The Case of Equality in the Dobrushin Deutsch Zenger Bound

Transcription

1 The Case of Equality in the Dobrushin Deutsch Zenger Boun Stephen J Kirklan Department of Mathematics an Statistics University of Regina Regina, Saskatchewan CANADA S4S 0A2 Michael Neumann Department of Mathematics University of Connecticut Storrs, Connecticut USA Deicate to Professor Shmuel Frielan on his 65 th birthay Abstract Suppose that A = (a i,j) is n n real matrix with constant row sums µ Then the Dobrushin Deutsch Zenger (DDZ) boun on the eigenvalues of A other than µ is given by Z(A) = nx 2 max a s,r a t,r When A s,t n r= a transition matrix of a finite homogeneous Markov chain so that µ =, Z(A) is calle the coefficient of ergoicity of the chain as it bouns the asymptotic rate of convergence, namely, max{ λ λ σ(a) \ {}}, of the iteration x T i = x T i A, to the stationary istribution vector of the chain In this paper we stuy the structure of real matrices for which the DDZ boun is sharp We apply our results to the stuy of the class of Research supporte in part by NSERC uner grant OGP03825 Research supporte by NSA Grant No 06G 232

2 graphs for which the transition matrix arising from a ranom walk on the graph attains the boun We also characterize the eigenvalues λ of A for which λ = Z(A) for some stochastic matrix A Key wors: Stochastic Matrices, Coefficient of Ergoicity, Graphs, Ranom Walks, Eigenvalues of Stochastic Matrices AMS Classification: 5A8, 5A48, 5A5, 60G50 Introuction an Preliminaries Let A = (a i,j ) be an n n real matrix with constant row sums, that is, there exists a number µ R, such that n a i,j = µ, for all i =,, n j= It is easily seen that µ is an eigenvalue of A corresponing to the n vector of all ones, Then an upper boun on the largest (in moulus) eigenvalue of A other than µ is given by λ Z(A) () where Z(A) := 2 max s,t n r= n a s,r a t,r The boun is ue to Eckart Deutsch an Zenger [7] In Seneta [9, p62 63] a self containe proof is given for this boun We shall return to elements of this proof later Now let A R n,n be a transition matrix for an ergoic homogeneous Markov chain on n states Then A is an n n nonnegative, row-stochastic, an irreucible matrix so that, by the Perron Frobenius theory, the spectral raius of A, which is an eigenvalue of A, is In this case the quantity γ(a) = max λ, (2) λ σ(a)\{} when it is smaller than, etermines the asymptotic rate of convergence of the iteration process zi T = zi T A to the stationary istribution vector of the chain In this context of transition matrices, Dobrushin [0] has shown that γ(a) Z(A) (3)

3 an calle Z(A) the coefficient of ergoicity of the chain In view of the aforementione history, we shall call Z(A) the Dobrushin Deutsch Zenger boun or the DDZ boun for short The main purpose of this paper is to stuy the properties an structure of nonnegative, stochastic, an irreucible matrices A for which equality hols in (3) an to apply these results to ranom walks for which the equality hols for the unerlying transition matrix We commence our investigation, however, in Section 2 by assuming only that A R n,n is a matrix whose row sums are a constant which we shall take to be Observe that there is no loss of generality in that assumption, since we can always a a suitable rank one matrix y T to A to put it in that form Throughout the paper we shall call an eigenvalue λ of A subominant if λ = γ(a) an usually enote this fact by writing λ as λ sub One application in which which there is equality in the DDZ boun is, in fact, in the Google matrix Suppose that the web has n pages an that for each i =,, n, page i has i > 0 outgoing links (The assumption that each page has at least one outgoing link oes not affect the valiity of the conclusion below) We now construct a stochastic matrix A = (a i,j ) R n,n as follows If i, then each link from page i to page j, we set a i,j = / i If there is no link from page i to page j, then we set a i,j = 0 Assume now that the web contains a union of k 2 isjoint strongly connecte components so that A has the form: A, A 2,2 0 A =, 0 A k,k A k+, A k+,2 A k+,k+ with A l,l R m l,m l, for l =,, k +, with m + m k+ = n, an with each A l,l being a stochastic matrix, l =,, k Let α (0, ) Then the Google matrix is given by G α = ( α)a + αev T, where v is a positive vector with v =, that is, v is a probability vector Then as shown in Ipsen an Kirklan [, Corollary 72], γ(g α ) = Z(G α ) = α 2

4 The DDZ eigenvalue boun in () has been applie in contexts other than transition matrices of Markov chains As an example, let G be an unweighte unirecte graph on n vertices v,, v n whose egrees are,, n, respectively Let M be the (0, ) ajacency matrix of G an D = iag(,, n ) Then L = D M is the Laplacian matrix associate with G It is easy to see that L has zero row sums an hence the DDZ boun is applicable to the eigenvalues of L We note in passing that L is also a positive semiefinite M matrix We comment that there is much interest in the literature in the eigenvalues of L an hence in fining goo bouns on them For example, the secon smallest eigenvalue of L is known as the algebraic connectivity of G In the situation we escribe here, we clearly have that ρ(l), the spectral raius of L, is boune above by Z(L) an several recent papers have investigate the structure of graphs G for which Z(L) = ρ(l), see, for example, Rojo, Soto, an Rojo [7] an Das [5, 6] As mentione above, we shall also seek to use the equality case in the DDZ boun to etermine the structure of certain graphs, but in a ifferent sense than in the papers [7] an [5, 6] Let G be an unirecte unweighte connecte graph on n vertices an let the matrices D an M be as above It is easy to see that the matrix A(G) = D M R n,n, which is nonnegative an irreucible, is the transition matrix for a ranom walk on G It is also straightforwar to see that A is iagonally similar to the symmetric matrix D 2 MD 2 so that, in particular, all the eigenvalues of A are real As an asie, we note that the so-calle normalize Laplacian matrix for G (see [4]) is given by L = I D 2 MD 2, so that eigenvalue bouns for A will generate corresponing eigenvalue bouns for L In Section 2 we evelop some preliminary results, while in Section 3 we characterize the complex numbers that can be attaine as an eigenvalue of a stochastic matrix yieling equality in () In Section 4 we stuy ranom walks on various families of graphs for which γ(d M) = Z(D M) Generally speaking the transition matrices for these ranom walks exhibit a certain nonzero zero block structure We close this introuctory section by giving two contrasting examples The first is a graph G whose Laplacian matrix yiels equality in the DDZ eigenvalue boun, but whose ranom walk transition matrix yiels strict inequality in the DDZ boun The secon example is a graph for which equality hols for the 3

5 DDZ eigenvalue boun for the transition matrix of the corresponing ranom walk, but not for the corresponing Laplacian matrix For the first example take: M =, in which case D = iag(3, 4, 4, 4, 4, 5) Then for the Laplacian L = D M, we fin that σ(l) = {0, 3, 4, 5, 6, 6} an Z(L) = 6, while for the associate transition matrix D M of the ranom walk we fin that: σ(d M) = {, 0059, 0, 02500, 02673, 05886} an Z(A) = 75 so that Z(A) > γ(a) For the secon example take the 4 4 ajacency matrix: M = Here σ(d M) = {, 05774, 05774, 05774, 0, 0, 0, 0, 0, 0, 05774, 05774, 05774, } so that γ(a) = = = Z(A) A computation now shows that for L = D A, ρ(l) = 7, while Z(L) = 8 4

6 2 The DDZ Boun Seneta s proof of the DDZ boun () rests on the following boun on the inner prouct of two vectors, one of which is orthogonal to the ones vector Lemma 2 (Paz [6, Chp IIa], Seneta [9, p63]) Let z = (z,, z n ) be an arbitrary row vector of complex numbers Then for any real vector δ 0 with δ T = 0, z T δ 2 max i,j n z i z j δ (24) To facilitate the stuy in this paper of the equality case in () we nee the characterization of the case of equality in (24) The following theorem comes from [3] Theorem 22 (Kirklan, Neumann, an Shaer [3, Theorem 2]) Let δ IR n be a vector such that δ T = 0 an let z C n Then equality hols in (24), viz z T δ = 2 max i,j n z i z j δ if an only if z an δ can be reorere simultaneously such that an where δ = δ δm δ m+ δ m+k 0 0 an z = a a b b c c n k m, (25) max z i z j = a b an δ i > 0, i =,, k + m (26) i,j n Throughout the remainer of this section A = (a i,j ) will always be an n n real matrix with row sums an subominant eigenvalue λ sub (A), in which case we can write that: Z(A) = { max (e T i,j n 2 i e T j )A } λ sub (A) (27) 5

7 We comment that in the case that A is also a nonnegative matrix, it reaily follows from the stochasticity of A an the efinition of Z(A) that Z(A) 2 (2 A ) (28) In our first lemma on the equality case of the DDZ boun on A we escribe some of the quantitative structure of the entries of A Lemma 23 Suppose that equality hols in (27) an let z be an eigenvector of A corresponing to λ sub Then for any pair of inices i, j n such that we have that z i z j = max p,q n { z p z q }, (e T i e T j )A 2 = Z(A) Further, there are entries a an b of z with a b = z i z j such that for any k, we have that an i) z k = a whenever a i,k a j,k > 0, ii) z k = b whenever a i,k a j,k < 0 In particular, if for some inex k we have z k a, b, then a i,k = a j,k Proof: For any vector v C n, let f(v) = max p,q, n { v p v q } We then have that λ sub (A) f(z) = λ sub (A) z i z j = (e T i e T j )Az 2 (e T i e T j )A f(z) Z(A)f(z) = λ sub (A) f(z) Consequently, it must be the case that 2 (e T i e T j )A = Z(A) Furthermore, we must also have that (e T i e T j )Az = (e T 2 i e T j )A f(z), an appealing to Theorem 22, we fin that conclusions (i) an (ii) follow 6

8 As an example consier the matrix A = Here Z(A) = 2 an the spectrum of A is given by σ(a) = {, 2,, 0, } Thus the (only) subominant eigenvalue of A is λ sub = 2 = λ sub (A) = Z(A) an the conitions of Lemma 27 are applicable The corresponing eigenvector to λ sub is given by z = 02236, in which case we see that a = 02336, b = 08944, an we observe that the inices i an j for which z i z j = max p,q n { z p z q }, are given by i =, 2, 3, an 4, an j = 5, respectively Taking, for example, the ifference of rows 2 an 5 of A, we get that it is given by the vector [0083, 00083, 00583, 00250, 02000] Notice that for k =,, 4, a 2,k a 5,k > 0 an we expect that z k = which we see is true, while for k = 5, a 2,5 a 5,5 < 0 an, as we expect from the lemma, z 5 = Base on Lemma 23 we can prove a further inequality on the entries of A when equality hols in (27) Theorem 24 Suppose that equality hols in (27) Let z be a λ sub eigenvector, an suppose that i an j are inices such that z i z j = max p,q n { z p z q } Then (a i,i a j,i )(a i,j a j,j ) 0 7

9 Proof: Suppose to the contrary that (a i,i a j,i )(a i,j a j,j ) > 0 Without loss of generality, we may assume that i =, j = 2, a, > a 2,, an a,2 > a 2,2 (otherwise we can simultaneously permute the rows an columns of A so that it has the esire form) We may also assume that the remaining rows an columns have been orere so that a,p > a 2,p,, for p = 3,, m, a,p < a 2,p, for p = m +,, m + q, an a,p = a 2,p, for p = m + q +,, n It follows from Lemma 23 that we have [ ] [ e T A = u T v T w T, e T 2 A u T 2 v T 2 w T ], an z = a b c, where the partitions are conformal an where we have u > u 2, v < v 2, an a b = max p,q n { z p z q } From the eigenequation Az = λ sub z we have λ sub a = au T + bv T + w T c an λ sub a = au T 2 + bv2 T + w T c, so that a(u u 2 ) T + b(v v 2 ) T = 0 Now w T = u T + v T = u T 2 + v2 T, so that (u u 2 ) T = (v2 T v T ) We conclue that (a b)(u u 2 ) T = 0, an hence that a = b, a contraiction A refinement of the results in Theorem 24 is given in the following lemma: Lemma 25 Suppose that equality hols in (27) Let z be a λ sub eigenvector, an suppose that i an j are inices such that i < j an z i z j = max p,q n { z p z q } Suppose further that (a i,i a j,i )(a i,j a j,j ) < 0, an efine the following sets of inices: Σ = {k a i,k > a j,k }, an Σ 2 = {k a i,k < a j,k }, Σ 3 = {k a i,k = a j,k } 8

10 Set σ i, := k Σ a i,k, σ i,2 := k Σ 2 a i,k, an σ j, := k Σ a j,k, σ j, := k Σ 2 a j,k Then an a) if a i,i > a j,i an a i,j < a j,j, then λ sub = σ i, σ j, b) if a i,i < a j,i an a i,j > a j,j, then λ sub = σ j, σ i, Proof: a) Without loss of generality, we assume that i = an j = 2 Further, we may simultaneously reorer inices 3,, n so that [ ] [ ] e T A = a, a,2 u T v T w T, e T 2 A = a 2, a 2,2 u T 2 v2 T w T, an z = where the partitions are conformal an where u > u 2, v < v 2, an a b = max p,q n { z p z q } From the eigenequation Az = λ sub z it follows that a b a b c, λ sub a = aσ, + bσ,2 + w T c an λ sub b = aσ 2, + bσ 2,2 + w T c Subtracting the two equations we fin that λ sub (a b) = a(σ, σ 2, ) + b(σ,2 σ 2,2 ) Now since we fin that w T = σ, + σ,2 = σ 2, + σ 2,2, σ,2 σ 2,2 = (σ, σ 2, ) 9

11 Hence λ sub (a b) = (σ, σ 2, )(a b), an conclusion (a) follows The proof of (b) is analogous The equality case in (27) allows us to prove results about the Joran block structure corresponing to the subominant eigenvalues of A We begin with the following lemma: Lemma 26 Suppose that equality hols in (27) for the matrix A Then for any k IN, equality also hols in (27) for the matrix A k, with λ sub (A k ) = Z(A k ) = (Z(A)) k Proof: From the proof of Proposition 4 on p70 of Paz [6] (see also Seneta [9, Lemma 43]) it follows reaily that Z(A k ) (Z(A)) k, for each k IN But then for any such k, we have (Z(A)) k = λ sub (A) k = λ sub (A k ) Z(A k ) (Z(A)) k Whence Z(A k ) = λ sub (A k ), for each k IN We can now prove: Theorem 27 Suppose that equality hols in (27) for the matrix A Then for any eigenvalue λ such that λ = Z(A), the geometric an algebraic multiplicities of λ coincie Proof: If λ = 0, the result follows reaily from the fact that in that case, A must have rank So, henceforth we take λ to be nonzero Suppose to the contrary that the geometric muliplicity of λ is less than the algebraic multiplicity of λ Then there are vectors x T an y T such that x T A = λx T, y T A = λy T +x T, an y T = Observe that necessarily y T = 0 = x T A straightforwar proof by inuction shows that y T A k = λ k y T + kλ k x T = λ k ( y T + k λ xt ) Note that ( λk y T + k ) λ xt ( k λ k λ λ x T ) y T ( k = (Z(A)) k x ) T λ In particular, we fin that for all sufficiently large k IN, y T A k > (Z(A)) k = Z ( A k) This last contraicts Lemma 26 We thus conclue that the geometric 0

12 an algebraic multiplicities of λ must be equal We comment that the converse of Theorem 27 oes not hol as the following example shows Let A = Then the spectrum of A is given by σ(a) = {, 2,, 0, 2}, so that the geometric an algebraic multiplicities of both subominant eigenvalues, ±2 are, yet Z(A) = 0247 > 2 = λ sub (A) Until now we have consiere the equality case in the DDZ boun for any real matrix Let us now assume that A is an n n nonnegative an irreucible matrix whose row sum is a constant In this case A is row stochastic an can be regare as a transition matrix of a finite homogeneous ergoic Markov chain on n states For such a Markov chain, Meyer [4] has shown that virtually any important parameter of the chain can be rea from the group generalize inverse Q # of the singular an irreucible M matrix 2 Q = I A Cleary Q = 0 an it is known that Q # = Q It is further known that σ(q # ) = {0} { λ λ σ(a) \ {} Thus, on applying the DDZ eigenvalue boun we can write that: λ sub min λ σ(a)\{} λ Z(Q # ) where the rightmost inequality is ue to Seneta, see [9] Z(A), (29) Suppose now that for A as above, the equality case in the DDZ boun () hols In this case we can write that λ sub (A) min λ σ(a)\{} λ Z(Q # ) λ sub (A), (20) for any λ sub (A) σ(a) It is now straight forwar to prove the following result: For comprehensive accounts on group generalize inverses of matrices, incluing when they exist, see Ben Israel an Greville [] an Campbell an Meyer [3] 2 For a comprehensive account on the Perron Frobenius theory for nonnegative matrices an on M matrices se see Berman an Plemmons [2]

13 Theorem 28 Suppose that A is an n n nonnegative, stochastic, an irreucible matrix for which equality hols in (3) eigenvalue λ sub (A) R +, then for Q = I A, we have Z(Q # ) = λ sub (A) If γ(a) < an A has an Let is give two examples First take: A = (2) Then σ(a) = {0000, i, i, i, i] Here Z(A) = 4 = λ sub (A), but we see that A coul not possibly fulfill the conitions of Theorem 28 Inee we fin that for Q = I A, Q # = for which max λ σ(a)\{} λ = 0467 < Z(Q # ) = 3240 < 6667 = As a secon example consier A = = Z(A) 2

14 Here σ(a) = [, 0375, 5 05, 5] so that λ sub (A) = 5 Furthermore we fin that Z(A) = 5 an hence Z(A) = λ sub (A) an so for this A the conitions of Theorem 28 are fulfille On computing the group inverse of Q = I A we obtain that: Z(Q # ) = an that Z(Q # ) = 2 = /( /2) = /( Z(A)) 3 The complex eigenvalues yieling equality in the DDZ inequality for stochastic matrices Much is known about the eigenvalues of stochastic matrices A For example, Dmitriev an Dynkin [8, 9], an Karpelevich [2] etermine the region within the unit circle in which the eigenvalues of an n n stochastic matrix must lie (see Minc [5]) for a more accessible acount of the result of Dmitriev an Dynkin) Romanovsky [8] (see also Varga [2, Corollary, p39]) showe that if A is an n n cyclic matrix of inex k 2, an so A is, in particular imprimitive, then its characteristic polynomial is given by φ(t) = λ m [ t k ρ k (A) ] [ t k δ 2 ρ(a) k] [t k δ r ρ k (A)], where δ i <, for < i r, if r > Observe for example, that Romanovsky s theorem oes not tell us about the nature of the eigenvalues other than when A is irreucible, but not k cyclic, that is, when A is primitive This is illustrate in the example given in (2), where the four eigenvalues other than of A continue to be the four non real roots of the equation t 5 = 4 3

15 In this section we show that if A is a stochastic matrix for which the equality case in the DDZ boun hols, then the subominant eignvalues of A satisfy equations of the form t k = α, where k n, with α R, α, an with further restrictions on k when α < 0 We begin with the following construction Suppose that we have n istinct complex numbers z,, z n, an let ρ = max i,j n { z i z j } The corresponing iameter graph for the vector z = z z 2 z n is the graph [ ] T Γ(z) on vertices,, n with i j in Γ(z) if an only if z i z j = ρ We begin with a useful lemma Lemma 3 Let λ C an n IN Suppose that there is an n n stochastic matrix A having eigenvalue λ for which Z(A) = λ Then there is a stochastic matrix M of orer at most n an an eigenvector z such that Mz = λz, Z(M) = λ, z has istinct entries, an the iameter graph of z has no isolate vertices Proof: Let x be an eigenvector of A corresponing to λ Suppose that x oes not have istinct entries; for concreteness we take x = x 2 without loss of generality Write A an x as A = a a 2 r T a 2 a 22 r2 T c c 2 A an x = Next, consier the matrix ˆB of orer n an the vector y given as follows: [ ] [ ] a + a 2 r ˆB T x = an y = c + c 2 A x Eviently ˆBy = λy, an it is reaily verifie that Z( ˆB) Z(A) Further, since Z(A) = λ Z( ˆB) Z(A), we see that in fact λ = Z( ˆB) Now, applying an inuction step on the orer of the matrix, it follows that we can fin a matrix B an vector u such that Bu = λu, Z(B) = λ an u has istinct entries If it happens that the iameter graph of u has no isolate vertices, then we are one So, suppose that the iameter graph [ of u has some ] isolate [ vertices ] Without loss of generality, we have B =, an u =, where B B 2 u B 2 B 22 u 2 x x 2 x 4

16 the subvector u 2 correspons to all of the isolate vertices in the iameter graph of u w T It follows from Lemma 23 that B 2 is rank an of the form for some nonnegative vector w T Note also that w T <, otherwise we have B = 0, from which it follows that u is multiple of, a contraiction From the eigenequation, we have B u + (w T u 2 ) = λu Next, consier the matrix M = B + wt w T et B Note that M is stochastic, an that Z(M) = Z(B ) = Z(A) A straightforwar computation reveals that the vector z = u + λwt e T u (λ )( w T ) is an eigenvector for M with corresponing eigenvalue λ Further, note that z has istinct entries, an that its iameter graph has no isolate vertices The following result will be applie to the iameter graph of a suitable eigenvector in Theorem 33 below Lemma 32 Suppose that G is a graph on n vertices with no isolate vertices an maximum egree at least two Let A be an n n real matrix such that Z(A) =, A has constant row sums, an all rows of A are istinct Suppose that for each pair of inices i, j =,, n such that i j in G, we have (e i e j ) T A = (e k e l ) T, for some k l in G Then A can be written as A = y T ± S, where S is a (0,, ) matrix with the properties thats has a single zero row, an for some inex i an every nonzero row of S is of the form (e i e j ) T for some suitable j Proof: Suppose without loss of generality that vertex of G has maximum egree, with ajacent to vertices 2, 3,, k Let S = A e T A, which has an all zero first row Then there are inices a, b, c, an, with a b, c in G, such that (e 2 e ) T S = (e a e b ) T an (e 3 e ) T S = (e c e ) T an hence e T 2 S = (e a e b ) T, e T 3 S = (e c e ) T Since Z(S) = Z(A) =, we fin that necessarily either a = c or b =, otherwise Z(S) > Without loss of generality, we suppose that a = b (in the case that b =, we consier S instea of S) Note that since S has istinct rows, necessarily b If k 4, we fin as above that for each j = 4,, k, there are inices p j an q j such that e T j S = (e p j e qj ) T Furthermore, for each such j, if p j a, then necessarily q j = b an q j =, a contraiction We conclue that p j = a for each j = 4,, k Suppose now that p q is an ege of G that is not incient with vertex 5

17 Let e T p S = x T so that for some inices i an j with i j in G, we have e T q S = x T +e T i et j For concreteness, we will henceforth take rows 2,, k of S to be e T e T 2,, e T e T k, respectively, an we will take p = k+ an q = k + 2, all without loss of generality Thus the first k + rows of S have the following form: while the row k + 2 of S has the form x x 2 x 3 x k x k+ x n, [x x 2 x 3 x k x k+ x n ] + e T i e T j for some i an j From the fact that Z(S) =, it follows that x 0, while x j 0, j = 2,, k; consequently we set x p = y p, for p = 2,, k Further, from the facts that each row sum of S is zero, e T S = 0 T, an Z(S) =, it follows that the sum of the positive elements in each row is boune above by Suppose first that i = Since both x, x + [0, ], we fin that necessarily x = 0 Also, by consiering row k + 2, we see that each of x k+,, x n must be nonpositive It now follows that row k + of S is 0 T, a contraiction since S has istinct rows Hence i 2 an a similar argument (reversing the roles of rows k + an k + 2) yiels j 2 Next, suppose that 2 i k an without loss of generality we take i = 2 Consiering row k + 2, we fin that y 2 0, which yiels x 2 = y 2 = Hence x p = 0 for p = 3,, k, an further, for each p = k +,, n, we have x p 0 All tol we have that 2 (e k+ e 3 ) T S = x x, which yiels x an hence x = It follows then that e T k+ S = et e T 2, a contraiction to the fact that S has istinct rows A similar argument (again, reversing the roles of rows k + an k + 2) shows that assuming that 2 j k leas to a contraiction 6

18 The last case is then n i, j k + Without loss of generality we take i = k + an j = k + 2 Note that necessarily x k+ 0 an x k+2 0, an we set y k+ = x k+ Fix an inex l between 2 an k We have that 2 (el e k+2 ) T S = x + y p + y l + y k+ + x k+2 p=2,,k,p l + p=k+3,,n x q 4 x y l y k+ x k+2, from which we fin that for each such l, (x + x k+2 ) + (y l + y k+ ) = 2 It then follows that x + x k+2 =, an that y k+ + y l =, l = 2,, k The latter conition, in conjunction with the fact that y 2 + +y k +y k+ easily yiels that y k+ = an y p = 0, for p = 2,, k Next, by consiering the fact that 2 (e 2 e k+ ) T S, it follows that 2 x x, from which we euce that x = Hence e T k+ S = et e T k+ an et k+2 S = e T e T k+2, which is of the esire form We conclue that each nonzero row of S is of the form e T e T p, for some suitable inex p We are now in a position to prove the main result of this section Theorem 33 Let λ C an n IN Then there is an n n stochastic matrix A having eigenvalue λ for which Z(A) = λ if an only if one of the following hols: i) there is a k IN with k n, a k th root of unity ω, an an r [0, ] such that λ = rω; ii) there is a smallest o number k 0 IN, with k 0 n, a k 0 th root of, α, an an r [0, k 0 ], such that λ = rα Proof: Fix k IN an let C k be a k k cyclic permutation matrix Observe that for each r [0, ], the stochastic matrix [ A = r n J + r Ck 0 0 I n k satisfies Z(A) = r an has, for each k th root of unity ω, the complex number rω as an eigenvalue Similarly, for each o k 0, with k 0 n, the stochastic ] 7

19 matrix satisfies Z(A) = number r k 0 [ B = r n J + r r k 0 k (J C 0 k 0 ) 0 k 0 J 0 an has, for each k 0 th root of, α, the complex α as an eigenvalue Thus we see that each λ C satisfying i) or ii) is realize as an eigenvalue of some stochastic matrix with the esire properties Now suppose that there is an n n stochastic matrix A having eigenvalue λ such that Z(A) = λ If λ =, then certainly λ is of the form escribe in i) Henceforth, we suppose that λ Let v be an eigenvector for A corresponing to λ Appealing to Lemma 3, we assume without loss of generality that v has istinct entries, that iameter graph of v has no isolate vertices, an that A is m m for some m n Consier the iameter graph Γ(v) First, suppose that every vertex of Γ(v) has egree one Then m is even, an Γ(v) is a collection of m 2 inepenent eges If i j in Γ(v), then from Lemma 23, there is an ege k l in Γ(v) such that (e i e j ) T A = (e k e l ) T, from which it follows that λ(v i v j ) = v k v l Consier the irecte graph D, whose vertices are orere pairs (i, j) such that i j in Γ(v), with an arc from (i, j) to (k, l) if an only if (e i e j ) T A = (e k e l ) T Observe that D has m vertices, an that each vertex of D has outegree Letting M be the ajacency matrix of D, we fin that λ is an eigenvalue of M, with an eigenvector whose entry in the position corresponing to (i, j) is v i v j, for each i an j Since λ is an eigenvalue of the (0, ) matrix M, each row of which contains a single one, it follows reaily that λ is a k th root of unity for some k m Suppose now that Γ(v) has maximum egree at least two We then fin that Z(A) A satisfies the hypotheses of Lemma 4 Hence, Z(A) A can be written as y T ± S, where S is of the form escribe in that lemma Since such an S can be written as e T i P, for some inex i an permutation matrix P, we see that for some vector x T, we have either A = x T + Z(A)P or A = x T Z(A)P In the former case, we fin that the eigenvalues of A istinct from are of the form Z(A)ω where ω is a k th root of unity for some k m On the other han, if we have A = x T Z(A)P, then note that each entry of x T is boune below by Z(A), an that x T = + Z(A) Since A is m m, then necessarily + Z(A) = x T mz(a), so that Z(A) m Further, ] 8

20 the eigenvalues of A ifferent from are either of the form Z(A)ω for some ω satisfying ω k = for some k m, (if P has an even cycle) or of the form Z(A)α, where α is a k th root of an k is o an at most m (if P has an o cycle) This latter case yiels eigenvalues of the form escribe in ii) Corollary 34 Let A be an irreucible stochastic matrix of orer n, with left stationary vector π T We have λ = Z(A) for each eigenvalue λ of A if an only if there is some k IN such that A k = (Z(A)) k I + ( (Z(A)) k )π T Proof: Suppose that λ = Z(A) for each eigenvalue λ From Theorem 33 it follows that there is a k IN such that λ k 0 for each eigenvalue λ We thus fin that λ k = (Z(A)) k for all such λ Further, by Theorem 27 for each such eigenvalue λ of A, the algebraic an geometric multiplicities coincie It now follows that the matrix A k has just two istinct eigenvalues: with algebraic multiplicity one, an (Z(A)) k with geometric multiplicity n It is now straightforwar to etermine that A k = (Z(A)) k I + ( (Z(A)) k )π T Conversely, if A k = (Z(A)) k I + ( (Z(A)) k )π T for some k IN, we fin that A k has two istinct eigenvalues, namely, an (Z(A)) k of algebraic multiplicity n Thus, if λ is an eigenvalue of A, then λ k = (Z(A)) k, yieling the esire conclusion Remark 35 By a slight moification of the techniques in this section, the following result can be establishe Let A be an n n matrix real with constant row sums µ such that Z(A) =, an equality hols in (27) for some eigenvalue λ µ Then either λ is a k th root of unity for some k =,, n, or λ is a k th root of for some o k between an n 4 Ranom Walk on a Graph Let G be a connecte graph on n vertices, conveniently labele i =,, n, an let,, n be their corresponing egrees Let M be the ajacency matrix of G an let D = iag(,, n ) Then, as explaine in the introuction, the matrix A = A(G) = D M R n,n is the transition matrix for a ranom walk on G In this section we shall stuy the structure of graphs G whose ranom 9

21 walk has a transition matrix A which satisfies the equality case in the DDZ boun, namely, that Z(A) = γ(a) We begin with the following lemma which can essentially be euce from the proof of [7, Theorem 4] an also from work in [5, 6] Lemma 4 Suppose that A R n,n is the transition matrix for the ranom walk on a graph G Let i, j n be two vertices of G, of egrees i an j, respectively, an suppose that i j Let N i an N j enote the neighbourhoos of vertices i an j, respectively Then (e i e j ) T A 2 = N i \ N j i Proof: Note that 2 (e i e j ) T A is given by the sum of the positive entries in (e i e j ) T A Since i j, we see that (e i e j ) T A has a positive entry in position k if an only if i k but vertices j k In that case, necessarily (e i e j ) T Ae k = i The result now follows Corollary 42 Let A be as in Lemma 4 Then { } Ni \ N j Z(A) = max i, j are vertices in G with j j i Corollary 43 Let G be a connecte graph with normalize Laplacian matrix LI /2 AD /2 If λ 0 is an eigenvalue of L, then { } Ni \ N j max i, j are vertices in G with j j i { } Ni \ N j λ + max i, j are vertices in G with j j i In the next lemma we obtain a block structure of a transition matrix of a ranom walk which satisfies the equality case in the DDZ eigenvalue boun Lemma 44 Let A be the transition matrix for the ranom walk on G Let z be an eigenvector corresponing to λ sub (A) an suppose that equality hols in (27) Let a an b a maximal an a minimal entries in z, respectively If z has entries that are strictly between a an b, then A an z can be partitione 20

22 conformally as A = A, A,2 0 J A 2, A 2,2 0 J 0 0 A 3,3 A 3,4 an z = b a c 0, (42) A 4, A 4,2 A 4,3 A 4,4 c where the entries of c 0 an c are strictly between a an b Further, A 3,4 J, an each of A 4,, A 4,2 an A 4,3 is positive b Proof: We begin by writing z as z = a, where b < c < a Applying c Lemma 23 we fin that for any pair of inices i an j such that z i = b an z j = a, an any k such that b < z k < a, we have that a i,k = a j,k It follows that for any inex p such that z p = b or z p = a, an each k such that b < z k < a, there is a w k such that a p,k = w k Since G is connecte, w k > 0, for some k n, an since every w k is an element in the p th row, it follows that there is some such that each nonzero w k is equal to It follows then that we may write A an z as A = A, A,2 0 J A 2, A 2,2 0 J 0 0 A 3,3 A 3,4 b an z = a c 0, A 4, A 4,2 A 4,3 A 4,4 c respectively, where A 4, an A 4,2 are positive matrices, an the elements of c 0 an c are strictly between a an b Let the subsets in the partitioning of A be S,, S 4, respectively, with carinalities m,, m 4, respectively Note that S 4, since G is connecte, but that S 3 may be empty Suppose that S 3 By consiering (e i e j ) T A for i S, j S 3 we fin that Z(A) m4, while by consiering (e i e k ) T A for i S, k S 2 we have, in light of Lemma 23, that Z(A) m4 Hence Z(A) = m4, an again by consiering (e i e j ) T A for i S, j S 3 it follows that A 3,4 The positivity of A 4,3 now follows from the combinatorial symmetry of A 2

23 Corollary 45 Suppose that G is as in Lemma 44 an that A an z have been partitione as in (42) Denote the corresponing subsets of the partitioning (in orer) as S,, S 4, with carinalities m,, m 4, respectively Note that necessarily S 4 Then: i) Suppose that S 3 = Let  enote the principal submatrix of A on the rows m4 an columns corresponing to S S 2 Then  can be written as  = A, where A is stochastic an yiels equality in (27) Further, there is a λ sub (A) eigenvector for A having only two istinct entries, with all entries corresponing to inices in S taking one value an all entries corresponing to inices in S 2 taking the other value ii) If S 3, then Z(A) = m4 Further, we have that either A, = 0, A 2,2 = 0, A,2 = ( m 4 ),, or A 2, = ( ) m4, λ sub ( = A,2 = 0, m 4 ), A 2, = 0, A, = A 2,2 = λ sub = ( ( ( m 4 m 4 m 4 ), ), Proof: i) From (42) an Lemma 23, it follows that if i S an j S 2, then 2 (e i e j ) T A = Z(A), from which we fin that Z(A) = Z(Â) From the [ ] [ ] b b eigenequation, we have that  + a Jc = λ sub (A) We claim that, a ) 22

24 in fact, [ λ sub ] (A) is an eigenvalue of  To see this note that if not, then it follows b ( Â) that = λ sub I a Jc, a contraiction since  has constant row sums Thus λ sub (A) is an eigenvalue of  It now follows that λ sub (Â) = Z(Â) = Z(A) Next let v = [ b a ] so that Âv + Jc m4 = λ sub v If λ sub (Â), set x = T c /(λ sub +m 4 ) It now follows that v +x is a λ sub (Â) eigenvector for  having two istinct entries, with all entries corresponing to S ientical an all entries corresponing to S 2 ientical m4 Finally, if λ sub (Â) =, it follows that  can be written as a irect sum of two nonnegative matrices, both necessarily with constant row sums m4, an the eigenvector conclusion now follows Setting A = m 4 Â, the esire conclusions are now evient ii) As in Lemma 44, we have Z(A) = m4 Further, from (42) it follows that each of A,, A,2, A 2,, an A 2,2 has constant rows sums, an eviently we have that A, + A,2 = ( m 4 ) an A 2, + A 2,2 = ( m 4 ) Letting x,, x,2, x 2,, an x 2,2 be the row sums of A,, A,2, A 2,, an A 2,2, respectively, we fin from the eigenequation that λ sub = x, x 2, = x 2,2 x 2, Thus if λ sub = m4, then x, = x 2,2 = m4 an x,2 = x 2, = 0, while if λ sub = ( ) m4, then x,2 = x 2, = m4 an x, = x 2,2 = 0 The conclusions on A,,, A 2,2 now follow Remark 46 Suppose that A is as in Corollary 45 an that S 3 = Set S S 2 = m an S 4 = k Let G(S S 2 ) an G(S 4 ) enote the inuce subgraphs of G on the vertex sets S S 2 an S 4, respectively From Lemma 44, it follows that G(S S 2 ) is regular, say of egree r Let r,, r k enote the egree sequence for the inuce subgraph G(S 4 ) Eviently = r + k In orer that Z(A) = Ni\Nj for some i S an j S 2, all of the following conitions must hol: i) for each p, q S 4 with r p r q, N p \ N q r p + m N i \ N j r + m 23

25 an ii) for each q S 4, either r + k m r q k N i \ N j, or { } (r + k)(m r) r q max r + k m, m N i \ N j Remark 47 Suppose that A is as in Corollary 45 an that S 3 Denote the egrees of the vertices in the subgraph inuce by S 3 by q i, i =,, m 3 an the egrees of the vertices in the subgraph inuce by S 4 by r j, j =,, m 4 Set p = m 4 In orer that Z(A) = N k\n l for some k S an l S 2, all of the following conitions must hol: i) For each i S 4, either m 4 p r i p + m 4 m m 2 m 3, or r i { max p + m 4 m m 2 m 3, m } 4 p (m + m 2 + m 3 ) (p + m 4 ) ; ii) For each i S 3 an j S 4, either q i + m 4 r j + m + m 2 + m 3 an or an m + m 2 + m 3 q i m + m 2 + m 3 + r j p p + m 4, q i + m 4 r j + m + m 2 + m 3 m 4 r j m 4 + q i p p + m 4 ; iii) For each i, j S 3 with q i q j, we have that N i \ N j q i + m 4 p p + m 4 ; iv) For each i, j S 4 with r i r j, we have that N i \ N j m + m 2 + m 3 + r i p p + m 4 Our next result gives the block structure of certain nonbipartite graphs which satisfy the equality case in the DDZ eigenvalue boun Theorem 48 Suppose that G is a nonbipartite, connecte graph for which equality hols in (27) Suppose further that there is a λ sub eigenvector z having 24

26 just two istinct entries a an b with a > b Suppose also that z = a, z 2 = b, that 2, which can be assume without loss of generality as G is connecte, an that > 2 Then A an z can be taken to have the following form, where the partitionings are conformal: A = 0 T 0 T T J 2 J 0 J A 4,4 A 4,5 J A 5,4 A 5,5 ] b an z = a b (43) [ A4,4 A 4,5 Further, the matrix is the ajacency matrix of a biregular A 5,4 A 5,5 graph with egrees N \ N 2 an N \ N 2 Proof: We partition the rows an columns of A, as well as z, as follows: S = {}, S 2 = {2}, S 3 = (N \ N 2 ) \ {2}, S 4 = (N 2 \ N ) \ {}, S 5 = N N 2, S 6 = {i z i = a, i, 2}, an S 7 = {i z i = b, i, 2} (We note that some subsets in this partitioning may be empty; however, S 5, since Z(A) is assume to be less than ) With this partitioning it follows that b 0 T 0 T T 0 T 0 T a T 2 T 2 T 0 T 0 T a A = A 3, A 3,2 A 3,7 an z = b b A 7, A 7,2 A 7,6 A 7,7 a b (44) From the eigenequation Az = λ sub z it is straightforwar to etermine that λ sub = N\N2 an that λ sub a = b In particular, we fin from this last observation that if i S 2 S 3 S 6, then a i,j = 0, for each j S 2 S 3 S 6 Applying that observation in conjunction with the combinatorial symmetry of b 25

27 A, it follows that 0 T 0 T T 0 T 0 T T 2 T 2 T 0 T 0 T A 3, 0 0 A 3,4 A 3,5 0 A 3,7 A = 0 A 4,2 A 4,3 A 4,4 A 4,5 A 4,6 A 4,7 A 5, A 5,2 A 5,3 A 5,4 A 5,5 A 5,6 A 5, A 6,4 A 6,5 0 A 6,7 0 0 A 7,3 A 7,4 A 7,5 A 7,6 A 7,7 Further, since λ sub b = a N \ N 2 ( + b N ) \ N 2, it follows that A 5,2 + A 5,3 + A 5,6 = N \ N 2 But then from the combinatorial symmetry of A, we see that A 5, > 0, so that if i S 5 an j S 6, then 2 (e i e j ) T A > N \ N 2, a contraiction an we conclue that S 6 = Thus we can take our matrix A to be written as 0 T 0 T T 0 T T 2 T 2 T 0 T A = A 3, 0 0 A 3,4 A 3,5 A 3,7 0 A 4,2 A 4,3 A 4,4 A 4,5 A 4,7 A 5, A 5,2 A 5,3 A 5,4 A 5,5 A 5,7 0 0 A 7,3 A 7,4 A 7,5 A 7,7 Note that A 4,2 +A 4,3 = N\N2 an that A 5,2 +A 5,3 = N\N2 Thus by only if consiering j S 4 or j S 5, it follows that 2 (e 2 e j ) T A N\N2 A 4,7 an A 5,7 are zero matrices Hence A 7,4 = 0 an A 7,5 = 0 by combinatorial symmetry But this last is a contraiction since then for any j S 7, e T 2 A an e T j A have isjoint support We conclue that S 7 = 26

28 Consequently, our matrix A can be written as 0 T 0 T T T 2 T 2 T A = A 3, 0 0 A 3,4 A 3,5, 0 A 4,2 A 4,3 A 4,4 A 4,5 A 5, A 5,2 A 5,3 A 5,4 A 5,5 b a with z partitione conformally as z = a By consiering 2 (e e j ) T A b b for any j S 3, it follows that A 3,5 > 0 (assuming that S 3 ) If S 4, we note that if A 3,4 contains a zero entry, say in the column corresponing to i S 4, then it follows that the columns of A 4,4 an A 5,4 corresponing to inex i must be zero columns Hence the rows of A 4,4 an A 4,5 corresponing to inex i must also be zero rows But then two rows of A have isjoint support, namely the secon row of A an the row corresponing to inex i S 4, a contraiction We conclue that if S 4, then A 3,4 > 0 It now follows that every row of A corresponing to an inex in S 3 is the same as row 2 of A as Collapsing S 2 an S 3 into a single set S 2, we fin that A an z can be written A = 0 T 0 T T J 2 J 0 A 4,2 A 4,4 A 4,5 A 5, A 5,2 A 5,4 A 5,5 b an z = a b From combinatorial symmetry, we see that A 4,2 an A 5,2 must be positive, as is A 5, Since A 4,2 = N\N2 an A 5,2 = N\N2, it follows that the vertices of S 4 an S 5 must all have egree In particular, A = 0 T 0 T T J 2 J 0 J A 4,4 A 4,5 J A 5,4 A 5,5 The conitions on A 4,4, A 4,5, A 5,4, an A 5,5 now follow reaily b 27

29 [ A4,4 A 4,5 Note that in Theorem 48, the matrix is the ajacency A 5,4 A 5,5 matrix of a biregular graph, say H, with the vertices in S 4 having egree N \ N 2 an the vertices in S 5 having egree N \ N 2 In orer that the matrix A of (43) satisfies Z(A) = N\N2, the following conitions on H must hol: ] i) each vertex in S 4 is ajacent to at most N \ N 2 vertices in S 4, an each vertex in S 5 is ajacent to at most N \ N 2 vertices in S 4 ; ii) for each pair of vertices i, j S 4, N i \ N j N \ N 2 ; iii) for each pair of vertices i, j S 5, N i \ N j N \ N 2 ; an iv) for each pair of vertices i S 4, j S 5, N i \ N j N \ N 2 Finally, we also note that if S 4 =, then necessarily H = K N \N 2 As an example of an ajacency matrix of a graph G satisfying the conitions of Theorem 48 we give the matrix M = Here D = iag([6, 4, 4, 4, 4, 6, 6, 6]) an the transition matrix for the ranom walk, A = D M inuce by G satisfies that γ(a) = 2/3 = 2/3 = Z(A) We observe that the eigenvector of A corresponing to λ sub = 2/3 is, inee, given by z = [ 02774, 0460, 0460, 0460, 0460, 02774, 02774, 02774] T In our next result we investigate the form of the transition matrix which satisfies the DDZ boun for a ranom walk inuce by a regular graph 28

30 [ ] A, A,2 Theorem 49 Suppose that A = is a n n transition matrix for A 2, A 2,2 a connecte regular graph [ G ] that satisfies equality in (27), with corresponing b λ sub eigenvector z =, where a > b Label the subsets of the partition a S an S 2, respectively Fix inices i an j with i S an j S 2, respectively, an with i j Then λ sub = N i \ N j Set α = N i N j S 2 an β = N i N j S Then A, = β, A,2 = N i \ N j + α, A 2, = N i \ N j + β, an A 2,2 = α Proof: From Lemma 25 b), we fin immeiately that λ sub = Ni\Nj From the equations Az = λ sub z an A =, we fin that each of the blocks A,,, A 2,2 must have constant row sums, say x,,, x 2,2, respectively It is straightforwar [ to see ] that, necessarily, the eigenvalues of the 2 2 matrix [ ] X = x, x,2 are an λ sub, with corresponing eigenvectors an x 2, x 2,2 [ ] b, respectively a Consiering (e i e j ) T A an applying Lemma 23 we see that one of two situations hols: either z k = a for each k N i \ N j, an z k = b, for each k N j \ N i, or z k = b for each k N i \ N j, an z k = a, for each k N j \ N i Suppose that the latter case occurs It then follows that ] X = [ Ni\N j +β β α N i\n j +α which fails to have λ sub as an eigenvalue Hence only the former case can occur, from which we fin that e T i A,2 = N i \ N j + α, an hence e T i A, = β A similar argument yiels e T j A 2, = N i \ N j + β, an hence e T j A 2,2 = α The conclusion now follows Remark 40 Suppose that A is as in Theorem 49 an let G(S ) an G(S 2 ) enote the subgraphs inuce by S an S 2, respectively It follows that G(S ) an G(S 2 ) are both regular, of egrees N i \ N j + α an N i \ N j + β, respectively Further, in orer that Z(A) = Ni\Nj, each of the following conitions 29

31 must hol: i) for each k, l S, N k \ N l N i \ N j ; an ii) for each k, l S 2, N k \ N l N i \ N j ; iii) for each k S, l S 2, N k \ N l = N i \ N j Accoring to Seneta [9, Definition 32], a stochastic matrix A is calle scrambling if any two rows of A have at least one positive element in a coincient position It is easy to see that for such matrices A, Z(A) < In a similar vein, we say that a graph G is scrambling if it has the property that each pair of vertices has a common neighbour Eviently G is scrambling if an only if the transition matrix for the corresponing ranom walk on G is a scrambling stochastic matrix This leas us to the following result: Theorem 4 Let G be a scrambling graph on n 4 vertices an let A be the transition matrix for the corresponing ranom walk on G Then λ sub (A) n 2 n Furthermore equality hols if an only if G = K 2 O n 2, where enotes the join of two graphs an O k enotes the empty graph on k vertices Proof: From (27) we know that λ sub (A) Z(A) an, on applying Lemma 4, it follows that { } Ni \ N j Z(A) = max i, j are vertices of G an i j i We thus reaily fin that { i λ sub (A) max i } i =,, n n 2 n Suppose now that λ sub (A) = n 2 n an note that necessarily G must have at least one vertex of egree n Let z be an eigenvector of A corresponing to λ sub, say, with maximum entry a an minimum entry b Suppose first that z has an entry strictly between a an b Let i an j correspon to entries in z equal to a an b, respectively Referring to (42) of Lemma 44, we fin that vertices i an j have the same egree, say Since n 2 30 n =, we fin that = n

32 But in this case, 2 (e i e j ) T A = n, a contraiction We conclue that z has no entries strictly between a an b Suppose next that G is regular, say of egree It follows that Z(A), from which we conclue that = n But then G = K n, the complete graph on n vertices, an again we have a contraiction It now follows that A satisfies the hypotheses of Theorem 48, necessarily with = n Referring to (43), we fin that A can be written as follows for some 2 : A = 0 n T n T 2 J 2 0 n n J n (J I) (45) Suppose that G has k vertices of egree 2 an n k vertices of egree n We fin reaily from (45) that Z(A) = k n k = n 2 It follows that 2 = 2 an that G = K 2 O n 2 Conversely, if G = K 2 O n 2, then A = 0 n n T n 0 n T so that necessarily we have that which is easily seen to have eigenvalues 0 (of multiplicity n 3),, n an n 2 n, Theorem 42 yiels the following corollary for the eigenvalues of the normalize Laplacian arising from scrambling graphs: Corollary 42 Let G be a scrambling graph on n 4 with normalize Laplacian matrix L If λ 0 is an eigenvalue of L, then n λ 2n 3 n Equality hols in either of the bouns on λ if an only if G = K 2 O n 2 Recall that G is a threshol graph on n vertices if it can be generate from a one vertex graph by repeate applications of the following two operations: (i) aition of a single isolate vertex to the graph, an (ii) aition of a single to the graph that is connecte to all other vertices Recall further that threshol graphs are characterize by the property that they contain no inuce subgraphs that are isomorphic to either P 4, C 4 or K 2 K 2 It is not ifficult to see that the only regular threshol graphs are either complete or empty Our final result in this paper is: 3

33 Theorem 43 Let G be a connecte threshol graph on n vertices an let A be the ajacency matrix for the corresponing ranom walk on G Then equality hols in (27) if an only if G can be written as G = O p K n p, for some p n Proof: Suppose first that A is of the form escribe in (43) in Theorem 48 an partition the rows an columns of A as S,, S 4 conformally with (43) Since 2 < n, we fin that p S 2 2 If S 3, then selecting vertices u, v S 2 an w S 3, we fin that the subgraph of G inuce by vertices, u, v, w is C 4, a contraiction Hence S 3 =, so that = n Hence the vertices in S 4 must also have egree n an it follows that G = O p K n p Suppose next that A is of the form (42) escribe in Theorem 44, an partition the rows an columns of A as S,, S 4 conformally with (42) Since the subgraph H of G inuce by the vertices of S S 2 is regular, it is either a complete subgraph or an empty subgraph The latter case then yiels Z(A) = 2 (e i e j ) T A = 0, for i S an j S 2, a contraiction Thus we see that S S 2 inuces a complete subgraph on + vertices from which it follows that Z(A) = Observe that in this situation, necessarily S 3 = for otherwise Z(A) Hence = n, so that λ sub(a) = n It now follows reaily that in fact G = K n Conversely, it is straightforwar to etermine that if G = O p K n p, then A yiels equality in (27) Acknowlegement: The first author extens his appreciation to J Chris Fisher for helpful conversations an a reference on iameter graphs References [] A Ben-Israel an T N E Greville, Generalize Inverses: Theory an Applications, 2n e, Springer, New York, 2003 [2] A Berman an R J Plemmons, Nonnegative Matrices in the Mathematical Sciences, SIAM, Philaelphia, 994 [3] SL Campbell, CDJ Meyer, Generalize Inverses of Linear Transformations, Dover Publications: New York, 99 32

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