An Alternative Proof of Primitivity of Indecomposable Nonnegative Matrices with a Positive Trace

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1 An Alternative Proof of Primitivity of Indecomposable Nonnegative Matrices with a Positive Trace Takao Fujimoto Abstract. This research memorandum is aimed at presenting an alternative proof to a well known theorem that an indecomposable nonnegative matrix with at least one positive diagonal element is primitive. Our proof is half topological and half combinatorial. Keywords: Indecomposability, Positive Diagonal Elements, Primitive Matrices 1 Introduction It is well known that a nonnegative matrix which is indecomposable and has at least one positive diagonal element is primitive. This is obvious from another well known characterization of indecomposable imprimitive matrices explained, e.g., in Nikaido[4] (and also [5]). This proposition may date back to A. Brauer. A direct and combinatorial proof is given in a forthcoming book by Krause[1], even in a nonlinear setting. However, it seems difficult to extend the method to spaces of an infinite dimension. In this note, a direct and simple proof is presented, which can be used in nonlinear generalizations as well as for infinite dimensional spaces. (See for the importance of primitivity [2] and [3].) Section 2 explains our definitions and notation, and then in Section 3 the theorem is proved in our own way with a help of two lemmata. In Section 4 some remarks are given when generalizing the theorem to nonlinear mappings and to spaces of an infinite dimension. 2 Notation and definitions Let R n be the Euclidean space of dimension n (n 2), R n + be the nonnegative orthant of R n,anda agivenn n nonnegative matrix. Considering future extensions, we denote the mapping Ax, x R n, by f(x) Ax, andinacomponentwiseway, 1

2 f i (x) (Ax) i nx a ij x j, where a ij is the (i, j) element of the matrix A, and x j is the j-th entry of the vector x. The symbol N stands for the index set j=1 N {1, 2,...,n}. For vector comparison, we use the following inequality signs. x y iff x i y i for all i N, x > y iff x y and x 6= y, x À y iff x i >y i for all i N. Now we define Definition 1. A nonnegative matrix A is decomposable iff there exist two nonempty subsets I and J of the index set N such that I J =, I J = N, and a ij =0 for i I and j J. Definition 2. A nonnegative matrix A is indecomposable iff it is not decomposable. Alternatively we may define the indecomposability as Definition 2a. A nonnegative matrix A is indecomposable iff for any two nonempty subsets I and J of the index set N such that we have I J =, I J = N, f i (x) >f i (y) in at least one i I for arbitrary two vectors x and y such that x i = y i for i I and x i >y i for j J. Definition 3. A nonnegative matrix A is primitive when there exists a positive integer k such that A k À 0. Otherwise it is imprimitive. 2

3 3 Propositions Let us consider the following 2 2 matrix A = µ This is indecomposable. When we make the second power of A, A 2 = µ which is decomposable. Thus, we need an additional condition to guarantee the indecomposability of the power of an indecomposable matrix. We first state a lemma. Lemma 1. If A is indecomposable and has at least one positive diagonal element, then A k is also indecomposable for any integer k>0. Proof. We prove this lemma using mathematical induction. First for k =1, the proposition is obvious. We assume this lemma holds for (k 1) with k 2. Let us also assume that f 1 (x) >f 1 (y) for any x and y such that x 1 >y 1 and x i = y i when i>1. (That is, a 11 > 0.) Suppose to the contrary that A k be decomposable and we have two nonempty subsets I and J of the index set N such that., I J =, I J = N, and a (k) ij =0 for i I and j J. Here, a (k) ij means the (i, j) element of A k. Among possible bipartitions between I and J, we adopt as I the one with the minimum number of elements. When 1 I, wesplit A k as A A k 1. Since a 11 > 0, a (k 1) 1j =0 for j J. Now if there is a positive entry among a (k 1) ij for i I and j J, say a h` > 0 for some ` J when h H & I, wewill consider the index set I H and J H. (The index subset H is assumed to be maximal in the sense that a (k 1) ij =0 for any pair (i, j) such that i (I H) and j J.) It is clear that a ih =0for any pair (i, h) such that i I and j H: otherwise we would have a (k) ij > 0 for that i I and some j J since A k = A A k 1, thus yielding a contradiction. Then, from another relationship that A k = A k 1 A, wehave a (k) ih =0 for any pair (i, j) such that i (I H) and h H. This implies that we have two nonempty subsets I 0 (I H) and J 0 (J H) of the index set N such that I 0 J 0 =, I 0 J 0 = N, and a (k) ij =0 for i I 0 and j J 0, is shown to be indecom- a contradiction to the minimality of the index set I. Hence,A k posable when 1 I. 3

4 When 1 J, wefirst split A k as A k 1 A, and in the second stage as A A k 1,looking for positive entries column-wise rather than row-wise. The above proof can proceed mutatis mutandis. Lemma 2. If A is indecomposable, Ax = λx with λ > 0 implies x À 0, i.e., strictly positive. Proof. Thisiswellknownandevidentfromthedefinition. Now we prove our main proposition. Theorem 1. A given nonnegative matrix A is primitive iff its power A k is indecomposable for any positive integer k. Proof. First, let us prove the only if part. Suppose that there is a positive integer p such that A p becomes decomposable. Then we can find a vector x with at least one zero element such that for any positive integer q, A pq x keeps zero elements in the same positions. This is a contradiction to the fact that A k À 0 for some integer k>0, and so A r À 0 for any integer r k. Now we proceed to the if part. Let us adopt the absolute sum norm k k in the n-dimensional Euclidean space R n, and we consider the subset S {x x R n + and kxk =1}. All we have to show is that starting from an arbitrary point x on the boundary of S, bd(s), a sequence of vectors X {x, Ax, A 2 x,...} comes into the interior of S, int(s). Suppose to the contrary that this sequence remains on the boundary for ever starting from a certain vector x. Indecomposability by Definition 2a requires that in the vector series X, there can be no two points on the same sub-simplex facet of the boundary bd(s). Otherwise x and A k x, for some integer k>0, has the same sign pattern, contradicting the indecomposability of A k, assured by Lemma 1. There are, however, only a finite number of facets on bd(s). Hence, the vector sequence X has to enter the interior of S. Corollary 1. If a given indecomposable matrix A has its trace positive, then it is primitive. Proof. It is clear from Theorem 1 and Lemma 1. 4 Remarks When we generalize the above propositions to nonlinear cases, it does not seem to cause much trouble. It is possible to redefine necessary concepts using f i (x) with the help of a basic assumption of monotonicity, i.e., f(x) f(y) if x y. Some difficulties turn up if we try to extend the results to spaces of an infinite dimension. We have in mind the space of piecewise continuous functions on a compact set, or the space L on a certain compact measurable space. To realize a finite number of facets 4

5 appearing in the proof of Theorem 1, we need to quantize the compact set by use of a neighborhood of a positive measure References [1] Krause, U.: Positive Discrete Dynamical Systems. Theory, Models, and Applications. (to appear in 2004). [2] Fujimoto, T., Krause, U.: Strong ergodicity for strictly increasing nonlinear operators. Linear Algebra and Its Applications 71, (1984) [3] Fujimoto, T., Krause, U.: Stable inhomogeneous iterations of nonlinear positive operators on Banach spaces. SIAM Journal on Mathematical Analysis 25, (1994) [4] Nikaido, H.: Convex Structures and Economic Theory. New York: Academic Press, 1968 [5] Nikaido, H.: Introduction to Sets and Mappings in Modern Economics. New York: Academic Press,