Geometry 2014: Notes and Exercises Sophie M. Fosson [email protected] June 5, 2014 ii These notes follow my exercise classes in chronological order; • are a support to student practice; • do not cover the whole course; • include some theory reminders; • include the exercises proposed during my exercise classes; • include further material and exercises; • are sure to contain several mistakes; • shall be updated every two-three weeks; • are expected to be enhanced by student feedback (that is, if you find any • mistakes please tell me! by email). Thank you, s.m.f. Contents 1 Week 1 1 1.1 Introduction to matrices . .1 1.2 Homogeneous linear systems . .2 1.3 Determinant . .4 1.4 Cramer's rule . .4 2 Week 2 7 2.1 Matrix reduction and rank . .7 2.2 Linear systems . .8 2.3 Matrix inversion . .9 3 Week 3 11 3.1 Rn as vector space . 11 3.2 Bases and dimension . 12 3.3 Products between vectors . 13 4 Week 4 15 4.1 Vector spaces . 15 4.2 Back to RN : straight lines and planes . 17 4.3 Lines and planes in R3: Cartesian Equations . 17 4.4 When a plane is a subspace? . 18 4.5 Geometric description of linear systems' solutions . 18 4.6 Geometric description of linear systems' solutions . 19 5 Week 5 21 5.1 Linear mappings . 21 5.2 Matrix associated with a linear mapping . 23 6 Week 6 25 6.1 Endomorphisms . 25 6.2 Eigenvectors, eigenvalues . 25 6.3 Multiple choice quizzes . 26 7 Week 7 29 7.1 Extra material . 29 7.2 Eigenvalues and multiplicities . 29 7.3 Change of basis . 31 iii iv CONTENTS 8 Week 8 33 8.1 Simple endomorphisms and diagonalizability . 33 9 Week 9 39 9.1 Circles and Spheres . 39 9.1.1 Circles in R2 ......................... 39 9.1.2 Spheres . 40 9.1.3 Circles in R3 ......................... 41 10 Week 10 43 10.1 Conics . 43 10.2 Quadrics . 43 10.3 Reduction of conics to canonical forms . 44 11 Week 11 47 11.1 Quadrics . 47 11.2 Critical points for functions of two variables . 48 11.3 Mixing quadrics and functions of two variables . 49 12 Week 12 51 Chapter 1 Week 1 1.1 Introduction to matrices A matrix is a table of numbers, called the entries of the matrix; in these notes will mainly consider matrices whose entries are real numbers. We write A Rm×n or A Rm;n to name a real matrix A, with m rows and n columns: 2 2 0 1 A11 A12 A1n ······ B A21 A22 A2n C B ······ C B .. C B A31 A32 . A3n C Aij R; i = 1; : : : ; m; j = 1; : : : ; n (1.1) B C 2 B . C @ . .. A Am1 Am2 Amn ······ The sum between two matrices in A; B Rn;m is the matrix in C = A+B n;m 2 2 R such that Ci;j = Ai;j + Bi;j, i = 1; : : : ; m, i = 1; : : : ; n. Two matrices must have the same dimensions m; n to be summed. Given a scalar α R, the product C = αA is given by Ci;j = αAi;j, i = 1; : : : ; m, i = 1; : : :2 ; n. The product between two matrices is a bit more complex. Given A Rm×n and B Rn×p, the entry (i; j) of the product AB is defined as 2 2 n X (AB)ij = aihbhj (1.2) h=1 Notice that the number of columns of A must be equal to the number of rows of B. Moreover, the product is not commutative! Exercise 1 Let 0 1=2 0 7 1 0 3 1 1 1 0 1 1 0 1 1 − A = @ 2 1 2 A B = @ 0 2 1 A C = @ 2 A D = @ 1 A 1 4 3 0 0 1 3=4 −6 − Evaluate: (a)A + 3B; 1 2 CHAPTER 1. WEEK 1 (b) B2; (c) BT B. (d) A(2C + D); Solution (a) 0 13=2 3 4 1 − A + 3B = @ 2 7 5 A 1 4 0 (b) 0 3 0 0 1 0 3 1 1 1 0 9 3 3 1 T B B = @ 1 2 0 A @ 0 2 1 A = @ 3 5 3 A 1 1 1 0 0 1 3 3 3 (c) 0 3 1 1 1 0 3 1 1 1 0 9 5 5 1 2 B = @ 0 2 1 A @ 0 2 1 A = @ 0 4 3 A 0 0 1 0 0 1 0 0 1 (d) 0 2 1 0 1 1 0 3 1 2C + D = @ 4 A + @ 1 A = @ 3 A 3=2 −6 15=2 0 3 1 0 51 1 − @ 3 A = @ 24 A : 15=2 15=2 − Exercise 2 Given xy x2 A = y2 −xy − where x; y R not both null, show that A2 = 0. 2 Solution xy x2 xy x2 A2 = = y2 −xy y2 −xy − − (xy)2 x2y2 x3y + x3y 0 0 = = : xy3 −xy3 −x2y2 + (xy)2 0 0 − − This exercise has shown that the power of a nonnull matrix may be null. 1.2 Homogeneous linear systems A linear system Ax = b, A Rm;n, x Rn = Rn;1, b Rm = Rm;1 is said to be homogeneous when b = 0.2 Homogeneous2 systems have2 at least one solution, namely x = 0. Performing the product, we get m equations for n unknowns. Given the equations, one could solve the system as learned in high school. The problem is 1.2. HOMOGENEOUS LINEAR SYSTEMS 3 that we aim at working even with large m; n, which makes such computations not feasible. This is why we work on the matrix in order to reduce the problem. More precisely, we are allowed to transform the matrix switching rows and performing any linear combinations between rows. In other terms, if rj is the generic jth row of the matrix, we can do: rj ri $ rj crj; c R (1.3) ! 2 rj rj + cri; c R ! 2 These operations can be used to reduce the matrix. A rigorous definition of reduction will be given next week, however the intuition is that we can transform A into a matrix Ae which has many zeros, which makes the linear system easier. Exercise 3 Solve the homogeneous system Ax = 0 where 0 1 2 0 1 A = @ 1 1 3 A : (1.4) 1 2 1 Solution T 3 First, notice that, even if not remarked, it is clear that x = (x1; x2; x3) R and 0 = (0; 0; 0)T R3. 2 2 Let us now perform some suitable operations from (1.3). But actually which operations? A more rigorous procedure will be explained next week, here just do operations that let you obtain some more zeros in the matrix! For example: 1. r3 r3 r1: ! − 0 1 2 0 1 @ 1 1 3 A (1.5) 0 0 1 2. r2 r2 3r1: ! − 0 1 2 0 1 @ 1 1 0 A (1.6) 0 0 1 3. r1 r1 r2: ! − 0 1 2 0 1 Ae = @ 0 1 0 A (1.7) 0 0 1 Given that Ax = 0 is equivalent to Axe = 0, we solve the second one. We have: 0 1 0 1 0 1 8 0 1 0 1 1 2 0 x1 0 < x1 + 2x2 = 0 x1 0 Axe = 0 @ 0 1 0 A @ x2 A = @ 0 A x2 = 0 @ x2 A = @ 0 A , , ) 0 0 1 x3 0 : x3 = 0 x3 0 (1.8) For exercise, one can try to solve Ax = 0 with no reduction. This is not impossible, but a little bit more tricky. 4 CHAPTER 1. WEEK 1 1.3 Determinant The determinant is a number associated with each square matrix. It is use- ful to infer many properties of a matrix, and also to solve systems (see next paragraph). The determinant can be computed as follows. In R2;2: a b a b det = = ad bc (1.9) c d c d − In R3;3: a b c e f d f d e d e f = a b + c (1.10) i j − h j h i h i j and then extend by induction to Rn;n with n 3. ≥ 1.4 Cramer's rule The Cramer's rule suggests a method to solve linear systems by using determi- nants. n;n T Given a square matrix A R and the system Ax = b, b = (b1; : : : ; bn) , T 2 x = (x1; : : : ; xn) , then det Ai xi = i = 1; : : : ; n: (1.11) det A where Ai is the matrix obtained from A by replacing its i-th column by the column vector b. Notice that we must have det A = 0. PROOF 6 First, let us recall the properties of the determinant: given a square matrix, if we perform the column transformation ci ci + αcj the determinant remains ! the same; otherwise, if we transform ci αci, α R, then the determinant of the new matrix is the determinant of the! old one2 multiplied by α. Now, consider 0 1 b1 a12 a13 : : : a1n B b2 a22 a23 : : : a2n C A = B C (1.12) 1 B . C @ . A bn an2 an3 : : : ann Pn and noting that bi = j=1 aijxj, i = 1; : : : ; n, we obtain that det A1 = x1 det A which proves the thesis for x1. Iterating the procedure for all the xi, the Cramer's rule is proved. Exercise 4 Given A of previous exercise, solve the linear system Ax = (0; 0; 1)T using the Cramer's rule.