
Lecture 1. A binary operation φ on a set A is a function from A × A to A, i.e., φ : A × A → A. φ((x, y)) is also denoted by xφy. Often we use a symbol for φ: +, ·, − ,? . A field F is a set with two operation, called addition and multiplication, which are denoted by + and · (often omitted), respectively, and which satisfy the following properties: 1. Both operations are commutative: a + b = b + a and ab = ba for all a, b ∈ F 2. Both operations are associative:(a + b) + c = a + (b + c) and (ab)c = a(bc) for all a, b, c ∈ F 3. There exists a unique identity element for each operation, denoted by 0 and 1, i.e., 0 + a = a + 0 = a and 1a = a1 = a for all a ∈ F 4. For every a ∈ F, there exists a unique b ∈ F such that a + b = b + a = 0. This element b is called the additive inverse of a in F, and is denoted by −a. × 5. For every a ∈ F := F \{0}, there exists a unique b ∈ F such that ab = ba = 1. This −1 element b is called the multiplicative inverse of a in F, and is denoted by a . 6. Multiplication and addition are related by the distributive laws: a(b + c) = ab + ac and (b + c)a = ba + ca. The axiom system above is not the ‘most economic’ one. Check that it implies that 0a = a0 = 0 for every a ∈ F. Examples of fields. Q – the field of rational numbers; R – the field of real numbers; C – the field of complex numbers; Fp – the finite field of p elements, p is prime. The field Fp is often denoted by Z/pZ or Zp, and is thought as the set of p elements {0, 1, . , p − 1} where addition and multiplication are done by modulo p. Math 672: Lecture Notes. Fall 2005. Felix Lazebnik. 1 We have Q ⊂ R ⊂ C, and the operations in the subfields are just restrictions of the corresponding operations in C. Suppose V is a set whose elements are called vectors and denoted byv ¯. Suppose there exists a binary operation on V , called addition and denoted by +, which is commutative, associative, there exists an identity element, denoted by 0¯ and called zero vector, and for every vectorv ¯, there exists an additive inverse which we denote by −v¯. Suppose F is a field and there exists a function µ from µ : F × V → V given by (k, v¯) 7→ kv¯, which satisfies the following axioms: 1. 1v = v for allv ¯ ∈ V 2. k(¯v +u ¯) = kv¯ + ku¯ for all k ∈ F and allv, ¯ u¯ ∈ V 3. (k + m)¯v = kv¯ + mv¯ for all k ∈ F and allv ¯ ∈ V 4. k(mv¯) = (km)¯v for all k, m ∈ F and allv ¯ ∈ V An ordered triple ((V, +), F, µ) is called a vector space. In a simpler manner, we just say that V is a vector space over the field F. The function µ is mentioned very rarely. We say that µ defines a ‘multiplication’ of elements of F by vectors from V . Often in this context the elements of F are called scalars, and we say that µ defines a ‘multiplication of vectors by scalars’. Note that this multiplication is not a binary operation on a set. Its result is an element of V , i.e., is always a vector. When we write k(mv¯) = (km)¯v, we mean that the scalars k and m are multiplied in F, and the result of this multiplication km, which is a scalar, is ‘multiplied’ by a vectorv ¯. Examples of vector spaces. • For a field F, and positive integer n, let n V = F := {v¯ = (v1, v2, . , vn): vi ∈ F for all i = 1, 2, . , n} n The addition on F and the multiplication by scalars are defined as follows: for everyv, ¯ u¯ ∈ V , and every k ∈ F, v¯ +u ¯ := (v1 + u1, . , vn + un) and kv¯ = k(v1, . , vn) := (kv1, . , kvn). n Then V = F is a vector space over F, which can be (and has to be) checked easily. For F = R, n we obtain the well familiar space R . Math 672: Lecture Notes. Fall 2005. Felix Lazebnik. 2 n n 3 3 If F = Fp, vector space Fp contains p vectors. E.g., F2 contains 2 = 8 vectors. • Let V = C(0, 1) be the set of all continuous real valued functions on (0, 1): f : (0, 1) → R. The addition on C(0, 1) and the scalar multiplication are defined as follows: for any f, g ∈ C(0, 1), and every k ∈ R, (f + g)(x) := f(x) + g(x) and (kf)(x) := kf(x). Again, C(0, 1) is a vector space over R, which has to be checked. Here the fact that + is a binary operation on C(0, 1), or more precisely, that it is ‘closed’, which means f + g ∈ C(0, 1), is not a trivial matter. The same for kf, though it is a little easier. Similarly we can consider 1 C(R) or C (0, 1) – the vector space over R of all real valued differentiable functions from (0, 1) to R with continuous first derivative. • V = R is a vector space over Q. V = C is a vector space over R. V = C is a vector space over Q. In all these examples the addition in V is the usual addition in the field, and the multiplication of vectors by scalars is the usual multiplication of two numbers in V . • V = F[x] – set of all polynomials of x with coefficients from a field F. V is a vector space over F with respect to the usual addition of polynomials and the multiplication of polynomials by numbers (elements of F). • V is the set of all functions f : R → R which satisfy the differential equation y00 − 5y0 + 6y = 0. V is a vector space over R. • V is the set of all sequences of real numbers (xn), n ≥ 0, defined by the recurrences: x0 = a, x1 = b, a, b ∈ R, and for all n ≥ 0, xn+2 = 2xn+1 − 3xn. V is a vector space over R. • V = Mm×n(F) – the set of all m × n matrices with entries from F with respect to usual addition of matrices and the multiplication of matrices by scalars. • Let A ∈ Mm×n, i.e., A is an m × n matrix over R. Then the set V of all solutions of the n homogeneous system of linear equations Ax¯ = 0,¯ i.e., the set of all vectorsx ¯ ∈ R such that Ax¯ = 0,¯ is a vector space over R with respect to usual addition of vectors and the scalar n n multiplication in R . Note that in this example elements of R are thought of as the column vectors ( n × 1 matrices). Proposition 1 Let V be a vector space over a field F. Then Math 672: Lecture Notes. Fall 2005. Felix Lazebnik. 3 (i) 0¯ is unique. (ii) for each a¯ ∈ V , the inverse of a¯ is unique. (iii) 0¯a = 0¯ for every a¯ ∈ V . (iv) (−1)¯a = −a¯ for every a¯ ∈ V . (v) −(−a¯) =a ¯ for every a¯ ∈ V . (vi) Cancellation Law: a¯ + ¯b =a ¯ +c ¯ if and only if ¯b =c ¯. Proof. (i) Indeed, if 0¯0 is a possible another identity element, then 0¯ + 0¯0 = 0¯ as 0¯0 is an identity, and 0¯ + 0¯0 = 0¯0 as 0¯ is an identity. So 0¯ = 0¯0. This justifies the notation 0.¯ (ii) Indeed, let ¯b, ¯b0 be inverses of a. Then consider the element (¯b +a ¯) + ¯b0. Since ¯b +a ¯ = 0,¯ then (¯b +a ¯) + ¯b0 = 0¯ + ¯b0 = ¯b0. Similarly, consider ¯b + (¯a + ¯b0). Sincea ¯ + ¯b0 = 0,¯ then ¯b + (¯a + ¯b0) = ¯b + 0¯ = ¯b. Due to the associativity, (¯b +a ¯) + ¯b0 = ¯b + (¯a + ¯b0). So ¯b0 = ¯b. This justifies the notation −a¯. (iii) (iv) (v) (vi) Math 672: Lecture Notes. Fall 2005. Felix Lazebnik. 4 From now on: We will NOT denote vectors by bars. When we write kv, we will mean that k ∈ F and v ∈ V . If we do not say otherwise , V will denote a vector space over an arbitrary field F. Math 672: Lecture Notes. Fall 2005. Felix Lazebnik. 5 Lecture 2. Let V be a vector space and k1, . , km ∈ F and v1, . , vm ∈ V . Then the vector k1v1 + ... + kmvm is called a linear combination of v1, . , vm. At this time we do not define infinite linear combinations. For a subset A of V the set of all (finite) linear combinations of vectors from A is called the span of A and is denoted by Span(A) or hAi. Clearly, A ⊆ Span(A). P Sometimes it is convenient to use a∈A kaa as a notation for a general linear combination of finitely many elements of A. In this notation we always assume that only finitely many coefficients ka are nonzero A subset W of a vector space V over a field F is called a subspace or a linear subspace of V if it is a vector space over the the operation on V restricted to W and the multiplication of elements of V by elements from F restricted to W . In order to check that a subset W is a subspace in V it is sufficient to check that it is “closed” with respect to the addition of V and with respect to the multiplication by scalars. All other axioms will be inherited from the ones in V . The existence of the zero vector in W and the additive inverses follows from the fact that in V 0a = 0 (= 0),¯ (−1)a = −a and that W is closed with respect to multiplication of vectors by scalars.
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