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HARVARD MATH 21B - Coordinates

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COORDINATES Math 21b, O. KnillB-COORDINATES. Given a basis ~v1, . . .~vn, define the matrix S =| . . . |~v1. . . ~vn| . . . |. It is invertible. If~x =Pici~vi, then ciare called the B-coordinates of ~v. We write [~x]B=c1. . .cn. If ~x =x1. . .xn, we have~x = S([~x]B).B-coordinates of ~x are obtained by applying S−1to the coordinates of the standard basis:[~x]B= S−1(~x) .EXAMPLE. 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.EXAMPLE. Let V be the plane x + y − z = 1. Find a basis, in which every vector in the plane has the formab0. SOLUTION. Find a basis, such that two vectors v1, v2are in the plane and such that a third vectorv3is linearly independent to the first two. Since (1, 0, 1), (0, 1, 1) are points in the plane and (0, 0, 0) is in theplane, we can choose ~v1=101~v2=011and ~v3=11−1which is perpendicular to the plane.EXAMPLE. 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.B-MATRIX. If B = {v1, . . . , vn} is a basis inRnand T is a linear transformation on Rn,then the B-matrix of T is defined asB =| . . . |[T (~v1)]B. . . [T (~vn)]B| . . . |COORDINATES HISTORY. Cartesian geometry was introduced by Fermat and Descartes (1596-1650) around1636. It had a large influence on mathematics. Algebraic methods were introduced into geometry. Thebeginning of the vector concept came only later at the beginning of the 19’th Century with the work ofBolzano (1781-1848). The full power of coordinates becomes possible if we allow to chose our coordinate systemadapted to the situation. Descartes biography shows how far dedication to the teaching of mathematics can go ...:(...) In 1649 Queen Christina of Sweden persuaded Descartes to go to Stockholm. However the Queen wanted todraw tangents at 5 a.m. in the morning and Descartes broke the habit of his lifetime of getting up at 11 o’clock.After only a few months in the cold northern climate, walking to the palace at 5 o’clock every morning, he diedof pneumonia.Fermat DescartesChristina BolzanoCREATIVITY THROUGH LAZINESS? Legend tells that Descartes (1596-1650) introduced coordinateswhile lying on the bed, watching a fly (around 1630), that Archimedes (285-212 BC) discovered a methodto find the volume of bodies while relaxing in the bath and that Newton (1643-1727) discovered New-ton’s law while lying under an apple tree. Other examples are August Kekul´e’s analysis of the Benzenemolecular structure in a dream (a snake biting in its tail revieled the ring structure) or Steven Hawkings discovery thatblack holes can radiate (while shaving). While unclear which of this is actually true, there is a pattern:According David Perkins (at Harvard school of education): ”The Eureka effect”, many creative breakthroughs have incommon: a long search without apparent progress, a prevailing moment and break through, and finally, atransformation and realization. A breakthrough in a lazy moment is typical - but only after long struggleand hard work.EXAMPLE. Let T be the reflection at the plane x + 2y + 3z = 0. Find the transformation matrix B in thebasis ~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.SIMILARITY. The B matrix of A is B = S−1AS, where S =| . . . |~v1. . . ~vn| . . . |. One says B is similar to A.EXAMPLE. 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.PROPERTIES OF SIMILARITY. A, B similar and B, C similar, then A, C are similar. If A is similar to B,then B is similar to A.QUIZZ: If A is a 2 × 2 matrix and let S =0 11 0, What is S−1AS?MAIN IDEA OF CONJUGATION S. The transformation S−1maps the coordinates from the standard basisinto the coordinates of the new basis. In order to see what a transformation A does in the new coordinates,we map it back to the old coordinates, apply A and then map it back again to the new coordinates: B = S−1AS.The transformation instandard coordinates.~vS← ~w = [~v]BA ↓ ↓ BA~vS−1→ B ~wThe transformation inB-coordinates.QUESTION. Can the matrix A which belongs to a projection from R3to a plane x + y + 6z = 0 be similar toa matrix which is a rotation by 20 degrees around the z axis? No: a non-invertible A can not be similar to aninvertible B: if it were, the inverse A = SBS−1would exist: A−1= SB−1S−1.PROBLEM. Find a clever basis for the reflection of a light ray at the line x +2y = 0. ~v1=12, ~v2=−21.SOLUTION. You can achieve B =1 00 −1with S =1 −22 1.PROBLEM. Are all shears A =1 a0 1with a 6= 0 similar? Yes, use a basis ~v1= a~e1and ~v2= ~e2.PROBLEM. You know A =3 0−1 2is similar to B =1 00 −1with S =0 −11 1. Find eA=1+A+A2+A3/3!+.... SOLUTION. Because Bk= S−1AkS for every k we have eA= SeBS−1=1/e 0e + 1/e


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HARVARD MATH 21B - Coordinates

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