3— Multiplicative Systems in Several Variables I, II, III: With C.J. Everett (LA-683, June 7, 1948) (LA-690, June 11, 1948) (LA-707, October 28, 1948)

### VII— Supercriticality Conditions

We give now necessary and sufficient conditions for supercriticality of the transformation G(x), namely that limGk(O) = I1 . By definition, mij = mi') = (Ogi/axj)i. Since we have assumed all (&gi/Oxj) non-vanishing for x :\$ 0 on I, clearly all mij > 0.

Theorem. The following conditions on G(x) are equivalent: (a) limGk(O) = x ° 1. (b) there exists an x such that 0 -< G(x) -< x 1, where gi(x) < ii, 0 < -i < 1 for all i. (c) there exists an x such that 0 -< G(x) -<x1, where 0 <Xi < I for all i. (d) there exists an i such that 0 -< G() -< x 1. (e) there exists an x such that _ mij(1 - Xj) > 1 - ji, where 0 < xi < 1, for all i. (f) there exists a 5 = (1i,..., St) such that E mijbj > 6i, where bi > 0 for all i. (g) the matrix [mij - ij]t = J(G(1)) -I contains at least one upper principal minor A, = [mij - bij] (1 < v <t) such that (-l1)v+ I[AI> 0 where (-l)"+l · IA, > 0 in case v = t.* (h) there exists a v = (vl,...,vt) such that Evimij > vj, where vj > 0 for all j. (i) there exists a characteristic root r > 1 of M = [mij] and a corresponding characteristic vector v = (v, .. ., vt) with all vi > 0, that is, Zvimij = rvj for such r and v.

In matrix notation m ii .l * mit (vi,..., vt) · =(rv, ..., rVt) mtl mtt or briefly, vM = rv.

* [aij]~ indicates a matrix whose indices i,j range from m to n. 6j is the Kronecker delta, 1 or 0 as i is or is not equal to j. I is the identity matrix [6ij]. IAI indicates determinant of A.

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1. The proof consists of the following implications: a e /\\/\d (c f-*g(- h-(i. \ / b The theorem has been amplified to permit proof in easy stages, but the essential content is the reduction to matrix theory in J(G)1 of the question of criticality of the (non-linear) transformation G(x).

2. (a-c) If x° 1, then 0 -G(x° ) = x° 1, 0 <x° < 1, all i, so that x° serves as x in (c).

3.(c> d) is of course trivial.

4. (d-a) If 0 -< G(x) - x 1, then Gk(0)-Gk(2) -< Gk-l(x) < ... -< G(.) < .l1, hence x° = limGk(O) __< 1.

5.(c —e) If 0 - G(x) -< a 1, 0 <xi < 1, all i, we have xi > gi(2) = gi(1) + - mij(xj - 1) + 1/2 E(igi/Ixj&xk)((i)(xj - 1)(xk - 1) where Xj <((i) < 1 for all j. Hence xi >1 -+ E mij(Xj - 1) and 1 - xi < Z^mij(1 - ;j) with 0 <Xj < 1, all j.

6. (e -- f) is trivial, since we may define (i = 1 - xi > 0 in terms of the x given in e.

7. (fb) Suppose t=' mij6j > 6i > 0, i = . .., t Since mij = (Ogi/9xj)l and Ogi/Oxj is continuous, there exists a 5, 0 < 1 < 1, such that (O9gi/Oxj)S6j > 6i, all i, whenever <Xj < 1, all j. Define p = min {(I1- )/Si;i j = 1,..., t} > 0. Then _(Ogi/9xj) p6 j > p6i all i, wherever I<Xj< 1, all j. Now define i = 1- Sip; i = 1,...,t. Then E(Ogi/Oxj)=(1 - Xj) > (1 - i) all i, whenever I<j< 1, all j. Since 0 < 1 - xi = 6iP< 1 (because of definition of p), we have 0 < < xi < 1, all i. Now define x = (1t,...,t) where, as we have seen, 0 < xi < 1.Then 1 = gi(1) = gi(x) + E(Ogi/0xj)(i)(1- xj)/x(i) where j < xj < 1, all i,j. x i < all i, j. K Since j< Xy < < , we have (O9gi/Oxj)=(i)(1 -j) > 1 - Zi and 1 > gi(x) + (1 - Xi). Hence gi(x) < xi, all i, and 0 -< G(x) -< x1 for an x of the type required by (b).

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8. (b , c) is trivial.

9. For the equivalence of (f) and (g) we need the following:

Lemma. Let A = [aij]l be a matrix in which aij > 0 for all i \$ j. Then there exists a point 6 = (6 1,...,6 t), with all 6i > 0, such that Eaij6j > 0, all i, if and only if, for some v, (1 <v < t) the upper principal minor A, = [aij]} satisfies the relation (-1)"+1A,Il> O, with (-1)V+1. IAVI > 0 in case v = t.

Proof. We first prove E aijSj > 0 implies the condition on principal minors. Proof is by induction on the order t. If t = 1, we are given that a116 1 > 0 for 61 > 0 so that all > 0, and (-1)2 1A1 1 > 0 as required. Assume the statement of the theorem for order t - 1, and suppose given j= 2aijSj > 0, i = 1,..., t, t > 2, all 6i > 0, aij > 0 for i 4 j. If all > 0, in the given system, we have at once (-1)2 All > 0 as required (since t > 2). Now suppose all < 0. Then taljSi > -a1161 and aii61 + -t 2aijSj > 0, i = 2,..., t. Since -all > 0 and ail > 0 (i = 2,...,t) we multiply to obtain t ailalj6j > ail(-aii)S1 and (-all)ai6lS + 2 aij (-all)6j > O, i = 2, ,t. .,tHence J (-a11) aij6j+ -ailaib6=Et(-allaij+ ailai)6 > 0; i = 2, ...,t.

Since, for i \$ j,aij > 0, and for i, j =2,...,t, ail > 0 and aij > 0 we have also (-allaij+ ailaij) > 0 for i 5 j. By the induction assumption on systems of order t - I there exists a minor of order v: AU = [-allaij + ailalj]2+l such that (-1)"+lA,Il > 0 with (-1)"+1 IAIl > 0 if v = t- 1. But for the original matrix we have lA+l = laijl+l = I...alj... = l/(-all)v ...-alj...a . . . . allai... l/(_ 1)'. i'alja( - alaij + ailaij ...) 1/(-ail) =ai/-a alla12 a) 1.alt

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Hence (-l1)+2. IA,+ =(-1)v+2ali/(- all)"IAv = (-1)"+l/( - all)-11A,l> 0 and > 0 in case v + = t (i.e., v = t - 1).

We now prove the converse, namely, the condition on minors implies the existence of a 6 such that E aijSj > 0. Proof is again by induction on order t.

In case t = 1, we are given that (-1)2 1All = all > 0, so for (say) =1 > 0 we have allb6 > 0.

Assume the theorem true for order t - 1, and suppose given A = [aij] with t> 2, having for a given v(1 < v < t) (-1)v+1 IA. > 0 (> 0 if v + t).

If all> 0, let 62 = ...= t = I and choose 61 > 0 so that 61 > max { - (1/al) t2aij; i = 2,..., t}. Then aijj = all6l+ E2 ali> 0 + 2 aij > 0 since alj > 0 for j \$ 1. Also, for i = 2,...,t, aij = al61= Et aij> - t aij aij= 0, by definition of 61. Thus the conclusion follows in case all = (_1)2. IA}l > 0.

Suppose then that all < 0. Then the given A, must have 2 < v< t. Now 0<(-1)v+l-A, = (-1)-+1aijv = (-1)V+1 . aij ... .. aij . ... (-l)v+ /(-aii)-. aj ..·.• (-_1)+1/ (-a11)v . (-allaij) ... ... a 13=(-1)v+l/(-al1)V-lallal2   . ait .. (-allaij + ailalj) l0 _l-1 (-1)"/(--all)-2·IAI-. Hence (-1)>"A,_1 > 0 for 1 <v-1<t -1 (> O if v-1=t-1) for the system of order t - 1:

A = [-allaij + ailaij]2 in which -allaij + ailalj > 0 for i 4 j just as before. By the induction hypothesis, 6i > 0, i = 2,... , t exist such that E ( - allaij+ aaiaj)6 j> 0, i 2,...,t. Hence ail ( aj6j/ - all) + t aij6j > 0, i = 2,..., t. Let 61 represent the parenthesis of the last inequality. Then el > 0, since aij > 0 for j \$ 1, and substituting, ailel + aij6j> 0,i = 2,..., t.

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Choose T so that 0 < T < el, and T < (1/ail)(ailel + 12 aijsj) for all i = 2,..., t. Finally, define 61 = e1 - T > 0. Then Etaij 6 j a11 61+ '2 alj6j = allel + 2- alj6+ (-all)T =(-ail)T> 0, and for i = 2,..., t,t aij0j = ai161+ aij6j = ailel + aijj - ailT > 0 because of the second inequality defining T. Hence the lemma is proved.

10.(f -9 g) If JEmijj > 5j > 0, all i, then Z(mij - 6ij)6j > 0, all i, and the matrix [mij- 6ij]t satisfied the condition of (g) because of the lemma.

11.(g -9 f) If the matrix [mij - bij]t satisfies the condition of (g), there exists by the lemma numbers 6i > 0 such that (mij - bij)6j > 0 and hence E nmijSj > 6i, all i.

12. (g --h) If [mij - 6j] has the property (-1)'+1 \IAI > 0 (> 0 if v = t) for some v, then the matrix [mij - 6ij]T = [mij - 6ij] has* (-1)+l' . IAI = (-1)>+ 1 . IA i > 0 (> 0 if v = t). Write mij = mji; then [mij - 6 j] has (-1)^+ · IlAj > 0 (> 0 if v = t). But we have seen that (g -+f) so that there exist vi > 0 such that Emijvj > vi. Hence _mjivj > vi, or changing notation, Zvimij > vj, all j.

13.(h -* g) If E vimij > vj > 0, all j, then E mjivj > v > 0, and, defining mij = mji, Emijvj> vi > 0. But we know that (f -+ g) so the matrix [imj - 6ij] has the property (-1)^+l · nij - —ijl _> 0 (> 0 if = t). But then (-1)v+l(mij - iji = (-1)V+l [mi - 6jil = (-1)V+lmij- 6ij\l> 0 (> 0 if v=t).

14. (i - h) If vimij = rvj for vj > 0, r > 1, we have vimi > v3, hence (h).

15.(ih i) Assume Ei=l vimij > vj > O. Since (Evimij)/lv > 1,

j = 1,...,t, we define ro so that ( vimij)/vj > ro> 1, j = 1,...,t. Then vimij> rovj, all j. Write _ Vimij = rovj + Ej.

* Recall that IAl = IAI .

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Define llvll = v2 > 0. Then E(vi/2 llvll)mi j = ro(vj/2. lvIl) + (ej/211vll ). Define vO= (vj/2 11vll) and ej = (e/211vll). We have now v?°mi = rov + ej, where ej > 0, vO > 0, all j, and Ilv°ll = 1/2. Define 1 = min l/2v°;mij/ /mij; all i,j= 1,...,t >0. Let A = {v; vj>t, llvll < 1, E vimij>rovj, all j}.

(A) The set A is non-void for v° E A and is a subset of euclidean t-space Et. For v.O> 2i > p by definition of u,Iv°0 l = 1/2 < 1, and v°mij > rov.

(B) The set A is bounded, since \lvll< I for all v e A.

(C) The set A is closed. For, let v = limv" where vaeA. We have to show v e A. Since vja > , all a, j, we have vj = lima vj > /p. Also, Ilvi = IIlimaaill = ([limvall < 1, since 11 I is continuous and all iva 11 < 1. Finally, vimij = E (limava) ij = lima Vmij> limara rv= ro lim a v=rovj Hence v E A.

(D) The set A is convex. That is, for every v,v' C A, and every a, a'> 0 with a+a' = 1, we have av+a'v' E A. For avj + a'v' > au+a' = (a + a')p = . Also llav + a'v'll<a lvl +'- lIv'il <a + a' = 1. And (avi + a'vD)mij = a Zvimij +a'_vimij>arovj + a'rovj = ro(avj + a'v).

(E) The set A has an inner point, e.g., v° . For v° E A there exists a neighborhood of v° contained in A. Let e = min vj- ; ej (ro + mij); j= l,...,t. .

Since vj > 2, > -, vO - > 0 and e > 0. Now let v be any point of the e- neighborhood of v° , that is, lIv - v°ll < e. We show that v e A. For Ivj -v°\< Z/ (vj - = v) 11 - lv- < e, so vj > vj, -e > /, by choice of e, and thus v satisfies the first requirement for membership in A.

Next, Illvll-IIvOlll<liv-v°0 1 < e, so llvll < e+llvll- = 1/2+e < 1. The latter inequality obtains since e < v -/ < vj< Z(v ) < v = ilv°ll = 1/2. Hence v satisfies the second requirement.

Finally, E vimj>E (v - e) mij = rov + ej- e Ei mij> rov. The latter inequality is true since it requires ro (v - vj)>e mij - ej, i.e.,

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o- j>(e Ei mij - j)/r . But we know v-j > -e and it suffices to prove -e>(e E mnij-ej) /ro or ej >e( i mij + ro). But e satisfies this requirement by definition.

The properties A-E show that A is a convex body in Et. It is well known, therefore, that any continuous transformation T of A into A has a fixed point T(v) = v, v E A. (Alexandroff-Hopf).

For every v E A, define T(v)= vM/llvMll, that is, T(v) = v' where V = Eivi, mij/ Ej(vimij)2 . Since for v c A, vi > > 0, and mij > 0, clearly IIvMll > 0 on A.

First we show that if v e A, then T(v) E A. For (vj)2 = (Evimij)2 /Ej(Zivimij)2 EviZ2mj/Ej(Ev2) (im) > 1/VII2 . Ei {mTjvi /Eimij }> (1/IIVI12 )2'EiVi2 = SO Vj > / as required for T(v) = v' E A.

Moreover, IIT(v)II = (IvM/IIvMlIII) = 1. Finally, EiVimjk = (1/I1vMll) Ej mjk Eivimij>(/ll|vMll) E mjkrovj = rov;, so T(v) e A, and T(A) c A. Clearly, T(v) is continuous on A. Hence, for some v e A, T(v) = v = vM/lIvM\l, and thus Vj · IIvMI = vimij>rovj, so I\vMII>ro > 1. Set r = llvMll and we have Evimij = rvj with vj > 0 and r > 1. Thus the theorem is complete.

16. Some general remarks on positive matrices are now appropriate. If [mij] is a matrix with all mij > 0, it can have at most one characteristic root r > 0 with a corresponding positive characteristic vector v, all vi > 0, that is, the equations vimij = rvj and Evjmij = r'vj; v j, vj, r, r' > 0 imply r = r'.

For, choose A, > > 0 so that A <(v'/vi)<j,i = 1,...,t. Then Avi<v i< p/i, and since vM = rv implies vMk = rkv and v'M = r'v' implies v'M = r'v', we have EAviMik<EV/m(k) < ,Virnm,k) and hence Arkvj<r'kvj < prkvj, where Mk = [m()] and clearly m( > 0. Then Avj <(r'/r)kv' < LiVj, an impossibility for high k, if the ratio r'/r is either less than or greater than one.

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17. We shall make use of the following

Lemma. If B = [bij] is a matrix with bij> 0 for all i \$j, and if Evibij = 0 v= Evibij for vi, v' > 0, then there exists an a > 0 such that v = avi, i = 1,...,t.

Proof. Induct on order t. Suppose t = 1. Then we have vlbll = 0 = vlbll, vl, v1> 0, so v' = avl where a = vl/vl. Assume the lemma for systems of order t - 1, and suppose given t vbij = 0 = tv'j with t > 2. Since vibil= -bll vi and bil > 0 (for i 4 1) and vi > 0, all i, we must have -b1l > 0. Note also vibij +2vib = 0, i = 2,...,t.

Hence T2vibibij = -blbiivl and vlblj(-bl) + Evi(-bli)bj = 0, j =2,...,t. Substituting, ' vi - bbj + bill) = 0, j = 2,...,t. For i 7 j, bij > 0 and - bl > 0, bil > 0, bi > 0 for i, j > 2. Thus the matrix [- bIbij+bilblj]t has the property of the lemma. In identical fashion we have also 2 v i( -b 11bij+ bilblj) = 0. Thus, by the induction hypothesis, I i v' = avj, j = 2,...,t, for some a > 0. But -bllv' i = vb il = a vibil = a(-bllvl) and thence also vi = avl.

18. It follows that, if mij > 0, all i, j, and E vimij = rvj, E v'mij= rv' for vj,v3 > 0, then v' = av. That is, a positive matrix can have at most one positive characteristic vector (up to the scalar multiples) corresponding to any one of its characteristic roots.

For, the above equations may be written vi (mij - ijr) = 0 = i v'(mij - bijr), where the non-diagonal elements of [mij - bijr] are the mij > 0.

19. If M is the moment matrix of any G, there is one and only one solution for r > 0, vi > 0 of the relation vM = rv. G is supercritical if and only if the solution r is greater than one. (See 20 for existence.)

20. The proof of property (i) for supercritical matrices was complicated by the fact that we wanted to prove r > 1. It is possible to show more easily the weaker result that a solution of the relation in 19 exists for arbitrary matrix M = [mij] with all mij > 0.

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Define = min{l/(4vt); mij/ ijm} > 0, A = {v; Vi ,>L, lvll< 1} and T(v) = vM/\\vMll for v in A. Then just as in the proof of (i) one verifies A is a convex body in Et, and T(v) is a continuous transformation of A into itself. As such, there exists a v in A such that T(v) = vM/IIvMIl = v, and for r = IlvMll > 0, vM = rv. Since v E A, vi> t > 0.

21. Note that, if a(v) = vi, o is a linear functional: ou(av + a'v') =Z(avi+ a'v') = a,vi + a' E via(Vv) + a'(v') . If we define T(v) = vM/a(vM) on the sector vi > 0 (v \$ 0), where M has all nij > 0, we see that a(T(v))u(vM)/a(vM) = 1 and T(v) may be regarded as the central projection of vM onto the hyperplane v 1+. . .+t = 1. Our operator T has some interest beyond that of the classical Markoff operators (see G. Birkhoff) since it need not be a contraction in the positive sector. For example, if M =[1 1 one finds that it is supercritical, has positive characteristic root r = 5, corresponding vector = (4/5,1/5) = v where o(v) = 1, and vM = rv. Hence T(t) = v. Let 6 = (1,0). Then IT(b) - T(v) I = /377/20 > 116 - vIl = /2/5. Hence T(6) is not closer to v than 6 was. Nevertheless we prove limTk(v) =v for all vectors v in the sector vi> 0 (v# 0). Our proof of this fact is lengthy but of a general topological character.

22. Let {yi} denote the simplex with vertices yl,...,y" in E, that is, the set of all points y = E aiyi , ai > 0, E ai = 1. If A is an arbitrary matrix, we have (Caiyi)A ai(yiA) so that the map under A of the simplex {yi} is the simplex with vertices yiA: {y'A= {yA)}

23. For arbitrary x with a(x) 0 define the transformation xS = x/ao(x). If the yi are all \$ 0 and have all components non-negative, clearly the same properties are possessed by all points of {yi}. Hence we may operate on {yi} with S and {yi}S = {yiS}. This says that the projection onto the hyperplane vi = 1 of the simplex {yi} is the simplex with vertices yi/ai where ci - ao(yi).

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For, {yi}S is the set of all aiyi/ aioi and {yiS} is that of all ,bi(yi/ai) where ai,bi> 0, and Eai = 1 = bi. We have to show that, given either the ai or the bi, the other set may be determined so that i aiyi/ E ai ri(bi/i)yi (1) Given the ai, definition of bi = aiai/ r aii implies bi> 0, E bi = 1 and validity of (1) so that {yi}S C {yiS}.

Given the bi, it suffices to find ai so that aiai = biajcj or E (biaj- bijoj)aj = O,i = 1,..., n. (2) J J

The determinant of the latter system is zero since addition of each row but the first successively to the first yields in the (1,j) position aj b bi - j =rj- aj-j = 0.

Hence a solution (al,...,an) \$ (0,...,0) exists, and we see from ai/i aii =bi/ai > 0 that all ai have the same sign. Hence we can choose a solution ai> 0 for all i, with Eai > 0, and then ai/Zai fulfills all requirements.

24. If T(v) = vM/oa(vM) then Tk(v), the k-th iterate of T, is given by Tk(v) = vMkl/a(vMk). Proof is trivial by induction on k. Hence Tk(v) = (vMk)S and Tk{yi} = ({yi}Mk)S = {yiMk}S = {yiMk/a(yi'M)} ={T (yi)}

25. Define j6 = (0,..., 1,. ., 0), the j-th unit vector of t-space, and A = {6i}, Ak = Tk (A) = {Tk(5i)}. Since mij > 0, clearly T(A) c A and hence A D T(A) D T2 (A) D ... is a "nest" of simplices. Define D = QAk, the intersection of all Ak. We prove that T(D) = D. Let de e D, i.e., de = T(dl) = T2 (d2) .... Then T(do) = T2 (dl) = T3 (d2) = ... is in D and hence T(D) c D.

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To obtain the other inclusion, we make use of the lemma. If {yi}1 is a simplex contained in A such that T(y) = T(y') for y 4y' in {yi}, then the linear dimension of the subspace spanned by the T(yi), i = 1,...,t is less than that of the space spanned by the yi, i = 1,..., t.

Proof. Let (say) yl,...,yd be linearly independent vertices of the simplex {yi} and the remaining yd+i dependent upon them.

Write y = Edciyi+ y'= d c'yi in {yi} C A, hence a(y) = a(y') = 1. Suppose that T(y) = ciyiM/l E ci&i = T(y') = E cyi M/ E cii where i = a(yiM). If we let C = ciai, C' = E ci then (ci/C c'/C')yiM = 0. Now not all coefficients here are zero, for if so, we should have c' = (C'/C)ci =Rci and hence y' = c'yi = R cyi = Ry. But then cr(y') = Ra(y), R = 1, and thus c i= ci, = = y', a contradiction. Hence the yiM are dependent (i = 1,...,d) and since the yd+i are dependent on the yi, i = 1,..., d so are the yd+iM dependent upon the yiM, i = 1,...,d. It follows that the space spanned by the yiM, i = 1,...,t is of lower linear dimension than d. But this space is identical with that spanned by the T(yi) = yiM/&i.

Since the dimension t of the space spanned by the 6i can suffer only a finite number of reductions, it follows that, for all k>K, for some K, T is one to one on Ak to Ak+l.

Now we prove D c T(D). Let d e D so that we have for all k, d =Tk(dk) for some dk E A. Now for all i > 0, T++l(dk+i+l) = Tk+i+2 (dk+i+2) or T(Tk+i(dk+i+l)) - T(Tk+i+(dki+2))

where the operands of T are in Ak+l C Ak. Since T is one to one on Ak, we have ek= Tk+i(dk+i+l) = Tk+i+l(dk+i+ 2) for all i. Hence d = T(ek) where ek e K +1 Akc C Q Ak = D, and thus D c T(D).

26. We now show D is itself a simplex. To this end, consider the sequences Tk(Si),i + 1,... ,t, of vertices of the Ak. There exists a common subsequence k, such that limTk(bi) = Ai all exist i = 1,...,t. We claim that the simplex {iV}t is indeed the set D.

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First, it is clear that D = I Tk(A) = T Tk"(A) since the Ak are nested. Hence it suffices to prove that {Ai } = TTkv(A).

Since the Ai = limTk"(6i), Ai is a limit of a sequence of points which are in Tk"(A) = Ak, for all v > N. Since Tk(A) is closed, Ai E TkN(A). Hence Ai c ; Tkv(A). Now all Tk"(A) are convex, hence so is their intersection D. Since all Ai are in the convex set n Tk"(A) = D, and {Ai} is the convex hull of the Ai , we have {Ai } c 0 Tkv(A) = D (Alexandroff-Hopf).

We prove now nTku(A) c {Ai}. Suppose d in the intersection but not in {Ai}. Then, since {Ai} is closed, there exists an e such that the e-neighborhood of d excludes {Ai}. However, since limTkv(6i) = Ai , we can find N so that, for i-+1,..., t, I TkN(6i) - Ai < e/2. Now d C TkN(A) hence write d = EaiTkN(Si). Then |ld - Eaiill = E ai(TkN(i) Ai)ll < Eaie/2 = e/2 and thus EaiAi of {Ai} is in the e-neighborhood of d, a contradiction.

27. Since {Ai} = n Tkv(A) = n Tk(A) =D, we have that T is one to one on the simplex {Ai} to all of itself. Now {Ai} is the convex hull of the finite set of points i,. . ., At. As such it is a convex polyhedron, and the set of all its geometric vertices* is a subset of the original, say Al,...,A, and {Ail = {Ai}. We know that {Ai}l T = {Ai} {T(Ai)}I. It follows that T(A1 ), T(A2 ),..., T(An ) is a permutation of the geometric vertices A1 ,..., A, and hence for some N, TN(Ai) = Ai = AMN /u(AiMN), i = l,...,n or AiMN = T(AiMN) . A. But MN is a matrix with all elements positive. By the uniqueness results obtained previously, it follows that n = 1, and D = n Tk(A) = {A'}, where AlMN =((AMN) ·-1 . But for M itself we have vM = rv so VMN = rNv and again by uniqueness, Al = v.

Hence limTk(v) = limvMk/a(vMk) = v for all v with vi> 0, (v y 0).

* A geometric vertex is the point of the polyhedron not an inner point of any segment {x, y} of the polyhedron, P, y, x in P.

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3— Multiplicative Systems in Several Variables I, II, III: With C.J. Everett (LA-683, June 7, 1948) (LA-690, June 11, 1948) (LA-707, October 28, 1948)