Unit 5: Change of Coordinates

1 LINEAR ALGEBRA AND VECTOR ANALYSISMATH 22 BUnit 5: Change of a basisBin a linear spaceX, we can write an elementvinXin a uniqueway as a sum of basis elements. For example, ifv=[34]is a vector inX=R2andB={v1=[1 1], v2=[16]}, thenv= 2v1+v2. We say that[21]Bare theBcoordinatesofv. Thestandard coordinatesarev=[34]are assumed if no otherbasis is specified. This meansv= 3e1+ {v1, v2, , vn}is a basis ofRn, then the matrixSwhich contains thevectorsvkas column vectors is called thecoordinate Change :IfSis the matrix ofB, thenS 1vare theBcoordinates the above example,S=[11 1 6]has the inverseS 1=[6 111]/7. WecomputeS 1[3,4]T= [2,1] [v]B= [a1, .. , an] are the new Coordinates ofv, this meansv=a1v1+ +anvn. But that meansv=S[v]B.

2 SinceBis a basis,Sis invertible and [v]B=S 1v. Theorem:IfT(x) =Axis a linear map andSis the matrix from a basischange, thenB=S 1 ASis the matrix ofTin the new The statement [y]B=B[x]Bcan be written using the last theoremasS 1y=BS 1xso thaty=SBS 1x. Combining withy=Ax, this givesB=S 1AS. two matricesA, BsatisfyB=S 1 ASfor some invertibleS, they are calledsimilar. The matricesAandBboth implement the transformationT, but they do itfrom a different perspective. It makes sense to adapt the basis to the situation. Forexample, here on earth, at a specific location, we use a coordinate system, wherev1 Linear Algebra and Vector Analysispoints east, wherev2points north and wherev3points straight up. The natural basishere in Boston is different than the basis in Z good coordinate system is adapted to the situation.

3 Whentalking about points on the globe, we can use a global coordinate systemwithe3in the earth axes. When working on earth say near Boston, weneed another a suitable basis is one of the main reasons why linear algebra is so idea will be a major one throughout the course. We will use eigen basis to diag-onalize a matrix, we will use good Coordinates to solve ordinary and partial us, the Change of Coordinates now is a way to figure out the matrix of atransformationTo find the matrixAof a reflection, projection or rotation matrix, finda good basis for the situation, then look what happens to the new basisvectors. This givesB. Now write down the matrixSand getA=SBS the matrixAwhich implements the reflectionTat the planeX={x+y+ 2z= 0}.

4 Take a basis adapted to the situation. Takev3= [1,1,2]Twhich is perpendicular to the plane, then choosev1= [1, 1,0]T, v2=[2,0, 1]Twhich are in the plane. Now, sinceT(v1) =v1, T(v2) =v2andT(v3) = v3,the transformation is described in that basis with the matrixB= 1 000 100 0 1 . The1 Figure (1) was rendered using creative commons Povray code by Gilles Tran authored Change transformationSisS= 1121 1020 1 . We can now getA=SBS 1= 1 2212 1110 . the matrix which rotates about the vector [3,4,0]Tby 90 degreescounter clockwise when looking from the tip (3,4,0) of the vector to the origin (0,0,0). build a basis adapted to the situation. Of course, we usev1= [3,4,0] need now two other vectors which are perpendicular to each other.

5 The vectorsv2= [ 4,3,0]Tandv3= [0,0,5] present themselves. It is good to have the two vectorswhich are moving to have the same length because then the matrixBis particularlysimple: sincev2 v3, v3 v2, v1 v1, we haveB= 100x0010 1 0 . WithS= 3 4 0430005 , we getA=SBS 1. This is now just matrix multiplication andcan be computed 912 10012 16754 30 /25. It would have been quite hard to find thecolumn vectors of this matrix by figuring out where each of the standard basis vectorsekgoes. Still, we have used that basic principle when figuring out the matrix of the projection on the line perpendicular to the hyperplanex+y+z+w= 0 inR4. Solution: there is a nice basis adapted to that situation. ItgivesB={v1= 1111 , v2= 11 1 1 v3= 1 11 1 v4= 1 1 11 }, S= 111111 1 11 11 11 1 11.

6 Sis invertible. In this caseS 1= 4S. Now, in the new basis, the transformationmatrix is very simple. Asv1goes tov1andv2andv3andv4all go to zero, we haveB= 1 0 0 00 0 0 00 0 0 00 0 0 0 andA=SBS 1= 1 1 1 11 1 1 11 1 1 11 1 1 1 /4. In this case, we might also have been able towrite down the matrix without going to a new coordinate system as the image of thefirst basis vector is the vector projection of [1,0,0,0] onto [1,1,1,1].Linear Algebra and Vector AnalysisHomeworkThis homework is due on Tuesday, 2/13 :What are theB- Coordinates of~vin the basisB.~v= 2345 ,B={ 3111 , 1110 , 1100 , 1000 }?Problem :What is the matrixBfor the transformationA=[31 1 4]in the basisB={[21],[10]}:Problem :Chose a suitable basis to solve:a) What matrixAimplements the reflection at the plane 3x+ 3y+ 6z= 0?

7 B) What matrixAimplements the reflection at the line spanned by[2,2,4]T?Problem :Find the matrixAcorresponding to the orthogonal pro-jection onto the plane spanned by the vectors 123 and 012 .Problem : Graphene are hexagonal planar structures. We canwork with them when using a good adapted basis. Assume the first isv=[10]. a) Findwso thatB={v, w}is the basis as seen in ) What are the standard Coordinates of[3 1]B?c) Is[2372]Ba vertex of a hexagon or the center of one?Figure are single layer hexagonal lattice carbon Knill, Math 22b, Harvard College, Spring 2019