UNCOUPLING THE PERRON EIGENVECTOR PROBLEM

UNCOUPLING THE PERRON EIGENVECTOR PROBLEM Carl D Meyer INTRODUCTION Foranonnegative irreducible matrix m m with spectral radius ρ,afundamental problem concerns the determination of the unique normalized Perron vector π m which satisfies m π = ρπ, π > 0, π i = For relatively small values of m, this is not a difficult problem to solve using adaptations of standard techniques for solving systems of linear equations However, there exist many applications for which m is too large to be comfortably handled by standard methods For large scale problems, it is only natural to attempt to somehow uncouple the original matrix into two or more smaller matrices say P, P 2,, P k oforder r,r 2,,r k, respectively, where k i= r i = m Ideally, this sequence of smaller matrices should have the following properties i= Each smaller matrix P i should also be nonnegative and irreducible so that each P i has a unique normalized Perron vector π (i) Each smaller matrix P i should have the same spectral radius ρ as the original matrix It should be possible to determine the π (i) s completely independent of each other For modern parallel computer architectures it is desirable to be able to execute the computation of the π (i) s in parallel Finally, it must be possible to easily couple the smaller normalized Perron vectors π (i) back together in order to produce the normalized Perron vector π for the original nonnegative matrix This paper is dedicated to showing how to accomplish the above four goals 2 PERRON COMPLEMENTTION The purpose of this section is to introduce the concept of a Perron complement in a nonnegative irreducible matrix and to develop some of the basic properties of Perron complementation The concept of Perron complementation will be the cornerstone for the entire development which follows Unless otherwise stated, will denote an m m nonnegative irreducible matrix with spectral radius ρ and will be partitioned as = 2 22 2k Center for Research in Scientific Computation & Department of Mathematics, North Carolina State University, Box 8205, Raleigh, NC 27695-8205 This work was supported in part by the National Science Foundation under Grant DMS-85254

C D Meyer Uncoupling Perron Vectors Page 2 where all diagonal blocks are square It is well known that ρi is a singular M -matrix of rank m and that every principal submatrix of ρi of order m orsmaller is a nonsingular M -matrix See Berman and Plemmons [979, pg 56] In particular, if i denotes the principal submatrix of obtained by deleting the i th row and i th column of blocks from the partitioned form of, then each ρi i is a nonsingular M -matrix Therefore, (ρi i ) 0 (2) so that the following terms are well defined DEFINITION 2 Let be an m m nonnegative irreducible matrix with spectral radius ρ and let have a k -level partition = 2 22 2k in which all diagonal blocks are square For a given index i, let i denote the principal block submatrix of obtained by deleting the i th row and i th column of blocks from and let i and i designate i =( i i2 i,i i,i+ ik ) and i = i i,i i+,i ki That is, i is the i th row ofblocks with ii removed and i is the i th column of blocks with ii removed The Perron complement of ii in is defined to be the matrix P ii = ii + i (ρi i ) i For example, the Perron complement of 22 in = 2 3 2 22 23 3 32 33 is given by ( ρi P 22 = 22 +( 2 23 ) 3 3 ρi 33 ) ( 2 32 ) The reason for the terminology Perron complement will become clear as later developments unfold lthough the concept of Perron complementation as defined above is not the same as the well known concept

C D Meyer Uncoupling Perron Vectors Page 3 of Schur complementation, there is an obvious connection For example, consider a 2 -level partition of a nonnegative irreducible matrix = 2 2 22 with spectral radius ρ ssuming square diagonal blocks, the Perron complement of is given by P = + 2 (ρi 22 ) 2 and the Perron complement of 22 is P 22 = 22 + 2 (ρi ) 2 For this case, it is easy to see that the Perron complement of is in fact [ρi Schur Complement(ρI 22 )] in the matrix ρi and the Perron complement of 22 is [ρi Schur Complement(ρI )] in the matrix ρi The purpose of the following observations stated as technical lemmas is to help prove theorems which will establish the fact that every Perron complement is also a nonnegative irreducible matrix LEMM 2 Let be a m m nonnegative irreducible matrix partitioned as = 2 22 2k in which all diagonal blocks are square and suppose that the m m identity matrix I 0 0 0 I I = 2 0 0 0 I k is partitioned accordingly so that each I j and the corresponding jj have the same size If I 0 0 0 0 0 I j 0 Q(i, j) = 0 I i 0 0 0 0 0 I k is the elementary permutation matrix obtained by interchanging the i th and j th rows (or columns) of blocks in I, define  to be the matrix  = Q(i, j)q(i, j) The matrix  is also a nonnegative irreducible matrix and if ˆP ii denotes the Perron complement of  ii in Â, then ˆP ii = P jj, ˆP jj = P ii, and ˆP hh = P hh when h i, j

C D Meyer Uncoupling Perron Vectors Page 4 PROOF: The proof is conceptually straightforward and follows from the fact that  and agree everywhere except in the i th and j th block rows and i th and j th block columns LEMM 22 If is a nonnegative irreducible matrix with spectral radius ρ and if Q = Q(,i) is the elementary permutation described in Lemma 2 which corresponds to an interchange of the st and i th block positions, let  be the matrix  = QQ If  is re-partitioned into a 2 2 block matrix à as shown below,   2  k - -  =  2  22  2k = à à 2 = à (22) à 2 à 22  k  k2  kk then  = à = ii and the Perron complement of ii in is given by P ii = P = à + à 2 (ρi à 22 ) Ã2 (23) PROOF: The Perron complement of  in  is easily seen to be given by ˆP = P = à + à 2 (ρi à 22 ) Ã2 The desired conclusion follows from Lemma 2 since P ii = ˆP LEMM 23 If = 2 22 2k is a nonnegative irreducible matrix with spectral radius ρ, then the Perron complement of the ii -block in T is the transpose of the Perron complement of the ii -block in Inother words, if P ii is the Perron complement of ii in, then P T ii is the Perron complement of T ii in T PROOF: This is evident from the definition of Perron complementation We are now in a position to develop some of the basic properties of Perron complementation following theorem is the first in the sequence The

C D Meyer Uncoupling Perron Vectors Page 5 THEOREM 2 If = 2 22 2k is a nonnegative irreducible matrix with spectral radius ρ, then each Perron complement P ii = ii + i (ρi i ) i is also a nonnegative matrix whose spectral radius is again given by ρ Moreover, if π () π (2) π = > 0 π (k) is a conformably partitioned positive eigenvector for associated with ρ, then P ii π (i) = ρπ (i) (24) That is, π (i) > 0 is a positive eigenvector for P ii associated with the spectral radius ρ PROOF: ssume that has been permuted and re-partitioned as described in Lemma 22 so that à à à = QQ = 2 à 2 à 22 where Q = Q(,i) is the permutation matrix associated with an interchange of the st and i th block positions ccording to (23), the Perron complement of ii in is the same as the Perron complement of à in à That is, P ii = P Since each principal submatrix of ρi (and ρi Ã) oforder m or less is a nonsingular M -matrix, it follows from (2) that (ρi à 22 ) 0 and hence ) P ii = P = à + à 2 (ρi à 22 Ã2 0 Remark: P ii need not be strictly positive an example is given after this proof To prove that P ii π (i) = ρπ (i), define π Q π and write π π = Q π () = π (2) where π () = π (i) Use the facts that π = ρπ and Q 2 = I to conclude that ρ π = ρq π = Q π = QQ 2 π = à π and hence à π () + à 2 π (2) = ρ π () à 2 π () + à 22 π (2) = ρ π (2)

C D Meyer Uncoupling Perron Vectors Page 6 It now follows that and therefore π (2) = ( ρi à 22 ) Ã2 π () P ii π (i) = P π () = à π () + à 2 ( ρi à 22 ) Ã2 π () = à π () + à 2 π (2) = ρ π () = ρπ (i) This proves that ρ is an eigenvalue for P ii with an associated positive eigenvector given by π (i) Tosee that ρ is indeed the spectral radius of P ii, suppose instead that r is the spectral radius of P ii and let v T 0 be anonzero left hand eigenvector of P ii associated with r so that v T P ii = ρv T Itfollows that Since v T π (i) 0,itmust be the case that ρ = r ρv T π (i) = v T P ii π (i) = rv T π (i) To see that a Perron complement need not be strictly positive, consider a 4 4 nonnegative irreducible matrix whose partitions and zero pattern are shown below For this configuration, + + 0 + + + = + + + 0 + + ----------------------------- ------ + + + + P = + 2 (ρi 22 ) 2 = + + 0 + + + + 0 + [+] ( + + +) = + + 0 + + + + + + + + + + lthough Perron complements can have zero entries, the zeros are always in just the right places so as to guarantee that each P ii is an irreducible matrix THEOREM 22 If = 2 22 2k is a nonnegative irreducible matrix with spectral radius ρ, then each Perron complement P ii = ii + i (ρi i ) i is also a nonnegative irreducible matrix with spectral radius ρ

C D Meyer Uncoupling Perron Vectors Page 7 PROOF: Suppose that ii and hence P ii is r r If π () π (2) π = > 0 π (k) is a conformably partitioned positive eigenvector for associated with its spectral radius ρ, then Theorem 2 guarantees that π (i) > 0 is a positive eigenvector for P ii associated with its spectral radius ρ IfD is the r r diagonal matrix D = π (i) π (i) 2 π (i) r then S = D P ii D ρ is a row stochastic matrix For all row stochastic matrices, it is known that the unit eigenvalue has index ie, the unit eigenvalue for S has only linear elementary divisors See Gantmacher [960, Vol 2, pg 84] Therefore, as an eigenvalue of P ii, Index(ρ) = and hence the Jordan form for ρi P ii is J ρi Pii = 0t t 0 0 X r r where t is the algebraic multiplicity of ρ as an eigenvalue of P ii and where X is nonsingular ssume now that has been permuted and re-partitioned as described in Lemma 22 so that à à à = 2 à 2 à 22, and P ii = P By observing ( ) ( I Ã2 (ρi à 22 ) ρi à à 2 I 0 ρi P 0 I à 2 ρi à 22 (ρi à 22 ) = 0 à 2 I 0 ρi à 22 it follows that Rank(ρI Ã) =Rank(ρI P )+Rank(ρI à 22 ) Suppose Ã, Ã, and à 22 have dimensions m m, r r, and q q, respectively, with r + q = m It is well known that à being nonnegative and irreducible implies that Rank(ρI Ã) = m and Rank(ρI à 22 )=q Thus Rank(ρI P ii )=Rank(ρI P )=m q = r This together with (25) produces the conclusion that ρ must in fact be a simple eigenvalue for P ii By virtue of Lemma 23, P T ii is the Perron complement of T ii in T so that Theorem 2 guarantees P T ii also has a positive eigenvector associated with its spectral radius ρ It is known Gantmacher [960, Vol 2, pg 79] that if the spectral radius ρ of a nonnegative matrix is a simple eigenvalue and if the matrix as well as its transpose each possess a positive eigenvector associated with ρ, then the matrix must in fact be irreducible Since this is precisely the situation for the Perron complement P ii, the conclusion is that P ii must be irreducible (25) ),

C D Meyer Uncoupling Perron Vectors Page 8 The above proof is purely algebraic in nature However, irreducibility can be viewed strictly as a combinatorial concept because of the fact that a matrix M is irreducible if and only if its directed graph G(M) is strongly connected It is therefore desirable to also argue the irreducibility of a Perron complement from a graph theoretic point of view The author is indebted to C R Johnson for suggesting the following combinatorial proof Combinatorial Proof of Theorem 22: Forany particular Perron complement P ii of size r r, it can be assumed without loss of generality that has been permuted and repartitioned to the form n n = 2 2 22 in which is r r and P ii becomes identified with P = + 2 (ρi 22 ) 2 To demonstrate that G(P ) is strongly connected, let h, k {, 2,,r}, and let E( ) denote the set of directed edges in the directed graph of a specified matrix Case : There is a path from node h to node k in G( ) In this case, there must also be a path from node h to node k in G(P ) because E( ) E(P ) Case 2: There is no path from node h to node k in G( ) For this situation, there must be a path from node h to node k in G() but which necessarily passes through at least one node That is, G() contains a sequence of directed edges leading from node h to node k such that i {r +,r+2,,n} h h h p i i q k k t k {(h h ), (h h 2 ),, (h p h p )} E( ), (26) {(k k 2 ), (k 2 k 3 ),, (k t k} E( ), (27) and where φ {i,i 2,,i q } {r +,r+2,,n} In particular, the existence of directed edges h p i i q k in G() guarantees that [ ] 2 q 22 2 0 h pk This, together with the fact that 2 (ρi 22 ) 2 = j=0 implies [ 2 (ρi 22 ) 2 ] 2 j 22 2 ρ j+, so that (h p k ) E(P ) Combining this with (26) and (27) together with the fact that E( ) E(P ) leads to the desired conclusion that there is a path from node h to node k in G(P ) h pk 0

C D Meyer Uncoupling Perron Vectors Page 9 3 UNCOUPLING ND COUPLING THE PERRON VECTOR dopt the following definition which is consistent with the terminology used in Horn and Johnson [985] DEFINITION 3 For anonnegative irreducible matrix with spectral radius ρ, the unique normalized eigenvector π satisfying the conditions π = ρπ, π > 0, and e T π = where e T =( ) is called the Perron vector for If = 2 22 2k is a nonnegative irreducible matrix with spectral radius ρ, then the results of the previous section guarantee that each Perron complement P ii = ii + i (ρi i ) i is also a nonnegative irreducible matrix which again has spectral radius ρ Furthermore, if π () π (2) π = π (k) is the conformably partitioned Perron vector for, then Theorem 2 guarantees that the vector defined by π (i) p i e T π (i) is the Perron vector for the associated Perron complement P ii Inwhat follows, the normalizing scalar ξ i e T π (i) > 0 will be called the i th coupling factor Observe that the Perron vector for the larger matrix can be written in terms of the Perron vectors of the smaller Perron complements as ξ p ξ 2 p 2 π = (3) ξ k p k This illustrates that at least in theory it is possible to uncouple the Perron eigenvector problem by using Perron complementation If matrix is partitioned to k levels so as to yield k Perron complements

C D Meyer Uncoupling Perron Vectors Page 0 one P ii for each diagonal block then the Perron vectors p i for the P ii s can be determined independent of each other and then combined in order to construct the Perron vector π for the larger matrix However, the expressions for the coupling factors ξ i have the form ξ i = e T π (i) = h π (i) h t first glance, this might seem to place the issue in a hopeless circle because prior knowledge of π in the form of the sums h π (i) h isnecessary in order to reconstruct π from the p i s Fortunately, this is not the case The following result shows that the coupling factors ξ i can very easily be determined without prior knowledge of the π (i) s This is a key feature in the uncoupling-coupling technique DEFINITION is partitioned as with square diagonal blocks and let p i 32 Let be a nonnegative irreducible matrix with spectral radius ρ Suppose that = 2 22 2k be the Perron vector for the Perron complement P ii = ii + i (ρi i ) i The coupling matrix associated with is defined to be the k k matrix C =[c ij ] whose entries are given by 3 c ij e T ij p j The reason for this definition willimmediately become apparent The utility of the coupling matrix is realized in the following theorem which demonstrates how the Perron vectors of the individual Perron complements can be easily coupled together by means of the coupling matrix C in order to construct the Perron vector for THEOREM 3 Foranonnegative irreducible matrix = 2 22 2k with spectral radius ρ and conformably partitioned Perron vector π () π π = (2), π (k) 3 Throughout, the size of the vector e T =( ) will always be defined by the context in which it appears

C D Meyer Uncoupling Perron Vectors Page the associated k k coupling matrix C is also a nonnegative irreducible matrix whose spectral radius is again given by ρ Furthermore, the Perron vector for C is precisely the vector ξ ξ ξ = 2 whose components are the positive coupling factors defined earlier as ξ k ξ i = e T π (i) The vector ξ is hereafter referred to as the coupling vector associated with the partitioned matrix PROOF: It is clear from the definition of the coupling matrix that C 0 To see that C is irreducible, note that because e T > 0, ij 0, and p j > 0, it must be the case that c ij =0 if and only if ij = 0 Since is irreducible, the preceding statement implies that C must also be irreducible otherwise, if C could be permuted to a block triangular form, then so could Toshow that the coupling vector ξ ξ ξ = 2 is the Perron vector for C, use the fact ξ k p j = π (j) ξ j and compute the i th component of the product C ξ to be (Cξ) i = k c ij ξ j = j= k e T ij p j ξ j j= k = e T ij π (j) = e T ρπ (i) = ρξ i j= Thus C ξ = ρξ so that ρ is a positive eigenvalue for C associated with a positive eigenvector ξ Since C is irreducible, ρ must therefore be the spectral radius of C Because π is a normalized vector, it follows that ξ is indeed the normalized positive eigenvector and hence the Perron vector for C by observing π () k k π ξ i = e T π (i) =(e T e T e T ) = et π = i= i= π (k) By combining the observation of (3) together with Theorem 3, we arrive at the major conclusion of this paper which is summarized below as a formal statement

C D Meyer Uncoupling Perron Vectors Page 2 THEOREM 32 (The Coupling Theorem) Let be a nonnegative irreducible matrix with spectral radius ρ and suppose that has a k -level partition = 2 22 2k with square diagonal blocks If p i is the Perron vector for the Perron complement P ii = ii + i (ρi i ) i and if ξ ξ ξ = 2 is the Perron vector for the k k coupling matrix C in which then ξ k c ij = e T ij p j, ξ p ξ π = 2 p 2 ξ k p k is the Perron vector for Conversely, if the Perron vector for is given in the form of a conformably partitioned vector π () π π = (2), π (k) then ξ i = e T π (i) and π (i) ξ i is the Perron vector for the Perron complement P ii The case of a 2 -level partition is of special interest The following corollary shows that it is particularly easy to uncouple and couple the Perron eigenvector problem for these situations COROLLRY partitioned as 3 Let be a nonnegative irreducible matrix with spectral radius ρ If is = 2 2 22 where and 22 are square, then the Perron vector for is given by ξ p π = ξ 2 p 2

C D Meyer Uncoupling Perron Vectors Page 3 where p and p 2 are the Perron vectors for the Perron complements P = + 2 (ρi 22 ) 2 and P 22 = 22 + 2 (ρi ) 2, respectively, and where the coupling factors ξ and ξ 2 are given by ξ = e T 2 p 2 ρ e T p + e T 2 p 2 and ξ 2 = ξ PROOF: Simply verify that the given values of ξ and ξ 2 are the two components of the Perron vector for the coupling matrix e C = T p e T 2 p 2 e T 2 p e T 22 p 2 There is always a balancing act to be performed when uncoupling the Perron eigenvector problem using Perron complementation s k increases and the partition of becomes finer, the sizes of the Perron complements become smaller thus making it easier to determine each of the Perron vectors p i t the same time however, the size of the coupling matrix C k k becomes larger thus making it more difficult to determine the coupling factors ξ i In the two extreme cases when either k = m or k =,there is no uncoupling whatsoever because C = when k = m and P = when k = Furthermore, as the size of a Perron complement becomes smaller, the order of the matrix inversion embedded in the Perron complement becomes larger One must therefore choose the partition which best suits the needs of the underlying application For example, if computation utilizing a particular parallel architecture is the goal, then the specific nature of the hardware and associated software may dictate the partitioning strategy Rather than performing a single uncoupling-coupling operation to a high level partition of,anal- ternative is to execute a fork-join procedure using only 2 -level partitions at each stage Starting with a nonnegative irreducible matrix (with spectral radius ρ ) ofsize m m, partition roughly in half as = 2 2 22 to produce two Perron complements, P and P 22, which are each nonnegative irreducible matrices (each having spectral radius ρ )oforder approximately m/2 The Perron complements P and P 22 may in turn be partitioned roughly in half to produce four Perron complements say (P ),(P ) 22,(P 22 ), and (P 22 ) 22 each of which is of order approximately m/4 (again, each has spectral radius ρ ) This process can continue until all Perron complements are sufficiently small in size so as to easily yield a Perron vector The small Perron vectors are then very easy to successively couple according to the rule given in Corollary 3 until the Perron vector π for the original matrix is produced t each stage, all Perron complements have spectral radius ρ so that no eigenvalue computation past the initial determination of ρ is necessary For example, consider the following 8 8 nonnegative irreducible matrix whose spectral radius is ρ 33248

C D Meyer Uncoupling Perron Vectors Page 4 8 6 3 5 7 0 7 0 7 3 8 5 6 4 2 6 3 8 8 7 2 8 4 0 7 7 8 2 = - - 2 4 6 2 4 0 4 3 6 6 0 6 5 7 6 5 8 4 8 2 4 5 5 0 7 0 3 4 For the indicated partition, the Perron complements of are given by 32 05 836 7639 9309 8738 66 03 7575 3058 989 775 274 P = - - 523 495 9 8862 and P 37 6592 72 306 22 = - - 8567 05 2880 640 6 038 73 9455 2842 5969 54 with coupling vector ξ = Now, the two Perron complements of P are 750 765 (P ) = 27 206 with coupling vector ξ = while the two Perron complements of P 22 are 689 856 (P 22 ) = 79 292 5469 453 and (P ) 22 = 503 4897 and (P 22 ) 22 = ( 60 ) 54 734 793 ( 2625 ) 9965 605 035 with coupling vector ξ 2 = 5595 4405 The Perron vector for (P ) is given by 5286 474 and the Perron vector for (P ) 22 so that the Perron vector for P is 5286 2697 503 p = ( 474 ) 4690 = 2406 2297 4897 530 2600 is 4690 530 32 The arithmetic indicated in this example is not exact Numbers have been rounded to four significant digits

C D Meyer Uncoupling Perron Vectors Page 5 Similarly, the Perron vector for (P 22 ) is given by 5878 so that the Perron vector for P 422 22 is 536 4684 536 2974 5595 p 2 = ( 4684 ) 5878 = 262 2589 4405 422 86 and the Perron vector for (P 22 ) 22 is Therefore, the Perron vector for must be 2697 475 2406 36 5469 2297 256 ξ p π = 2600 422 = = ξ 2 p 2 2974 347 262 88 453 2589 73 86 0823 In addition to the fork-join process illustrated above, there are several other variations and hybrid techniques eg, iterative methods which are possible It is clear that the remarks of this section can serve as the basis for a fully parallel algorithm for computing the Perron vector for a nonnegative irreducible matrix However, there are several substantial computational issues which must be addressed and therefore a more detailed discussion concerning parallel implementations will be considered in separate papers 4 PRIMITIVITY ISSUES Primitivity is, of course, an important issue ccordingly, it is worthwhile to make some observations concerning the degree to which primitivity or lack of primitivity in a partitioned matrix is inherited by the smaller Perron complements P ii The first observation to make is that being a primitive matrix is not sufficient to guarantee that all Perron complements are primitive For example, the matrix = 0 2 0 0 2 - - 0 0 is irreducible and primitive because 5 > 0 However, for the indicated partition, the Perron complement is not primitive P = 0 0 Nevertheless, it is true that if a particular diagonal block in is primitive, then the corresponding Perron complement must also be primitive

C D Meyer Uncoupling Perron Vectors Page 6 THEOREM 4 Let be anonnegative irreducible matrix partitioned as = 2 22 2k in which all diagonal blocks are square If a particular diagonal block ii Perron complement P ii must also be primitive is primitive, then the corresponding PROOF: Since ii 0 and i (I i ) i 0,itfollows that for each positive integer n, P n ii = [ ii + i (I i ) ] n i = n ii + N where N 0 Therefore, P n ii > 0 whenever n ii > 0 The following theorem explains why all but a special class of Perron complements must be primitive THEOREM 42 If ii has at least one nonzero diagonal entry, then the corresponding Perron complement P ii must be primitive PROOF: If ii has at least one nonzero diagonal entry, then so does P ii Theorems 2 and 22 guarantee that each P ii is always nonnegative and irreducible It is well known see Berman and Plemmons [979, pg 34] that an irreducible nonnegative matrix with a positive trace must be primitive Therefore, P ii must be primitive The converse of Theorem 4 as well as the converse of Theorem 42 is false The matrix 0 0 0 0 0 0 0 0 0 0 0 = 2 0 0 0 0 - - 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 (4) is irreducible but notice that neither, 22, nor itself, is primitive Nevertheless, the corresponding Perron complements P = 0 2 0 0 and P 22 = 0 0 2 0 2 0 2 0 0 0 are each primitive

C D Meyer Uncoupling Perron Vectors Page 7 This example indicates another important advantage which Perron complementation can provide The matrix in (4) is not primitive and hence its Perron vector π cannot be computed by the simple iteration 4 x n+ = x n where x 0 is arbitrary ρ However, P and P 22 are each primitive and therefore each P ii yields a Perron vector p i by means of the straightforward iterations p (n+) i = ( Pii ρ ) p (n) i where p (0) i is arbitrary Take note of the fact that the two iterations represented here are completely independent of each other and consequently they can be implemented simultaneously By using the coupling factors described in Corollary 3, it is easy to couple p with p 2 in order to produce the Perron vector for the larger imprimitive matrix 5 SUMMRY Foranonnegative irreducible matrix which is partitioned as a matrix with square diagonal blocks m m = 2 22 2k which has spectral radius ρ and conformably partitioned Perron vector π () π π = (2), π (k) the concept of the Perron complement of ii is introduced to be P ii = ii + i (ρi i ) i, i =, 2,,k Perron complementation is shown to possess the following properties: Each P ii is also a nonnegative and irreducible matrix Each P ii also has spectral radius ρ The Perron vector p i for P ii is the normalized i th segment of π That is, p i = π (i) ξ i normalizing factors (also referred to as the coupling factors) are ξ i = e T π (i) ξ ξ 2 The vector ξ k = ξ k where the is the Perron vector for the k k coupling matrix C which also is a nonnegative irreducible matrix with spectral radius ρ whose entries are defined to be c ij = e T ij p j 4 The spectral radius ρ is included here for the sake of generality Notice that the matrix in (4) is column stochastic so that ρ =

C D Meyer Uncoupling Perron Vectors Page 8 These results allow the Perron eigenvector problem to be completely uncoupled in the sense that the Perron vector p i of each small Perron complement P ii can be determined independently from the Perron vectors of the other Perron complements By using the Perron vector ξ for the coupling matrix C, the smaller Perron vectors p i associated with the individual Perron complements can be coupled together to form the Perron vector π for the original large matrix by taking ξ p ξ π = 2 p 2 ξ k p k This work represents a generalization of the results of Meyer [987] in which the concept of stochastic complementation was introduced in order to uncouple finite Markov chain problems

C D Meyer Uncoupling Perron Vectors Page 9 REFERENCES Berman and R J Plemmons [979], Nonnegative Matrices In The Mathematical Sciences, cademic Press, New York F R Gantmacher [960], Matrix Theory Vol II, Chelsea, New York R Horn and C Johnson [985], Matrix nalysis, Cambridge University Press, New York C D Meyer [987], Stochastic complementation with applications to uncoupling Markov chains and the Simon-ndo theory of nearly reducible systems, NCSU Center Res Sci Comp Tech Report # 00870