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MATH2201 Lecture Notes MATH2201 Lecture Notes Andrei Yafaev (based on notes by Richard Hill, John Talbot and Minhyong Kim) December 8, 2009 If you find a mistake then please email me ([email protected]). Contents 1 Number Theory 2 1.1 Euclid’salgorithm ............................... .... 2 1.2 Factorizationintoprimes . ...... 8 1.3 Congruences..................................... 12 2 Polynomial Rings 16 2.1 IrreducibleelementsinRings . ....... 16 2.2 Euclid’s algorithm in k[X]............................... 19 3 Jordan Canonical Form 23 3.1 Revisionoflinearalgebra . ...... 23 3.2 Matrix representation of linear maps . ......... 27 3.3 Minimalpolynomials.............................. 33 3.4 GeneralizedEigenspaces . ...... 36 3.5 Jordan Bases in the one eigenvalue case . ........ 40 3.6 Jordan Canonical (or Normal) Form in the one eigenvalue case .......... 44 3.7 Jordancanonicalformingeneral. ........ 49 4 Bilinear and Quadratic Forms 52 4.1 MatrixRepresentation . ..... 52 4.2 Symmetric bilinear forms and quadratic forms . ........... 56 4.3 Orthogonality and diagonalization . ......... 58 4.4 ExamplesofDiagonalising. ...... 60 4.5 Canonical forms over C ................................ 63 4.6 Canonical forms over R ................................ 63 5 Inner Product Spaces 66 5.1 GeometryofInnerProductSpaces . ...... 66 5.2 Gram–Schmidt Orthogonalization . ........ 70 5.3 Adjoints........................................ 73 5.4 Isometries...................................... 74 5.5 OrthogonalDiagonalization . ....... 78 1 Lecture 1 Sketch of the course: Number theory (prime numbers, factorization, congruences); • Polynomials (factorization); • Jordan canonical form (generalization of diagonalizing); • Quadratic and bilinear forms; • Euclidean and Hermitian spaces. • Prerequisites (you should know all this from first year algebra courses) Fields and vector spaces (bases, linear independence, span, subspaces). • Linear maps (rank, nullity, kernel and image, matrix representation). • Matrix algebra (row reduction, determinants, eigenvalues and eigenvectors, diagonaliza- • tion). 1 Number Theory For this chapter and the next (Polynomials), I recommend the book ‘A concrete introduction to higher algebra’, by Lindsay Childs. Number theory is the theory of Z = 0, 1, 2,... Recall also the notation N = 1, 2, 3, 4,... { ± ± } { } 1.1 Euclid’s algorithm We say that a Z divides b Z iff there exists c Z such that b = ac. We write a b. • ∈ ∈ ∈ | A common divisor of a and b is d that divides both a and b. • The greatest common divisor or highest common factor of a and b is a common factor • d of a and b such that any other common factor is smaller than d. This is written d = gcd(a, b) = hcf(a, b) = (a, b). Note that every a Z is a factor of 0, since 0 = 0 a. Therefore every number is a common ∈ × factor of 0 and 0, so there is no such thing as hcf(0, 0). However if a, b Z are not both zero, ∈ then they have a highest common factor. In particular if a > 0 then hcf(a, 0) = a. Euclid’s algorithm is a method for calculating highest common factors. Note that if a divides b, then also a divides b or a divides b or a divides b. To remove − − − − this ambiguity, we will usually work with positive integers. The following obvious remark : any divisor of a 0 is smaller or equal to a, is often used in ≥ the proofs. 2 1.1.1 Euclidean divison Let a b 0 be two integers. There exists a UNIQUE pair of ≥ ≥ integers (q,r) satisfying a = qb + r and 0 r < b. ≤ Proof. Two things need to be proved : the existence of (q,r) and its uniqueness. Let us prove the existence. Consider the set S = x,x integer 0: a xb 0 { ≥ − ≥ } The set S is not empty : 1 belongs to S. The set S is bounded : any element x of S satisfies x a . THerefore, S being a bounded set of positive integers, S is finite and hence contains a ≤ b maximal element. Let q be this maximal element and let r := a qb. − We need to prove that 0 r < b. By definition r 0. To prove that r < b, let us argue by ≤ ≥ contradiction. Suppose that r b. Then, replacing r by a qb, we get ≥ − a (q + 1)b 0 − ≥ This means that q + 1 S but q + 1 > q. This contradicts the maximality of q. Therefore r < b ∈ and the existence is proved. Let us now prove the uniqueness. Again we argue by contradiction. Suppose that there exists a pair (q′,r′) satisfying a = q′b+r′ with 0 r < b and such that q = q. By subtracting the inequality 0 r < b to this inequality, ≤ ′ ′ 6 ≤ we get b<r r < b i.e. − ′ − r r′ < b | − | Now by subtracting a = q′b + r′ to a = qb + r and taking the modulus, we get r r′ = q q′ b | − | | − | By assumption q = q , hence q q 1 and we get the inequality 6 ′ | ′ − |≥ r r′ b | − |≥ The two inequalities satisfied by r r contradict each other, hence q = q . And now the equality − ′ ′ r r = q q b gives r = r . The uniqueness is proved. 2 | − ′| | − ′| ′ We now prove the following property which lies at the heart of Euclid’s algorithm. 1.1.2 Let a b 0 be two integers and (q,r) such that ≥ ≥ a = bq + r, 0 r < b ≤ Then hcf(a, b) = hcf(b, r). Proof. Let A := hcf(a, b) and B := hcf(b, r). As r = a bq and A divides a and b, A divides − r. Therefore A is a common factor of b and r. As B is the highest common factor of b and r, A B. ≤ 3 In exactly the same way, one proves (left to the reader), that B A and therefore A = B. ≤ 2 This proposition leads to the following algorithm (Euclids algorithm). Let a b 0 be two integers. We wish to calculate hcf(a, b). ≥ ≥ The method is this: Set r1 = b and r2 = a. If ri 1 = 0 then define for i 3: − 6 ≥ ri 2 = qiri 1 + ri, 0 ri <ri 1. − − ≤ − The fact that ri < ri 1 (strict inequality !!!) implies that there will be an integer n such that − rn = 0. Then by the theorem we have hcf(a, b) = hcf(r2,r3)= ... = hcf(rn 1, 0) = rn 1. − − 1.1.3 Remark When performing Euclid’s algorithm, be very careful not to divide qi by ri. This is a mistake very easy to make. 1.1.4 Example Take a = 27 and b = 7. We have 27 = 3 7 + 6 × 7 = 1 6 + 1 × 6 = 6 1 + 0. × Therefore hcf(27, 7) = hcf(7, 6) = hcf(6, 1) = hcf(1, 0) = 1. Euclid’s algorithm is very easy to implement on a computer. Suppose that you have some standard computer language (Basic, Pascal, Fortran,...) and that it has an instruction r := a mod b which returns the remainder of the Euclidean division of a by b. The implementation of the algorithm would be something like this: Procedure hcf(a, b) If a < b then Swap(a, b) While b = 0 6 Begin r := a mod b a := b b := r End Return a End The following lemma is very important for what will follow. It is essentially ‘the Euclid’s algorithm’ run backwards. 4 1.1.5 Bezout’s Lemma As usual, let a b 0 be integers. Let d = hcf(a, b). Then there ≥ ≥ are integers h, k Z such that ∈ d = ha + kb. Note that in this lemma, the integers h and k are not positive, in fact exactly one of them is negative or zero. Prove it ! Proof. Consider the sequence given by Euclid’s algorithm: a = r1, b = r2,r3,r4,...,rn = d. In fact we’ll show that each of the integers ri may be expressed in the form ha+kb. We prove this by induction on i: it is certainly the case for i = 1, 2 since r = 1 a+0 b and r = 0 a+1 b. 1 × × 2 × × For the inductive step, assume it is the case for ri 1 and ri 2, i.e. − − ri 1 = ha + bk, ri 2 = h′a + k′b. − − We have ri 2 = qri 1 + ri. − − Therefore r = h′a + k′b q(ha + kb) = (h′ qh)a + (k′ qk)b. i − − − 2 1.1.6 Example Again we take a = 27 and b = 7. 27 = 3 7 + 6 × 7 = 1 6 + 1 × 6 = 6 1 + 0. × Therefore 1 = 7 1 6 − × = 7 1 (27 3 7) − × − × = 4 7 1 27. × − × So we take h = 1 and k = 4. − We now apply the Euclid’s algorithm and B´ezout’s lemma to the solution of linear diophan- tine equations. Let a,b,c be three positive integers. A linear diophantine equation (in two variables) is the equation ax + by = c A solution is a pair (x, y) of integers (not necessarily positive) that satisfy this relation. Such an equation may or may not have solutions. For example, consider 2x + 4y = 5. Quite clearly, if there was a solution, then 2 will divide the right hand side, which is 5. This is not the case, therefore, this equation does not have a solution. 5 On the other hand, the equation 2x + 4y = 6 has many solutions : (1, 1), (5, 1),.... This − suggests that the existence of solutions depends on whether or not c is divisible by the hcf(a, b) and that if such is the case, there are many solutions to the equation. This indeed is the case, as shown in the following theorem. Before we prove the theorem, let us prove a couple of preliminary lemmas. 1.1.7 Lemma Let a and b be two positive integers. Let d := hcf(a, b). Then hcf(a/d, b/d) = 1. Proof. Use B´ezout’s lemma : there exist h, k such that ah + kb = d.
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