2/28/2002, COORDINATES Math 21b, O. KnillHOMEWORK: 3.3: 22,24,32,36,40,56*, 3.4: 4,14,16,22,32*,48SUMMARY LAST HOUR:LINEAR SPACE. X linear space:~0 ∈ X closedunder addition and scalar multiplication.Examples: X = ker(A), X = im(A) are linearspaces.BASIS. B = {~v1, . . . , ~vn} ⊂ XB linear independent: a1~v1+ ... + an~vn= 0, ⇒a1= ... = an= 0.B span X: ~v ∈ X then ~v = a1~v1+ ... + an~vn.B basis: both linear independent and span.DIMENSION. The number of elements in a basis inX is independent of the choice of the basis and calleddimension of X.UNIQUE REPRESENTATION. ~v1, . . . , ~vn∈ X ba-sis ⇒ every ~v ∈ X can be written uniquely as a sum~v = a1~v1+ . . . + an~vn.BASIS DEFINES INVERTIBLE MATRIX . v1, . . . , vnform a basis in Rn⇔ S =| . . . |~v1. . . ~vn| . . . |is invertible.BASIS OF IMAGE A basis of the image of a matrixA is obtained by the pivotal columns of A.BASIS OF KERNEL. rewrite rref(A) as a systemof linear equations and introduce a free variable forevery nonpivotal column.B-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). In other words:B-coordinates of ~x are obtained by applying S−1to the coordinates of the standard basis.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. The beginningof the vector concept came only later at the beginning of the 19’th Century with the work of Bolzano (1781-1848). The full power of coordinates becomes possible if we allow to chose our coordinate system adapted tothe situation.Decartes 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 DecartesBolzanoEXAMPLE. Find the coordinates of ~v =23with respect to the basis B = {~v1=10, ~v2=11}.We have S =1 10 1and S−1=1 −10 1. Therefore [v]B= S−1~v =−13. Indeed −1~v1+ 3~v2= ~v.CREATIVITY THROUGH LAZINESS? Decartes (1596-1650) introduced coordinates while lying on the bed,watching a fly (around 1630). Archimedes (285-212 BC) discovered a method to find the volume of bodieswhile relaxing in the bath. Newton (1643-1727) discovered Newton’s law while lying under an apple tree.According to Harvards David Perkins: ”The Heureka effect”, many creative breakthroughs have in common:• Long search without apparent progress.• Prevailing moment and break through.• Transformation and realisation.According to this pattern, having a breakthrough in a lazy moment is typical - but only after a long struggleand hard work.B-MATRIX. If B = {v1, . . . , vn} is a basis in Rnand T is a linear transformation on Rn, then the B-matrix ofT is B =| . . . |[T (~v1)]B. . . [T (~vn)]B| . . . |.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 S =0 11 0, then what is S−1AS?MORE ON THE 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 in”bad” coordinates.~vS← ~wA ↓ ↓ BA~vS−1→ B ~wThe transformation in”good” coordinates.QUESTION. Can the matrix A belonging to a projection from R3to a plane x + y + 6z = 0 be similar to amatrix which is a rotation by 20 degrees around the z axis? No: an invertible matrix can not be similar to ainvertible matrix.PROBLEM. Find a clever basis for the reflection of a light ray at the line x + 2y = 0. ~v1=12, ~v2=−21.SOLUTIon. Yo can achive B =1 00 −1. In the standard with S =1 −22 1.PROBLEM. All shears A =1 a0 1with a 6= 0 are similar. SOLUTION. Use a basis ~v1= a~e1and ~v2=
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