Math 19b: Linear Algebra with Probability Oliver Knill, Spring 2011Lecture 16: CoordinatesA basis B = {~v1, . . . ~vn} of Rndefines the matrix S =| . . . |~v1. . . ~vn| . . . |. It is calledthe coordinate transformation matrix of the basis.By definition, the matrix S is invertible: the linear independence of the column vectors implies Shas no kernel. By the rank-nullety theorem, t he image is the entire space Rn.If ~x is a vector in Rnand ~x = c1~v1+· · ·+cnvn, then ciare called the B-coordinatesof ~v.We have seen that such a representation is unique if the basis is fixed.We write [~x]B=c1. . .cn. If ~x =x1. . .xn, we have ~x = S([~x]B).The B-coordinates of ~x are obtained by applying S−1to the coordinates of the standard basis:[~x]B= S−1(~x)This just rephrases that S[~x]B= ~x. Remember the column picture. The left hand side is justc1~v1+ · · · + cn~vnwhere the vjare the column vectors of S.1 If ~x =4−23, then 4, −2, 3 are the standard coordinates. With the standard basis B ={e1, e2, e3} we have ~x = 4e1− 2e2+ 3e3.2 If ~v1="12#and ~v2="35#, then S ="1 32 5#. A vector ~v ="69#has the coordinatesS−1~v ="−5 32 −1#"69#="−33#.Indeed, as we can check, −3~v1+ 3~v2= ~v.3 Find the coordinates of ~v ="23#with respect to the basis B = {~v1="10#, ~v2="11#}.We have S ="1 10 1#and S−1="1 −10 1#. Therefore [v]B= S−1~v ="−13#. Indeed−1~v1+ 3~v2= ~v.If B = {v1, . . . , vn} is a basis in Rnand T is a linear transformation on Rn, t hen theB-matrix of T isB =| . . . |[T (~v1)]B. . . [T (~vn)]B| . . . |.4 Find a clever basis for the reflection of a light ray at the line x + 2y = 0. Solution: Useone vector in the line and an other o ne perpendicular to it: ~v1="12#, ~v2="−21#. Weachieved so B ="1 00 −1#= S−1AA with S ="1 −22 1#.If A is the matrix of a transformation in the standard coordinates thenB = S−1ASis the matrix in the new coordinates.The transformation S−1maps the coordinates from the standard basis into the coo r dina t esof the new basis. In order to see what a transformation A does in the new coordinates, wefirst map it back to the old coordinates, apply A and then map it back again to the newcoordinates.~vS← ~w = [~v]BA ↓ ↓ BA~vS−1→ B ~w5 Let T be the reflection at the plane x + 2 y + 3z = 0. Find the transformation matrix B inthe basis ~v1=2−10~v2=123~v3=03−2. Because T (~v1) = ~v1= [~e1]B, T (~v2) = ~v2=[~e2]B, T (~v3) = −~v3= −[~e3]B, the solution is B =1 0 00 −1 00 0 1.Two matrices A, B which are related by B = S−1AS are called similar.6 If A is similar to B, then A2+ A + 1 is similar to B2+ B + 1. B = S−1AS, B2=S−1B2S, S−1S = 1, S−1(A2+ A + 1)S = B2+ B + 1.7 If A is a general 2 × 2 matrix"a bc d#and let S ="0 11 0#, What is S−1AS? Solution:"d cb a#. Both the rows and columns have switched. This example shows that the matrices"1 23 4#,"4 32 1#are similar.Homework due March 9, 20111 Find the B-matrix B of the linear transformation which is given in standard coordinatesasT"xy#="1 11 1#"xy#if B = {"21#,"12#}.2 Let V be the plane spanned by122and012. Find the matrix A o f reflection atthe plane V by using a suitable coordinat e system.3 a) Find a basis which describes best the points in the following lattice: We aim todescribe the lattice points with integer coordinates (k, l).b) Once you find the basis, draw all the points which have (x, y) coordinates in the discx2+ y2≤