Math 1b Practical — February 25, 2011The Leontief input-output model in economicsWe are given a square matrix C of nonnegative real numbers called the consumptionmatrix. The rows and columns are indexed by ‘industries’ 1, 2,...,n.Theentrycijrep-resents the number of units of material that industry j must purchase from industry i inorder to manufacture one unit of j’s own product.If the economy produces xjunits of the product of industry j,andx is the columnvector (x1,...,xn), then Cx gives the am ounts of the various products that are used orinternally consumed by the industries. Then x − Cx =(I − C)x gives the amounts of thevar i ous products available for external use.Here is an example I found athttp://www.krellinst.org/AiS/textbook/unit7/f77_7.11_leontief.html.The coordinates of the column vector x on the right are the economy’s productions ofp etroleum, textiles, transportation, and chemicals measured in millions of dollars. WehaveCx =⎛⎜⎝880110550825.50⎞⎟⎠and (I − C)x =⎛⎜⎝20190300−25.50⎞⎟⎠.So e.g. $190 million mor e dollars’ worth of textiles were produced than consumed in theeconomy’s pro duction. Unfortunately, it seems that $25.5 million more dollars’ worthof chemicals were consumed than produced. I guess these chemicals h ad to come from‘outside’. But the economy still made money.Anyway, whether you like or understand this mo del or not, here is a mathematicalquestion that arises in this theory. Supp ose C ≥ O and that d ≥ 0 is a column vector(whose coordinates diare the ‘demands’ for the product of industry i from outside thesystem). Can we find a nonnegative vector x so that(I − C)x = d, or (I − C)x ≥ d ?We have discussed finding nonnegative solutions of systems of linear equations in thefirst week of class—and the question of whether the system of linear inequalities (I −C)x ≥d has nonnegative solutions x is equivalent to the question of whether the system of linearequations (I − C)x = d + y,with2n variables, has n onnegative solutions (x, y). So maybethere is nothing more to say?Of course, if I − C is nonsingular, there is a unique solution x to (I − C)x = d,whichmay or may not be nonnegative.The following theorem, and its corollary, answers our question in many cases (likethat of our example), without calculations (unless you really need the actual solution).Theorem 1. If the spectral radius of a nonnegative matrix C is strictly less than 1, i.e.if all eigenvalues μ of C, real or complex, satisfy |μ| < 1,thenI − C is nonsingular and(I − C)−1≥ O.Then (I − C)x = dhas a unique nonnegative solution for any nonnegative vector d,namely x =(I − C)−1d.Proof: (Partial.) The eigenvalues of I − C are 1 minus the eigenvalues of C,andthesearenever 0 under our hypothesis. So I − C is nonsingular.We have not talked about infinite series of matrices b efore now, but we claim that(I − C)−1= I + C + C2+ C3+ ...for matrices C of spectral radius < 1. Of course, this notation means thatlimm→∞(I + C + C2+ ...+ Cm)=(I − C)−1(∗)(in ev ery entry). Since C ≥ O, it is clear that the limit (I − C)−1is nonnegative.To prove (∗), we need a fact about matrices C of spectral radius < 1, namely thatCk→ O as k →∞. (By Problem 2 of Set 7, this is true when C is diagonalizable. Theproof when C is not diagonalizable is har d er and is not given here.) Note that(I − C)−1− (I + C + C2+ ...+ Cm)=Cm+1(I − C)−1(check this by multiplying both sides by I − C), and so(I − C)−1− (I + C + C2+ ...+ Cm) → O as m →∞.By Pr oblem 4 of Set 7, if the sum of every column of a square nonnegative matrix Mis at most s,then|λ|≤s for every eigenvalue λ of M. Thesameistruewhen‘column’isreplaced by ‘row’. Thus a consequence of the theorem is the following.Corollary. If all column sums of a square nonnegative matrix C are l ess than 1,orifallrow sums of C are less than 1,then(I − C)x = dhas a unique nonnegative s olution forany nonnegative vector d.The sumni=1cijof column j of C is the total cost of materials industry j needsto produce one dollar’s worth of pro ducts, so normally one would expect the sum of thecolum n j to be < 1; otherwise industry j makes no profit.In our numerical example, the column sums of C are all less than 1, so (I − C)x = dwill have nonnegative solutions for every d ≥ 0.HereC =⎛⎜⎝0.10.40.60.200.100.10.20.15 0.10.30.40.30.25 0.2⎞⎟⎠,I− C =⎛⎜⎝0.9 −0.4 −0.6 −0.200.90−0.1−0.2 −0.15 0.9 −0.3−0.4 −0.3 −0.25 0.8⎞⎟⎠,(I − C)−1=⎛⎜⎝2.16971 1.75826 1.85101 1.456340.161815 1.30439 0.183522 0.2723240.994573 1.18796 2.1036 1.185991.45634 1.73952 1.6517 2.45091⎞⎟⎠.For example, the solution of(I − C)x =⎛⎜⎝100100100100⎞⎟⎠is x =⎛⎜⎝723.532192.205547.213729.847⎞⎟⎠.(I do not know how to get the industries to modify their productions to match x.)One fact we used in the proof of Theorem 1 (though the fact wasn’t proved completely)is the following assertion, which we r efer to again in the handout on probability matrices.Theorem 2. If C is any squar e matrix, real or complex, or sp ectral radius < 1,thenCk→ O as k