New version page

Stanford MATH 51 - Lecture Notes

Upgrade to remove ads
Upgrade to remove ads
Unformatted text preview:

MATH 51TA Section Notes for Thu 23 Oct 08Jason LoToday’s topics:• systems of co ordinates• Gram-Schmidt pro c es s1 Systems of CoordinatesGiven a basis B = {v1, · · · , vk} for a subspace V ⊂ Rn, recall that[v]B=c1...ckdenotes the vector whose entries are coordinates of v = c1v1+ · · · + ckvkwithrespect to the basis B.The diagram in the blue box on p.150 encapsulates most of the importantinformation in se ction 21: ifT : Rn→ Rn: v 7→ Avis the linear transformation defined by multiplication by the matrix A, andB = {w1, · · · , wn} is any basis for Rn, thenvA//T (v)C−1[v]BCOOB//[T (v)]B(here C = [w1| · · · |wn]).The following example illustrates the use of the diagram above:Example 1. Ex 21.6(a)(ii), (b)(ii)Note that, if we are given a basis B = {v1, · · · , vn} of Rn, and we aretold what happens to the basis elements viunder the linear transformation T ,then we do not need to use B = C−1AC to work out B (which would involvecomputing the inverse of C, which is quite some work): we simply have (bluebox on p.153)B = [[T (v1)]B| · · · |[T (vn)]B]Example 2. First part of Ex 21.912 Orthonormal Bases and Gram-Schmidt ProcessA basis B = {v1, · · · , vk} for a subspace V ⊂ Rnis called orthonormal if:• the viare unit vectors• the viare mutually orthogonal (i.e. for all i 6= j, vi· · · vj= 0)Given an orthonormal basis B = {v1, · · · , vk}, it is easy to work out thecoordinates [v]Bof any vector v ∈ V with resp ect to B:[v]B=v · v1...v · vk(quickly prove this if time permits).Example 3. Consider the vectorsv1=2−20, v2=−330, v3=001• These vectors are mutually orthogonal and linearly independent. Do theyform an orthonormal basis for R3? Turn them into an orthonormal basis.• Write the following vectors with respect to this orthonormal basis:x =100−100100(draw a diagram to illustrate this)2.1 Gram-Schmidt Orthogonalisation ProcessGiven any basis B = {v1, · · · , vk} for any subspace V ⊂ Rn, the Gram-Schmidtprocess turns it into an orthonormal basis. This process is explained on p.167;here is another take at explaining how it works:We want to obtain an orthonormal basis {w1, · · · , wk} for the same subspaceV . Define Vr= span{v1, · · · , vr}.Then we do:1. Define w1:=v1kv1k. (If we want to, we could relabel the viso that v1,hence w1, is particularly simple.)2. Computey2:= v2− ProjV1(v2)(draw diagram to illustate this)= v2− (v2· · · w1)w123. Define w2=y2ky2k. This will be the second vector in the orthonormalbasis.4. Computey3:= v3− ProjV2(v3)= v3− (v3· w1)w1− (v3· w2)w25. Define w3=y3ky3k. This will be the third vector in the orthonormal basis.6. etc.Example 4. Ex 22.10Example 5. Apply the Gram-Schmidt process to the basis in Example 3.(Draw pictures to illustrate what is happening.)Extra Examples Ex


View Full Document
Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?