A Cycle-Based Bound for Subdominant Eigenvalues of Stochastic Matrices

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1 A Cycle-Based Bound for Subdominant Eigenvalues of Stochastic Matrices Steve Kirkland Department of Mathematics and Statistics University of Regina Regina, Saskatchewan, Canada, S4S 0A2 August 3, 2007 Abstract Given a primitive stochastic matrix, we provide an upper bound on the moduli of its non-perron eigenvalues The bound is given in terms of the weights of the cycles in the directed graph associated with the matrix The bound is attainable in general, and we characterize a special case of equality when the stochastic matrix has a positive row Applications to Leslie matrices and to Google-type matrices are also considered Keywords: subdominant eigenvalue, stochastic matrix, directed graph AMS Classifications: 15A42, 15A51 Abbreviated Title: A Cycle-Based Bound for Stochastic Matrices 1 Introduction and preliminaries A square, entrywise nonnegative matrix S is stochastic if each of its row sums is 1 ie if S1 = 1, where 1 is the all ones vector of the appropriate order Stochastic matrices arise in the study of discrete time Markov chains that are Research partially supported by NSERC under grant number OGP

2 time homogenous and have a finite state space Evidently 1 is an eigenvalue of S, and it follows from Perron-Frobenius theory that for any eigenvalue λ of a stochastic matrix S, we have λ 1 Further, if S is primitive ie some power of S has all positive entries, then in fact if λ 1 is an eigenvalue of S then λ < 1 In the case that S is primitive, the sequence of powers S k, k = 1, 2, 3, converges as k That convergence is a central component of a corresponding result on the convergence of the Markov chain associated with S Specifically, when S primitive, the sequence of iterates in the corresponding Markov chain converges to the stationary vector for S, ie the left Perron vector w T for S, normalized so that w T 1 = 1 We say that λ is a subdominant eigenvalue of a primitive stochastic matrix S if λ is an eigenvalue of S having second-largest modulus after 1 Through a slight abuse of notation, we use λ 2 (S) to denote a subdominant eigenvalue for S (the abuse arising from the fact that S may have several subdominant eigenvalues) Evidently the rate of convergence of the sequence S k is governed by the moduli of the subdominant eigenvalues of S We refer the reader to [8] for further background, terminology and results on stochastic matrices and Markov chains For a given stochastic matrix S (or indeed any square entrywise nonnegative matrix), the question of whether S is primitive is purely combinatorial, as it depends only on the positions of the positive entries in S, not on their size Specifically, let (S) be the directed graph associated with S That is, if S is n n, say, the vertices of (S) are labeled 1,, n, and there is an arc i j in (S) if and only if s i,j > 0 Then S is primitive if and only if (S) is strongly connected and, in addition, the greatest common divisor of the lengths of the directed cycles in (S) is equal to 1 We refer the reader to [1] for these and other results on the relationship between a matrix and its directed graph Since the structure of (S) is influential on the primitivity of S, it is natural to wonder whether the influence of that combinatorial structure can be detected in the moduli of the subdominant eigenvalues of S In the special case that the Markov chain associated with S is time reversible, ie w i s i,j = w j s j,i for all i, j = 1,, n, there is an existing body of work relating the combinatorial features of (S) to the modulus of λ 2 (S) The reader is referred to [3] and the references therein for results in that direction We note in passing that the time reversible setting is highly structured: (S) can be thought of as an undirected graph (since in the time reversible case, for each i, j = 1,, n, (S) contains the arc i j if and only if it contains 2

3 the arc j i), while S itself is diagonally similar to a symmetric matrix, and so has all real eigenvalues For the case of a general irreducible stochastic matrix S, there are rather fewer results relating (S) to λ 2 (S) There is some work in that area, however; for instance, in both [6] and [7], hypotheses on (S) are used to produce lower bounds on λ 2 (S) The present paper continues in a similar spirit Specifically, we combine the combinatorial information in the cycles of (S) with the quantitative information in the entries of S to produce an upper bound on λ 2 (S) We note that there are a number of bounds on λ 2 (S) in the literature The following bound, which can be found in [8], will be particularly useful for our purposes Proposition 11 Suppose that S is a stochastic matrix of order n Let τ(s) = max{1 n min{s ik, s jk } i, j = 1,, n} k=1 Then for each eigenvalue λ 1 of S, we have λ τ(s) Observe that τ(s) = 1 if and only if S has a pair of rows with disjoint support - that is, a pair of rows i, j such that min{s i,k, s j,k } = 0 for all k We note in passing that if S is of order 2, then in fact we have λ 2 (S) = τ(s) However, as the following example shows, the bound provided by Proposition 11 may not perform so well, even for matrices of order 3 Example 12 Suppose that x (0, 1), and consider the matrix x x(1 x) (1 x) 2 S x = Then the eigenvalues of S x are 1, (1 x)e 2πi 3 and (1 x)e 2πi 3, but τ(s x ) = 1 for all x (0, 1) 2 A bound on subdominant eigenvalues Consider a stochastic matrix S, and let C be a cycle in (S) We use w(c) to denote the product of the entries in S that correspond to the arcs of C 3

4 We begin with a lower bound on the ratio of two entries in the stationary vector for S Lemma 21 Suppose that S is an irreducible n n stochastic matrix, and that i and j are distinct integers in {1,, n} Suppose further that (S) contains the path i k 0 k 1 k m+1 j For each p = 1,, m such that j k p, let C p denote the cycle k p k m+1 k p in (S) Then w j w i Π m p=0s kp,k p+1 Π m+1 p=1 (1 s kp,k p ) w(c 1 ) m p=2 j kp Π p 1 l=1 (1 s k l k l )w(c p ) (21) If equality holds in (21), then there is just one path from i to j in (S), namely the path k 0 k 1 k m+1 Proof: We suppose without loss of generality that j < i Let S i denote the principal submatrix of S formed by deleting row i and column i, and let r T denote the row vector of order n 1 formed by deleting the i th entry of e T i S Let ŵ T denote the vector formed from w T by deleting the i th entry It now follows from the eigenequation w T S = w T that ŵ T = w i r T (I S i ) 1, so that w j w i = r T (I S i ) 1 e j Since (S) contains the path k 0 k 1 k m+1, we have r T s k0,k 1 e T k 1, while the principal submatrix of S i on rows and columns k 1,, k m+1 dominates (entrywise) the matrix s k1 k 1 s k1 k s k2 k 2 s k2 k A = s kmkm s kmkm+1 s km+1 k 1 s km+1 k m s km+1 k m+1 Consequently, r T (I S i ) 1 e j s k0,k 1 e T k 1 (I A) 1 e km+1, and from the adjoint formula for the inverse of I A, we have s k0,k 1 e T k 1 (I A) 1 e km+1 = Π m p=0s kp,k p+1 Π m+1 p=1 (1 s kp,k p ) w(c 1 ) m p=2 j kp Π p 1 l=1 (1 s k l k l )w(c p ) 4

5 Thus inequality (21) holds Note that (I S i ) 1 = q=0 Sq i, from which we find that for each a, b i, the entry of (I S i ) 1 corresponding to row a column b of S can be written as the sum of the weights of all walks in (S) that start at vertex a, end at vertex b and that do not go through vertex i (here the weight of a walk is the product of the corresponding entries in S) In particular, if equality holds in (21), the only walks in (S) that start at i, end at j, and do not contain i as an intermediate vertex are the walks already accounted for in the expression for s k0,k 1 e T k 1 (I A) 1 e km+1 above It now follows that the only path in (S) from i to j is the path k 0 k 1 k m+1 Keeping the notation as above, we make the following definition For each i = 1,, n, we let α(i, i) = s i,i, and for any pair of distinct indices i, j = 1,, n, such that j i in (S), we let w(c α(i, j) = max{ 0 ) i = k Π m+1 p=1 (1 s kp,kp ) w(c 1) Pm Π p 1 l=1 (1 s 0 k m+1 = k l k l )w(c p) p=2 j kp j k 0 is a cycle in (S)} We take the convention if (S) does not contain the arc j i, then α(i, j) = 0 Here is one of our main results Theorem 22 Suppose that S is an irreducible n n stochastic matrix Then n λ 2 (S) max{1 min{α(i 1, j), α(i 2, j)} i 1, i 2 = 1,, n} (22) j=1 Proof: Denote the stationary vector for S by w T Let W = diag(w) and consider the stochastic matrix T given by T = W 1 S T W For any i j we have t i,j = w j w i s j,i, and applying Lemma 21, we find that for any cycle i = k 0 k m+1 = j k 0 in (S), we have w j w i s j,i w(c 0 ) Π m+1 p=1 (1 s kp,k p ) w(c 1 ) m p=2 j kp Π p 1 l=1 (1 s k l k l )w(c p ) It follows that for each i, j = 1,, n, t i,j α(i, j) Since T is similar to S T, we have λ 2 (S) = λ 2 (T ) max{1 n j=1 min{t i 1,j, T i2,j} i 1, i 2 = 1,, n} max{1 n j=1 min{α(i 1, j), α(i 2, j)} i 1, i 2 = 1,, n} 5

6 Remark 23 Observe that Theorem 22 provides a nontrivial bound on λ 2 (S) if and only if no pair of columns in S has disjoint support, or, equivalently, if and only if for each pair of vertices i, j of (S), there is a vertex k such that k i and k j Remark 24 Note that the cycle structure of (S) is reflected the quantities α(i, j), and hence in (22) Remark 25 The right hand side of (22) can be shown to be nondecreasing in each diagonal entry of S Example 26 Here we revisit the matrix of Example 12 Fix x (0, 1), and let x x(1 x) (1 x) 2 S x = Then we have α(1, 1) = x, α(2, 1) = x and α(3, 1) = 1, while for all other pairs (i, j), we have α(i, j) = 0 From (22) we find that λ 2 (S x ) 1 x As noted in Example 12, the eigenvalues of S x are 1, (1 x)e 2πi 3 and (1 x)e 2πi 3 Thus, not only does Theorem 22 provide a nontrivial upper bound on λ 2 (S x ), but in fact equality holds in (22) for S x Note that for any irreducible stochastic matrix that has a row with all positive entries, the upper bound of Theorem 22 is strictly less than 1, by Remark 23 Motivated in part by that observation, we next present a bound on λ 2 for irreducible stochastic matrices having a positive row The proof is immediate from Theorem 22 Corollary 27 Suppose that S is an irreducible stochastic matrix of order n, and that its first row has all positive entries Let β 1 = s 1,1, and for each i = 2,, n, let β i = w(c 0 ) Π m+1 p=1 (1 s kp,kp ) w(c 1) P m p=2 Πp 1 max{ i = k l=1 (1 s k l k l )w(c 0 k m+1 = 1 is p) a path in (S), where C p denotes the cycle k p k m+1 k p } Finally let µ(s) = min{β i, i = 1,, n} Then The following is a consequence of Corollary 27 λ 2 (S) 1 µ(s) (23) 6

7 Corollary 28 Suppose that S is an irreducible stochastic matrix, and that its first row has all positive entries Let a be the minimum positive entry in S, and let d be the maximum distance from a vertex i (S) to vertex 1 Then λ 2 (S) 1 2a+ad+2 1 2a+a d+1 Proof: Note that for any cycle C of length m in (S), w(c) a m It now follows that if there is a path of length l from vertex i 1 to vertex 1 in (S), then β i l+1 Since 0 < a 1, we find that a 1 a a 2 a l 2 a the expression l+1 a is nonincreasing in l, and that a l+1 1 a a 2 a l 1 a a 2 a l a Hence we have µ(s) d+1 It now follows from Corollary 27 that 1 a a 2 a d a λ 2 (S) 1 d+1 = 1 2a+ad+2 1 a a 2 a d 1 2a+a d+1 3 Characterization of equality in (23) Having established the upper bound (22) in Theorem 22, and seen via Example 12 that equality can hold, it is natural to wonder about which matrices yield equality in the bound That problem appears to be quite difficult, in part because it requires a characterization of equality in the bound of Proposition 11 In this section, we deal with a somewhat less daunting problem: that of characterizing the case of equality in (23) of Corollary 27 We begin with four different classes of examples for which equality holds in (23) Example 31 Suppose that for some z (0, 1), the stochastic matrix S has the form [ ] z y T S =, (34) 1 0 where I(i) y i z(1 z), i = 1,, n 1, and I(ii) y T 1 = 1 z Note that β 1 = z, and that for each i = 1,, n 1, β i+1 = y i z 1 z Hence we have µ(s) = z, so that by (23), λ 2 (S) 1 z Observe that S has rank 2, and so has just two nonzero eigenvalues Since S is stochastic, it has 1 as an eigenvalue, and by considering the trace of S, we deduce that λ 2 (S) = (1 z) In particular, we see that for the matrix S, equality holds in (23) The set of matrices S of the form (34) satisfying conditions I(i) and I(ii) will be referred as Class I 7

8 Example 32 Suppose that for some 1 p n 2, S is a stochastic matrix with a positive first row having the form x 1 x 2 x p 1 x p x p+1 x p+2 x n 1 x n y 2 1 y y 3 1 y y S = p 1 y p 0 0, y p y p y p y p y p y p y n 1 y n (35) where II(i) y i > 0, i = 2,, n, y p+1 < 1, and II(ii) m = gcd{n j p + 1 y p+j < 1} 2 Suppose further that for some z (0, 1), the following conditions hold: II(iii) x 1 z; II(iv) y j x j 1 P j 1 i=1 x i z for j = 2,, p; II(v) x j+p y p+1 y p+j = z(1 z) j 1 (1 p i=1 x i) for j = 1,, n p; and II(vi) z n p (1 z) j 1 j=1 y p+1 y p+j = 1 We note in passing that condition II(vi), in conjunction with condition II(v), ensures that the first row sum for S is 1 First, we claim that µ(s) = z To see the claim, first note that β 1 = x 1 z For each j = 2,, p, we have β j = y 2 y j x j y 2 y j 1 (1 P j 1 = y jx j i=1 x i) 1 P j 1 i=1 x i, which, by condition II(iv), is bounded below by z Similarly, we find that β p+1 = y p+1 x p+1 1 P, and that for j = 2,, n p, β y p i=1 x p+j = p+1 y p+j x p+j i 1 P p i=1 x i P j 1 i=1 y p+1y p+i x p+i It now follows from condition II(v) that β p+j = z for j = 1,, n p In particular, we find that µ(s) = z, and hence λ 2 (S) 1 z Next, we claim that equality holds in (23) To see this, we first note that the left stationary vector for S, w T, say, has the property that for each j = 1,, n, β j = w 1x j w j, so that in fact equality holds in (21) of Lemma 21 Letting T = W 1 S T W, where W = diag(w), we see that T can be written as T = z1e T 1 + (1 z)m, where M is a stochastic matrix Since β j+p = z for j = 1,, n p, we find that the principal submatrix of M on rows and columns p + 1,, n is stochastic Further, from condition II(ii), it follows 8

9 that that principal submatrix is periodic with period m Consequently, e 2πi m is an eigenvalue of M, so that (1 z)e 2πi m is an eigenvalue of S Hence λ 2 (S) = 1 z, as claimed The set of matrices S of the form (35) satisfying II(i)-II(vi) will be referred as Class II Example 33 Suppose that for some n m 2 we have n = rm 1 for some integers r, m 2, and that S is a stochastic matrix with a positive first row having the form S = x 1 x 2 x n 1 x n y y 2 0 y y y n 1 y n where III(i) y i > 0, i = 2,, n, and III(ii) y i < 1 only if i is a multiple of m, (36) Suppose also that for some z (0, 1), the following conditions hold: III(iii) x i = z(1 z) i 1, i = 1,, m 1; ( ) 1 β III(iv) x lm+i = x lm lm β lm z(1 z) i 1 for i = 1,, m 1, and l = 1,, r 1 It is then straightforward to determine that β m = ymxm and that for (1 z) m 1 l = 2,, r 1, β lm = y m y 2m y lm x lm (1 z) m 1 [(1 z) (m 1)(l 1) l 1 j=1 (1 z)(m 1)(l j 1) y m y 2m y jm x jm ] Suppose further that III(v) β lm z, l = 1,, r 1, and III(vi) (1 z) m 1 = r 1 q=1 x qm[(1 z) m (1 z)m 1 β lm ] First, we claim that µ(s) = z Note that by condition III(v), β lm z for l = 1,, r 1 Since y i = 1 for i = 1,, m 1, we have β j = x j 1 P j 1 i=1 x i for j = 1,, m 1, and so condition III(iii), combined with induction on j, shows that β j = z for j = 1,, m 1 Next, fix l between 1 and r 1, and note that y lm+i = 1 for i = 1, m 1 We have β lm+i = 9

10 y 2 y lm+i x lm+i 1 x 1 P lm+i 1 = j=2 y 2 y j x j y 2 y lm x lm+i 1 x 1 P lm 1 j=2 y 2 y j x j P lm+i 1 j=lm y 2y lm x j = x lm+i x lm P lm+i 1 β lm j=lm Applying condition III(iv), it now follows by induction on i that β lm+i = z for l = 1,, r 1 and i = 1,, m 1 Finally, we note that condition III(vi) simply ensures that when conditions III(iii) and III(iv) hold, the top row of S sums to 1 In particular, it now follows that µ(s) = z, and so by (23), we find that λ 2 (S) 1 z Next, we claim that equality holds in (23) As in Example 32, note that the left stationary vector for S, w T, say, has the property that for each j = 1,, n, β j = w 1x j w j, so that in fact equality holds in (21) of Lemma 21 Setting W = diag(w) and T = W 1 S T W, we see that T can be written as T = z1e T 1 +(1 z)m, where M is a stochastic matrix Note that M j,1 = β j z 1 z for each j = 1,, n In particular, since β lm+i = z for i = 1,, m 1, l = 0,, r 1, we have M j,1 > 0 only if m j; note also that M n,i > 0 only if m i by condition III(ii) It now follows that in (M), the length of each cycle is divisible by m We thus find that e 2πi m (1 z)e 2πi m is an eigenvalue of M, so that is an eigenvalue of S Hence λ 2 (S) = 1 z, as claimed The set of matrices S of the form (36) satisfying conditions III(i)-III(vi) will be referred as Class III Example 34 Suppose that for some n m 2, and some a, r IN, we have n = am 1 + r, and that S is a stochastic matrix with a positive first row having the form S = x 1 x 2 x n r 1 x n r x T y v 2 u T 2 0 y 3 0 v 3 u T y n r v n r u T n r where IV(i) x T = [ x n r+1 x n ], IV(ii) y i > 0, i = 2, n r, and IV(iii) y i + v i + u T i 1 = 1, i = 2, n r Suppose further that IV(iv) v i > 0 only if m i, and that IV(v) u T i 0 T only if m (i 1) x j, (37) 10

11 If a = 1, suppose also that for some z (0, 1), IV(vi) x i = z(1 z) i 1, i = 1,, m 1; IV(vii) x n r+j z(1 z) m 1, j = 1,, r; and IV(viii) r j=1 x n r+j = (1 z) m 1 Observe that from the hypotheses on v i and u T i, we find that in fact y i = 1 for i = 2,, n r Hence we have β j = for j = 2,, n r, while for j = n r + 1,, n, β j = x j 1 P n r i=1 x i x j 1 P j 1 i=1 x i Applying conditions IV(vi) and IV(vii), we see that β j = z, j = 1,, n r, while for j = n r + 1,, n, β j = x j z Thus we have µ(s) = z, so that by (23), λ (1 z) m 1 2 (S) 1 z Next, we claim that equality holds in (23) As in the previous example, note that equality holds in (21) of Lemma 21 for the left stationary vector w T of S Letting W = diag(w) and T = W 1 S T W, we see that T can be written as T = z1e T 1 + (1 z)m, where M is a stochastic matrix From the fact that β j = z for j = 1,, n r, it then follows from conditions IV(iv) and IV(v) that the length of every cycle in (M) is divisible by m We thus find that e 2πi m is an eigenvalue of M, so that (1 z)e 2πi m is an eigenvalue of S Hence λ 2 (S) = 1 z, as claimed On the other hand, if a 2, suppose, in addition to conditions IV(i)- IV(iii), that for some z (0, 1), IV(vi) x i = z(1 z) i 1, i = 1,, m 1; IV(vii) x lm+1 y lm+1 = z 1 β lm β lm x lm ; IV(viii) x lm+i = x lm+1 (1 z) i 1 for i = 1,, m 1; and IV(ix) x n r+j x n r (1 z), j = 1,, r It is straightforward to show that for each l = 1,, a 1, β lm = y 2 y lm x lm (1 z) m 1 [(1 z) (m 1)(l 1) P l 1 j=1 (1 z)(m 1)(l j 1) y 2 y jm x jm ] Suppose further that IV(x) β lm z for l = 1,, a 1; and IV(xi) (1 z) m 1 = a 1 l=1 x lm(1 + ( 1 β lm β lm y lm+1 )(1 (1 z) m 1 )) + r j=1 x n r+j Note that from the hypotheses on v i and u T i, we find that for each l = 0,, a 1, and i = 2,, m 1, y lm+i = 1 Also, for each j = 2,, n r, y we have β j = 2 y j x j 1 x 1 P j 1, while for j = 1,, r, we have β i=2 y 2y i x n r+j = i y 2 y n r x n r+j 1 x 1 P n r In particular, for each i = 1,, n r 1, we find that i=2 y 2y i x i β i+1 = y i+1x i+1 β i, while for each j = 1, r, we have β x i (1 β i ) n r+j = x n r+jβ n r x n r (1 β n r ) From conditions IV(vi) - IV(viii), it now follows that for each j = 1,, n r 11

12 such that j is not a multiple of m, β j = z Condition IV(ix) also yields the fact that β n r+j z, j = 1,, n r From condition IV(x), we have β lm z, l = 1,, a 1 Hence we find that µ(s) = z, so that from (23), λ 2 (S) 1 z Next, we claim that equality holds in (23) As above, note that the left stationary vector for S, w T, say, has the property that for each j = 1,, n r, β j = w 1x j w j, while for j = 1,, r, w 1x n r+j w n r+j β n r+j Letting W = diag(w) and T = W 1 S T W, we see that T can be written as T = z1e T 1 + (1 z)m, where M is a stochastic matrix Note that for j = 1,, n r, M j,1 = β j z, 1 z so that in particular, for such j, M j,1 > 0 only if m j Note also that for j = 1,, r, M n r+j,i > 0 only if m (i 1), while M n r,i > 0 only if m i It now follows that in (M), the length of each cycle is divisible by m We thus find that e 2πi m is an eigenvalue of M, so that (1 z)e 2πi m is an eigenvalue of S Hence λ 2 (S) = 1 z, as claimed The set of matrices S of the form (37) satisfying conditions IV(i)-IV(v) and either IV(vi)-IV(viii) or IV(vi) -IV(xi) will be referred as Class IV The rest of this section is devoted to providing a converse to Examples 31-34, namely that the matrices of Classes I-IV are, up to permutation similarity, the only ones yielding equality in (23) Throughout the remainder of this section, we assume that S is an irreducible stochastic matrix whose first row has all positive entries, and such that equality holds in (23) Let w T be the left stationary vector for S, set W = diag(w), and let T = W 1 S T W Finally, let M be given by M = 1 T µ(s) 1 µ(s) 1 µ(s) 1eT 1, where µ(s) is defined in Corollary 23 Observe that M is stochastic; further, since left non-perron eigenvectors for T are also eigenvectors for M, we find that because λ 2 (T ) = λ 2 (S) = 1 µ(s), it must be the case that λ 2 (M) = 1 Throughout, we let J = {j 1 β j = µ(s)}, and observe that J may be empty Lemma 35 Suppose that i J Then there is just one path from i to 1 in (S), say i = k 0 k m+1 = 1 Further, each of vertices k 1,, k m+1 has indegree two in (S) Proof: Since β i = µ(s), then necessarily equality holds in (21); the fact that there is a unique path from i to 1 now follows from Lemma 21 Denote that path by i = k 0 k m+1 = 1 Suppose that for some j = 1,, m + 1, vertex k j has indegree at least 3 Then there is a vertex p 1, k j 1 such that 12

13 p k j Observe that p i, otherwise there is more than one path from i to 1 Considering the walk k 0 k m+1 p k j k m+1, and referring to the proof of Lemma 21, we see that this walk makes a contribution to w 1 w i that is unaccounted for in the right side of (21); in particular, inequality (21) must be strict, a contradiction We conclude that each of vertices k 1,, k m+1 has indegree two, as desired Corollary 36 Suppose that we have distinct indices i, j J Then either the path from i to 1 contains the path from j to 1, or the path from j to 1 contains the path from i to 1 Proof: Suppose that the unique paths from i to 1 and j to 1 are i = a 0 a p a p+1 = 1 and j = b 0 b q b q+1 = 1, respectively Without loss of generality, we assume that p q; we claim that b q+1 k = a p+1 k for k = 0, 1,, q + 1 We establish the claim by induction on k, and note that certainly the claim holds for k = 0, since b q+1 = 1 = a p+1 Suppose now that for some k q we have b q+1 k = a p+1 k By Lemma 35, b q+1 k has indegree two, so that only vertex 1 and vertex b q k have arcs into b q+1 k However, since a p+1 k = b q+1 k, a similar argument shows that only vertex 1 and vertex a p k have arcs into a p+1 k We conclude then that a p k = b q k, completing the induction step The claim now follows, and hence we see that the path from i to 1 contains the path from j to 1 Corollary 37 If J, then there is a unique index k J such that for each j J, the path from k to 1 in (S) contains the path from j to 1 in (S) Proof: For each vertex i 1, let d(i, 1) be the distance in (S) from i to 1 If J has cardinality one, the result is immediate, so suppose that J has at least two elements Let k be a vertex in J such that d(k, 1) d(j, 1) for each j J Fix a j J such that j k; by Corollary 36, either the path from j to 1 contains the path from k to 1, or vice versa If the former holds, then d(j, 1) > d(k, 1), a contradiction It now follows that for each j J, the path from k to 1 contains the path from j to 1, and that k is the unique vertex in J with that property 13

14 Lemma 38 One of the following holds: i) M is reducible with a single periodic essential class; ii) M is irreducible and periodic Proof: Since λ 2 (M) = 1, then it is well-known (see [8], for example) that either M has two or more essential classes, or M is reducible with a single periodic essential class, or M is irreducible and periodic We claim that the first case cannot arise To see the claim, let C 1 be the (possibly empty) set of inessential indices for M, and let C 2,, C k be the classes of essential indices for M (note that k 3) It follows that M can be permuted and partitioned as M = M 1,1 M 1,2 M 1,3 M 1,k 0 M 2, M 3, M k,k where the partitioning of M corresponds to the sets C 1,, C k Note that if index 1 is not in C 1, then it follows that S is reducible, contrary to hypothesis Hence 1 C 1 Observe that necessarily, it must be the case that if i C 2 and j C 3, we have β i = µ(s) = β j However, it is not the case that in (S), either the path from i to 1 contains the path from j to 1, or vice versa, contradicting Corollary 36 Hence M has at most one essential class, as claimed The conclusion now follows, Corollary 39 If J =, then S has the form [ ] µ(s) y T S =, (38) 1 0 where y i > µ(s)(1 µ(s)), i = 1,, n 1 and y T 1 = 1 µ(s) In particular, S is in Class I Proof: Since J =, s 1,1 = β 1 = µ(s) < β i for i = 2,, n 1 Observe that (M) differs from (S T ) only in the fact that the former has no loop at vertex 1 In particular, M is irreducible, and, from Lemma 38, M is 14

15 also periodic For each i = 2,, n, (M) contains the arc i 1, so it follows that the index of periodicity for M is 2 We now deduce that for some positive vector y, S has the form [ ] µ(s) y T S = 1 0 Evidently it must be the case that y T 1 = 1 µ(s) Further, since β i = y i 1 1 µ(s) for i = 2,, n, we must have y i > µ(s)(1 µ(s)) for each such i Thus, S is in Class I Lemma 310 Suppose that M is reducible with a single periodic essential class, say C 2, and let C 1 denote the set of inessential indices for M Then 1 C 1, and there is a single pair of vertices u, v such that u C 2, v C 1, and (S) contains the arc u v There is a unique path from v to 1 in (S) Let k be the unique vertex in J whose path to 1 contains the path from j to 1 for each j J Then the subgraph of (S) induced by C 2 consists of a single Hamilton path from k to u, along with arcs into k so as to yield a strongly connected periodic directed graph Proof: Observe that M can permuted and partitioned to the form [ ] M1,1 M M = 1,2, 0 M 2,2 where the partitioning corresponds to C 1 and C 2 Note that necessarily 1 C 1, otherwise S is reducible Further, we have C 2 J, and the principal submatrix of M on the rows and columns corresponding to C 2 is irreducible, periodic, and stochastic From the fact that the subgraph of (S) induced by C 2 is strongly connected, and the fact that each index in C 2 has a unique path to 1, we find that in (S) there is a single arc from a vertex in C 2 to a vertex in C 1, say u v Necessarily, there is a unique path from v to 1 in (S) Since the path from k to 1 contains the path from u to 1, we find that necessarily k C 2 and that the subgraph of (S) induced by C 2 contains a Hamilton path from k to u From Lemma 35, each vertex on that Hamilton path that is distinct from k has indegree two in (S), so the only possible other arcs in that subgraph are arcs into vertex k 15

16 Remark 311 It is straightforward to show, using the structure described in Lemma 310, that if S satisfies the hypotheses of that lemma, then it is permutationally similar to a matrix of the form x 1 x 2 x p 1 x p x p+1 x p+2 x n 1 x n y 2 1 y y 3 1 y y p 1 y p 0 0, y p y p y p y p y p y p y n 1 y n (39) where a) y i > 0, i = 2,, n, y p+1 < 1, and b) m gcd{n j p + 1 y p+j < 1} 2 Here m is the period of the principal submatrix of S on the rows and columns corresponding to C 2 Lemma 312 Suppose that principal submatrix of M on its last n p rows and columns is irreducible, stochastic, and periodic with period m Suppose also that S has the form (39) and satisfies conditions a) and b) of Remark 311 Then x 1 µ(s), and for j = 2,, p we have y j x j 1 P j 1 i=1 x i µ(s) For j = 1,, n p we have x j+p y p+1 y p+j = µ(s)(1 µ(s)) j 1 (1 p i=1 x i) Finally, we have µ(s) n p j=1 = 1 In particular, S is in Class II (1 µ(s)) j 1 y p+1 y p+j Proof: Necessarily we have µ(s) = min j β j, and from the structure of S we see that necessarily β p+j = µ(s) for j = 1,, n p It is straightforward to see that for j = 2,, p, β j =, and so the first two conclusions y jx j 1 P j 1 i=1 x i follow from the fact that β j µ(s) for j = 1,, p We find that β p+1 = y p+1x p+1 1 P, and that for j = 2,, n p, β p i=1 x p+j = i y p+1 y p+j x p+j 1 P p i=1 x i P j 1 Since β i=1 y p+1y p+i x p+j = µ(s), j = 1,, n p, a straightforward induction proof shows that x j+p y p+1 y p+j = µ(s)(1 µ(s)) j 1 (1 p+i p i=1 x i) for each such j The last equation follows from the fact that n i=1 x i = 1 Hence S is in Class II 16

17 Lemma 313 Suppose that M is irreducible and periodic with period m Partition the vertex set of (M) into the classes C 1, C 2,, C m, so that the only arcs in (M) are of the form u v, where for some i = 1,, m, u C i and v C (i 1) mod m Suppose without loss of generality that 1 C 1 Then (M) has one of the following two forms: i) a Hamilton path from a vertex k C m 1 to vertex 1, along with possible arcs into k from vertices in C m and possible arcs from vertex 1 to one or more of the vertices in C m ; ii) a collection of vertices v 1,, v r C m each with a single arc out to a vertex k C m 1, along with a path from k to 1 through all remaining vertices, possible arcs into k from vertices in C m, and possible arcs from vertices in C 1 into v 1,, v r Proof: Note that m 1 i=1 C i J, and let k be the vertex in J whose path (in (S)) to 1 contains all other paths to 1 that start from vertices in J If it were the case that k C i for some i = 1,, m 2, then there would be a vertex l C i+1 J such that l k, a contradiction to the fact that the path from k to 1 contains that from l to 1 We conclude that necessarily k C m 1 or k C m Suppose that k C m 1 and consider the path P in (S) (and hence in (M)) from k to 1 Observe that each vertex in m 1 i=1 C i sits on P If P is a Hamilton path, then the form described in i) follows readily from the fact that only vertex k can have indegree greater than two in (S) If P is not a Hamilton path, then the only vertices not on P are necessarily in C m Label these vertices as v 1,, v r The form described in ii) now follows upon noting that the only vertices that can have indegree larger than two in (S) are k and v 1,, v r Finally, if k C m, then in fact the path from k to 1 must be a Hamilton path, otherwise there is a vertex l C 1, l 1 such that l k, a contradiction The form described in ii) (with r = 1 and v 1 = k) now follows Remark 314 Suppose that M satisfies the hypotheses of Lemma 313 If form i) holds, then necessarily n = rm 1 for some integer r 2 Further, it is straightforward to show that in that case, S is permutationally similar 17

18 to a matrix of the form x 1 x 2 x n 1 x n y y 2 0 y y y n 1 y n, (310) where a) y i > 0, i = 2,, n and b) y i < 1 only if i is a multiple of m Lemma 315 Suppose M is irreducible and periodic with period m Suppose also that S has the form (310) and satisfies conditions a) and b) of Remark 314 Write n = rm 1 for some r 2 Then x i = µ(s)(1 µ(s)) i 1, i = 1,, m 1, and x lm+i = x lm ( 1 β lm β lm ) µ(s)(1 µ(s)) i 1 for i = 1,, m 1, and l = 1,, r 1, where β lm = y m y 2m y lm x lm (1 µ(s)) m 1 [(1 µ(s)) (m 1)(l 1) l 1 j=1 (1 µ(s))m 1)(l j 1) y m y 2m y jm x jm ] y for l = 2,, r 1 Further, β m = mx m µ(s), and β (1 µ(s)) m 1 lm µ(s), l = 1,, r 1 Finally, we have (1 µ(s)) m 1 = r 1 q=1 x qm[(1 µ(s)) m (1 µ(s)) m 1 β lm ] In particular, S is in Class III Proof: Since M is irreducible and periodic with period m, we find that β lm+i = µ(s) for each i = 1,, m 1, and l = 0,, r 1 Now β 1 = x 1, and for y each j 2, β j = 2 y j x j 1 x 1 P j 1 Note also that y i=2 y 2y i x j < 1 only if m divides i j Since β j = µ(s) for j = 1,, m 1, an induction proof shows that x i = µ(s)(1 µ(s)) i 1, i = 1,, m 1 It now follows that β m = which must necessarily be bounded below by µ(s) For each l 1, we have β lm+1 = ymy lmx lm+1 1 x 1 P lm = x lm+1 i=2 y x 2y i x lm i x β lm lm y mx m (1 µ(s)) m 1, = x lm+1β lm x lm (1 β lm ) Since β lm+1 = µ(s), we find that x lm+1 = x lm µ(s) 1 β lm β lm Further, since β lm+i = µ(s) for i = 1,, m 1, it follows by induction that x lm+i = x lm 1 β lm β lm µ(s)(1 µ(s)) i 1, i = 1,, m 1 From that we deduce that for each l 1, m 1 i=0 y 2 y lm+i x lm+i = y m y lm x lm [(1 µ(s)) m (1 µ(s))m 1 β lm ]

19 An argument by induction on l now shows that β lm = y m y 2m y lm x lm (1 µ(s)) m 1 [(1 µ(s)) (m 1)(l 1) l 1 j=1 y my 2m y jm x jm ] for l = 2,, r 1 Evidently β lm µ(s) for l = 1,, r 1 Finally, the fact that (1 µ(s)) m 1 = r 1 q=1 x qm[(1 µ(s)) m (1 µ(s))m 1 now follows from the above considerations, and the condition that n i=1 x n = 1 We conclude that S is in Class III β lm ] Remark 316 Suppose that M satisfies the hypotheses of Lemma 313 If form ii) holds, then necessarily n = am 1 + r for some a, r IN It is straightforward to show that in that case, S is permutationally similar to a matrix of the form x 1 x 2 x n r 1 x n r x T y v 2 u T 2 0 y 3 0 v 3 u T y n r v n r u T n r , (311) where a) x T = [ x n r+1 x n ], b) yi > 0, i = 2, n r, and c) y i + v i + u T i 1 = 1, i = 2, n r Further, d) v i > 0 only if i is a multiple of m, and e) u T i 0 T only if i 1 is a multiple of m The proof of the following result is analogous to that of Lemma 315, and is omitted Lemma 317 Suppose that M is irreducible and periodic with period m Suppose also that S has the form (311), and satisfies conditions a)-e) of Remark 316 Write n r = am 1 for some a, r IN i) If n r = m 1, then x i = µ(s)(1 µ(s)) i 1, i = 1,, m 1, x n r+j µ(s)(1 µ(s)) m 1, j = 1,, r, and r j=1 x n r+j = (1 µ(s)) m 1 ii) If n r = am 1 for some a 2, then x i = µ(s)(1 µ(s)) i 1, i = 1,, m 1 For each l = 1,, a 1, β lm = y 2 y lm x lm (1 µ(s)) m 1 [(1 µ(s)) (m 1)(l 1) l 1 j=1 (1 µ(s))(m 1)(l j 1) y 2 y jm x jm ], 19

20 x lm+1 y lm+1 = µ(s) 1 β lm β lm x lm, and x lm+i = x lm+1 (1 µ(s)) i 1 for i = 1,, m 1 Also, x n r+j x n r µ(s), j = 1,, r, and finally, we have (1 µ(s)) m 1 = a 1 l=1 x lm(1 + ( 1 β lm β lm y lm+1 )(1 (1 µ(s)) m 1 )) + r j=1 x n r+j In either case, S belongs to Class IV Here is the main result of this section Theorem 318 Suppose that S is an irreducible stochastic matrix whose first row is positive Equality holds in (23) if and only if S is permutationally similar to a matrix in one of Classes I - IV Proof: If S is permutationally similar to a matrix in one of Classes I - IV, then the fact that equality holds in (23) follows from Examples Conversely, suppose that equality holds in (23) for S By Lemma 38, either M is reducible with a single periodic essential class, or M is irreducible and periodic In the former case, we find from Lemma 310, Remark 311, and Lemma 312 that S is permutationally similar to a matrix in Class II Suppose next that M is irreducible and periodic If J =, then from Corollary 39, we find that S is in Class I Now suppose that J, so that M satisfies Lemma 313 i) or ii) If i) holds, then from Remark 314 and Lemma 315, we find that S is permutationally similar to a matrix in Class III On the other hand, if ii) holds, it follows from Remark 316 and Lemma 317 that S is permutationally similar to a matrix in Class IV 4 Applications and examples In this section we consider some applications of our results to certain special classes of matrices Example 41 Suppose that we have positive numbers x i, i = 1,, n, with n i=1 x i = 1 Consider the stochastic companion matrix C = x 1 x 2 x n 1 x n

21 Observe that τ(c) = 1, so that Proposition 11 yields no new information on the modulus of λ 2 (C) However, from Corollary 27, we find that λ 2 (C) P n j=i+1 1 min{x 1, x j P n i = 1,, n} j=i x j x i 1 P i 1 i = 2,, n} = max{ j=1 x j Equality can hold in this bound on λ 2 (C) For instance, if t (0, 1), and we have the parameters x i = t(1 t) i 1, i = 1,, n 1, x n = (1 t) n 1, we find that C is in Class IV (with m = n and a = r = 1) so that equality holds in (23) with µ(c) = t Indeed when C is constructed with this collection of parameters, then for each n-th root of unity ω 1, we find that λ = ω(1 t) is an eigenvalue of C A square nonnegative matrix is a Leslie matrix if it has the property that its only positive entries lie in the first row and on the first subdiagonal (and typically that subdiagonal is taken to have all positive entries) Such matrices arise in a discrete-time, age-dependent model of population growth, known as the Leslie model (see [5] for a brief description of the model, and [2] for a more comprehensive treatment) In the event that we have a primitive Leslie matrix L, it turns out that the age distributions in the corresponding Leslie model converge to a Perron vector of L Further, if the Perron value of L is ρ, and λ is an eigenvalue of L of next largest modulus after ρ, then the rate of convergence of the age distributions in the Leslie model is given by λ ρ (Note the natural parallel with the behaviour of a Markov chain having a primitive transition matrix) The next sequence of results applies the ideas of Section 2 to the eigenvalues of certain Leslie matrices Theorem 42 Let L be a Leslie matrix of order n, given by m 1 m 2 m n 1 m n p L = 0 p 2 0 0, 0 0 p n 1 0 where m i > 0, i = 1,, n and p i > 0, i = 1,, n 1 Denote the Perron value of L by ρ Then for any eigenvalue λ ρ of L, we have n j=i+1 λ ρ max{ p 1 p j 1 m j ρ n j n j=i p i = 1,, n} (412) 1 p j 1 m j ρ n j 21

22 Proof: Let x 1 = m 1, and for each i = 2,, n, let x ρ i = p 1p i 1 m i It is known ρ i (see [5]), and not so difficult to show, that 1 L is diagonally similar to the ρ matrix x 1 x 2 x n 1 x n P Applying the bound of Example 41 yields λ max{ n j=i+1 x j P ρ n i = 1,, n}, j=i x j which is equivalent to (412) Our next result deals with a certain class of stochastic companion matrices Theorem 43 Suppose that 2 k n, and that we have positive numbers 2 x k,, x n such that n j=k x j = 1 Consider the companion matrix 0 0 x k x n S = Define γ by γ = min{x k,, x 2k 1, x 2k+x 2 k x k x Pn n k+1,, Pn x k x n 1 } j=n k+1 x j j=n 1 x j Then λ 2 (S) (1 γ) 1 k+1 P, x 2k+1+x k x n j=k x Pn k+1 j j=k+1 x j,, xn+x kx n k Proof: It is straightforward to show that x k+j, 1 j k 1 e T 1 S k+1 e j = x k+j + x k x j, k j n k x k x j, n k + 1 j n, Pn j=n k x j, so that the first row of S k+1 is positive Letting the left stationary vector of S be w T, we find from Lemma 21 that w 1 w j P 1 n (as it happens, equality i=j x i 22

23 holds) Letting W = diag(w), we find that the matrix T = W 1 (S k+1 ) T W is stochastic, and its first column is positive Indeed, T j,1 = Sk+1 1,j P n ; the result i=j x i now follows from the fact that λ 2 (S) = λ 2 (T ) 1 k+1 1 τ(t ) k+1 1 (1 γ) k+1 We have the following application of Theorem 43 to a particular subclass of Leslie matrices Corollary 44 Suppose that 2 k n, that m 2 i > 0, i = k,, n, and that p i > 0, i =,, n 1 Let L be the Leslie matrix given by 0 0 m k m n p p L = p n 1 0 Let ρ be the Perron value for L and let ˆγ = min{ p 1p k 1 m k,, p 1p 2k 2 m 2k 1, ρ k ρ 2k 1 p k p n 1 m n+p 1 p n k 1 m n k m k Pn i=n k p kp i 1 m i ρ n i, p k p 2k 1 m 2k +p 1 p k 1 m k m k P 2k i=k p kp i 1 m i ρ 2k i + P n p k p i 1 m i i=2k+1 p 1 p n k m n k+1 m k,, ρ i 2k Pn i=n k+1 p kp i 1 m i,, Pn p 1 p n 2 m n 1 m k ρ n+1 i i=n 1 p kp i 1 m i } ρ n 1+k i Then for any eigenvalue λ ρ of L, we have λ ρ(1 ˆγ) 1 k+1 Proof: For i = k,, n, let x i = p 1p i 1 m i As in the proof of Theorem 42, ρ i we find that 1 L is diagonally similar to the companion matrix S of Theorem 43 The conclusion now follows from Theorem 43 after substituting for ρ x k,, x n in terms of p 1,, p n 1, m k,, m n, and then simplifying We close with an example discussing an application of Theorem 22 to a certain iterative method that has been proposed for the computation of Google s PageRank Example 45 Suppose that we have an n n stochastic matrix P, and a positive row vector v T of order n, normalized so that v T 1 = 1 Fix a parameter c (0, 1), and consider the stochastic matrix G given by G = cp + (1 c)1v T A matrix G of this form is a Google type matrix, so named because 23

24 a matrix of that type is used in Google s PageRank algorithm, which uses the stationary vector of G in order to rank web pages on the internet A typical approach to computing the stationary vector for G is via the power method, and it is well known that in the case that P has at least two essential classes (which is certainly the case in practice), or has at least one periodic essential class, the convergence of the power method is geometric, with convergence rate c In order to frame our discussion, we impose some assumptions on P We take P to have essential classes C 1,, C k, and assume that for any essential index j, p j,j = 0 (this happens to be the case in the PageRank application) We suppose also that the rows and columns of P have been simultaneously permuted so that for each i = 1,, k, i C i Finally, we assume that the labelling of the rows and columns of P is such that P has the form P = 0 P 12 P 13 P 21 P 22 P 23 0 P 32 P 33 here we have partitioned out the first k rows and columns of P, while the second subset of the partition corresponds to those indices i such that p i,j > 0 for some j = 1,, k, and the third subset of the partition consists of the remaining indices Partition v T conformally with P as v T = [ ] v1 T v2 T v3 T In [4], the authors consider an approach to computing the stationary vector of G via a certain iterative aggregation/disaggregation technique, and show that if aggregation is performed on the first k rows and columns, then the convergence is also geometric, and the rate of convergence is given by c λ 2 (H), where H is the following (necessarily irreducible) stochastic matrix: [ P22 + cp H = 21 P 12 P 23 + cp 21 P 13 P 32 P [ ] 33 P21 1 [ ] cβv T 1 P (1 c)(1 + βδ)v2 T cβv1 T P 13 + (1 c)(1 + βδ)v3 T ; here δ = v1 T 1 and β = 1 c Observe that for each index i such that (P ) 1 (1 c)δ contains an arc of the form i j for some j = 1,, k, the corresponding row of H has all positive entries Note that if H is of massive order (as is the case in the PageRank setting), then λ 2 (H) may be difficult to compute, and so an upper bound on λ 2 (H) would then be of some use in estimating the rate of convergence of the iterative method discussed in [4] Since H has at least one positive row, Theorem 22 can be used to bound λ 2 (H) 24 ; ] +

25 Let σ = P 21 1, and suppose that σ has order p; let u T = cβv1 T P 12 + (1 c)(1 + βδ)v2 T (also of order p) and let x T = cβv1 T P 13 + (1 c)(1 + βδ)v3 T (of order q, say) Fix an index j {1,, p} It is straightforward to determine σ that if j i {1,, p}, then α(i, j) σ j u i u i j 1 σ i u i, while α(j, j) = σ j u j Similarly, if i {p + 1,, p + q} and (P ) contains the path i i 0 i 1 i m+1 j (observe that such a path always exists), then p i0,i α(i, j) 1 p im,im+1 x i0 1 σ j u j m l=1 p i l,i l+1 p im,im+1 x il Thus, these estimates can be used in the bound of Theorem 22 to provide an upper bound on λ 2 (H) Consequently, this approach not only provides an alternate proof of the fact (shown in [4]) that λ 2 (H) < 1, but also yields quantitative information (in terms of the entries in P, the entries in v T, and the parameter c) on how close λ 2 (H) can be to 1 As an illustration of the above, suppose that the minimum positive entry in P is a, and that the minimum positive entry in v T is b, and that q 1 We find that δ bk, that β 1 c, that σ a1, that 1 (1 c)bk ut (1 c)b 1 (1 c)bk 1T, and that x T (1 c)b 1 (1 c)bk 1T It now follows that for each j {1,, p}, for the matrix H we have α(j, j) (1 c)ab, while for distinct i, j {1,, p}, then 1 (1 c)bk (1 c) 2 a 2 b 2 (1 (1 c)bk)(1 (1 c)b(k+a)) for the matrix H, α(i, j) Similarly, if j {1,, p} and i {p + 1,, p + q}, and if there is a path in (P ) from i to j of (1 c)(1 a)ba length d, then for the matrix H, α(i, j) d It (1 a)(1 (1 c)b(k+a)) (1 c)ab(1 a d 1 ) is not difficult to see that this last quantity, considered as a function of d, is nonincreasing if 1 a b(1 c)(k (k 2)a a 2 ), and nondecreasing in d otherwise Setting d j = max i {p+1,,p+q} d(i, j), it follows that for each j {1,, p} and any i {1,, p + q}, we have for the matrix H that Since α(i, j) min{ (1 c)ab 1 (1 c)b(k+a), (1 c) 2 a 2 b 2, (1 c)ab, (1 (1 c)bk)(1 (1 c)b(k+a)) 1 (1 c)bk (1 c)(1 a)ba d j (1 a)(1 (1 c)b(k+a)) (1 c)ab(1 a d j 1 ) } (1 c) 2 a 2 b 2 (1 (1 c)bk)(1 (1 c)b(k+a)) (1 c)ab 1 (1 c)bk (1 c)ab 1 (1 c)b(k+a) α(i, j) min{ (1 c) 2 a 2 b 2, (1 (1 c)bk)(1 (1 c)b(k+a)), we thus find that (1 c)(1 a)ba d j (1 a)(1 (1 c)b(k+a)) (1 c)ab(1 a d j 1 ) } γ j Consequently, we have λ 2 (H) 1 k j=1 γ j, thus providing a upper bound on the rate of convergence of the iterative aggregation/disaggregation method discussed in [4] Observe that this bound, while crude, depends only on the number of essential classes for P, the minimum entries in P and v T, and upon the lengths of certain paths in (P ) 25

26 We note that in certain (highly artificial) examples, in fact the bound of Theorem 22 can actually be attained by λ 2 (H) For instance, suppose that n 3 and that P is the adjacency matrix of a directed n cycle, say P = In terms of the discussion above, we have k = 1, and it turns out that the stochastic matrix H has the form x 1 x 2 x n 2 x n H = , where [ x1 x 2 x n 2 x n 1 ] = 1 c 1 (1 c)v 1 [ v2 v 3 v n ] + c 1 (1 c)v 1 e T n 2 ) i 2 Ob- ( Suppose further that for i = 3,, n 1, v i = v 1 (1 c)(v1 +v 2 ) 2 1 (1 c)v 1 serve that in order for this last condition to hold, it must be the case that c < (1 (1 c)(v 1+v 2 )) n 2 (1 (1 c)v 1, otherwise v ) n 3 n = 1 n 1 i=1 v i 0 (it is sufficient to take v 2 is close to zero in order to ensure that 1 n 1 i=1 v i > 0) If all of ) i 1 these conditions hold, it follows that x i = (1 c)v 2 1 (1 c)v 2 1 (1 c)v 1, i = 1,, n 2 As noted in Example 41, equality then holds in (23), with λ 2 (H) = 1 (1 c)v 2 1 (1 c)v 1 = 1 µ(h) References 1 (1 c)v 1 ( [1] R Brualdi and H Ryser Combinatorial Matrix Theory Cambridge University Press, Cambridge, 1991 [2] H Caswell Matrix Population Models: Construction, Analysis, and Interpretation (2nd Edition) Sinauer, Sunderland, Massachusetts,

27 [3] P Diaconis and D Stroock Geometric bounds for eigenvalues of Markov chains Annals of Applied Probability 1: 36-61, 1991 [4] I Ipsen and S Kirkland Convergence analysis of a PageRank updating algorithm by Langville and Meyer SIAM Journal on Matrix Analysis and Applications 27: , 2006 [5] S Kirkland An eigenvalue region for Leslie matrices SIAM Journal on Matrix Analysis and Applications, 13: , 1992 [6] S Kirkland A note on the eigenvalues of a primitive matrix with large exponent Linear Algebra and its Applications, 253: , 1997 [7] S Kirkland Girth and subdominant eigenvalues for stochastic matrices Electronic Journal of Linear Algebra, 12:25-41, 2005 [8] E Seneta Non-negative Matrices and Markov Chains Springer-Verlag, New York,