EC 521 MATHEMATICAL METHODS FOR ECONOMICS Lecture 1: Preliminaries Murat YILMAZ Boğaziçi University In this lecture we provide some basic facts from both Linear Algebra and Real Analysis, which are going to be useful for later lectures. These lecture notes are mostly based on Chapter 1 in Advanced Mathematical Economics by R.V. Vohra. 1 Facts from Linear Algebra Let S = {x 1, x 2,..., x n,...} be a set of vectors and also denote the index set of vectors. Definition 1 A vector y can be expressed as a linear combination of vectors in S = {x 1, x 2,...} if there are real numbers {λ j } j S such that y = j S λ jx j. The set of all vectors that can be expressed as a linear combination of vectors in S is called the span of S and denoted by span(s). Definition 2 A finite set S = {x 1, x 2,..., x n } of vectors is said to be linearly independent (LI) if for all sets of real numbers {λ j } j S, if j S λ jx j = 0 then λ j = 0, j S. S is linearly dependent if it is not LI. That is, there exist real numbers {λ j } j S not all zero such that j S λ jx j = 0. Equivalently, one of the vectors in S can be expressed as a linear combination of the other vectors in S. Example 1 S 1 = {(0, 1, 0), ( 2, 2, 0)} is LI, but S 2 = {(0, 1, 0), ( 2, 2, 0), ( 2, 3, 0)} is LD. Definition 3 The rank of a (not necessarily finite) set S of vectors is the size of the largest subset of linearly independent vectors in S. 1
Example 2 rank(s 2 ) = 2. Definition 4 Let S be a set of vectors and B S be finite and linearly independent. The set B of vectors is said to be a maximal LI set if the set B {x} is linearly dependent for all vectors x S\B. A maximal LI subset of S is called a basis of S. Theorem 1 Every S R n has a basis. If B is a basis for S, then span(s) = span(b). If B and B are two bases of S, then B = B. (The proof uses the Exchange Lemma) Remark 1 If S has a basis B, then rank(s) = B. Definition 5 Let S be a set of vectors. The dimension of span(s) is the rank(s). Example 3 span{(1, 0), (0, 1)} = R 2. So dimension of R 2 is 2. Likewise, we can generalize and see that the dimension of R n is n. Let A be a matrix. We write span(a) to denote the span of the columns of A and span(a T ) the span of the rows of A. Definition 6 The kernel or nullspace of A is the set {x R n : A m n x n 1 = 0 m 1 }. Theorem 2 If A is an m n matrix, then the dimension of the span(a) plus the dimension of the kernel of A is n. That is, dim(span(a)) + dim(ker(a)) = n Or, rank(a) + dim(ker(a)) = n Similarly, rank(a T ) + dim(ker(a T )) = m The column rank of a matrix is the dimension of the span of its columns. Similarly, the row rank of a matrix is the dimension of the span of its rows. Theorem 3 For any A m n, the column rank of A and A T are the same. Thus, column rank of A and row rank of A are the same, called rank(a). rank(a) is simply the dimension of span(a). 2
2 Facts from Real Analysis Definition 7 Given a subset S of real numbers, the supremum of S, sup(s), is the smallest number that is larger than every number in S. The infimum of S, inf(s), is the largest number that is smaller than every number in S. Example 4 Let S = {x R : 0 < x < 1}. Then sup(s) = 1 and inf(s) = 0 Let x, y R n. Let d(x, y) denote the Euclidean distance between x and y, that is, d(x, y) = n j=1 (x j y j ) 2. And x denotes d(x, 0). The dot product of two vectors x and y is denoted by x y = n j=1 x jy j = d(x, 0)d(y, 0) cos θ where θ is the angle between vectors x and y. Note that d(x, 0) 2 = x x. And, x and y are orthogonal if x y = 0. Real Sequences: A sequence (x n ) R is convergent if there is a real number x such that the later terms of the sequence get arbitrarily close to x. Formally; Definition 8 (x n ) converges to x if for any given ɛ > 0 there exists a N N such that x n x < ɛ for all n N. Then we say (x n ) is convergent and x is the limit of (x n ) and we write lim n x n = x, or x n x. ***INSERT FIGURE 1 HERE*** All but finitely many terms of (x n ) are contained in the open interval (x ɛ, x + ɛ)!!! If (x n ) does not converge to a real number, then it is called divergent. If, for every number y, there exists an M N with x n y for each n M, then (x n ) diverges to. That is, lim x n =, or x n. If x n diverges to then we say x n diverges to. Divergent does not mean divergent to (or )!!! For instance, ***INSERT FIGURE 2 HERE*** The whole idea is about the tail of the sequence and how it behaves. Note also that N in the definition may depend on ɛ > 0. Example 5 lim 1 n = 0. 3
To show this, pick an arbitrary ɛ > 0 and ask if there is an N N large enough to guarantee that 1 n 0 < ɛ for all n N. Here the choice of N is clear. Just pick N something strictly larger than 1 ɛ. Example 6 limx n = 0 where x n = n if n 100 1 n 100 if n > 100 Here we need to pick N large enough, larger than the one in Example 5. In fact, we need to pick N > (1/ɛ) + 100. Example 7 Let (x n ), (y n ) and (z n ) be real sequences such that x n y n z n for each n. Let s show that if limx n = limz n = a, then y n a. Proof. For an arbitrary ɛ > 0 we have N x, N z such that x n a < ɛ 3, n N x and z n a < ɛ 3, n N z. Now, note that y n a = y n x n + x n a y n x n + x n a z n x n + x n a = z n a + a x n + x n a z n a + x n a + x n a < ɛ 3 + ɛ 3 + ɛ 3 = ɛ where the first inequality follows from the triangular inequality. Picking, N y = max{n x, N z } we get y n a < ɛ for all n > N y Definition 9 (x n ) is bounded from above if there exists s K R with x n K for all n N. (x n ) is bounded from below if there exists s K R with x n K for all n N. Finally, (x n ) is bounded if it is bounded from both above and below. Proposition 1 Every convergent real sequence is bounded. 4
Proof. Take any (x n ) R N with x n x for some real number x. Pick ɛ = 1. Then there exist a N N such that x n x < 1 n N, and hence (from triangle inequality) x n < x + 1 for n N. But then x n max{ x + 1, x 1, x 2,..., x N } for all n N. Note that converse does not hold. Check ( 1) n. Definition 10 (x n ) is increasing if x n x n+1 for all n N. (x n ) is strictly increasing if x n < x n+1 for all n N. (x n ) is decreasing if x n x n+1 for all n N. (x n ) is strictly decreasing if x n > x n+1 for all n N. (x n ) is monotonic if it s either increasing or decreasing. We write x n x when x n is increasing and converges to x. We write x n x when x n is decreasing and converges to x. Proposition 2 Every increasing real sequence that is bounded from above converges. Every decreasing real sequence that is bounded from below converges. Proof. Let (x n ) be an increasing real sequence that is bounded from above. Let S := {x 1, x 2,...}. Since for every non-empty subset S of R that is bounded from above there is a supremum in R, we can write x := sups, x R. (Here supremum is the least upper bound.) We claim that x n x. To show this, pick an arbitrary ɛ > 0. Since x is the l.u.b. (least upper bound) of S, x ɛ can not be an upper bound of S, so x N > x ɛ for some N N. Since x n is increasing, we must have x + ɛ > x x n x N > x ɛ for all n N. So, x n x < ɛ for all n N. Proposition 3 Every real sequence has a monotonic subsequence. Proof. See Efe Ok s text, page 52 or check Thurston. Theorem 4 (Bolzano - Weierstrass Theorem) Every bounded real sequence has a convergent subsequence. Proof. Follows directly from Proposition 2 and Proposition 3 above. 5
Open and Closed Sets: Definition 11 A set S R n is said to be closed if it contains all of its limit points. S is open if S C is closed. Example 8 [0, 1] is closed. To see this, let {x n } n 1 [0, 1] be a convergent subsequence with lim x n = x 0. Let x 0 / [0, 1]. Say, x 0 > 1. Pick ɛ = x 0 1 2 > 0. Since lim x n = x 0, ɛ > 0 N such that x n x 0 ɛ n > N. For our choice of ɛ, this implies x n > 1 n > N, contradicting {x n } [0, 1]. Definition 12 A set S R is said to be closed if it contains all of its accumulation points. S is open if S C is closed. x R is an accumulation point of S R if [(x ɛ, x + ɛ)\{x}] S, ɛ > 0. Example 9 [0, 1] is closed, since any x [0, 1] is an accumulation point of [0, 1]. [a, b] is closed. [a, b) is neither closed nor open. {a,b,c,d} is closed. (a,b) is open. is closed and open (clopen). R is closed and open (clopen). Remark 2 Not closed open! Not open closed! Using the definition of closed and open sets with accumulation points, we can prove the Lemma below. Lemma 1 S R is open iff x S δ > 0 such that (x δ, x + δ) S. Proof. ( ): Let S R be an open subset. Suppose the conclusion is false. So x 0 S such that δ > 0 (x 0 δ, x 0 + δ) S. In particular, for δ = 1 a 1 / S but a 1 (x 0 1, x 0 + 1) with a 1 x 0. For δ = min{ 1 2, x 0 a 1 }, a 2 / S but a 2 (x 0 δ, x 0 +δ). Inductively construct (a n ) such that a n / S, x 0 a n 1 n < ɛ. So ɛ > 0 choosing n > 1 ɛ we find some a n such that a n (R\S) ((x 0 ɛ, x 0 + ɛ)\{x 0 }). So, x 0 is an accumulation point of R\S by definition. 6
Since R\S is closed, we must have x 0 R\S which contradicts x 0 S. ( ): Let x 0 be any accumulation point of R\S. Suppose x 0 S. Then δ > 0 such that (x 0 δ, x 0 + δ) S. Since x 0 is an accumulation point of R\S, we have, for this δ, [(x 0 δ, x 0 + δ)\{x 0 }] (R\S). So y S (R\S) since (x 0 δ, x 0 + δ) S. But S (R\S) = which is a contradiction. Thus, x 0 R\S, hence R\S is closed, hence S is open. We will later generalize distance concept and define metric spaces. Just as a sneak peek, look at the following definition of an open set: Definition 13 A set S R n is called open if x S, ɛ > 0 such that {y R n : d(x, y) < ɛ} S. And S is closed if S C is open. We also define interior, boundary and closure of a set: Definition 14 A point x S is an interior point of S if {y R n : d(x, y) < ɛ} S for all sufficiently small ɛ > 0. A point x S is a boundary point of S if {y R n : d(x, y) < ɛ} S and {y R n : d(x, y) < ɛ} S C for all sufficiently small ɛ > 0. Interior of S, denoted Int(S), is the set of all interior points of S. Boundary of S, denoted Bd(S), is the set of all boundary points of S. Definition 15 The closure of a set S is the set S combined with all points that are the limits of sequences of points in S. Definition 16 A set S R n is bounded if a finite positive number r such that x r x S. S is compact if it is closed and bounded. Remark 3 In R n compactness is equivalent to being closed and bounded. However, in general that s not the case: A compact set is always closed and bounded, but a set may be closed and bounded but not compact. We will talk about this in detail later. Continuity: Definition 17 A real valued function f : R n R is continuous at x if ɛ > 0, δ > 0 such that y R n with x y < δ, we have f(x) f(y) < ɛ. And f is continuous if it is continuous at all x R n. 7
Theorem 5 (Weierstrass Maximum Theorem) Let S R n be compact and f be a continuous real valued function on S. Then argmax x S f(x) and argmin x S f(x) exist and are both in S. Proof. (This proof is for later!) Suppose the set T = {y R : x S such that y = f(x)} is compact. Then sup(t ) = sup x S f(x) and inf(t ) = inf x S f(x). Since T is closed, it follows that sup(t ) and inf(t ) are contained in T. So x S such that f(x ) = sup x S f(x) and x S such that f(x ) = inf x S f(x). To prove that T is compact, let {K i } i M be a collection of open sets that cover T. Since f is continuous, f 1 (K i ) = {x S : f(x) K i } are open. (Why?) Then the collection {f 1 (K i ) i M } forms an open cover for S. Since S is compact, by Heine Borel Theorem, there exists a finite subcover, say indexed by M. If T i M K i, we are done. If y T and x such that f(x ) = y, then j M such that x f 1 (K j ). This implies y K j, i.e. y j M K j. 8