Topics in Economic Sciences
Topics in Economic Sciences EconS 521
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Date Created: 09/17/15
Differential Calculus in Matrix Notation 1 Consider the simple linear function aTx 2a x1 If we take the partial derivative with 11 respect to each Xi we obtain BaTx a a1 x1 Balx a a2 x2 BaTx an 6x Arranging the 7 partial derivatives as an ngtltl vector we have T MW 6x 2 Consider the set of m linear functions Ax where A is an mgtltn matrix so that n a x 24 1 J 1x 11 all 1quot x1 n a a a x Z a x a x 21 22 Zn 2 1 2 2o Ax 39 39 39 j aml aml amn xn Z a x amox 1 W J where the quotquot for the second subscript on the far righthandside indicates rows of A For each 139 lm that is each row of A we have from the previous case that an ngtltl column vector Since Ax is an mgtltl column vector to keep track of rows and columns properly we should have 6Ax6xT de ned as an mgtltn matrix That is the number of rows of 6Ax6xT equals the number of rows of A while the number of columns of 6Ax6xT equals the number of elements of x and columns of A Therefore we arrange the rows 6aiox6xT q vertically to obtain the de nition a1 6Ax a2 A 6xT I a Similarly if we arrange the columns of 6 a1 xT 6x a side by side we have 0AxT 6W ax ax a1T azf amTAT Remarks l Ax is an mgtltl column vector and xT an ngtltl row vector so we must have 6Ax6xT as an mgtltn matrix 2 xTAT AxT is a lgtltm row vector and x is an ngtltl column vector so we must have 6AxT 6x as an ngtltm matrix 3 This is precisely what these de nitions give us 4 However note that it is essential to keep track of and account for the relative positions and orientation of the function ie Ax and the variables ie x 3 Consider xTA where A is an ngtltm matrix and x is an ngtltl column vector all all alm a a a T 21 22 int x A x1 x2 xn aVll aVKZ arm n n n 211a11x1 1461 le 211almxl 39 For eachj lm we have 6aIx6x a an ngtltl column vector xTA is a lgtltm row 1 vector so we must have 6xTA6x as an ngtltm matrix We arrange the columns aj side by side to obtain the de nition a TA m 02 MFA x 4 Consider the quadratic form xTAx where A is a symmetric ngtlt matrix and x is an ngtltl column vector xTAxx1 x2 n 2 2 2 aux1 26112x1x2 2a1nx1xn anx2 261xe3 amxn Taking partial derivative of xTAx with respect to the individual elements of x gives 6 xTAx Q 2a11x1 alzxz 39 39 39 alnxn 2alox 6x1 6xTAx 2anx1 anx2 aux max xz 6xTAx 6 2an1x1 anlxl 39 39 39 arian zanox xquot where 61 aji for all 11 ln has been used eg an an Since xTAx is a scalar we must have 6xTAx6x as an ngtltl column vector Therefore we arrange the elements of 6xTAx6xj vertically to obtain the de nition 6xTAx 2Ax 6x 5 Consider y f x where y is an mgtltl vector with each element of y a function of the ngtltl vector x that is M 00 y2 ym fmx Each y can be partially differentiated with respect to each Xj giving a total of mn partial derivatives which we arrange as an mgtltn matrix a x a x 51306 6x1 6x 6x 2 n 62 62 62 Lei ix ix 1 2 V1 6xT 6xT afmx afmx afmx 6x 6x 6x 1 Z n Note that this is consistent with the de nition in 2 above for linear functions 6 To distinguish between minimum and maximum points we need to look at second order derivatives Let y f x be a scalar function of the ngtltl vector x and set the ngtltl vector of rstorder partial derivatives By 6x 6fx6x equal to 0 the ngtltl column vector of zeros gives a stationary point for y That is y is neither increasing nor decreasing in any of the elements of x at the point x say The point y f x will be a local maximum if n n 2 0 Z Zde 61x lt 0 11 1 6x 6x for every vector dx such that the individual was are arbitrarily small not all are zero and the secondorder partial derivatives are evaluated at the stationary point x In matrix notation we de ne the Hessian matrix for y f x as 62f 62f 62f 6x12 6x16x2 6x16xn 62 62 62 62ya 6 6x3 xt39ixT xt39ixT x1 2 2 2 391 3Zf 62f 62f Bxl xn 6x26xn 6x Note The Hessian matrix of a twice continuously differentiable function is symmetric This is known as Young s Theorem Using matrix notation the sufficient secondorder condition for a local maximum is de yy xdxlt0 V dx 0dx E lzr dxf lt8 Bx x where 8 gt 0 is arbitrarily small Similarly the sufficient secondorder condition for a local minimum is dxldxgt0 V dx 0dxquotE 421de lt8 x 6xT Therefore if the Hessian matrix 62yx 6x6xT is positive definite negative definite at the point x then x is a local minimum maximum Discussion Because the Hessian matrix is evaluated at the point x the quadratic form 2 o Qdx de dx is a function of the ngtltl vector dx and the Hessian matrix is x x an 71er matrix of constants with respect to dx for any given value of x Note that Qdx is homogeneous of degree 2 in dx that is Qtdx E t2 Qdx V t2 0 6 As a result if we can nd a vector dx with length ethat causes Q to become positive in the case where we are checking to see if x maximizes y then we can also nd a vector z E tdx of any length we desire that also causes Q to be positive This is why the conditions on the Hessian eg negative definiteness can be stated in terms of any z e Rquot not just for those z s that are contained in an ndimensional ball with radius 8 Jacobians Let y1 y2 y fix be n functions of n variables Suppose each f is differentiable in each xj The Jacobian determinant or Jacobian is defined by 51306 51306 51306 6x1 6x2 6x aflee Bax 6120 M 6x1 6x2 6xquot egg x 5206 5206 31206 6x 6x 6x 1 Z n The n functions f1 q are functionally independent in the ndimensional ball around the point x defined by x Ax xquot quotAxquot quot231ij lt e for some small 8 gt 0 if and only if M at 0 at x Note For all x Ax close enough to x the functions f x Ax are given by where lim 0 0 V i ln by Taylor s theorem Therefore for small enough 8 gt0 8 we can ignore the term 08 and write quot 6 y E xAx xm f xij i l n F1 6x In matrix notation we have 31106 31106 31306 6x1 6x2 6x Axl Ayl 622m 622m Bax Ax A 6x1 6x2 6x 2 y2 Z Z Axn Ayn 61206 61206 afnx 6x1 6x2 6xquot which is a linearized representation of the local transformation from changes in x to changes in y This relationship is well defined globally for each given x0 f x pair As 8 becomes arbitrarily small the representation becomes an exact of the function space xy e Rquot X Rquot Functional Independence From linear algebra we know that the matrix 6fx6xj spans n dimensions if and only if M at 0 Keep in mind that oflx6xj is independent of ie is not a function of Axkfor anyk ln Suppose that f2x g f1x where g is an arbitrary differentiable function of y1 Without loss of generality WLOG we only need to consider f1 and f2 since we can always reorder the rows off Then x aJi3 39 6x gf1x a by the chain rule That is B oxj is proportional to Bfloxj at each point x We know that this implies that M 0 from basic results in linear algebra Note that the factor of proportionality g39flx generally changes with x But remember that this factor is constant with respect to Ax in the linearized representation Now suppose again WLOG that fnx gf1xa f2xquot 11106 for some function g of y1 y2 y 1 Then we have amt agfxfx am j 1 n 6x 11 6y 6x J 3fquot 00 6x J Clearly this implies that for each x is an exact linear combination of the j1n 6x J first nil column vectors with coefficients given by x J 11 n j1n agmm agmx J which in general vary with x 6y1 ByH The implication is that at each point x if any is a differentiable function of the remaining k at i then M 0 and the matrix of partial derivatives Bflx6xj can not span nspace We have the following result Theorem The vector valued function f Rquot gt Rquot is functionally independent at the point x if and only if lJl 0 at x If lJl 0 Vx in the domain off then f Rquot gt Rquot is functionally independent on its domain Functional independence is equivalent to the full dimension of the linear space locally spanned by a system of n functions in n variables This concept plays a fundamental role in virtually all qualitative analyses of comparative statics and dynamics properties in economics