on W is b ▯ v1 b ▯ 2 b ▯ vk b W = v 1 v2+ ... + v k v1▯ v1 v2▯ v2 vk▯ vk ▯k ▯k b ▯ i Note that this is the sum vi= b vi 1 vi▯ vi 1 2 EXAMPLE: From our previous example: Let b = [22,11,11,33] ∈ R and 4 W = sp([1,0,2,−1],[−1,1,−2,2]). We cannot use the theorem above since {[1,0,2,−1],[−1,1,−2,2]} is not an orthogonal set. But W = sp([1,0,2,−1],[1,6,2,5]) as well since [1,6,2,5] ∈ W and the two vectors are linearly independent. Also, they are orthogonal since [1,0,2,−1] ▯ [1,6,2,5] = 1 + 4 − 5 = 0 Thus [22,11,11,33] ▯ [1,0,2,−1] [22,11,11,33] ▯ [1,6,2,5] b W = [1,0,2,−1] + [1,0,2,−1] ▯ [1,0,2,−1] [1,6,2,5] ▯ [1,6,2,6] This gives b = 11[1,0,2,−1] + 25[1,6,2,5] = [6,25,12,19] W 6 6 as expected. In this example, I did not explain how I found the vector [1,6,2,5]. In n general, given any basis for a subspace, W of R we want to be able to ﬁnd a basis for W which is orthogonal so that we can apply the above theorem. Further more, we want each basis vector to be a unit vector. We call such a set an orthonormal basis for W DEFINITION: Let W be a subspace of R . A basis {q ,q ,..1q 2 forkW is orthonormal if it is orthogonal and each vector i is a unit vector for all i = 1,2...,k. Thus, the theorem in the previous page now gives the following: THEOREM: The projection of b on W with orthonormal basis {q ,q ,...q } 1 2 k is ▯k b = (b ▯ q )q + (b ▯ q )q + ... + (b ▯ q )q = b W 1 1 2 2 k k qi 1 3 HOW TO FIND SUCH AN ORTHONORMAL BASIS OF W: THEOREM: The Gram-Schmidt Theorem. Let W be a subspace of R n with basis a 1a 2...,a k Let W = sp(a ,a ,...,a ) j 1 2 j for j = 1,2,...,k. Then there is an orthonormal basis {q ,q ,1..q2} fok W such that W =jsp(q ,q 1..2q ) j proof. To prove this theorem I will use the following two theorems done earlier: n n THEOREM A : If W is a subspace of R and b ∈ R , then there are ⊥ unique vectors b W ∈ W and b W ⊥ ∈ W such that b = b + b ⊥ or that b = b − b . W W W⊥ W n THEOREM B: If {v ,v ,...v1} 2e an krthogonal basis for W in R , then n for any b ∈ R the projection of b on W is b·v1 b·2 b·k b W = b W = v1·1 v1+ v2·v2v2+ ... + vk·vkv k We start oﬀ by ﬁnding an orthogonal basis {v ,v ,...v } of W. Then we 1 2 k v i can ﬁnd {q ,1 ,2..q }kby letting q =i for i = 1,2,...,k ||vi|| (i) We let v1= a a1d W = s1(v ) = s1(a ). 1 n (ii) To get v2: By theorem A applied on the vector a ∈ R 2nd the ⊥ subspace W we1get unique vectors a2W1 ∈ W 1 and a2W ⊥ ∈ W 1 such 1 that: a2W ⊥ = a 2 a 2W 1. 1 a2▯ v1 Let v2= a 2 ⊥ . By theorem B we get that a 2W = v1. Thus: W1 1 v1▯ v1 a ▯ v v = a = a − a = a − 2 1 v 2 W 1 2 W 1 2 v ▯ v 1 1 1 Finally, v ∈ W ⊥ so v ▯ v = 0. 2 1 2 1 Since v1= a ,1then v ∈ 2p(a ,a 1 an2 {v ,v } 1s a2 orthogonal set by construction. Now let W = sp(v ,v ) = sp(a ,a ). 2 1 2 1 2 n (iii) To get v3: By theorem A applied on the vector a ∈ R a3d the 4 ⊥ subspace W we2get unique vectors a 3W2 ∈ W a2d a 3 ⊥ ∈ W 2 such that: W2 a3 ⊥ = a 3 a 3W . Let v 3 a 3 ⊥ W 2 2 W 2 By theorem B , since v ,v is an orthogonal basis for W we get that 1 2 2 v 1 a 3 v2▯ a3 a3W 2= v1+ v2 v 1 v 1 v 2 v 2 ▯ ▯ v1▯ a 3 v 2 a 3 v3= a 3W ⊥ = a 3 a 3W2 = a 3 v 1 v 2 2 v1▯ v 1 v2▯ v 2 ⊥ Finally, v3∈ W 2 and since v 1v ∈2W then2 v ▯ v =30, v1▯ v = 0.3 2 Also, v ∈ sp(a ,v ,v ) = sp(v ,v ,v ). Now let W = sp(v ,v ,v ) = 3 3 1 2 1 2 3 3 1 2 3 sp(a 1a 2a )3 Then, {v ,v 1v }2is 3n orthogonal basis for W by const3uc- tion. (iv) Continuing the process, if we have W i−1 = sp(v ,v1,.2.,v i−1) = sp(a 1a 2...,a i−1) such that {v ,1 ,2..,v i−1} is an orthogonal basis for W i−1, constructed as above, then we can get v by litting ▯ ▯ v 1 a i v 2 a i vi−1 ▯ ai v i a iW⊥ = a ia W i−1 = a i v 1 v 2 ... + vi−1 i−1 v1▯ v 1 v 2 v 2 v i−1v i−1 And W = ip(v ,...1 ) = ip(a ,a ,.1.,2 ) . i Eventually, we get the orthogonal basis {v ,v ,.1.,v2} of Wksuch that W = sp(v ,..v ) = sp(a ,...,a ). i 1 i 1 i vi By letting q =i for i = 1,2,...,k we get an orthonormal set of vectors ||vi|| {q 1q 2...,q k such that W = sp(v ,...1 ) = kp(q ,q ,..1,q2) withkthe prop- erty that W =isp(v ,..1v ) =isp(q ,q ,1..,2 ) i EXAMPLE: Let {a 1 [1,1,1,0],a = [021,1,0],a = [0,0,130]} be a n basis for the subspace W of R . Find an orthogonal basis for
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