On the Measurement of Intangibles IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 268 Vol. 5 Issue 2 2013 ISSN 1936-6744 THE ANALYTIC HIERARCHY PROCESS WITHOUT THE THEORY OF OSKAR PERRON Thomas L. Saaty University of Pittsburgh Pittsburgh, PA 15260 USA saaty@katz.pitt.edu ABSTRACT It is known and has been mathematically proven that the principal eigenvector is necessary for deriving priorities from judgments in the Analytic Hierarchy Process (AHP). According to the work of Oskar Perron, the principal eigenvector can be obtained as the limiting power of a positive matrix. In this paper we show that the principal eigenvector does not need the theory of Perron for its existence based on the fact that the principal eigenvalue and corresponding principal eigenvector are transparently obtained for a consistent matrix. By perturbation theory the result is obtained for a near consistent matrix. Keywords: Theory of Perron, AHP eigenvector, pairwise comparison matrix solution 1. Introduction We begin by honoring the name of Oskar Perron who proved a very powerful theorem in mathematics about real positive matrices. I regret not meeting Perron (born in 1880) before he died in Munich, Germany, in 1975. My work on the Analytic Hierarchy Process (AHP) with real positive reciprocal matrices leads to the principal eigenvalue and eigenvector without the need for invoking any of Perron’s results as will be shown in this paper. He proved that if ( ) is an ijA a n n= × real positive matrix: 0 for 1 , ,ija i j n> ≤ ≤ then there is a positive real number maxλ , called the Perron root or the Perron–Frobenius eigenvalue, such that maxλ is an eigenvalue of A and any other eigenvalue λ (possibly, complex) is strictly smaller than maxλ in absolute value, |λ| < maxλ . There exists an eigenvector w = (w1,…,wn) of A with eigenvalue maxλ such that all components of w are positive: max , 0 for 1 i .iAw w w nλ= > ≤ ≤ There also exists a positive left eigenvector max: , 0. T T iv v A v vλ= > Perron also proved that the principal eigenvector w corresponding to maxλ can be obtained by raising the matrix A to infinite powers. mailto:saaty@katz.pitt.edu http://en.wikipedia.org/wiki/Complex_number Rob Typewritten Text http://dx.doi.org/10.13033/ijahp.v5i2.191 IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 269 Vol. 5 Issue 2 2013 ISSN 1936-6744 His theorem was later modified by Frobenius whose theory assures us of a similar result except that he proved the root may no longer be simple if there are zero entries in the matrix. In the Analytic Hierarchy Process the pairwise comparison judgment matrices are real, positive and reciprocal 1 ( ( ), 0, for all and ) ,ij ji ji ij A a a a i j a = > = and their order is not much larger than 7x7. The entries values lie between 9 and 1/9. Interestingly, as we shall show here, the principal eigenvalue and its principal eigenvector can be found for a real reciprocal positive matrix of small order without Perron’s theory. The principal eigenvalue and eigenvector can be obtained from the solution of a system of equations without using the powers of the matrix as does Perron. We observe that if we know either maxλ or w , we also know the other. If, for example, we know maxλ , we get w by solving, in the familiar way, the homogenous system of linear equations: max 1 , 1,..., . n ij j i j a w w i nλ = = =∑ If we know w then because of the normalization condition 1 1 n i i w = =∑ in our case, after taking the sum on both sides of the equation with respect to i and interchanging the sums on the left we obtain: max max 1 1 1 . n n n j ij i j i i w a wλ λ = = = = =∑ ∑ ∑ In other words we obtain maxλ as the scalar product of the vector w with the vector of column sums of the matrix A. If the matrix has real coefficient that are positive, and if w is real and positive then maxλ is real and positive. But, we have not yet established that it is a simple eigenvalue and that it dominates all other eigenvalues in modulus. We follow two routes to obtain the principal eigenvector :w one is by using the general idea of perturbation of the coefficients of a consistent matrix, which also involves reciprocal values in the transpose position, and the other is by considering the graph- theoretic concept of dominance along paths of different lengths leading to Cesaro summability. For us, applied to AHP reciprocal matrices, Cesaro summability says that the average value of the normalized vector of the row sums of the powers of a positive reciprocal matrix is equal to the normalized vector of the row sums of the limiting power of that matrix, which, of course, according to Perron, is the principal eigenvector of that matrix. The latter again gives the same answer as the theory of Perron does without the need for Perron’s logic, whether that matrix is consistent (trivial because Ak = nk-1A) or inconsistent, by perturbation arguments from the work of J. H. Wilkinson that it yields the principal eigenvalue and eigenvector in the limit. IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 270 Vol. 5 Issue 2 2013 ISSN 1936-6744 2. Consistent Positive Reciprocal Matrices Assume that one is given n stones, 1,..., nA A , that we will refer to as alternatives, having known weights ,...,i nw w , respectively. We form a matrix A of pairwise ratios whose ith row gives the ratios of the weights of the ith stone with respect to all the others. 𝐴1 𝐴2 ⋯ 𝐴𝑛 𝐴1 𝑤1/𝑤1 𝑤1/𝑤2 ⋯ 𝑤1/𝑤𝑛 𝐴 = 𝐴2 𝑤2/𝑤1 𝑤2/𝑤2 ⋯ 𝑤2/𝑤𝑛 ⋮ ⋮ ⋮ ⋯ ⋮ 𝐴𝑛 𝑤𝑛/𝑤1 𝑤𝑛/𝑤2 ⋯ 𝑤𝑛/𝑤𝑛 We note that we can recover the vector of weights ,...,i nw w w= by solving the system of equations defined below: Aw = 𝑤1/𝑤1 𝑤1/𝑤2 ⋯ 𝑤1/𝑤𝑛 𝑤2/𝑤1 𝑤2/𝑤2 ⋯ 𝑤2/𝑤𝑛 ⋮ ⋮ ⋯ ⋮ 𝑤𝑛/𝑤1 𝑤𝑛/𝑤2 ⋯ 𝑤𝑛/𝑤𝑛 𝑤1 𝑤2 ⋮ 𝑤𝑛 = 𝑛 𝑤1 𝑤2 ⋮ 𝑤𝑛 = 𝑛𝑤 Solving this homogeneous system of linear equations Aw nw= to find w is a trivial eigenvalue problem, because the existence of a solution depends on whether or not a solution exists if n is an eigenvalue of the characteristic equation of .A But the matrix A has rank one and thus all its eigenvalues but one are equal to zero. The sum of the eigenvalues of a matrix is equal to its trace, the sum of its diagonal elements, which in this case is equal to n. Thus n is the largest, or the principal, eigenvalue of A and w is its corresponding principal eigenvector and it is positive and unique to within multiplication by a constant, and thus belongs to a ratio scale. We now know what must be done to recover the weights iw , whether they are known in advance or not. Definition: An n by n matrix ( )ijA a= is consistent if , , 1,...,ij jk ika a a i j k n= = holds among its entries. A consistent matrix always has the form ( )i j w A w = and we know that 1k kA n A−= . The consistent case has no need for the theorem of Perron to prove the existence of a largest real eigenvalue and its corresponding positive eigenvector, nor to prove that this vector is the limit to which powers of the matrix converge. Of course, real world IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 271 Vol. 5 Issue 2 2013 ISSN 1936-6744 reciprocal pairwise comparison matrices are very unlikely to be consistent unless they use actual measurement data. Now, we give a mathematical discussion to show why when a matrix is inconsistent we still need the principal right eigenvector for our priority vector. It is clear that no matter what method we use to derive the weights ,iw we need to get them back so they are proportional to the expression 1 1,..., n ij j j a w i n = =∑ , that is, we must solve 1 = for 1,..., n ij j i j a w cw i n = =∑ . Otherwise, 1 for 1,..., n ij j j a w i n = =∑ would yield another set of different weights and they in turn could be used to form new expressions 1 1,..., n ij j j a w i n = =∑ , and so on ad infinitum. Unless we solve the principal eigenvalue problem, our quest for priorities becomes meaningless. We learn from the consistent case that what we get on the right is proportional to the sum on the left that involves the same scale used to weight the judgments that we are looking for. Thus we have the proportionality constant c. A better way to see this is to use the derived vector of priorities to weight each row of the matrix and take the sum. This yields a new vector of priorities (relative dominance of each element) represented in the comparisons. This vector can again be used to weight the rows and obtain still another vector of priorities. In the limit (if one exists), the limit vector itself can be used to weight the rows and get the limit vector back perhaps proportionately. Our general problem, possibly with inconsistent judgments, takes the form: 𝐴𝑤 = 1 𝑎12 ⋯ 𝑎1𝑛 1 𝑎12 1 ⋯ 𝑎2𝑛 ⋮ ⋮ ⋮ ⋮ 1 𝑎1𝑛 1 𝑎2𝑛 ⋯ 1 𝑤1 𝑤2 ⋮ 𝑤𝑛 = 𝑐𝑤 This homogeneous system of linear equations Aw cw= has a solution w if c is the principal eigenvalue of .A That this is the case can be shown using an argument that involves both left and right eigenvectors of .A Two vectors 1 1( ,..., ), ( ,..., )n nx x x y y y= = are orthogonal if their scalar product 1 1 ... n nx y x y+ + is equal to zero. It is known that any left eigenvector of a matrix corresponding to an eigenvalue is orthogonal to any right eigenvector corresponding to a different eigenvalue. This property is known as biorthogonality. IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 272 Vol. 5 Issue 2 2013 ISSN 1936-6744 Theorem 1: For a given positive matrix ,A the only positive vector w and only positive constant c that satisfy ,Aw cw= is a vector w that is a positive multiple of the principal eigenvector of ,A and that the only such c is the principal eigenvalue of .A Proof: We know that the right principal eigenvector and the principal eigenvalue satisfy our requirements. We also know that the algebraic multiplicity of the principal eigenvalue is one, and that there is a positive left eigenvector of A (call it z ) corresponding to the principal eigenvalue. Suppose there is a positive vector y and a (necessarily positive) scalar d such that .Ay dy= If d and c are not equal, then by biorthogonality y is orthogonal to ,z which is impossible since both vectors are positive. If d and c are equal, then y and w are dependent since c has algebraic multiplicity one, and y is a positive multiple of .w Thus the proof is complete. Let ija be the relative dominance of iA over .jA In order to simplify the notation let the matrix corresponding to the reciprocal pairwise relation be denoted by ( )ijA a= . The relative dominance of iA over jA along paths of length k is given by ( ) 1 ( ) 1 1 n k ij j n n k ij i j a a = = = ∑ ∑ ∑ where ( )kija is the (i,j) entry of the k th power of the matrix ( )ija . The total dominance ( )iw A , of alternative i over all other alternatives along paths of all lengths is given by the infinite series ( ) 1 ( )1 1 1 ( ) n k i j j i n n kk i j i j a w A a ∞ = = = = = ∑ ∑ ∑∑ which coincides with the Cesaro sum in (Equation 1). (Equation 1) ( ) 1 ( )1 1 1 1 lim n k i jM j n nx kk i j i j a M a = →∞ = = = ∑ ∑ ∑∑ IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 273 Vol. 5 Issue 2 2013 ISSN 1936-6744 But this limit of weighted averages (the Cesaro sum) can be evaluated; we have for an n by n consistent matrix ( )where ( ) ( ) , 1,... ,iij ij j w A a a i j n w = = = that 1k kA n A−= and the limit in Equation 1 is simply the eigenvector w normalized. In general, it is also true that the Cesaro sum converges to the same limit as does its thk term /k T kA e e A e that yields k step dominance. Here we see that the requirement for rank takes on the particular form of the principal eigenvector. We will not assume it for the inconsistent case, but will prove its necessity again for that more general case. We now develop a necessary and sufficient condition for rank preservation. For emphasis, recall from graph theory that an element ( ) of Am mija gives the cumulative dominance of the ith element over the jth element along all chains of length m. That is precisely how one measures the consistency relation between that row and each column. In fact, when A is consistent we have from 1m mA n A−= that the entries of mA and those of A differ by a constant thus maintaining consistency. In general we have 𝐴𝑚 = (𝑎𝑖𝑗 (𝑚)) . Theorem 2: For a positive reciprocal matrix A lim 𝑚→∞ 𝑎𝑖𝑘 (𝑚) ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖−1 = lim 𝑚→∞ 𝑎𝑖𝑠 (𝑚) ∑ 𝑎𝑖𝑠 (𝑚)𝑛 𝑖−1 , 𝑘, 𝑠 = 1, 2, … , 𝑛. Proof. Let 𝐵 = 𝑁𝐴𝑁−1 (where N and N-1 are non-singular matrices that we will define later) be the Jordan canonical form of A given by: 𝐵 = 𝜆1 𝐵2 ⋱ 𝐵𝑟 , where 𝜆1 ≡ 𝜆𝑚𝑎𝑥 , and 𝐵𝑝 , 𝑝 = 2, 3, … , 𝑟 is the 𝑛𝑝 × 𝑛𝑝 Jordan block defined by 𝐵𝑝 = 𝜆𝑝 0 0 ⋯ 0 0 1 𝜆𝑝 0 0 0 0 1 𝜆𝑝 0 0 ⋮ ⋱ ⋮ 0 0 0 ⋯ 1 𝜆𝑝 , IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 274 Vol. 5 Issue 2 2013 ISSN 1936-6744 where 𝜆𝑝 , 𝑝 = 2, … , 𝑟 are distinct eigenvalues with multiplicities 𝑛2, … , 𝑛𝑟 respectively, and ∑ 𝑛𝑝 = 𝑛 − 1𝑟𝑝=2 . We have 𝐴 = 𝑁−1𝐵𝑁 and 𝐴𝑚 = 𝑁−1𝐵𝑚𝐵, where 𝐵𝑚 is given by: 𝐵𝑚 = 𝜆1 𝑚 0 ⋯ 0 0 𝐵2 𝑚 0 ⋮ ⋱ ⋮ 0 0 ⋯ 𝐵𝑟𝑚 , Let us denote 𝑁−1 ≡ 𝐷 = �𝑑𝑖𝑗� and 𝑁 = (𝑛𝑖𝑗). We have 𝐴𝑚𝐷𝐵𝑚𝑁 = 𝑛11𝑑11𝜆1 𝑚 + ⋯ , 𝑛12𝑑11𝜆1 𝑚 + ⋯ , … , 𝑛1𝑛𝑑11𝜆1 𝑚 + ⋯ 𝑛11𝑑21𝜆1 𝑚 + ⋯ , 𝑛12𝑑21𝜆1 𝑚 + ⋯ , … , 𝑛1𝑛𝑑21𝜆1 𝑚 + ⋯ ⋮ ⋱ ⋮ 𝑛11𝑑𝑛1𝜆1 𝑚 + ⋯ , 𝑛12𝑑𝑛1𝜆1 𝑚 + ⋯ , … , 𝑛1𝑛𝑑𝑛1𝜆1 𝑚 + ⋯ Let 𝑒 = (1, 1, … , 1)𝑇 = 𝑎1𝑤1 + ⋯ + 𝑎𝑟𝑤𝑟 , where 𝑤𝑝 is the principal right eigenvector corresponding to 𝜆𝑝. We have, 𝑒𝑇𝐴𝑚 = 𝑎1𝜆1 𝑚𝑤1𝑇 + ⋯ + 𝑎𝑟𝜆1 𝑚𝑤1𝑇 = �𝑛11 �𝑑11𝜆1 𝑚 + ⋯ , … , 𝑛1𝑛 �𝑑11𝜆1 𝑚 + ⋯ 𝑛 𝑖=1 𝑛 𝑖=1 �. Given two columns of 𝐴, 𝑘 and 𝑠 we have, 𝑎𝑖𝑘 (𝑚) ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 = 𝑛1𝑘𝑑𝑖1𝜆1 𝑚+⋯ 𝑛1𝑘 ∑ 𝑑𝑖1𝜆1 𝑚+⋯𝑛𝑖=1 and 𝑎𝑖𝑠 (𝑚) ∑ 𝑎𝑖𝑠 (𝑚)𝑛 𝑖=1 = 𝑛1𝑠𝑑𝑖1𝜆1 𝑚+⋯ 𝑛1𝑠 ∑ 𝑑𝑖1𝜆1 𝑚+⋯𝑛𝑖=1 Since both numerators and denominators are polynomials in 𝜆𝑝𝑚, 𝑝 = 1, 2, … , 𝑟, and 𝜆1 = 𝜆𝑚𝑎𝑥 > �𝜆𝑝�, 𝑝 ≠ 1, we have for the ith entries of two arbitrary columns k and s: lim 𝑚→∞ 𝑎𝑖𝑘 (𝑚) ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 = lim 𝑚→∞ 𝑎𝑖𝑠 (𝑚) ∑ 𝑎𝑖𝑠 (𝑚)𝑛 𝑖−1 = 𝑑𝑖1 ∑ 𝑑𝑖1𝑛𝑖−1 . Definition: A positive matrix 𝐴 is said to be m-dominant if there exists 𝑚0 such that for 𝑚 ⩾ 𝑚𝑛 either 𝑎𝑖𝑘 (𝑚) ⩾ 𝑎𝑖′𝑘 (𝑚) or 𝑎𝑖𝑘 (𝑚) ⩽ 𝑎𝑖′𝑘 (𝑚)for all k and for any pair i and i'. IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 275 Vol. 5 Issue 2 2013 ISSN 1936-6744 Corollary: A positive reciprocal matrix is asymptotically m-dominant. Proof. We have from Theorem 2 that the normalized columns of mA are the same in the limit. Since the elements in each row are identical, the result follows by choosing 0m to be the maximum of its values for each pair of rows. We now show that the rank of an inconsistent matrix A is determined in terms of the powers of .A To do this, we demonstrate that there is a method of estimating w which coincides with the normalized limiting columns of .A This method is precisely the eigenvalue method. Theorem 3: lim 𝑚→∞ � 𝑎𝑖𝑘 (𝑚) ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 � = 𝑤𝑖, 𝑖 = 1,2, … , 𝑛. Proof: From lim 𝑚→∞ � 𝐴 𝑚𝑒 �|𝐴𝑚|� � = 𝑤, we have 𝑤𝑖 = lim𝑚→∞ � 1 �|𝐴𝑚|� �∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑘=1 . Multiplying and dividing 𝑎𝑖𝑘 (𝑚) by ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑘=1 , and then re-grouping the terms, we have, on distributing the limit with respect to the finite sum, 𝑤𝑖 = � lim𝑚→∞ � 𝑎𝑖𝑘 (𝑚) �|𝐴𝑚|� , ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 � 𝑛 𝑘=1 = � � lim 𝑚→∞ 𝑎𝑖𝑘 (𝑚) ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 � 𝑛 𝑘=1 � lim 𝑚→∞ ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 �|𝐴𝑚|� �. By Theorem 2 lim 𝑚→∞ 𝑎𝑖𝑘 (𝑚) ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 is the same constant for all 𝑘 hence we have 𝑤𝑖 = lim𝑚→∞ 𝑎𝑖𝑘 (𝑚) ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 � � lim 𝑚→∞ ∑ 𝑎𝑖𝑘 (𝑚)𝑛 𝑖=1 �|𝐴𝑚|� � 𝑛 𝑘=1 . Since �|𝐴𝑚|� = ∑ ∑ 𝑎𝑖𝑘 (𝑚),𝑛𝑘=1 𝑛 𝑖=1 the proof is complete. There is a natural way to derive the rank order of a set of alternatives from a pairwise comparison matrix A. The rank order of each alternative is the relative proportion of its dominance over the other alternatives. This is obtained by adding the elements in each row in A and dividing by the total. However, A only captures the dominance of one alternative over each other in one step. But an alternative can dominate a second by first dominating a third alternative, and then the third dominates the second. Thus, the first alternative dominates the second in two steps (along a path of length two). It is known that the result for dominance in two steps is obtained by squaring the pairwise comparison matrix. Similarly dominance can occur in three steps, four steps and so on, IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 276 Vol. 5 Issue 2 2013 ISSN 1936-6744 the value of each obtained by raising the matrix to the corresponding power. The rank order of an alternative is the average of the relative values for dominance in one step, two steps, and so on. We show below that when we take this infinite series of dominance along paths of length one, two, three, and so on and calculate its limiting value we obtain precisely the principal right eigenvector of the matrix A. This demonstrates that the eigenvector is derived deductively to obtain a relative scale among n alternatives from their matrix of pairwise comparisons. It is the desired solution because it preserves rank order rather than because it is a convenient criterion introduced for minimization purposes. We have Theorem 4 below from Saaty and Vargas (1984). Theorem 4: The relative dominance of an alternative is given by the solution of the eigenvalue problem 𝐴𝑤 = 𝜆𝑚𝑎𝑥𝑤. Proof: The relative dominance of an alternative along all paths of length 𝑘 ⩽ 𝑚 is given by 1 𝑚 ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 . Let 𝑠𝑘 = 𝐴𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 and 𝑡𝑚 = 1 𝑚 ∑ 𝑠𝑘𝑚𝑘=1 . Note that lim mm t→∞ < ∞ . This a consequence of a theorem due to G. H. Hardy (1949) which gives the necessary and sufficient conditions for a transformation of a convergent sequence to also be convergent. Let 𝑇 be such a transformation mapping (𝑠1, … , 𝑠𝑚) → 𝑡𝑚 = � 𝑐𝑚,𝑘𝑠𝑘 ∞ 𝑘=1 . 𝑇 is regular if 𝑡𝑚 → 𝑠 as 𝑚 → ∞ whenever 𝑠𝑘 → 𝑠 as k → ∞. It is known (Hardy, 1949) that 𝑇 is regular if and only if the following conditions hold: (1) ∑ �𝑐𝑚,𝑘� < 𝐻∞𝑘=1 (independent of 𝑚), (2) 𝑐𝑚,𝑘 → 𝛿𝑘 for each 𝑘, when 𝑚 → ∞, (3) ∑ 𝑐𝑚,𝑘 → 𝛿∞𝑘=1 when 𝑚 → ∞, (4) 𝛿𝑘 = 0 for each 𝑘, (5) 𝛿 = 1. Here, 𝑐𝑚,𝑘 = � 1 𝑚 𝑓𝑜𝑟 1 ⩽ 𝑘 ⩽ 𝑚 0 𝑓𝑜𝑟 𝑘 > 𝑚. , Thus, we have (1) ∑ �𝑐𝑚,𝑘� = ∑ � 1 𝑚 � = 1𝑚𝑘=1 ∞ 𝑘=1 , IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 277 Vol. 5 Issue 2 2013 ISSN 1936-6744 (2) 𝑐𝑚,𝑘 = 1 𝑚 → 0 as 𝑚 → ∞. Hence (4) 𝛿𝑘 = 0 for each 𝑘, (3) ∑ 𝑐𝑚,𝑘 = ∑ � 1 𝑚 � = 1 ∞𝑘=1 ∞ 𝑘=1 and hence (5) 𝛿 = 1. It follows that 𝑇 is regular. Since 𝑠𝑘 = 𝐴𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 → 𝑤 as 𝑘 → ∞ (Saaty, 1980), where 𝑤 is the principal right eigenvector of 𝐴 we have, 𝑡𝑚 = 1 𝑚 ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 → 𝑤𝑚𝑘=1 as 𝑚 → ∞ In input/output analysis in economics, multipliers are traced by raising the input/output matrix to higher and higher powers and taking their sums to obtain the overall impact of each sector of the economy on every other sector. Still, another argument can be constructed from Theorem 4 because for large m the normalized columns of Am are the same and converge to the principal eigenvector. Theorem 5: The Cesaro sum, 𝑙𝑖𝑚 𝑚→∞ 1 𝑚 ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 ,is the principal right eigenvector of A. Proof: By Theorem 4 of Saaty and Vargas (1984) we know that, lim 𝑚→∞ 1 𝑚 ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 = lim𝑚→∞ ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 Multiplying lim 𝑚→∞ 1 𝑚 ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 by 𝐴 on the left we have, 𝐴� lim 𝑚→∞ 1 𝑚 � 𝐴𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 � = 𝐴� lim 𝑘→∞ 𝐴𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 � = � lim 𝑘→∞ 𝑒𝑇𝐴𝑘+1𝑒 𝑒𝑇𝐴𝑘𝑒 �� lim 𝑘→∞ 𝐴𝑘+1𝑒 𝑒𝑇𝐴𝑘+1𝑒 � = � lim 𝑘→∞ 𝑒𝑇𝐴𝑘+1𝑒 𝑒𝑇𝐴𝑘𝑒 �� lim 𝑚→∞ 1 𝑚 � 𝐴𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 � There is a vector 𝑦 = lim 𝑚→∞ 1 𝑚 ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 and a constant 𝑑 = lim𝑘→∞ 𝑒 𝑇𝐴𝑘+1𝑒 𝑒𝑇𝐴𝑘𝑒 such that 𝐴𝑦 = 𝑑𝑦. Under the assumption that A has r distinct eigenvalues 1,..., rλ λ with multiplicities 1,..., rn n , respectively, by using the Jordan canonical form of A we can write 1A N BN−= where N is an invertible matrix and B is shown below: IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 278 Vol. 5 Issue 2 2013 ISSN 1936-6744 𝐵 = � 𝐵1 0 ⋯ 0 0 𝐵2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 𝐵𝑟 � and 𝐵𝑝 = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 𝜆𝑝 0 0 0 ⋯ 0 1 𝜆𝑝 0 0 ⋯ 0 0 1 𝜆𝑝 0 ⋯ 0 0 0 1 𝜆𝑝 ⋱ 0 ⋮ ⋮ ⋮ ⋱ ⋱ ⋮ 0 0 0 ⋯ 1 𝜆𝑝⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ , 𝑝 = 1, … , 𝑟. We have: 𝐴𝑘 = 𝑁−1𝐵𝑘𝑁, 𝐵𝑘 = ⎣ ⎢ ⎢ ⎡𝐵1 𝑘 0 ⋯ 0 0 𝐵2 𝑘 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 𝐵𝑟𝑘⎦ ⎥ ⎥ ⎤ The matrix kpB is shown in Equation 2 below. Letting 𝑁−1 = 𝐷 = �𝑑𝑖𝑗� and 𝑁 = �𝑛𝑖𝑗� we write: 𝐴𝑘 = 𝑁−1𝐵𝑘𝑁 = 𝐷𝐵𝑘𝑁. The first 1n columns of kA are given by 11 k nD B , as shown in Equation 3 below, and 𝐷𝑛1+1,𝑛2𝐵1 𝑘 as shown in Equation 4 below. IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 279 Vol. 5 Issue 2 2013 ISSN 1936-6744 (Equation 2) 𝐵𝑝𝑘 = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 𝜆𝑝 𝑘 0 0 0 ⋯ 0 𝑘𝜆𝑝𝑘−1 𝜆𝑝𝑘 0 0 ⋯ 0 𝑘(𝑘−1) 2! 𝜆𝑝𝑘−2 𝑘𝜆𝑝𝑘−1 𝜆𝑝𝑘 0 ⋯ 0 𝑘(𝑘−1)(𝑘−2) 3! 𝜆𝑝𝑘−3 𝑘(𝑘−1) 2! 𝜆𝑝𝑘−2 𝑘𝜆𝑝𝑘−1 𝜆𝑝𝑘 ⋱ 0 ⋮ ⋮ ⋮ ⋱ ⋱ ⋮ 𝑘(𝑘−1)⋯(𝑘−𝑛𝑝+1) (𝑛𝑝−1)! 𝜆𝑝 𝑘−𝑛𝑝+1 𝑘(𝑘−1)⋯(𝑘−𝑛𝑝+2) (𝑛𝑝−2)! 𝜆𝑝 𝑘−𝑛𝑝+2 𝑘(𝑘−1)⋯(𝑘−𝑛𝑝+3) (𝑛𝑝−3)! 𝜆𝑝 𝑘−𝑛𝑝+3 ⋯ 𝑘𝜆𝑝𝑘−1 𝜆𝑝𝑘⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ (Equation 3) 𝐷𝑛1𝐵1 𝑘 = ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡𝑑11𝜆1 𝑘 + 𝑑12 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑1𝑛1 � 𝑘 𝑛1 − 1 �𝜆1 𝑘−𝑛1+1 𝑑21𝜆1 𝑘 + 𝑑22 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑2𝑛1 � 𝑘 𝑛1 − 1 �𝜆1 𝑘−𝑛1+1 ⋮ 𝑑𝑛1𝜆1 𝑘 + 𝑑𝑛2 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑𝑛𝑛1 � 𝑘 𝑛1 − 1 �𝜆1 𝑘−𝑛1+1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡𝑑12𝜆1 𝑘 + 𝑑13 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑1𝑛1 � 𝑘 𝑛1 − 2 �𝜆1 𝑘−𝑛1+2 𝑑22𝜆1 𝑘 + 𝑑23 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑2𝑛1 � 𝑘 𝑛1 − 2 �𝜆1 𝑘−𝑛1+2 ⋮ 𝑑𝑛2𝜆1 𝑘 + 𝑑𝑛3 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑𝑛𝑛1 � 𝑘 𝑛1 − 2 �𝜆1 𝑘−𝑛1+2 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ , … , ⎣ ⎢ ⎢ ⎢ ⎡𝑑1𝑛1𝜆1 𝑘 𝑑2𝑛1𝜆1 𝑘 ⋮ 𝑑𝑛𝑛1𝜆1 𝑘⎦ ⎥ ⎥ ⎥ ⎤ ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 280 Vol. 5 Issue 2 2013 ISSN 1936-6744 (Equation 4) 𝐷𝑛1+1,𝑛2𝐵1 𝑘 = ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡𝑑11𝜆1 𝑘 + 𝑑12 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑1𝑛1 � 𝑘 𝑛1 − 1 �𝜆1 𝑘−𝑛1+1 𝑑21𝜆1 𝑘 + 𝑑22 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑2𝑛1 � 𝑘 𝑛1 − 1 �𝜆1 𝑘−𝑛1+1 ⋮ 𝑑𝑛1𝜆1 𝑘 + 𝑑𝑛2 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑𝑛𝑛1 � 𝑘 𝑛1 − 1 �𝜆1 𝑘−𝑛1+1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡𝑑12𝜆1 𝑘 + 𝑑13 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑1𝑛1 � 𝑘 𝑛1 − 2 �𝜆1 𝑘−𝑛1+2 𝑑22𝜆1 𝑘 + 𝑑23 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑2𝑛1 � 𝑘 𝑛1 − 2 �𝜆1 𝑘−𝑛1+2 ⋮ 𝑑𝑛2𝜆1 𝑘 + 𝑑𝑛3 � 𝑘 1 �𝜆1 𝑘−1 + ⋯ + 𝑑𝑛𝑛1 � 𝑘 𝑛1 − 2 �𝜆1 𝑘−𝑛1+2 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ , … , ⎣ ⎢ ⎢ ⎢ ⎡𝑑1𝑛1𝜆1 𝑘 𝑑2𝑛1𝜆1 𝑘 ⋮ 𝑑𝑛𝑛1𝜆1 𝑘⎦ ⎥ ⎥ ⎥ ⎤ ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ Thus, finally, we have: 𝐴𝑘 = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡�𝑑1𝑗𝑛𝑗1𝜆1 𝑘 + � 𝑑1𝑗𝑛𝑗1𝜆2 𝑘 + ⋯ 𝑛2 𝑗=𝑛1+1 𝑛1 𝑗=1 ⋯ �𝑑1𝑗𝑛𝑗𝑛𝜆1 𝑘 + � 𝑑1𝑗𝑛𝑗𝑛𝜆2 𝑘 + ⋯ 𝑛2 𝑗=𝑛1+1 𝑛1 𝑗=1 ⋮ ⋱ ⋮ �𝑑𝑛𝑗𝑛𝑗1𝜆1 𝑘 + � 𝑑𝑛𝑗𝑛𝑗1𝜆2 𝑘 + ⋯ 𝑛2 𝑗=𝑛1+1 𝑛1 𝑗=1 ⋯ �𝑑𝑛𝑗𝑛𝑗𝑛𝜆1 𝑘 + � 𝑑𝑛𝑗𝑛𝑗𝑛𝜆2 𝑘 + ⋯ 𝑛2 𝑗=𝑛1+1 𝑛1 𝑗=1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ . IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 281 Vol. 5 Issue 2 2013 ISSN 1936-6744 Let us assume 1 2 rλ λ λ≥ ≥ ≥ . Then we have the ratio: (Equation 5) lim 𝑘→∞ 𝑒𝑇𝐴𝑘+1𝑒 𝑒𝑇𝐴𝑘𝑒 = lim 𝑘→∞ ∑ 𝑑𝑖𝑗𝑛𝑖𝑗𝜆1 𝑘+1 + ∑ 𝑑𝑖𝑗𝑛𝑖𝑗𝜆2 𝑘+1 + ⋯𝑛2𝑖,𝑗=𝑛1+1 𝑛1 𝑖,𝑗=1 ∑ 𝑑𝑖𝑗𝑛𝑖𝑗𝜆1 𝑘 + ∑ 𝑑𝑖𝑗𝑛𝑖𝑗𝜆2 𝑘 + ⋯𝑛2𝑖,𝑗=𝑛1+1 𝑛1 𝑖,𝑗=1 Since 1λ is the principal eigenvector of A and 1Ay yλ= , then 𝑦 = lim𝑚→∞ 1 𝑚 ∑ 𝐴 𝑘𝑒 𝑒𝑇𝐴𝑘𝑒 𝑚 𝑘=1 is the principal right eigenvector of A. The ratio given in Equation 5 shows how to calculate the principal eigenvalue without solving the characteristic equation or using sophisticated mathematical software. We apply it to the following inconsistent pairwise comparison matrix A : 𝐴 = � 1 2 3 4 1/2 1 5 6 1/3 1/5 1 7 1/4 1/6 1/7 1 � 𝐴2 = � 1 4 9 16 1/4 1 25 36 1/9 1/25 1 7 1/16 1/36 1/49 1 � 𝐴10 = � 923637. 1.02324 × 106 2.52682 × 106 7.73336 × 106 850356. 941992. 2.32613 × 106 7.11986 × 106 369456. 409271. 1.01053 × 106 3.09298 × 106 119539. 132434. 327011. 1.00078 × 106 � 𝐴11 = � 4.21087 × 106 4.66478 × 106 1.15187 × 107 3.52551 × 107 3.87669 × 106 4.29457 × 106 1.06043 × 107 3.24561 × 107 1.68418 × 106 1.86579 × 106 4.60711 × 106 1.41002 × 107 544954. 603711. 1.49077 × 106 4.56261 × 106 � Computing the ratio below using the 10th and 11th power of the matrix A we obtain a first estimate of the eigenvalue. 𝑒𝑇𝐴11𝑒 𝑒𝑇𝐴10𝑒 = 136340459 29907404 = 4.558753 IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 282 Vol. 5 Issue 2 2013 ISSN 1936-6744 Going all the way to the 20th and 21st powers of the matrix and forming the ratio as shown below gives us a second somewhat different estimate of the eigenvalue that is quite close to the value computed by a presumably quite accurate commercial mathematical software package. 𝑒𝑇𝐴21𝑒 𝑒𝑇𝐴20𝑒 = 5.28609 × 1014 1.15953 × 1014 = 4.558805 The four eigenvalues of ,A obtained from the commercial mathematical software, are shown below: 1 2 3 4 4.558805319078529 0.03996796194816203 1.583542975555991 0.03996796194816203 1.583542975555991 0.4788693951822055 i i λ λ λ λ = = − + = − − = − We see that our final ratio has converged to the same value, 4.558805, to six decimal places, and we arrived at it using our method of raising the matrix to a sufficiently high power and computing the eigenvalue as a ratio of two successive powers. 3. Conclusion So, in conclusion, we have shown that for AHP pairwise comparison matrices, which are positive and reciprocal, we do not need to use the beautiful and general theorem of Perron. IJAHP Article: Saaty/ The Analytic Hierarchy Process without the theory of Oskar Perron International Journal of the Analytic Hierarchy Process 283 Vol. 5 Issue 2 2013 ISSN 1936-6744 REFERENCES Hardy, G.H. (1949) Divergent Series, New York: Oxford University Press. Horn, R.A. & C.R. Johnson (2012) Matrix Analysis, Cambridge: Cambridge University Press. Saaty, T.L. & L.G. Vargas (1984) Inconsistency and rank preservation, Journal of Mathematical Psychology, 28(2), 205-214. Saaty, T.L. (1980). The Analytic Hierarchy Process, New York: McGraw Hill International. Saaty, T.L. (1985). New Light on the Theorem of Perron, Homenaje Al Professor Sixto Rios, Trabajos De Estadistica Y De Investigacion Operativa, 36(3), 253-257. ABSTRACT 1. Introduction 2. Consistent Positive Reciprocal Matrices