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Econ 509, Introduction to Mathematical Economics IProfessor Ariell ReshefUniversity of VirginiaLecture notes based mostly on Chiang and Wainwright, Fundamental Methods of MathematicalEconomics.1Mathematical economicsWhy describe the world with mathematical models, rather than use verbal theory and logic? Afterall, this was the state of economics until not too long ago (say, 1950s).1. Math is a concise, parsimonious language, so we can describe a lot using fewer words.2. Math contains many tools and theorems that help making general statements.3. Math forces us to explicitly state all assumptions, and help preventing us from failing toacknowledge implicit assumptions.4. Multi dimensionality is easily described.Math has become a common language for most economists. It facilitates communication betweeneconomists. Warning: despite its usefulness, if math is the only language for economists, thenwe are restricting not only communication among us, but more importantly we are restricting ourunderstanding of the world.Mathematical models make strong assumptions and use theorems to deliver insightful conclusions. But, remember the A-A’C-C’Theorem:Let C be the set of conclusions that follow from the set of assumptions A. Let A’ be a smallperturbation of A. There exists such A’ that delivers a set of conclusions C’ that is disjointfrom C.Thus, the insightfullness of C depends critically on the plausibility of A.The plausibility of A depends on empirical validity, which needs to be established, usuallyusing econometrics. On the other hand, sometimes theory informs us on how to look at existingdata, how to collect new data, and which tools to use in its analysis. Thus, there is a constantdiscourse between theory and empirics. Neither can be without the other (see the inductivism vs.deductivism debate).Theory is an abstraction of the world. You focus on the most important relationships thatyou consider important a priori to understanding some phenomenon. This may yield an economicmodel.1

2Economic modelsSome useful notation: 8 for all, 9 exists, 9! exists and is unique. If we cross any of these, or pre xby : or2.1, then it means "not": e.g., @, :9 and9 all mean "does not exist".Ingredients of mathematical models1. Equations:De nitions/Identities : RC: Y C I G X: Kt 1 (1M) Kt It: Mv PYBehavioral/Optimization : q d p: MC MR: MC PEquilibrium : q d q s2. Parameters: e.g., ,from above.3. Variables: exogenous, endogenous.Parameters and functions govern relationships between variables. Thus, any complete mathematicalmodel can be written asF ( ; Y; X) 0 ;where F is a set of functions (e.g., demand, supply and market clearing conditions),is a setof parameters (e.g., elasticities), Y are endogenous variables (e.g., price and quantity) and X areexogenous, predetermined variables (e.g., income, weather). Some models will not have explicitX variables. Moving from a "partial equilibrium" model closer to a "general equilibrium" modelinvolves treating more and more exogenous variables as endogenous.Models typically have the following ingredients: a sense of time, model population (who makesdecisions), technology and preferences.2.2From chapter 3: equilibrium analysisOne general de nition of a model’s equilibrium is "a constellation of selected, interrelated variables so adjusted to one another that no inherent tendency to change prevails in the model2

which they constitute".Selected: there may be other variables. This implies a choice of what is endogenous andwhat is exogenous, but also the overall set of variables that are explicitly considered in themodel. Changing the set of variables that is discussed, and the partition to exogenous andendogenous will likely change the equilibrium.Interrelated: The value of each variable must be consistent with the value of all othervariables. Only the relationships within the model determine the equilibrium.No inherent tendency to change: all variables must be simultaneously in a state of rest,given the exogenous variables and parameters are all xed.Since all variables are at rest, an equilibrium is often called a static. Comparing equilibria is calledtherefore comparative statics (there is di erent terminology for dynamic models).An equilibrium can be de ned as Y that solvesF ( ; Y; X) 0 ;for givenand X. This is one example for the usefulness of mathematics for economists: see howmuch is described by so little notation.We are interested in nding an equilibrium for F ( ; Y; X) 0. Sometimes, there will be nosolution. Sometimes it will be unique and sometimes there will be multiple equilibria. Each ofthese situations is interesting in some context. In most cases, especially when policy is involved,we want a model to have a unique equilibrium, because it implies a function from ( ; X) to Y(the implicit function theorem). But this does not necessarily mean that reality follows a uniqueequilibrium; that is only a feature of a model. Warning: models with a unique equilibrium areuseful for many theoretical purposes, but it takes a leap of faith to go from model to reality— as ifthe unique equilibrium pertains to reality.Students should familiarize themselves with the rest of chapter 3 on their own.2.3NumbersNatural, N: 0; 1; 2::: or sometimes 1; 2; 3; :::Integers, Z: :::2; 1; 0; 1; 2; :::Rational, Q: n d where both n and d are integers and d is not zero. n is the numerator andd is the denominator.Irrational numbers: cannot be written as rational numbers, e.g., , e,3p2.

Real, R: rational and irrational. The real line: ( 1; 1). This is a special set, because it isdense. There are just as many real numbers between 0 and 1 (or any other two real numbers)as on the entire real line.Complex: an extension of the real numbers, where there is an additional dimension in whichpwe add to the real numbers imaginary numbers: x iy, where i 1.2.4SetsWe already described some sets above (N, Q, R, Z). A set S contains elements e:S fe1 ; e2 ; e3 ; e4 g ;where ei may be numbers or objects (say: car, bus, bike, etc.). We can think of sets in terms ofthe number of elements that they contain:Finite: S fe1 ; e2 ; e3 ; e4 g.Countable: there is a mapping between the set and N. Trivially, a nite set is countable.In nite and countable: Q. Despite containing in nitely many elements, they are countable.Uncountable: R, [0; 1].Membership and relationships between sets:e 2 S means that the element e is a member of set S.Subset: S1S2 : 8e 2 S1 ; e 2 S2 . Sometimes denoted as S1S2 . Sometimes a strict subsetis de ned as 8e 2 S1 ; e 2 S2 and 9e 2 S2 ; e 2 S1 .Equal: S1 S2 : 8e 2 S1 ; e 2 S2 and 8e 2 S2 ; e 2 S1 .The null set, ?, is a subset of any set, including itself, because it does not contain any elementthat is not in any subset (it is empty).Cardinality: there are 2n subsets of any set of magnitude n jSj.Disjoint sets: S1 and S2 are disjoint if they do not share common elements, i.e. if @e suchthat e 2 S1 and e 2 S2 .Operations on sets:Union (or): A [ B feje 2 A or e 2 Bg.4

Intersection (and): A \ B feje 2 A and e 2 Bg.Complement: de neMinus: for Bas the universe set. Then A or Ac feje 2and e 2 Ag.A, AnB feje 2 A and e 2 Bg. E.g., A nA.Rules:Commutative:A[B B[AA\B B\AAssociation:(A [ B) [ C A [ (B [ C)(A \ B) \ C A \ (B \ C)Distributive:A [ (B \ C) (A [ B) \ (A [ C)A \ (B [ C) (A \ B) [ (A \ C)Do Venn diagrams.2.5Relations and functionsOrdered pairs: whereas fx; yg fy; xg because they are sets, but not ordered, (x; y) 6 (y; x)unless x y (think of the two dimensional plane R2 ). Similarly, one can de ne ordered triplets,quadruples, etc.Let X and Y be two sets. The Cartesian product of X and Y is a set S that is given byS XY f(x; y) jx 2 X; y 2 Y g :For example, Rn is a Cartesian productRn RR:::R f(x1 ; x2 ; :::xn ) jxi 2 Rg :Cartesian products are relations between sets:8x 2 X; 9y 2 Y such that (x; y) 2 X5Y ;

so that the set Y is related to the set X. Any subset of a Cartesian product also has this trait.Note that each x 2 X may have more than one y 2 Y related to it (and vice versa). Thus therelation may assign to any x 2 X a set of values in Y , Sx 2 Y . (Analysis of the shape of these setsin the context of relations will be useful when discussing dynamic programming.)If8x 2 X; 9!y 2 Y such that (x; y) 2 SXY ;then y is a function of x. We write this in shorthand notation asy f (x)orf :X!Y :The second term is also called mapping, or transformation. Note that although for y to be afunction of x we must have 8x 2 X; 9!y 2 Y , it is not necessarily true that 8y 2 Y; 9!x 2 X. Infact, there need not exist any such x at all. For example, y a x2 , a 0.In y f (x), y is the value or dependent variable; x is the argument or independentvariable. The set of all permissible values of x is called domain. For y f (x), y is the imageof x. The set of all possible images is called the range, which is a subset of Y .2.6Functional formsStudents should familiarize themselves with polynomials, exponents, logarithms, "rectangular hyperbolic" functions (unit elasticity), etc. See Chapter 2.5 in CW.2.7Functions of more than one variablez f (x; y) means that8 (x; y) 2 domainXY; 9!z 2 Z such that (x; y; z) 2 SXYZ :This is a function from a plane in R2 to R or a subset of it. y f (x1 ; x2 ; :::xn ) is a function fromthe Rn hyperplane or hypersurface to R or a subset of it.3Equilibrium analysisStudents cover independently. Conceptual points are reported above in Section 2.2.6

44.1Matrix algebraDe nitionsMatrix:Amn266 64a11a21.a12a22:::3a1na2n.777 [aij ] i 1; 2; :::m; j 1; 2; :::n :5am1 am2 : : : amnNotation: usually matrices are denoted in upper case; m and n are called the dimensions.Vector:xm1266 64x1x2.xm3777 :5Notation: usually lowercase. Sometimes called a column vector. A row vector isx0 4.2x1 x2xm:Matrix operationsEquality: A B i aij bij 8ij. Clearly, the dimensions of A and B must be equal.Addition/subtraction: AB C i aijbij cij 8ij.Scalar multiplication: B cA i bij c aij 8ij.Matrix multiplication: Let Am– if n k then the product Ammand Bklbe matrices.n Bn lexists and is equal to a matrix Cmlof dimensionsm Am nexists and is equal to a matrix Cknof dimensionsl.– if m l then the product Bkknn.7

– If product exists, thenAmn Bn l2666 664 Transpose: Let Amn"!a11a21.a12a22:::a1na2n.32b11 b12 : : : b1l77 6 b21 b22b2l767 6# .74.5bn1 bn2 : : : bnlam1 am2 : : : amn#nXcij aik bkji 1; 2; :::m; j 1; 2; :::l :37775k 1 [aij ]. Then A0nm [aji ]. Also denoted AT . Properties:– (A0 )0 A– (A B)0 A0 B 0– (AB)0 B 0 A0Operation rules– Commutative addition: A B B A.– Distributive addition: (A B) C A (B C).– NON commutative multiplication: AB 6 BA, even if both exist.– Distributive multiplication: (AB) C A (BC).– Association: premultiplying A (B C) AB AC and postmultiplying (A B) C AC BC.4.3Special matricesIdentity matrix:266I 6431 0 ::: 00 10 77:. . . 7. . 5.0 0 ::: 1AI IA A (dimensions must conform).Zero matrix: all elements are zero. 0 A A, 0A A0 0 (dimensions must conform).Idempotent matrix: AA A. Ak A ; k 1; 2; ::.Example: the linear regression model is yn1 Xn k k 1 "n 1 .0(X X) 1 X 0 y. Therefore weThe estimated modelby OLS is y Xb e, where b have predicted values8

1yhb Xb X (X 0 X)X 0 y and residuals e y yb y Xb y X (X 0 X) 1 X 0 y iI X (X 0 X) 1 X 0 y. We can de ne the projection matrix as P X (X 0 X) 1 X 0 and theresidual generating matrix as R [IP ]. Both P and R are idempotent. What does itmean that P is idempotent? And that R is idempotent? What is the product P R, and whatdoes that imply?Singular matrices: even if AB 0, this does NOT imply that A 0 or B 0. E.g.,A 2 41 2; B 2142:Likewise, CD CE does NOT imply D E. E.g.,C 2 36 91 11 2; D ; E 2 13 2:This is because A, B and C are singular: there is one (or more) row or column that is a linearcombination of the other rows or columns, respectively. (More on this to come).Nonsingular matrix: a square matrix that has an inverse. (More on this to come).Diagonal matrix266D 64d11 0 : : : 00 d220.00 : : : dnn3777 :5Upper triangular matrix. Matrix U is upper triangular if uij 0 for all i j, i.e. allelements below the diagonal are zero. E.g.,32a b c4 0 e f 5 :0 0 iLower triangular matrix. Matrix W is lower triangular if wij 0 for all i j, i.e. allelements above the diagonal are zero. E.g.,23a 0 04 d e 0 5 :g h iSymmetric matrix: A A0 .Permutation matrix: a matrix of 0s and 1s in which each row and each column contains9

exactly one 1. E.g.,230 0 14 1 0 0 5 :0 1 0Multiplying a conformable matrix by a permutation matrix changes the order of the rows orthe columns (unless it is the identity matrix). For example,0 11 01 23 43 41 2 and1 23 40 11 02 14 3 :Partitioned matrix: a matrix with elements that are matrices themselves, e.g.,AgDkhhBgEkihCgFkjh:(g k) (h i j)Note that the dimensions of the sub matrices must conform.4.4Vector productsScalar multiplication: Let xm1be a vector. The scalar product cx is23cx16 cx2 767cxm 1 6 . 7 :4 . 5cxmInner product: Let xm1and ym1be vectors. The inner product is a scalarx0 y mXx i yi :i 1This is useful for computing correlations.Outer product: Let xmand yn 1 be vectors. The outer product is a matrix23x1 y1 x1 y2 : : : x1 yn6 x2 y1 x2 y2x 2 yn 767:xy 0 6 . 74 . 51xm y1 xm y2 : : : xm ynThis is useful for computing the variance/covariance matrix.10m n

Geometric interpretations: do in 2 dimensions. All extends to n dimensions.– Scalar multiplication.– Vector addition.– Vector subtraction.– Inner product and orthogonality (xy 0 means x?y).4.5Linear independenceDe nition 1: a set of k vectors x1 ; x2 ; :::xk are linearly independent i neither one can be expressedas a linear combination of all or some of the others. Otherwise, they are linearly dependent.De nition 2: a set of k vectors x1 ; x2 ; :::xk are linearly independent i :9 a set of scalars c1 ; c2 ; :::ckPsuch that ci 6 0 for some or all i and ki 1 ci xi 0. Otherwise, they are linearly dependent. I.e.,if such set of scalars exists, then the vectors are linearly dependent.Consider R2 :All vectors that are multiples are linearly dependent. If two vectors cannot be expressed asmultiples then they are linearly independent.If two vectors are linearly independent, then any third vector can be expressed as a linearcombination of the two.It follows that any set of k 2 vectors in R2 must be linearly dependent.4.6Vector spaces and metric spacesThe complete set of vectors of n dimensions is a space, a vector space. If all elements of thesevectors are real numbers (2 R), then this space is Rn .Any set of n linearly independent vectors is a base for Rn .A base spans the space to which it pertains. This means that any vector in Rn can beexpressed as a linear combination of the base (it is spanned by the base).Bases are not unique.Bases are minimal: they contain the smallest number of vectors that span the space.Example: unit vectors. Consider the vector space R3 . Then2 32 32 310045454e1 0; e2 1; e3 0 500111

is a base. Indeed, e1 ; e3 ; e3 are linearly independent.Distance metric: Let x; y 2 S, some set. De ne the distance between x and y by a functiond: d d (x; y), which has the following properties:– d (x; y)0.– d (x; y) d (y; x).– d (x; y) 0 , x y.– d (x; y) 0 , x 6 y.– d (x; y)d (x; z) d (z; y) 8x; y; z (triangle inequality).A metric space is given by a vector space distance metric. The Euclidean space is givenby Rn the following distance functionvu nuX(xid (x; y) tqyi ) (x2i 1y)0 (xy) :Other distance metrics give rise to di erent metric spaces.4.7Inverse matrixDe nition: if for some square (nn) matrix A there exists a matrix B such that AB I, then Bis the inverse of A, and is denoted A1,1i.e. AA I.Properties:1Not all square matrices have an inverse. If Adoes not exist, then A is singular. Otherwise,A is nonsingular.A is the inverse of A1and vice versa.The inverse is square.The inverse, if it exists, is unique. Proof: suppose not, i.e. AB I and B 6 AA1 AB A1 I,IB B A1,1.Thena contradictionOperation rules:A1I )1 A.AA1a contradictionProof: suppose not, i.e.1 I1)A11AA11211 B and B 6 A. I ) BA1AThen AA1 IA ) BI B A,

(AB)1(AB)1C B B1A 1,1but only if both Band A(AB) I C (AB) CAB ) CABB1exist. Proof: Let (AB)1 CA IB1 B11A 1Note that in the linear regression model above P X (X 0 X)we CANNOT write (X 0 X)1 X1X 0 1.11 C. Then) CAA1 X 0 , but unless X is square(If we could, then P I but then there are nodegrees of freedom: The model ts exactly the data, but the data are not very informative,because they are only one sample drawn from the population).(A0 )1 AI0 I ) A1 0.Proof: Let (A0 )1 AB 0 A1I1 B. Then (A0 )) B0 A11) B AA0 I BA0 ) (BA0 )0 AB 0 1 0Conditions for nonsingularity:Necessary condition: matrix is square.Given square matrix, a su cient condition is that the rows or columns are linearly independent. It does not matter whether we use the row or column criterion because matrix issquare.A is square linear independence , A is nonsingular , 9A{z} 1necessary and su cient conditionsHow do we nd the inverse matrix? Soon. Why do we care? See next section.4.8Solving systems of linear equationsWe seek a solution x to the system Ax cAnn xn 1 cn1) x cA1;where A is a nonsingular matrix and c is a vector. Each row of A gives coe cients to the elementsof x:row 1 :row 2 :nXi 1nXa1i xi c1a2i xi c2i 1Many linear (or linearized) models can be solved this way. We will learn clever ways to computethe solution to this system. We care about singularity of A because (given c) it tells us somethingabout the solution x.13

4.9Markov chainsWe introduce this through an example. Let x denote a vector of employment and unemploymentrates: x0 e u , where e u 1 and e; u0. De ne the matrix P as a transition matrix thatgives the conditional probabilities for transition from the state today to a state next period,P pee peupue puu;where pij Pr (state j tomorrowjstate i today). Each row of P sums to unity: pee peu 1 andpue puu 1; and since these are probabilities, pijx0t et ut .0 8ij. Now add a time dimension to x:We ask: What is the employment and unemployment rates going to be in t 1 given xt ? Answer:x0t 1 x0t P et utpee peupue puu et pee ut pue et peu ut puu:What will they be in t 2? Answer: x0t 2 x0t 1 P x0t P 2 . More generally, x0t0 k x0t0 P k .A transition matrix, sometimes called stochastic matrix, is de ned as a square matrixwhose elements are non negative and all rows sum to 1. This gives you conditional transitionprobabilities starting from each state, where each row is a starting state and each column is thestate in the next period.Steady state: a situation in which the distribution over the states is not changing over time.How do we nd such a state, if it exists?Method 1: Start with some initial condition x0 and iterate forward x0k x00 P k , taking k ! 1.Method 2: de ne x as the steady state value. Solve x0 x0 P . Or P 0 x x.14

5Matrix algebra continued and linear models5.1RankDe nition: The number of linearly independent rows (or, equivalently, columns) of a matrix A isthe rank of A: r rank (A).If Amnthen rank (A)If a square matrix Anmin fm; ng.nhas rank n, then we say that A is full rank.Multiplying a matrix A by a another matrix B that is full rank does not reduce the rank ofthe product relative to the rank of A.If rank (A) rA and rank (B) rB , then rank (AB) min frA ; rB g.Finding the rank: the echelon matrix method. First de ne elementary operations:1. Multiply a row by a non zero scalar: c Ri , c 6 0.2. Adding c times of one row to another: Ri cRj .3. Interchanging rows: Ri Rj .All these operations alter the matrix, but do not change its rank (in fact, they can all beexpressed by multiplying matrices, which are all full rank).De ne: echelon matrix.1. Zero rows appear at the bottom.2. For non zero rows, the rst element on the left is 1.3. The rst element of each row on the left (which is 1) appears to the left of the row directlybelow it.The number of non zero rows in the echelon matrix is the rank.We use the elementary operations in order to change the subject matrix into an echelon matrix,which has as many zeros as possible. A good way to start the process is to concentrate zeros at thebottom. Example:20A 4 24116132442 5 R1 R3 : 4 200151611231012 5R1 : 4 240414611302 54

R223201 142 5 R3 2R2 : 4 0 5 1240 0145 2112R1 : 4 0011There is a row of zeros: rank (A) 2. So A is singular.5.232301 41022 5R2 : 4 0 1 4 11 51100 00Determinants and nonsingularityDenote the determinant of a square matrix as jAnis zero then the matrix is singular.1. jA11j a11 .2. jA22j a11 a22n j.This is not absolute value. If the determinanta12 a21 .3. Determinants for higher order matrices. Let Akis the determinant of the matrix given2a bA 4 d eg hkbe a square matrix. The i-j minor jMij jby erasing row i and column j from A. Example:3ce ff 5 ; jM11 j :h iiThe Laplace Expansion of row i gives the determinant of A:jAkkj kXi j( 1)j 1aij jMij j kXaij Cij ;j 1where Cij ( 1)i j jMij j is called the cofactor of aij (or the i-j th cofactor). Example:expansion by row 1a b cd e fg h i aC11 bC12 cC13 a jM11 j b jM12 j c jM13 jd ed fe f cb ag hg ih i a (ei f h) b (di f g) c (dh eg) :In doing this, it is useful to choose the expansion with the row that has the most zeros.Properties of determinants1. jA0 j jAj2. Interchanging rows or columns ‡ips the sign of the determinant.16

3. Multiplying a row or column by a scalar c multiplies the determinant by c.4. Ri cRj does not change the determinant.5. If a row or a column are multiples of another row or column, respectively, then the determinantis zero: linear dependence.6. Changing the minors in the Laplace expansion by alien minors, i.e. using jMnj j instead ofjMij j for row i 6 n, will give zero:kXj 1aij ( 1)i j jMnj j 0 ; i 6 n :This is like forcing linear dependency by repeating elements.Pkj 1 aij( 1)i j jMnj j is thedeterminant of some matrix. That matrix can be reverse engineered from the last expression.If you do this, you will nd that that reverse-engineered matrix has linear dependent columns(try a 33 example).Determinants and singularity: jAj 6 0, A is nonsingular, columns and rows are linearly independent, 9A1, for Ax c ; 9!x A1c, the column (or row) vectors of A span the vector space.5.3Finding the inverse matrixLet A be a nonsingular matrix,AnThe cofactor matrix of A is CA :n266 64266CA 64a11a21.a12a22: : : a1na2n.an1 an2 : : : annC11C21.C12C22: : : C1nC2n.Cn1 Cn2 : : : Cnn173777 :53777 ;5

0 :where Cij ( 1)i j jMij j. The adjoint matrix of A is adjA CA32C11 C21 : : : Cn16 C12 C22Cn2 7760adjA CA 6 . 7 :4 . 5C1n C2n : : : Cnn0 :Consider ACA0ACAPnPn2 Pna1j C1ja1j C2j : : :a1j Cnjj 1j 1Pj 1PnPnn6j 1 a2j Cnjj 1 a2j C2jj 1 a2j C1j6 6.4Pn .PnPn .j 1 anj C1jj 1 anj C2j : : :j 1 anj Cnj2 Pn0:::0j 1 a1j C1j Pn600j 1 a2j C2j6 6.4.Pn .00:::j 1 anj Cnj32jAj 0 : : : 06 0 jAj0 776 6 . 7 jAj I ;4 . 5003777537775: : : jAjwhere the o diagonal elements are zero due to alien cofactors. It follows that0ACA jAj I10 IACAjAjadjA0 1 :A 1 CAjAjjAjExample:A 1 23 4; CA 4231430; CA 21And you can verify this.5.4Cramer’s ruleFor the system Ax c and nonsingular A, we havex A1c 18adjAc:jAj; jAj 2; A1 232112:

Denote by Aj the matrix A with column j replaced by c. Then it turns out thatxj jAj j:jAj0 c is c times row of C 0 , i.e. each row r isTo see why, note that each row of CAAPjCjr cj , which isa Laplace Expansion by row r of some matrix. That matrix is Aj and the Laplace expansion givesthe determinant of Aj .5.5Homogenous equations: Ax 0Let the system of equations be homogenous: Ax 0.If A is nonsingular, then only x 0 is a solution. Recall: if A is nonsingular, then its columnsPare linearly independent. Denote the columns of A by Ai . Then Ax ni 1 xi Ai 0 impliesxi 0 8i by linear independence of the columns.If A is singular, then there are in nite solutions, including x 0.5.6Summary of linear equations: Ax cFor nonsingular A:1. c 6 0 ) 9!x 6 02. c 0 ) 9!x 0For singular A:1. c 6 0 ) 9x, in nite solutions 6 0.If there is inconsistency— linear dependency in A, the elements of c do not follow thesame linear combination— there is no solution.2. c 0 ) 9x, in nite solutions, including 0.One can think of the system Ax c as de ning a relation between c and x. If A is nonsingular,then there is a function (mapping/transformation) between c and x. In fact, when A is nonsingular,this transformation is invertible.19

5.7Inverse of partitioned matrix (not covered in CW)Let A be a partitioned matrix such thatA A11 A12A21 A22;Su cient conditions for nonsingularity of A are that A11 and A22 are square, nonsingular matrices. In that caseAwhere B11 A11B11B11 A12 A22111A22 A21 B11 A22 A221 A21 B11 A12 A221 A12 A221 A21Awhere B22 A2211 11(1);(2), or alternativelyA111 A111 A12 B22 A21 A111B22 A21 A111A21 A111 A12;A111 A12 B22B22. (This is useful for econometrics.)To prove the above start with AB I and gure out what the partitions of B need to be. Toget (1) you must assume (and use) A22 nonsingular; and to get (2) you must assume (and use) A11nonsingular.Note that A11 and A22 being nonsingular are not necessary conditions in general. For example,A 0 11 0is nonsingular but does not meet the su cient conditions. However if A is positive de nite (we willde ne this below; a bordered Hessian is not positive de nite), then A11 and A22 being nonsingularis also a necessary condition.5.8Leontief input/output modelWe are interested in computing the level of output that is required from each industry in aneconomy that is required to satisfy nal demand. This is not a trivial question, because output of allindustries (depending on how narrowly you de ne an industry) are inputs for other industries, whilealso being consumed in nal demand. These inter-industry relationships constitute input/outputlinkages.Assume1. Each industry produces one homogenous good.2. Inputs are used in xed proportions.20

3. Constant returns to scale.This gives rise to the Leontief ( xed proportions) production function. The second assumptioncan be relaxed, depending on the interpretation of the model. If you only want to use the frameworkfor accounting purposes, then this is not critical.De ne aio as the unit requirement of inputs from industry i used in the production ofoutput o. I.e., in order to produce one unit of output o you need aio units of i. If someindustry o does not require its own output for production, then aoo 0.For n industries Ann [aio ] is a technology matrix. Each column tells you how much ofeach input is required to produce one unit of all outputs. Alternatively, each row tells youthe input requirements to produce one unit of the industry of that row.If all industries were used as inputs as well as output, then there would be no primary inputs(i.e. time, labor, entrepreneurial talent, natural resources, land). To accommodate primaryinputs, we add an open sector. If the aio are denominated in monetary values— i.e., inPorder to produce 1 in industry o you need aio of input i— then we must have ni 1 aio 1,because the revenue from producing output o is 1. And if there is an open sector, then wePmust have ni 1 aio 1. This means that the cost of intermediate inputs required to produce 1 of revenue is less than 1. By CRS and competitive economy, we have the zero pro tcondition, which means that all revenue is paid out to inputs. So primary inputs receivePn(1i 1 aio ) dollars from each dollar produced by industry o.Equilibrium impliessupply demand demand for intermediate inputs nal demand .In matrix notationx Ax d :And soxAx (I21A) x d :

Let A0o be the oth row vector of A. Then for some output o (row) we havexo A0o x donX aoi xi doi 1 ao1 x1 ao2 x2 ::: aon xn do : {z} {z}intermediate inputs nalFor example, ao2 x2 is the amount of output o that is required by industry 2, because you need ao2units of o to produce each unit of industry 2 and x2 units of industry 2 are produced. This impliesao1 x1ao2 x2 ::: (1aoo ) xoIn matrix notation2(1 a11 )a12a136a21(1 a22 )a2366aa(1a33 )313266.4.an1an2an3ao;o 1 xo 1.a1na2na3n.(1ann )Or(I(I:::32767676767654aon xn do :x1x2x3.xn327 67 67 67 67 65 4d1d2d3.dn37777 :75A) x d :A) is the Leontief matrix. This implies that you need to produce more than just naldemand because some x are used as intermediate inputs (loosely speaking, "IIf (IA I").A) is nonsingular, then we can solve for x:x (IA)1d:But even then the solution to x may not be positive. While in reality this must be triviallysatis ed in the data, we wish to nd theoretical restrictions on the technology matrix to satisfy anon-negative solution for x.5.8.1Existence of non negative solutionConsiderDe ne23a b cA 4 d e f 5 :g h iPrincipal minor: the determinant of the matrix that arises from deleting the i-th row and22

i-th column. E.g.jM11 j e fh ia cg i; jM22 j ; jM33 j a bd ek-th order principal minor: is a principal minor of dimensions kof the original matrix are nthe same n:k. If the dimensionsn, then a k-th order principal minor is obtained after deletingk rows and columns. E.g., the 1-st order principal minors of A arejaj ; jej ; jij :The 2-nd order principal minors are jM11 j, jM22 j and jM33 j given above.Leading principal minors: these are the 1st , 2nd , 3rd (etc.) order principal minors, wherewe keep the upper most left corner of the original matrix in each one. E.g.a bd ejM1 j jaj ; jM2 j ; jM3 j a b cd e fg h i:Simon-Hawkins Condition (Theorem): consider the system of equations Bx d. If(1) all o -diagonal elements of Bn(2) all elements of dnThen 9x

Lecture notes based mostly on Chiang and Wainwright, Fundamental Methods of Mathematical Economics. 1 Mathematical economics Why describe the world with mathematical models, rather than use verbal theory and logic? After all, this was the state of economics until not too long ago (say, 1950s

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