Lecture15-16f.pdf

10 Pages
192 Views
Unlock Document

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
More Less

Related notes for MATB24H3

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit