We have referred at several instances to the problem of "solving" the implicit first-order equations
for the explicit
relations fli x a) 0 X x 2 , a ) 0 xx -- = !*( x2 = ? (5-22) (5-23)
where x\ and x2 are the choice variables and a represents the
parameters of the model.
Sufficient conditions under which this procedure is valid are known as the implicit function theorem. One should be wary, incidently, of a "nontheorem" that appears every now and then. This nontheorem asserts that if there are n equations and n unknowns, a unique solution results. This proposition is valid only in the case of linear equations whose coefficient matrix has a nonzero determinant. Figures 5-la, b, and c demonstrate why the theorem cannot be applied to nonlinear functions. fix) fix) fix) 0 x Two solutions (a) 0 x One solution (b) 0 x No solutions (c) FIGURE 5-1
In general, with nonlinear functions, no general assertions are possible regarding the number of solutions to n equations and n unknowns.
106 THE STRUCTURE OF ECONOMICS
FIGURE 5-2
The Implicit Function Theorem. Around any point where the circle is not vertical, a unique y exists for any x. However, around x — +1 or x — — 1, two values of y are associated with anyjc, no matter how small the interval is made around that x value. If the function is not vertical, an explicit solution y = /(JC) exists for an explicit relation g{x, y) = 0. However, dg/dy j= 0, while sufficient, is not necessary.
The implicit function theorem is narrower in scope than the above nontheorem. Suppose Eqs. (5-22) have a unique
simultaneous solution at some point (x®, x°, a0). Under what
conditions can the implicit relations (5-22) be written as the explicit relations (5-23)?
To answer this, consider first the simplest case of one equation in two unknowns, e.g., the unit circle, depicted in Fig. 5-2,
JC2 + y2 = 1 (5-24)
For this function to be written as some explicit function, y = /(JC), a unique y must be associated with any JC, around a certain point. Of course, (5-24) can be solved for y as
y = ±(\-x2)x'2
The function as written here is technically not a function at all; for each JC, two values of y are given, instead of a unique y. However, such is not the case for solutions around individual points on the unit circle. Consider some point, A, x = 1 /V2, y = l/\/2. In some neighborhood around x = 1/V2, a unique value of y is associated. That is, around x = 1/V2, y = l/\/2, the explicit functional relation
_ r2xl/2 (5-25)
is valid. The implicit function (5-24) admits an explicit solution around the point A, not necessarily for all JC.
The situation is different, however, at the intercepts of the unit circle and the x axis, the points (—1, 0) and (1, 0). At either of these two points, no matter how small the interval is made around the point, any value of JC will be associated with two values of y. The implicit relation (5-24) does not admit of an explicit solution y = /(JC). An explicit solution of x on _y, that is, x = g(y), does exist—for any value of y around (1, 0) or (—1, 0), a unique value of JC is implied [however, not at the points (0, 1) and (0,-1)].
MATRICES AND DETERMINANTS 107
It can be seen that the reason why the implicit equation x2 + y2 =
1 does not admit of a unique solution _y = f(x) at (1, 0) and (—1, 0) is that at these points, the function turns back on itself. Moving counterclockwise around the circle, as y increases through the value 0, on the right semicircle, x first increases and then decreases. (On the left semicircle, moving clockwise, x decreases and then increases.) At the points (1,0) and (—1,0) the implicit function x2 + y2 = 1 is vertical, that is, dy/dx -> ±oo. As long as the function is not vertical, the implicit relation yields a well-defined explicit solution y = f(x).
We can see how the preceding analysis relates to the ability to do comparative statics, in one-variable models. Consider one implicit choice equation, which might be the first-order equation of some objective function:
h(y,a)=0 (5-
26)
To find dy/da, an explicit solution of (5-26) must be assumed:
y = y*(a) (5-
27)
Substituting (5-27) into (5-26), the identity
h(y*(a),a)=0 (5-
28)
results. Differentiating with respect to a, dy*
hy~ ^ - + ha= 0 (5-
29) da In order to solve (5-29) for dy*/da,
hy=/=0 (5-
30)
must be assumed. This amounts to assuming that the function h(y,a) is not vertical (a plotted horizontally, _y vertically).
In maximization models, the sufficient second-order conditions guarantee the existence of the explicit solutions (5-27). In these models, the implicit relation (5-26) is already the first partial of some objective function, f(y,a). That is, (5-26) is
fy(y, ot) = h(y,a) =0
The condition that hy j= 0 is guaranteed by the sufficient second-order
condition for a maximum,
fyy = h y< 0
It should be noted that whereas hy^0is sufficient to be able to
write y = y*(a), it is not necessary. There are some functions for
which hy = 0 at some point, and it is still possible to write y = y*(a).
For example, consider the function
The solution to this equation, depicted in Fig. 5-3, is y = al/3
108 THE STRUCTURE OF ECONOMICS
FIGURE 5-3
The Function y = a1//3. This function
illustrates why the condition hy j= 0
is sufficient but not necessary for writing an implicit function in explicit form. This function becomes vertical at the origin, yet it is still possible to define y as a single-valued function of a, because
a1//3 does not turn back on itself. If h
y ^ 0, the explicit formulation is
always possible; if hy = 0, it may
not be.
Although dy /da —>• oo as a — > 0, it is still the case that a unique y is associated with any a around a = 0; the function, while vertical at a = 0, does not turn back on itself there.
In models with two equations and two choice variables, the situation is algebraically more complicated, but conceptually similar. Consider the system (5-22) again, but let us just assume that these are
just two equations in three unknowns, X\,x2, and a, without assuming
for the moment that there exists an f(x\, x2, a) for which /i = df/dxi, f2
= df/dx2. A sufficient condition that Eqs. (5-22) admit the explicit
solution (5-23) at some point is that neither of the explicit functions (5-23) become vertical, for any a, if a is one of many parameters. Let us try to solve for dx*/da and 3*2/30:.
Differentiating Eqs. (5-22), we get
9/L 9/L 3XI 3x2 9/2_ Bh_ dx\ dx2 da dx*2 \~da — f\a ~fla (5-31)
A necessary and sufficient condition for solving for dx*/da and dx^/da uniquely is that the determinant
J = L dx2
ML dx2
(5-32)
This determinant, whose rows are the first partials of the equations to be solved, is called a Jacobian determinant. If / ^ 0, the partials dx*/da are well defined, and in fact are so because the explicit
MATRICES AND DETERMINANTS 109 is precisely the sufficient condition that allows solution of the simultaneous equations (5-22) for the explicit equations (5-23). This is the generalization of relation (5-30) for one equation.
Condition (5-32) is implied by the sufficient second-order
conditions for a maximum. In maximization models, f\{x\, x2, a) and
f2(xi, x2, «) are df/dxi, df/dx2. Therefore, dfi/dxi = fu, etc., and the
Jacobian is
J = /ll
From the sufficient second-order conditions, 7^0, since J > 0. For models with n equations
fl( xu. . . , xn, a ) =0
(5-33) fn( x i , . . . , xn, a )
= O a sufficient condition for explicit solutions
xt=x*{a) (5-34)
to exist at some point is that the Jacobian of (5-33) be nonvanishing there: dxn (5-35) Bfn Bfn dxn PROBLEMS
1. Evaluate the following determinants 1 2 -1 -3 -2 -1 -4 -3 1 2 0 1 -1 0 1.206 1 1 1.207 0 1 1 1 0 ( a ) ( b
110 THE STRUCTURE OF ECONOMICS
2. Suppose a square matrix is triangular; i.e., all elements below the diagonal are 0:
/flu a 12 ■ ■ • a] n
0 a22 a2n
A =
V 0 0 aj
Show that |A| = a\\a21, ..., ann, the product of the diagonal elements.
3. Consider the system of n equations in n unknowns Ax = b
where the vector b consists of 0s in all entries except a 1 in some rowj. Assuming |A| =/= 0, show that the solutions can be represented as
* = ]A[ U=1..."
where A;i is the cofactor of element a,,-.
APPENDIX SIMPLE MATRIX OPERATIONS
A matrix, again, is any rectangular array of numbers: fan • • • c
am n/
This matrix has m rows and n columns. Suppose some other matrix B has n rows and r columns (B must have the same number of rows as A has columns):
/bn • • • b
B = ;
\bni bnrl
The matrix product C = AB is defined to be the m x r matrix c \r\ I ' an ■ ■ ■
ai n\ / bn •
dm\
where any element c,y of C is defined to be ykj k=l
Schematically, each element of any row of A is multiplied, term by term, by the elements of some column of B (as shown by the direction of the arrows above) and
MATRICES AND DETERMINANTS 111 the result is summed. Note that while the product AB may be well defined, BA may not be, since the number of columns in the left-hand matrix must equal the number of rows in the right-hand matrix in a matrix product. In general, even for square matrices, matrix multiplication is not commutative, i.e., in general,
AB^BA
The associative and distributive laws do hold, however. If A is m x n, B is n x r, and C is r x p, then ABC is ra x p and the following laws are valid:
Associative law (AB)C = A(BC) If A is m x n, B and C are n x p, then: Distributive law A(B + C) = AB + AC For the associative law, we simply note
h=\ k=\ k=\ h=\
For the distributive law,
aik(bkj + ckj) =
k=\ k=\ k=\
The transpose of any matrix, A', is the matrix A with its rows and columns interchanged. That is,
The transpose of a product is the product of the transposed matrices, in the reverse order:
(AB)' = B A To
prove this, let c,7 be an element of (AB)'. By
definition,
k=l An element of B'A' is
k=\ identical to the former sum.
A matrix is called symmetric if it equals its transpose; that is, A = A'
That is, for every element a{j, atj = a^. The rows and columns can be
interchanged leaving the same matrix. This is a very important class of matrices in economics. The matrices encountered in maximization models are the second partials of some
112 THE STRUCTURE OF ECONOMICS
objective function, f ( x i , . . . , xn) . By Young's theorem, fa = /},-.
Hence, these matrices are symmetric. The Rank of a Matrix
Consider anmxn matrix (a,-y) and consider each of its rows, Ai, ...,
Am, separately. Each row /,
represents a point in Euclidean n-space. It is important to discuss the "dimensionality" of these m points; i.e., do they all lie on a single line (one dimension), a plane (two dimensions), etc.? Algebraically, if these m vectors lie in an ra-dimensional space, then it is not possible to write any vector A as a linear combination of the others. In other words, if
fcjAi H- - -h kmAm = 0
where the kt are scalars (ordinary numbers), then all the kt 's must be
zero. In this case, Ai, ..., Am are said to be linearly independent.
For any given matrix A, the maximum number of linearly independent row vectors in A is called the rank of A. If A has m rows and n columns, and n > m, then the maximum possible rank of A is m. It is not obvious, but true that the number of linearly independent column vectors of A equals the number of linearly independent row vectors. Thus the rank of a matrix is the maximum number of linear independent vectors in A, formed from either the rows or the columns of A.
Example 1. The vectors Ai = (1, 0, 0), A2 = (0,1, 0), A3 = (0, 0, 1)
are linearly independent.
fc,Ai + k2A2 + £3A3 = (fci, k2, k3) = 0 if and only if k} = k2 = &3 = 0. The matrix
therefore has rank 3.
Example 2. Let Ai = (1, 1, 0), A2 = (1, 0, 1), A3 = (1,-1, 2). These
vectors are linearly dependent. Here, A3 = 2A2 — A,, or
A, - 2A2 + A3 = 0
Any one of these vectors can be written as a linear combination of the other two, but not less than two. The matrix
/I 1 0\ A= 1 0 1
\1 -1 2/ therefore has rank 2.
MATRICES AND DETERMINANTS 113
A set of m linearly independent vectors Ai, ..., Am is said to form
a basis for Euclidean ra-space. Any vector b in that space can be
written as a linear combination of Ai, ..., Am, that is,
where the fc/'s are scalars.
Consider a system of n equations in n unknowns,
or, in matrix notation, Ax = b. If the rank of A is less than n, then some row of A is a linear combination of other rows. But this is the procedure for solving the above system for the JC'S. If rank (A) < n, then at least one equation is derivable from the others, i.e., there are really less than n independent equations in n unknowns. In this case, no unique solution exists. We saw in the chapter that simultaneous equations admitted a unique solution if the determinant of A, |A|, was nonzero. An important result of matrix theory is thus:
Theorem. If A is a square n x n matrix, then the rank of A is n if and only if |A| ^ 0.
Discussion.This algebra is the basis of the nonvanishing Jacobian determinant of the implicit function theorem. Briefly, if rank (A) < n, then some row (or column) is a linear combination of the other rows (columns). By repeated application of the corollary to Theorem 5 in the chapter proper, |A| — 0. Conversely, if |A| = 0, some row of A is either 0 or a linear combination of the other rows,
and hence A],..., An are linearly dependent. A more formal proof of
this part can be found in any standard linear algebra text.
A square nxn matrix A that has a rank n is called nonsingular. If rank (A) < n, A is called singular.
The Inverse of a Matrix
In ordinary arithmetic, the inverse of a number x is its reciprocal, l/x. The inverse of a number x is that number y which makes the product xy = 1. In matrix algebra, the unity element for square n x n matrices is the identity matrix I where,
/I 0 • • • 0\
0 1 0
1/
That is, I is a square n x n matrix with Is on the main diagonal, and
0s elsewhere. Formally, if (atj) is the identity matrix, then atj = 1 if/
114 THE STRUCTURE OF ECONOMICS be verified that for any square matrix A,
AI = IA = A
Thus the identity matrix I corresponds to the number 1 in ordinary arithmetic. Is there a reciprocal matrix B, for some matrix A such that
AB = I
If so, we call B the inverse of A, denoted A"1.
The problem of finding the inverse of a matrix is equivalent to solving
Ax = b
for a unique x, where A is an n x n square matrix. If A"1 exists,
premultiply the equation by A"1, yielding
x = A Jb
This is the simultaneous solution for x. This solution exists if and
only if |A| ^= 0. This is correspondingly the condition that A"1 exists,
that is, A must be nonsingular, or have rank n.
Assuming |A| ^= 0, consider the following matrix, A*, called the adjoint of A:
A* = (An A2 l ■ ■ ■ An l\
An A22 An2
Ain A2n AnnJ
The adjoint, A*, is formed from the cofactors of the a,-/s, transposed. Consider the matrix product AA*:
an
\ i \ i \
= |A|I Any element of AA* off the main diagonal is formed by the product of the elements of some row of A and the cofactors of some other row; these products sum to zero by the theorem on alien cofactors. The diagonal elements of A A*, however, are formed from the sums of products of a row of A and the cofactors of that row; this sums to |A|. Hence
MATRICES AND DETERMINANTS 115
The inverse of A, A"1, is thus (1/|A|)A*, or
A"1 =
\|A| |A
By inspection, it can be seen that if AA"1 = I, then A"1 A = I also;
that is, the left or right inverse of A is the same A"1. Also, A"1 is unique. Suppose there exists some B such that
AB =
Premultiplying by A "1
A AB = IB = B = A It is also true that
= B The proof of this is left as an exercise. Orthogonality
Two vectors are called orthogonal if their scalar product is 0.
Example 1. The vectors Ei = (1, 0, 0), E2 = (0, 1,0), and E3 = (0,
0, 1) are all mutually orthogonal.
Example 2. Let a = (2, -1, 1), b = (-1,-1, 1). Then ab = 0; thus a and b are orthogonal.
Orthogonal vectors must be linearly independent. Suppose a square matrix A is made up of row vectors ai, ..., an which are mutually orthogonal, and whose Euclidean "length" is unity:
A is called an orthogonal matrix. It can be quickly verified that the transpose of A, A', is the inverse of A, i.e.,
AA = I -l
116 THE STRUCTURE OF ECONOMICS PROBLEMS
1. Find the rank of the following matrices. For which does | A| ^ 0? /-I 1 2\ / 1 0 -1\ /-I 0 1\ A = ( 1 — 1 — 2 I B = ( — 1 1 1 C = (
1 - 1 1
V-2 2 4/ V 1 -1 -1/ \ 0 —1
3/
1.208 Prove that (AB)"1 = B^A"1, if A and B are two square
nonsingular matrices.
1.209 Prove that (A"1)"1 = A, that is, the inverse of the inverse is the original matrix.
1.210 Show that (A)"1 = (A"1)', i.e., the transpose of the inverse is the inverse of the transpose.
1.211 Show that if A is n x n, and h is an n x 1 column vector, then h'Ah =
The expression h'Ah is called a quadratic form. These expressions appear in the theory of maxima and minima.
1.212 Show that if h'Ah < 0 for any vectors h ^ 0, then (among other things) the diagonal
elements of A are all negative; that is, an < 0, i = 1,..., n.
1.213 Prove that if A is an orthogonal matrix, A'A = I; that is, A' = A"1.
1.214 Prove that if the rows of a square matrix A are orthogonal and have unit length, the columns
likewise have these properties. SELECTED REFERENCES
The implicit function theorem can be found in any advanced calculus text. Classic references are:
Apostol, T.: Mathematical Analysis, Addison-Wesley Publishing Co., Inc., Reading, MA, 1957. Courant, R.: Differential and Integral Calculus, 2d ed., vols. 1 and 2, Interscience Publishers, Inc., New York, 1936.
Matrices and determinants are the subject of any linear, or matrix, algebra text. Perhaps the clearest and most useful for economists is:
Hadley, G.: Linear Algebra, Addison-Wesley Publishing Co., Inc., Reading, MA, 1961.
Samuelson, P. A.: Foundations of Economic Analysis, Harvard University Press, Cambridge, MA, 1947.
The first systematic exposition of the application of the implicit function theorem in economic
CHAPTER
6
COMPARATIVE STATICS: THE TRADITIONAL
METHODOLOGY
6.1 INTRODUCTION; PROFIT MAXIMIZATION ONCE MORE