# Moreover this binary operation satisfies the following properties

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1 Contents 1 Algebraic structures Group Definitions and examples Subgroup Group homomorphism Ring Definitions and examples The rings Z/nZ Field Polynomials The ring K[X] Definition and operations Degree Polynomial arithmetic Polynomial maps Definition Derivative polynomial Root Vector spaces The structure of vector space First examples Definitions More examples Subspace Definition and characterization Linear span Sum and direct sum Linear map Definition and examples Linear map and subspaces Isomorphic vector spaces Sets of linear maps Finite-dimensional vector spaces Family of vectors Linearly independent family Spanning family and basis

2 2 CONTENTS 4.2 Finite dimension Finite-dimensional vector space Dimension Finite-dimensional subspace Linear map on finite-dimensional vector space Matrices Definition and operations The vector space M p,n (K Multiplication of matrices The ring M n (K Matrices associated to vectors and linear maps Coordinate vector Matrix associated to a linear map Change of basis Rank Lecture notes

3 Chapter 1 Algebraic structures 1.1 Group Definitions and examples Definition 1.1 (Binary operation A binary operation (or binary law on a nonempty set S is a map from S S to S. Such a binary operation is usually denoted : S S S (a, b a b Example : The set N of all natural numbers comes with the binary operation addition of two numbers. + : N N N (a, b a + b Example : Similarly, the addition provides a binary operation on the set Z of all integers. + : Z N Z (a, b a + b Moreover this binary operation satisfies the following properties i (a, b, c Z 3, (a + b + c = a + (b + c = a + b + c ii 0 Z is such that a Z, a + 0 = 0 + a = a iii a Z, b Z/ a + b = b + a = 0 (actually b = a, the opposite number of a Example : Let B be the set of all bijective functions from the segment [0, 1] to itself. B = {f : [0, 1] [0, 1] / f bijective} Notice this set is nonempty (for instance Id [0,1] = (x x B. The composition of functions provides a binary operation on B. : B B B (f, g f g Indeed f g is well in B since f g : [0, 1] [0, 1] is bijective as composition of bijective functions. Moreover this binary operation satisfies the following properties

4 4 CHAPTER 1. ALGEBRAIC STRUCTURES i (f, g, h B 3, (f g h = f (g h = f g h ii Id [0,1] B is such that f B, f Id [0,1] = Id [0,1] f = f iii f B, g B/ f g = g f = Id [0,1] (actually g = f 1, the bijective inverse function of f Definition 1.2 (Group Let (G, be a nonempty set with a binary operation. (G, is said to be a group if it satisfies each of the following group axioms i Associativity: (a, b, c G 3, (a b c = a (b c = a b c ii Identity element: e G/ a G, a e = e a = a (e is called identity element iii Inverse element: a G, a G/ a a = a a = e (a is called inverse element of a Examples : (Z, + and (B, are groups whose identity elements are respectively 0 Z and Id [0,1] B. Definition 1.3 (Abelian group A binary operation on a nonempty set S is said commutative if (a, b S 2, a b = b a An abelian group (or commutative group is a group (G, for which its binary operation is commutative. Example : (Z, + is an abelian group since the addition of numbers is commutative. Example : (B, is not abelian since (f, g B 2 / f g g f. For instance, consider f = (x x 2 B (its associated inverse element is f 1 = (y y B and g = (x 1 x B (its associated inverse element is g 1 = (x 1 x = g B. Then x [0, 1], { f g(x = f(1 x = (1 x 2 = 1 2x + x 2 g f(x = g(x 2 = 1 x 2 So f g g f (for instance at x = 1 2. Further examples : a Additive groups. (Z, + (Q, + (R, + (C, + are abelian groups. b Multiplicative groups. (Q, (R, (C are abelian groups. The same goes for (R +, (but not for R because the multiplication is not a binary operation on this set. Proof : For instance for (R, : 1. is a binary operation on R since the multiplication of two real numbers not equal to 0 remains a real number not equal to is associative: (a, b, c (R 3, (abc = a(bc = abc 3. 1 R is identity element: a R, a 1 = 1 a = a 4. Every real number a not equal to 0 has a multiplicative inverse 1 a R which is inverse element of a (since a 1 a = 1 a a = Finally the group is abelian since the multiplication of numbers is commutative. Lecture notes

5 1.1. GROUP 5 c Counterexamples. (N, + is not a group since 1 N but there is no positive integer a N such that 1 + a = a + 1 = 0 (actually 1 / N. The same goes for (Z {0}, since 2 Z but 1 2 / Z. Proposition 1.4 Let (G, be a group. Then the following properties hold 1. The identity element e G is unique. 2. For any element a G, the inverse element of a in (G, is unique, denoted a Simplifications are possible: (a, b, c G 3, { a b = a c = b = c b a = c a = b = c 4. (a, b G 2, (a b 1 = b 1 a 1 Proof : 1. By contradiction, assume e 1 and e 2 in G are identity elements for. Then a G, { a e1 = e 1 a = a a e 2 = e 2 a = a a = e 2 in the first line gives e 2 e 1 = e 1 e 2 = e 2. a = e 1 in the second line gives e 1 e 2 = e 2 e 1 = e 1. Consequently e 1 = e 1 e 2 = e 2, that is all identity elements must be equal. In other words there is only one identity element. 2. Similarly, given any element a G, assume a and a in G are inverse elements of a. Then { a a = a a = e a a = a a = e Using the identity element e and the associativity of binary operation, we get: a = a e = a (a a = (a a a = e a = a Thus, there is only one inverse element of a. 3. Using the inverse element of a and the associativity, we get: a b = a c = a 1 (a b = a 1 (a c = (a 1 a b = (a 1 a c = e b = e c = b = c The same goes for the second simplification. 4. We need to check the axiom of inverse element in Definition 1.2. { (a b (b 1 a 1 = a (b b 1 a 1 = a e a 1 = a a 1 = e (b 1 a 1 (a b = b 1 (a 1 a b = b 1 e b = b 1 b = e So b 1 a 1 is the inverse element of the element a b. Example : If f and g are two bijective functions from the segment [0, 1] to itself then (f g 1 = g 1 f 1 (which is not necessary equal to f 1 g 1 since composition of functions is not a commutative binary operation. Sébastien Godillon

6 6 CHAPTER 1. ALGEBRAIC STRUCTURES Subgroup Example : Consider the abelian group (Z, + and the following subsets E = {even numbers} = {2k, k Z} O = {odd numbers} = {2k + 1, k Z} We have (a, b E, a + b E (if a = 2k and b = 2k then a + b = 2k with k = k + k Z but it is no longer true in O. In other words, + is a binary operation on E but not on O. Definition 1.5 (Subgroup Let (G, be a group and H G be a nonempty subset. (H, is a subgroup of (G, if it satisfies each of the following subgroup axioms i is a binary operation on H ii (H, is a group Example : For any group (G, with identity element e, ({e}, is a subgroup of (G, called the trivial subgroup. Proposition 1.6 Let (G, be a group and H G be a nonempty subset. (H, is a subgroup of (G, if and only if it satisfies the following condition (a, b H 2, a b 1 H Proof : Necessary. Let a and b be two elements in H. The axiom of inverse element in Definition 1.2 gives b 1 H. And because is a binary operation on H, we get a b 1 H. Sufficient. We need to check that each axiom from Definition 1.2 is satisfied for (H, and is a binary operation on H. 1. The axiom of associativity for (H, comes directly from that one satisfied by (G,. 2. Denote e G the identity element of (G,. Since H is nonempty, there exists an element a H. It follows e = a a 1 H. Consequently there exists an identity element in (H, or equivalently, the axiom of identity element is checked for (H,. 3. Now the inverse element in (G, of any a H satisfy a 1 = e a 1 H (since e H. In particular any element a H has an inverse element in (H, or equivalently, the axiom of inverse element is checked for (H,. 4. Finally for every elements a and b in H, we get a b = a (b 1 1 H (since b 1 H. In other words, is well a binary operation on H. Remark : In practice, it is more convenient to use Proposition 1.6 to show that (H, is a subgroup of (G, than Definition 1.5 Examples : (Z, + is a subgroup of (Q, +, (R, + and (C, +. (Q, + is a subgroup of (R, + and (C, +. (R +, is a subgroup of (R, and (C,. Lecture notes

7 1.1. GROUP 7 Proposition 1.7 For every positive integer n N, consider the following subset of Z nz = {multiples of n} = {nk, k Z} Then (nz, + is a subgroup of (Z, +. Proof : nz is well a nonempty subset of Z (for instance 0 nz. Let a = nk and b = nk be two elements in nz. Then Hence, the result follows from Proposition 1.6. a + ( b = (nk + ( (nk = n(k k nz Example : E = {even numbers} = 2Z is a subgroup of Z. Further examples : a Consider the following set of all complex numbers with absolute value 1 (that is the unital circle in the complex plane T = {z C / z = 1} Then (T, is a subgroup of (C, called the circle group. Proof : If z = 1 for a complex number z C then z 0. Moreover 1 T. So T is a nonempty subset of C. Now using the claims that the absolute value of the multiplicative inverse of a complex number is the multiplicative inverse of its absolute value and the absolute value of the product of two complex numbers is equal to the product of their absolute values, we get The conclusion follows with Proposition 1.6. (z, w T 2, z w 1 = z w = 1 b For every positive integer n N, consider the following subset of C R n = {n th root of unity} = {z C / z n = 1} Then (R n, is a subgroup of (T, (and of (C, as well. It is finite with cardinality n. Proof : If z n = 1 for a complex number z C then z n = z n = 1 and consequently z = 1 (because z R +. Moreover 1 is a root of unitaly for every power n N. So R n is a nonempty subset of T. Actually R n may be written as follows R n = {z = e ı 2kπ n, k {0, 1,..., n 1} } where θ R, e ıθ = cos(θ + ı sin(θ In particular, R n is a finite set with cardinality n. We conclude as in the previous proof for (T, since the exponentiation function z z n satisfies the same required properties as the absolute value for complex numbers. Sébastien Godillon

8 8 CHAPTER 1. ALGEBRAIC STRUCTURES Group homomorphism Definition 1.8 (Group homomorphism Let (G, and (H, be two groups. function ϕ : G H such that Moreover A group homomorphism from (G, to (H, is a (a, b G 2, ϕ(a b = ϕ(a ϕ(b If a group homomorphism ϕ : G H is a bijection, then it is called a group isomorphism. If ψ : G G is a group homomorphism from (G, to itself, then it is called a group endomorphism. If furthermore ψ is bijective and hence a group isomorphism, it is called a group automorphism. Remark : If ϕ : G H is a group isomorphism, then its bijective inverse ϕ 1 : H G is also a group homomorphism. Proof : Let a and b be two elements in H. Since ϕ is a surjective function, there exist two elements a and b in G such that a = ϕ(a and b = ϕ(b. Then ϕ 1 (a b = ϕ 1 (ϕ(a ϕ(b = ϕ 1 (ϕ(a b = a b = ϕ 1 (a ϕ 1 (b Some examples : a The constant function ϕ : a ϕ(a = e H equal to the identity element of (H, is a group homomorphism from (G, to (H, (even onto the trivial subgroup ({e H },. b The identity function ϕ = Id G = (a a is a group automorphism of (G,. c Given n Z, ϕ n : Z Z, k nk is a group endomorphism from (Z, + to itself and a group isomorphism from (Z, + onto its subgroup (nz, +. d Given λ R, ϕ λ : R R, x λx is a group endomorphism from (R, + to itself. Moreover it is a group automorphism if and only if λ 0. Proof : (x, y R 2, ϕ λ (x + y = λ(x + y = λx + λy = ϕ λ (x + ϕ λ (y And ϕ λ is a bijective function over R if and only if λ 0 (in this case, the associated bijective inverse is (x x/λ = ϕ 1/λ. e exp : R R +, x exp(x is a group isomorphism from (R, + to (R +,. Proof : (x, y R 2, exp(x + y = exp(x exp(y And exp : R R + is a bijective function (whose bijective inverse is ln : R + R. Lecture notes

9 1.2. RING Ring Definitions and examples Definition 1.9 (Ring Let (R, +, be a nonempty set with two binary operations denoted + and. (R, +, is said to be a ring if it satisfies each of the following ring axioms i (R, + is an abelian group ii the binary operation is associative: (a, b, c R 3, (a b c = a (b c = a b c iii the binary operation is distributive over the binary operation +: (a, b, c R 3, { a (b + c = (a b + (a c (b + c a = (b a + (c a Moreover a ring (R, +, is said unital if there exists an identity element for the binary operation : e R/ a R, a e = e a = a commutative if the binary operation is commutative: We denote (a, b R 2, a b = b a 0 R the identity element for the binary operation +, called the additive identity 1 R the identity element for the binary operation in case (R, +, is unital, called the multiplicative identity a the inverse element of an element a R for the binary operation +, called the additive inverse (or opposite Example : (Z, +, (Q, +, (R, +, (C, +, are commutative unital rings. Proof : For instance for (Q, +, : 1. + and are binary operations on Q: (a, b Q 2, a + b Q and ab Q 2. (Q, + is an abelian group. 3. is associative on Q: (a, b, c Q 3, (abc = a(bc = abc 4. is distributive over +: (a, b, c Q 3, a(b + c = (ab + (ac 5. has a multiplicative identity in Q: 1 Q = 1 since a Q, a 1 = 1 a = a 6. is commutative on Q: (a, b Q 2, ab = ba Sébastien Godillon

10 10 CHAPTER 1. ALGEBRAIC STRUCTURES Example : Let F be the set of all functions from R to itself. F = {f : R R} For every functions f and g in F, the function f +g F is defined by x R, (f +g(x = f(x+g(x (using addition of real numbers and the function f g F by x R, (f g(x = f(xg(x (using multiplication of real numbers. Then (F, +, is a commutative unital ring. Proof : 1. + and are binary operations on F by definition. 2. (F, + is an abelian group: the associativity and the commutativity come from those ones of (R, +, the additive identity is the constant function 0 F : x 0 and the additive inverse of a function f F is the function f defined by f : x x. 3. is associative on R, so the same does on F as well. 4. is distributive over + on R, so the same does on F as well. 5. The constant function 1 F : x 1 is a multiplicative identity for in F. 6. is commutative on R, so the same does on F as well. Proposition 1.10 Let (R, +, be a ring. Then the following properties hold 1. If (R, +, is unital, then the multiplicative identity 1 R R is unique. 2. a R, a 0 R = 0 R a = 0 R 3. If (R, +, is unital, then a R, ( 1 R a = a ( 1 R = a 4. (a, b R 2, ( a b = a ( b = (a b Proof : 1. Actually the same proof as for the first property from Proposition 1.4 still holds. 2. Using the distributivity of + over and the additive identity 0 R, we get: 0 R a + 0 R a = (0 R + 0 R a = 0 R a = 0 R + 0 R a Now a simplification on each side by 0 R a (see Proposition 1.4 gives 0 R = 0 R a as needed. The same goes for a 0 R. 3. We have a + ( 1 R a = 1 R a + ( 1 R a = (1 R + ( 1 R a = 0 R a = 0 R Consequently ( 1 R a is the additive inverse of the element a (since (R, + is an abelian group whose its additive identity is 0 R. The same goes for a ( 1 R. 4. We have The conclusion follows. a b + ( a b = (a + ( a b = 0 R b = 0 R Remark : In particular, the second point of the previous proposition shows that the additive identity 0 R of a ring (R, +, has no multiplicative inverse: there is no element a R such that 0 R a = a 0 R = 1 R. Lecture notes

11 1.2. RING 11 Definition 1.11 (Subring Let (R, +, be a ring and T R be a nonempty subset. (T, +, is a subring of (R, +, if it satisfies each of the following subring axioms i (T, + is a subgroup of (R, + ii is a binary operation on T In this case, (T, +, is a ring. Trivial example : For any ring (R, +,, ({0 R }, +, is a subring called the trivial subring. Examples : (Z, +, is a subring of (Q, +,, (R, +, and (C, +,. For every positive integer n N, (nz, +, is a subring of (Z, +, which is not unital as soon as n 2 (since 1 / nz. Proof : For instance for (nz, +, : 1. (nz, + is a subgroup of (Z, + (see Proposition If a = nk and b = nk are two elements in nz then a b = (nk (nk = n(knk nz. Hence, is a binary operation on nz. Definition 1.12 (Ring homomorphism Let (R, +, and (T,, be two rings. A ring homomorphism from (R, +, to (T,, is a function ϕ : R T such that (a, b R 2, { ϕ(a + b = ϕ(a ϕ(b ϕ(a b = ϕ(a ϕ(b Moreover If a ring homomorphism ϕ : R T is a bijection, then it is called a ring isomorphism. If ψ : R R is a ring homomorphism from (R, +, to itself, then it is called a ring endomorphism. If furthermore ψ is bijective and hence a ring isomorphism, it is called a ring automorphism. Remark : The same remark as for group isomorphism holds: the bijective inverse of any ring isomorphism is also a ring homomorphism The rings Z/nZ Remind the following result: Theorem 1.13 (Division algorithm For any given two integers a and d with d 0, there exist unique integers q and r such that { a = qd + r 0 r < d The integer q is called the quotient, r the remainder, d the divisor and a the dividend. Sébastien Godillon

12 12 CHAPTER 1. ALGEBRAIC STRUCTURES Fix a positive integer n N. Definition 1.14 (Congruence modulo n Two integers a and b are said congruent modulo n if n divides their difference a b, that is if there exists an integer q such that a b = qn (equivalently if the remainder of the division algorithm with dividend a b and divisor n is equal to 0. In this case, we write a b[n]. Remark : In particular, any integer a is congruent modulo n to its associated remainder r of the division algorithm with dividend a and divisor n (since a r = qn. Moreover if two integers a and b have the same remainder r from the division algorithm with divisor n, then they are congruent modulo n (since a = qn + r and b = q n + r with r = r imply a b = (q q n. Definition 1.15 (Congruence class For any given integer a Z, the congruence class modulo n of a is the following set a = {integers congruent modulo n to a} = {b Z / b a[n]} = {..., a 2n, a n, a, a + n, a + 2n, a + 3n,... } = a + nz Proposition 1.16 Let a and b be two congruence classes modulo n associated to two integers a and b. Then a and b are equal if and only if a and b are congruent modulo n. (a, b Z 2, a = b a b[n] Proof : Necessary. If a = b then in particular b a that is b is congruent modulo n to a. Sufficient. If b is congruent modulo n to a then there exists an integer q such that a = b + qn. Consequently a = a + nz = b + qn + nz = b + n(q + Z = b + nz = b Remark : In particular, for any given integer a Z, if r denotes the remainder of the division algorithm with dividend a and divisor n then a = r. Definition 1.17 (Z/nZ The set of all congruence classes modulo n is denoted by Z/nZ (read Z over nz. Z/nZ = {congruence classes modulo n} = {a, a Z} = {congruence classes modulo n of the remainders = {r, r Z and 0 r < n} = {0, 1,..., n 1} It is a finite set with cardinality n. from the division algorithm with divisor n} Lecture notes

13 1.2. RING 13 Example : For n = 2, Z/2Z = {0, 1}. For instance 2 = 0 and 5 = 1 (equivalently we may write 2 0[2] and 5 1[2]. Actually { a even = a = 0 a odd = a = 1 Example : For n = 3, Z/3Z = {0, 1, 2}. For instance 6 = 0 (or 6 0[3] and 13 = 1 (or 13 1[3] because 13 = Definition 1.18 (Addition and multiplication in Z/nZ Two binary operations on Z/nZ denoted by + and are defined as follows (using addition and multiplication of integers (r, r {0, 1,..., n 1} 2, { r + r = r + r r r = rr Proposition 1.19 (a, b Z 2, { a + b = a + b a b = ab Proof : From Theorem 1.13, a = qn + r and b = q n + r for some integers q, q, r, r with 0 r, r < n. We have (a + b (r + r = (q + q n. So (a + b and (r + r are congruent modulo n. It follows from Proposition 1.16 that We have a + b = r + r = r + r = a + b ab = (qn + r(q n + r = qq n 2 + (qr + q rn + rr Thus ab rr = (qq n + qr + q rn that is ab and rr are congruent modulo n. It follows from Proposition 1.16 that ab = rr = r r = a b Corollary 1.20 (Z/nZ, +, is a commutative unital ring whose identity elements are respectively 0 for the binary operation + and 1 for the binary operation. Proof : Proposition 1.19 gives everything we need from the fact that (Z, +, is a commutative unital ring whose identity elements are respectively 0 and 1. Example : Some computations in Z/6Z = {0, 1, 2, 3, 4, 5}: a 15 = 3 since 15 = b = = 9 = = 3 c = 6 = 0 thus 4 = 2 d 5 4 = 5 4 = 20 = = 2 Remark : In Z/6Z, we have 3 2 = 6 = 0 but 3 0 and 2 0. Sébastien Godillon

14 14 CHAPTER 1. ALGEBRAIC STRUCTURES 1.3 Field Definition 1.21 Let (F, +, be a nonempty set with two binary operations. (F, +, is said to be a field if it satisfies each of the following field axioms i (F, +, is an unital ring ii every element except 0 F has an inverse element in F for the binary operation : a F {0 F }, a 1 F/ a a 1 = a 1 a = 1 F Moreover a field (F, +, is said commutative if the binary operation is commutative: (a, b F 2, a b = b a The same notations as for ring (0 F, 1 F and a are used and we denote a 1 the inverse element of an element a F {0 F } for the binary operation (called the multiplicative inverse. Remark : Equivalently (F, +, is a field if and only if it satisfies each of the following conditions i (F, + is an abelian group (whose its additive identity is denoted by 0 F ii (F {0 F }, is a group iii the binary operation is distributive over the binary operation + Examples : (Q, +, (R, +, (C, +, are commutative fields. Counterexample : (Z, +, is not a field since 2 Z {0} has no multiplicative inverse in Z: there is no integer a Z such that 2a = 1 (equivalently (Z {0}, is not a group. Proposition 1.22 Let (F, +, be a field. Then (a, b F 2, a b = 0 F = either a = 0 F or b = 0 F Proof : Assume a 0 F. Consequently, there exists a multiplicative inverse a 1 F and we get: b = 1 F b = (a 1 a b = a 1 (a b = a 1 0 F = 0 F Remark : In particular, if there exist two elements a and b in a ring (R, +, such that a b = 0 R but a 0 R and b 0 R (such elements are called zero divisors then (R, +, is not a field. More precisely, any zero divisor does not have multiplicative inverse. Example : (Z/6Z, +, is not a field (since 3 2 = 0 but 3 0 and 2 0. Theorem 1.23 Let n N be a positive integer. Then (Z/nZ, +, is a commutative field if and only if n is a prime number. Lecture notes

15 1.3. FIELD 15 Proof : Necessary. The proof is the same as the previous example. By contradiction, assume that n N is not a prime number. In other words, there exist two integers p and q such that 1 < p, q < n and n = pq. Then p q = pq = n = 0 but p 0 and q 0. That is a contradiction with Proposition Sufficient. Assume n is a prime number and let r Z/nZ be a congruence class not equal to the congruence class 0. We may assume that 0 < r < n. In particular, r and n are relatively prime. Remind the following result: Theorem 1.24 (Bézout s identity If two integers a and b are relatively prime then there exist integers x and y such that ax + by = 1 Here we get two integers x and y such that rx + ny = 1. Consequently r x = rx = 1 ny = 1 In other words, x is the multiplicative inverse of r. So, any congruence class in Z/nZ not equal to 0 has a multiplicative inverse. Example : In Z/7Z, the multiplicative inverse of 2 is 4 since 2 4 = 2 4 = 8 = 1 (and then the multiplicative inverse of 4 is 2. Moreover, we have 3 1 = 5 (since 3 5 = 15 = = 1 and 6 1 = 6 (since 6 6 = 36 = = 1. Sébastien Godillon

16 Chapter 2 Polynomials In this chapter, fix a commutative field (K, +, (for instance K = Q, R or C. We denote 0 the additive identity a the additive inverse of an element a K 1 the multiplicative identity a 1 the multiplicative inverse of an element a K = K {0} 2.1 The ring K[X] Definition and operations Definition 2.1 (Polynomial Let (a n n N = (a 0, a 1, a 2,..., a n,... be an infinite sequence of elements in K which are eventually equal to zero, that is d N/ n > d, a n = 0 The polynomial with coefficients (a n n N is the following formal expression Moreover P (X = a 0 + a 1 X + a 2 X a d X d = the formal symbol X is called the variable d a n X n = n=0 + n=0 a n X n the formal symbols X 0 = 1, X 1 = X, X 2,..., X n,... are called the powers of X for any n N, a n is called the coefficient of the term with degree n a 0 is called the constant term We denote K[X] the set of all polynomials with coefficients in K. Examples : P (X = X + X 3 + X 5 = 0 + 1X + 0X 2 + 1X 3 + 0X 4 + 1X 5 is a polynomial in K[X] but also Q(X = X 2 or R(X = 0. S(X = 1 2 2X 2 + 5X 4 is a polynomial in R[X] or C[X] but not in Q[X].

17 2.1. THE RING K[X] 17 Definition 2.2 (Addition in K[X] Let P (X and Q(X be two polynomials in K[X] with coefficients respectively (a n n N and (b n n N. We define the sum of P (X and Q(X, denoted (P + Q(X, to be the polynomial in K[X] with coefficients (a n + b n n N. That provides a binary operation + on K[X]. P (X + Q(X = (P + Q(X = + n=0 (a n + b n X n Example : For instance, we have in R[X]: (1 + 2X + 3X 3 + (4 X + 5X 4 = ( (2 1X + (0 + 0X 2 + (3 + 0X 3 + (0 + 5X 4 Definition 2.3 (Multiplication in K[X] = 5 + X + 3X 3 + 5X 4 Let P (X and Q(X be two polynomials in K[X] with coefficients respectively (a n n N and (b n n N. We define the product of P (X and Q(X, denoted (P Q(X, to be the polynomial in K[X] with coefficients ( n k=0 a kb n k n N. That provides a binary operation on K[X]. P (XQ(X = (P Q(X = + n=0 ( n a k b n k X n = k=0 + n=0 k,l N k+l=n a k b l X n Remark : The previous definition is natural with respect to the following property Example : For instance, we have in R[X]: (k, l N 2, X k X l = X k+l (3 X 2X 2 (2 + 6X + 4X 2 = 6 + (18 2X + (12 6 4X 2 + ( 4 12X 3 8X 4 Proposition 2.4 = X + 2X 2 16X 3 8X 4 (K[X], +, is a commutative unital ring whose identity elements are respectively the constant polynomials 0 for addition and 1 for multiplication. Proof : At first, K[X] is nonempty (for instance the constant polynomial 0 is in K[X] and are binary operations on K[X] by definition. 2. We have: (a + is associative on K, so the same does on K[X] as well. (b The constant polynomial 0 K[X] is the additive identity for + in K[X]. (c For any polynomial in K[X] with coefficients (a n n N, the polynomial in K[X] with coefficients ( a n n N is its additive inverse. (d + is commutative on K, so the same does on K[X] as well. Consequently (K[X], + is an abelian group. Sébastien Godillon

18 18 CHAPTER 2. POLYNOMIALS 3. Let P (X, Q(X and R(X be three polynomials in K[X] with coefficients respectively (a n n N, (b n n N and (c n n N. We have: [P (XQ(X] R(X = = = = = + n=0 + N=0 + N=0 + N=0 + k=0 Thus, is associative on K[X]. k,l N k+l=n n,j N n+j=n a k b l k,l,j N k+l+j=n k,m N k+m=n a k X k Xn k,l N k+l=n a k b l c j a k + m=0 + j=0 c j X j a k b l c j XN l,j N l+j=m l,j N l+j=m XN b l c j b l c j XN Xm = P (X [Q(XR(X] 4. Let P (X, Q(X and R(X be three polynomials in K[X] with coefficients respectively (a n n N, (b n n N and (c n n N. We have: + n=0 k,l N k+l=n a k (b l + c l Xn = + n=0 k,l N k+l=n a k b l Xn + + n=0 k,l N k+l=n a k c l Xn Equivalently, P (X (Q(X + R(X = P (XQ(X + P (XR(X. And the same goes for (Q(X + R(X P (X = Q(XP (X + R(XP (X. Thus, is distributive over + on K[X]. 5. Let P (X be a polynomial in K[X] with coefficients (a n n N. Since every coefficient of the constant polynomial 1 K[X] is equal to zero except the constant term equal to 1, we have: P (X1 = + n=0 And the same goes for 1P (X = P (X. multiplicative identity for in K[X]. (a a a n a n 1 X n = + n=0 a n X n = P (X Thus, the constant polynomial 1 K[X] is the 6. + and are commutative on K, so the same goes for on K[X] as well. Finally, K[X] satisfies all conditions to be a commutative unital ring. Remark : But (K[X], +, is not a field. For instance, the polynomial P (X = X K[X] {0} has no multiplicative inverse in K[X]: there is no polynomial Q(X K[X] such that XQ(X = 1. Proof : For any polynomial Q(X = a 0 + a 1 X + a 2 X a d X d K[X], the constant term of XQ(X = a 0 X + a 1 X 2 + a 2 X a d X d K[X] is 0 but that one of the constant polynomial 1 K[X] is 1. So the equality can not hold. Lecture notes

19 2.1. THE RING K[X] Degree Definition 2.5 (Degree The degree of a polynomial P (X K[X], denoted deg(p (X or shortly deg(p, is the highest exponent for terms with non zero coefficient. More precisely if (a n n N are the coefficients of P (X then deg(p is an element of N { } defined by { deg(p = if P (X = 0 max{n N / a n 0} otherwise Moreover, the coefficient a deg(p K (in case P (X 0 is called the leading coefficient. Some examples : a The (non zero constant polynomials P (X = a 0 K[X] where a 0 0 are of degree 0. b The linear polynomials P (X = a 0 + a 1 X K[X] where a 1 0 are of degree 1. c The quadratic polynomials P (X = a 0 + a 1 X + a 2 X 2 K[X] where a 2 0 are of degree 2. d The cubic polynomials P (X = a 0 + a 1 X + a 2 X 2 + a 3 X 3 K[X] where a 3 0 are of degree 3. etc. Remark : It is usefull to define the degree of the zero constant polynomial to be (furthermore it is convenient to take max = as a convention. In the following, we introduce the rules: Proposition 2.6 k N { }, k max{k, } = k k + ( = ( + k = Let P (X and Q(X be two polynomials in K[X]. Then the following properties hold 1. deg(p + Q max{deg(p, deg(q} with equality if deg(p deg(q 2. deg(p Q = deg(p + deg(q Proof : We write P (X = deg(p n=0 a n X n and Q(X = deg(q n=0 b n X n. 1. In case deg(p < deg(q we get (P + Q(X = deg(p n=0 (a n + b n X n + deg(q n=deg(p +1 b n X n And b deg(q 0 implies that deg(p + Q = deg(q = max{deg(p, deg(q}. The same goes when deg(p > deg(q. Now if deg(p = deg(q = d then (P + Q(X = d (a n + b n X n n=0 Consequently, we have deg(p + Q d = max{deg(p, deg(q}. Sébastien Godillon

20 20 CHAPTER 2. POLYNOMIALS 2. We have: (P Q(X = + n=0 k,l N k+l=n a k b l Xn = + n=0 k deg(p l deg(q k+l=n a k b l X n = deg(p +deg(q Moreover the coefficient of the term with degree n = deg(p + deg(q is a k b l = a deg(p b deg(q 0 The conclusion follows. k deg(p l deg(q k+l=deg(p +deg(q n=0 k deg(p l deg(q k+l=n a k b l X n Remarks : The results remain true if P (X or Q(X (or both is the zero constant polynomial. In the first property, the sufficient condition for equality is not necessary. For instance deg(x + X = deg(2x = 1 = deg(x Actually if P (X and Q(X are two polynomials of same degree d N whose their leading coefficients are respectively a d and b d then Polynomial arithmetic Theorem 2.7 (Polynomial division algorithm deg(p + Q = d a d b d For any given two polynomials P (X and D(X with D(X 0, there exist unique polynomials Q(X and R(X such that { P (X = Q(XD(X + R(X deg(r < deg(d The polynomial Q(X is called the quotient, R(X the remainder, D(X the divisor and P (X the dividend. Proof : Existence. Fix a polynomial D(X K[X] with D(X 0 and call d 0 its degree. Remark that if d = 0, that is D(X = b 0 0, then Q(X = b 1 0 P (X and R(X = 0 are suitable. So we may assume that d 1. We will prove by induction the following property for every k 0 P k = the existence part is true for every P (X K[X] with deg(p k At first, if deg(p d 1 then Q(X = 0 and R(X = P (X are suitable. Hence, P k is satisfied for every integer k such that 0 k d 1 (and at least for k = 0 since d 1. Now assume P k is satisfied for a given integer k d 1. Let P (X be a polynomial in K[X] of degree deg(p = k + 1 d. We write: P (X = a 0 + a 1 X + a 2 X a d X d + + a k+1 X k+1 with a k+1 0 D(X = b 0 + b 1 X + b 2 X b d X d with b d 0 Lecture notes

21 2.1. THE RING K[X] 21 Consider the polynomial P 1 (X = P (X a k+1 b 1 d Xk+1 d D(X K[X]. From Proposition 2.6, we have: deg(a k+1 b 1 d Xk+1 d D(X = deg(a k+1 b 1 d Xk+1 d + deg(d = (k + 1 d + d = k + 1 and { deg(p 1 max deg(p, deg(a k+1 b 1 d Xk+1 d D(X } = max{k + 1, k + 1} = k + 1 Moreover the coefficient of the term in P 1 with degree k + 1 is a k+1 a k+1 b 1 d b d = 0. Then we have deg(p 1 k. By inductive hypothesis P k applied to P 1 (X, there exist two polynomials Q 1 (X and R 1 (X such that { P1 (X = Q 1 (XD(X + R 1 (X deg(r 1 < d Now take Q(X = a k+1 b 1 d Xk+1 d + Q 1 (X and R(X = R 1 (X then we get { P (X = ak+1 b 1 d Xk+1 d D(X + P 1 (X = Q(XD(X + R(X deg(r < d Consequently P k+1 is satisfied and the conclusion follows by induction. Uniqueness. By contradiction, assume Q 1 (X, R 1 (X and Q 2 (X, R 2 (X are such that { P (X = Q1 (XD(X + R 1 (X deg(r 1 < deg(d and { P (X = Q2 (XD(X + R 2 (X deg(r 2 < deg(d Then Q 1 (XD(X + R 1 (X = Q 2 (XD(X + R 2 (X or equivalently (Q 1 (X Q 2 (X D(X = R 2 (X R 1 (X From Proposition 2.6, we have: { deg ((Q1 Q 2 D = deg (Q 1 Q 2 + deg(d deg(r 2 R 1 max {deg(r 1, deg(r 2 } < deg(d So we get deg(q 1 Q 2 < 0 (since D(X 0 implies deg(d 0 that is deg(q 1 Q 2 = and hence Q 1 (X Q 2 (X = 0. It follows Q 1 (X = Q 2 (X and hence R 1 (X = R 2 (X. Finally the quotient and the remainder of the polynomial division algorithm are unique. Examples : In order to compute the quotient and the remainder of a polynomial division algorithm, one may use a long division algorithm as follows a For P (X = X 3 12X 2 42 and D(X = X 3 X 3 = X 2 (X 3 +3X 2 12X 2 +3X 2 = 9X (X 3 27X 0 27X = 27 (X = 0 (X and the sum of all these equalities gives after simplifications X 3 12X 2 42 = (X 2 9X 27(X that is Q(X = X 2 9X 27 and R(X = 123 Sébastien Godillon

22 22 CHAPTER 2. POLYNOMIALS b For P (X = X 4 + 7X 3 3X 2 11X + 5 and D(X = X 2 2X 3 X 4 = X 2 (X 2 2X 3 +2X 3 + 3X 2 7X 3 +2X 3 = 9X (X 2 2X 3 +18X X 3X 2 +3X X 2 = 18 (X 2 2X 3 +36X X +27X + 36X = 0 (X 2 2X 3 +52X = 0 (X 2 2X and the sum of all these equalities gives after simplifications X 4 + 7X 3 3X 2 11X + 5 = (X 2 + 9X + 18(X 2 2X X + 59 that is Q(X = X 2 + 9X + 18 and R(X = 52X + 59 c For P (X = X 5 + X 4 X 1 and D(X = X 2 1 X 5 = X 3 (X 2 1 +X 3 X 4 = X 2 (X 2 1 +X 2 0 +X 3 = X (X 2 1 +X 0 +X 2 = 1 (X X +X = 0 (X = 0 (X 2 1 and the sum of all these equalities gives after simplifications X 5 + X 4 X 1 = (X 3 + X 2 + X + 1(X 2 1 that is Q(X = X 3 + X 2 + X + 1 and R(X = 0 Definition 2.8 (Multiple and divisor Let A(X and B(X be two polynomials in K[X]. We say B(X divides A(X or equivalently A(X is a multiple of B(X if Q(X K[X]/ A(X = Q(XB(X In this case, we write B(X A(X or shortly B A. Remark : A polynomial B(X K[X] divides a polynomial A(X K[X] if and only if either B(X = 0 and A(X = 0 or B(X 0 and the remainder from the polynomial division algorithm with dividend A(X and divisor B(X is R(X = 0 Proposition 2.9 Let A(X and B(X be two polynomials in K[X]. If B A with A(X 0 then deg(b deg(a Proof : There exists a polynomial Q(X K[X] such that A(X = Q(XB(X. Moreover Q(X 0 since A(X 0. In particular deg(q 0 and from Proposition 2.6 deg(a = deg(qb = deg(q + deg(b deg(b Lecture notes

23 2.2. POLYNOMIAL MAPS 23 Proposition 2.10 The ring (K[X], +, has no zero divisor. That is (A(X, B(X (K[X] 2, A(XB(X = 0 = either A(X = 0 or B(X = 0 Proof : Assume A(XB(X = 0 with B(X 0. The polynomial division algorithm with dividend 0 and divisor B(X is 0 = 0 B(X + 0 But we have 0 = A(X B(X + 0 Consenquently the uniqueness part of Theorem 2.7 gives A(X = Polynomial maps Definition Definition 2.11 (Polynomial map Let P (X be a polynomial in K[X] with coefficients (a n n N. The polynomial map associated to P (X is the following map P : K K x P (x = + n=0 a n x n Some examples : a The polynomial map associated to P (X = 0 is the constant map x 0. b The polynomial map associated to P (X = X is the identity map x x. c The polynomial map associated to P (X = 1 + 2X + X 3 R[X] is the cubic map P : R R x P (x = x 3 + 2x 1 d Recall that Z/2Z = {0, 1} is a commutative field since 2 is a prime number. Moreover we have = 0 and = 2 = 0 Consequently the polynomial map associated to P (X = X + X 2 Z/2Z[X] is the constant map P : Z/2Z Z/2Z x P (x = 0 But notice P (X 0. Sébastien Godillon

24 24 CHAPTER 2. POLYNOMIALS Proposition 2.12 Denote F(K the set of all functions from K to itself. F(K has a natural ring structure coming from that one of K. Then the following map K[X] F(K P (X P is a ring homomorphism. In particular for any given α K, the following map is a ring homomorphism as well. K[X] K P (X P (α Proof : Everything comes from the definitions of addition and multiplication of two polynomials and from the ring homomorphism F(K K, f f(α Derivative polynomial Definition 2.13 (Derivative polynomial Let P (X be a polynomial in K[X] with coefficients (a n n N. The derivative polynomial of P (X is the following polynomial P (X = a 1 + 2a 2 X + 3a 3 X 2 + = + n=0 (n + 1a n+1 X n = + n=1 na n X n 1 By induction over k 1, we define the k th order polynomial derivative, denoted P (k (X, to be the polynomial derivative of P (k 1 (X with the notation P (0 = P (and then P (1 = P. Remarks : Notice that any integer n N may be considered in K if we write it as follows n = } {{ } K n times with 1 K In particular the derivative polynomial of any polynomial with coefficients in K is well in K[X]. The polynomial map P associated to the derivative polynomial P (X of a polynomial P (X is the derivative of the map P as expected. But limit, differentiation or any calculus tool are not needed here to define the derivative of a polynomial map. Example : Consider P (X = 7 + 8X 5X 2 + 2X 3 X 5 R[X]. Then P (X = P (1 (X = 8 10X + 6X 2 5X 4 P (2 (X = X 20X 3 P (3 (X = 12 60X 2 P (4 (X = 120X P (5 (X = 120 P (6 (X = 0 etc. Lecture notes

25 2.2. POLYNOMIAL MAPS 25 Proposition 2.14 The following properties hold 1. (P (X, Q(X (K[X] 2 { (P + Q (X = P (X + Q (X (P Q (X = P (XQ(X + P (XQ (X 2. Consider the polynomial P (X = X n with n N. Then where n! (n k! k 1, (P (k (X = = n(n 1(n 2... (n k + 1 { n! (n k! Xn k if 1 k n 0 if k n P (X K[X] is a constant polynomial if and only if P (X = 0 4. If P (X K[X] is a non constant polynomial then deg(p = deg(p 1 More generally, if P (k (X 0 for some k 1 then deg(p (k = deg(p k Proof : 1. Denote respectively (a n n N and (b n n N the coefficients of P (X and Q(X. Then (P + Q (X = + n=1 n(a n + b n X n 1 = + n=1 na n X n 1 + And, using new indices n = n 1, k = k 1 and l = l 1, we get (P Q (X = = = = + n=1 + n=1 + n =0 n k,l N k+l=n k,l N k+l=n a k b l Xn 1 (k + la k b l k,l N k +l=n Xn 1 (k + 1a k +1b l Xn + + n =0 + n=1 nb n X n 1 = P (X + Q (X k,l N k+l =n a k (l + 1b l +1 Xn ( + ( + ( + ( + (k + 1a k +1X k b l X l + a k X k (l + 1b l +1X l k =0 = P (XQ(X + P (XQ (X l=0 2. An induction over the order k 1 gives the result. 3. If P (X = a 0 K then P (X = 0 by definition of the polynomial derivative. Conversely, P (X = + n=1 na nx n 1 = 0 implies that a n = 0 for every n 1. The result follows. 4. Let P (X be a non constant polynomial in K[X] and denote by a deg(p its leading coefficient. By definition of the polynomial derivative, we have deg(p deg(p 1. Moreover the coefficient of the term with degree n = deg(p 1 is deg(p a deg(p which is not zero since deg(p 1 (P is non constant and a deg(p 0 (as leading coefficient of P (X. Thus, we have deg(p = deg(p 1. The remain follows by induction over the order k 1. k=0 k =0 Sébastien Godillon

26 26 CHAPTER 2. POLYNOMIALS Corollary 2.15 Let P (X be a polynomial in K[X] of degree d = deg(p. Then k d + 1, P (k (X = 0 Theorem 2.16 (Exact Taylor s formula Let P (X be a polynomial in K[X] of degree d = deg(p. Then P (X = P (0 + P (0X + P (2 (0 2 where 1 = (1 2 3 n 1 n! More generally, for any a K we have P (X = d n=0 X P (d (0 X d = d! P (n (a (X a n n! d n=0 P (n (0 X n n! Proof : Call R(X the following polynomial R(X = P (X d n=0 P (n (a (X a n n! We will prove by induction for every integer k with 0 k d that R (d k (X = 0. In particular, k = d will give the result. At first, using Proposition 2.6, we have: { ( d } P (n (a det(r max deg(p, deg (X a n d n! Then from Corollary 2.15 and Proposition 2.14, we get (R (d (X = R (d+1 (X = 0 that is R (d (X is a constant polynomial. Consequently Proposition 2.14 gives R (d (X = R (d (a = P (d (a d 1 n=0 P (n (a n! n=0 0 P (d (a d! = P (d (a P (d (a = 0 d! So the inductive hypothesis is true for k = 0. Now assume the inductive hypothesis is satisfied for a given integer k with 0 k d 1. Then (R (d k 1 (X = R (d k (X = 0 that is R (d k 1 (X is a constant polynomial. Consequently Proposition 2.14 gives R (d k 1 (X = R (d k 1 (a d k 2 = P (d k 1 (a d n=d k P (n (a n! n=0 P (n (a n! 0 P (d k 1 (a (d k 1! (d k 1! n! (a an (n (d k 1! = P (d k 1 (a 0 P (d k 1 (a 0 = 0 Finally the inductive hypothesis is still true for k + 1. The result follows by induction. Lecture notes

27 2.2. POLYNOMIAL MAPS Root Definition 2.17 (Root α K is said to be a root of a polynomial P (X K[X] if P (α = 0. Example : 1 and 3 are roots of P (X = 3 4X +X 2 since P (1 = = 0 and P (3 = = 0. Actually P (X = (X 1(X 3 in order that (X 1 P (X and (X 3 P (X. Example : 1 and 1 are roots of P (X = 1 2X 2 + X 4 = (X = (X 1 2 (X Proposition 2.18 α K is a root of P (X K[X] if and only if (X α P (X. Proof : Sufficient. If (X α P (X then there exists a polynomial Q(X K[X] such that P (X = (X αq(x and then P (α = (α αq(α = 0. Necessary. From Theorem 2.7, we get two polynomials Q(X and R(X such that { P (X = (X αq(x + R(X deg(r < deg(x α = 1 In particular, R(X is a constant polynomial that is R(X = r K. But α is a root of P (X implies 0 = P (α = (α αq(α + r = r Consequently R(X = 0 and P (X = (X αq(x as needed. Further example : a The polynomial P (X = 1 + X 2 R[X] has no root since x R, P (x = 1 + x 2 1 > 0. In particular, P (X can not be written as a product of two linear polynomials in R[X]. b But ı and ı are roots of Q(X = 1 + X 2 C[X] since Q(X = (X ı(x + ı. Definition 2.19 (Root of higher multiplicity Let k 1 be a positive integer. α K is said to be a root of multiplicity k of a polynomial P (X K[X] if (X α k P (X and (X α k+1 P (X, or equivalently if Q(X K[X]/ P (X = (X α k Q(X and Q(α 0 Furthermore, the multiplicity of a root α K of a polynomial P (X 0 is the following positive integer k α = max { k 1 / (X α k P (X } Proposition 2.20 α K is a root of multiplicity k 1 of the polynomial P (X 0 if and only if P (α = P (α = P (2 (α = = P (k 1 (α = 0 and P (k (α 0 Sébastien Godillon

28 28 CHAPTER 2. POLYNOMIALS Proof : Sufficient. The Taylor s formula (see Theorem 2.16 gives P (X = = with Q(α = P (k (α k! + + n=1 + n=0 k 1 n=0 P (n (α (X α n n! P (n (α n! = 0 + (X α k + = (X α k Q(X (X α n + n=0 + n=k P (n (α (X α n n! P (n+k (α (X α n (n + k! P (n+k (α (n+k! (α α n = P (k (α k! since P (k (α 0 Necessary. If P (X = (X α k Q(X with Q(α 0 then Proposition 2.14 gives P (α = P (α = P (2 (α = = P (k 1 (α = 0 and P (k (α = k!q(α 0 Proposition 2.21 Let P (X 0 be a polynomial in K[X]. Denote by k 1, k 2,..., k n the multiplicities of the roots of P (X. Then k 1 + k k n deg(p In particular P (X has at most deg(p roots. Proof : Denote by α 1, α 2,..., α n the roots of P (X associated to the multiplicities k 1, k 2,..., k n. Then the polynomial (X α 1 k 1 (X α 2 k 2... (X α n kn divides P (X. But from Proposition 2.6 we have: deg ((X α 1 k 1 (X α 2 k 2... (X α n kn = k 1 + k k n Hence, the conclusion follows from Proposition 2.9. To conclude, just state the following important and powerful result without proof. Theorem 2.22 (Fundamental theorem of algebra Every non constant polynomial in C[X] has at least one root. Remark : In particular, the inequality of Proposition 2.21 becomes an equality in C[X]. More precisely, any non constant polynomial P (X C[X] may be written as a product of linear polynomials: where C C is the leading coefficient of P (X α 1, α 2,..., α n are the roots of P (X P (X = C(X α 1 k 1 (X α 2 k 2... (X α n kn k 1, k 2,..., k n are their associated multiplicities Lecture notes

29 Chapter 3 Vector spaces In this chapter, fix a commutative field (K, +, (for instance K = R or K = C. 3.1 The structure of vector space First examples Example of the real plane R 2 : The set of all vectors in the plane consists of the set of all arrows starting at one fixed point in the plane. We may write it as follows R 2 = { v = (x, y / x R and y R} The real numbers x and y are called the coordinates of the vector v. Given two arrows v and w starting at one fixed point in a plane, the parallelogram spanned by these two arrows contains one diagonal arrow which starts at the same fixed point. This new arrow defines the sum of v and w.

30 30 CHAPTER 3. VECTOR SPACES More precisely, the addition of two vectors in R 2 is defined by the following binary operation + : R 2 R 2 R 2 ( v, w v + w = (x + x, y + y where { v = (x, y w = (x, y Notice that this binary operation provides an abelian group structure on R 2. Moreover, any arrow v starting at one fixed point in a plane may be scaled: given any positive real number λ, the scaling of v by λ is the arrow whose diection is the same as v but is dilated or shrunk by multiplying its length by λ. When λ is negative, the scaling of v by λ is defined as the arrow pointing in the opposite direction, instead. More precisely, the multiplication of a vector in R 2 by a scalar is defined as follows. : R R 2 R 2 But notice that. is not a binary operation. Example of the real space R 3 : (λ, v λ. v = (λx, λy where v = (x, y Similarly to the plane, it is the same for the space. Indeed the set of all vectors in the space consists of the set of all arrows starting at one fixed point in the space. R 3 = { v = (x, y, z / x R, y R and z R} Each vector in the space R 3 has three coordinates (instead of two for a vector in the plane R 2. The addition of two vectors in R 3 is defined by the following binary operation + : R 3 R 3 R 3 ( v, w v + w = (x + x, y + y, z + z where { v = (x, y, z w = (x, y, z The multiplication of a vector in R 3 by a scalar is defined as follows. : R R 3 R 3 (λ, v λ. v = (λx, λy, λz where v = (x, y, z We get a similar algebraic structure as for the real plane R 2 : a binary operation + such that (R 3, + is an abelian group and a multiplication by a scalar which is not a binary operation. Lecture notes

31 3.1. THE STRUCTURE OF VECTOR SPACE Definitions Definition 3.1 (External binary operation An external binary operation (or external binary law over K on a nonempty set S is a map from K S to S. Such a binary operation is usually denoted. : K S S (λ, x λ.x Definition 3.2 (Vector space Let (V, +,. be a nonempty set with a binary operation denoted + and an external binary operation over K denoted. and called the scalar multiplication. (V, +,. (or simply V is said to be a vector space over K (or a K-vector space if it satisfies each of the following vector space axioms i (V, + is an abelian group ii the scalar multiplication. is distributive over the binary operation + on V : λ K, (v, w V 2, λ.(v + w = λ.v + λ.w iii the scalar multiplication. is distributive over the binary operation + on K: (λ, µ K 2, v V, (λ + µ.v = λ.v + µ.v iv the scalar multiplication. is compatible with the binary operation on K: (λ, µ K 2, v V, (λ µ.v = λ.(µ.v v 1 K K is the identity element for the scalar multiplication.: v V, 1 K.v = v In this case, the elements of K are called scalars and those of V are called vectors. Furthermore if λ 1, λ 2,... λ n are scalars and v 1, v 2,..., v n are vectors, then the linear combination of those vectors with those scalars is given by λ 1.v 1 + λ 2.v λ n.v n = n λ k.v k V k=1 Examples : The real plane R 2 and the real space R 3 are vector spaces over R. Proof : Notice that the scalar multiplication. on R 2 or R 3 corresponds for every coordinate to the multiplication on R. Consequently, the distributivity of the scalar multiplication. over the addition + on V or R follows from the distributivity of the multiplication over the addition + on R, the compatibility of the scalar multiplication. with the multiplication on R follows from the associativity of the multiplication on R, and 1 R R is the identity element for the scalar multiplication. since it is the identity element for the multiplication on R. Sébastien Godillon

32 32 CHAPTER 3. VECTOR SPACES Proposition 3.3 Let (V, +,. be a vector space over K. Then the following properties hold 1. λ K, λ.0 V = 0 V 2. v V, 0 K.v = 0 V 3. λ K, v V, λ.v = 0 V = either λ = 0 K or v = 0 V 4. λ K, v V, ( λ.v = (λ.v Proof : 1. From the distributivity of the scalar multiplication over + on V we get λ.0 V + λ.0 V = λ.(0 V + 0 V = λ.0 V = 0 V + λ.0 V Now, a simplification on each side by λ.0 V gives λ.0 V = 0 V as needed. 2. From the distributivity of the scalar multiplication over + on K we get 0 K.v + 0 K.v = (0 K + 0 K.v = 0 K.v = 0 V + 0 K.v Now, a simplification on each side by 0 K.v gives 0 K.v = 0 V as needed. 3. Assume λ 0 K. Consequently, there exists a multiplicative inverse λ 1 K and using the compatibility of the scalar multiplication with on K and the first point, we get Finally either λ = 0 K or v = 0 V. v = 1 K.v = (λ 1 λ.v = λ 1.(λ.v = λ 1.0 V = 0 V 4. From the distributivity of the scalar multiplication over + on K and the second point we get λ.v + ( λ.v = (λ + ( λ.v = 0 K.v = 0 V More examples Definition 3.4 (Coordinate space Fix a positive integer n N. The coordinate space of dimension n is the set of all n-tuples of elements of K which is denoted K n = {v = (x 1, x 2,..., x n / x 1 K, x 2 K,..., and x n K} It is a K-vector space for the following operations + : K n K n K n (v, w v + w = (x 1 + x 1, x 2 + x 2,..., x n + x n where { v = (x1, x 2,..., x n w = (x 1, x 2,..., x n. : K K n K n (λ, v λ.v = (λx 1, λx 2,..., λx n where v = (x 1, x 2,..., x n The elements x 1, x 2,..., x n in K are called the coordinates of the vector v = (x 1, x 2,..., x n. Lecture notes

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