An Inequality for Nonnegative Matrices and the Inverse Eigenvalue Problem

Transcription

1 An Inequality for Nonnegative Matrice and the Invere Eigenvalue Problem Robert Ream Program in Mathematical Science The Univerity of Texa at Dalla Box 83688, Richardon, Texa Abtract We preent two verion of the ame inequality, relating the maximal diagonal entry of a nonnegative matrix to it eigenvalue We demontrate a matrix factorization of a companion matrix, which lead to a olution of the nonnegative invere eigenvalue problem (denoted the nniep) for matrice of trace zero, and we give ome ufficient condition for a olution to the nniep for 5 5 matrice of trace zero We alo give a neceary condition on the eigenvalue of a 5 5 trace zero nonnegative matrix in lower Heenberg form Finally, we give a brief dicuion of the nniep in retricted cae Keyword nonnegative matrice, maximal diagonal entry, nniep, companion matrice AMS ubject claification A18, A23, A2 1

2 An Inequality for Nonnegative Matrice An n n matrix with real entrie i aid to be nonnegative if all of it entrie are nonnegative A nonnegative matrix i aid to irreducible if under imilarity with a permutation matrix, it cannot be written in the form ( ) A11, A 21 A 22 where A 11 and A 22 are quare matrice of order le than n The Perron-Frobeniu theorem tate that if A i a nonnegative matrix, then it ha a real eigenvalue r (known a the Perron root) which i greater than or equal to the modulu (or abolute value) of each of the other eigenvalue, and alo A ha an eigenvector v aociated with r uch that each of it entrie are nonnegative Further, if A i irreducible then r i poitive and the entrie of v are poitive [L-T] If A i irreducible, then knowing that the vector v = (v 1, v 2,, v n ) T jut mentioned ha poitive entrie, we can write v = Dw, where D = diag(v 1,, v n ) and w = (1, 1,, 1) T, and then Av = rv can become D 1 ADw = rw, ie under imilarity an r-eigenvector i w = (1, 1,, 1) T It wa Brauer [B] in 1952 who firt howed that if the pectrum of an arbitrary n n matrix A i (the arbitrary et) σ = {r, λ 2,, λ n }, if v = (v 1, v 2,, v n ) T i an eigenvector aociated with r, and we take the n n matrix B, the i th column of which i α i v for each i, 1 i n, then A + B ha eigenvalue r + n i=1 α iv i, λ 2,, λ n To ee thi, let P be an n n invertible matrix which upper triangularize A, and chooe the firt column of P a v, o that r λ P 1 2 AP =, λ n then and we re done α 1 α 2 α n P 1 (A + B)P = P 1 AP + P, r + n i=1 α iv i λ 2 =, 2

3 We come now to our firt reult Theorem 1: Let A = (a ij ) be an n n nonnegative matrix with pectrum σ = {r, λ, λ 3,, λ n }, where r i the Perron root, λ i real and n 3, then max a ii 1 λ j 1 i n n 2 Proof: Suppoe (for now) that A i irreducible We will aume without lo of generality that A ha eigenvector w = (1, 1,, 1) T correponding to the Perron root r Under permutation imilarity we can alo aume that a nn i maximal among the diagonal entrie of A Now conider the matrix C = A + B, where the j th column of B i β j w for each j, 1 j n We know from Brauer theorem that C ha eigenvalue r = r + n j=1 β j, λ, λ 3,, λ n Take now β j = a nn a jj, 1 j n, o that C i nonnegative, ha all it diagonal entrie equal, and ha eigenvector w (till) correponding to the Perron root r From Geršgorin theorem applied to C = (c ij ) we have that λ c ii c ij = r c ii, for ome i, 1 i n j=1 j i But C ha all it diagonal entrie equal to a nn, o rewriting thi inequality a j=3 r + a nn λ a nn r a nn, then jut uing the left-hand inequality, and knowing r, we obtain 2a nn r + λ = r + (a nn a jj ) + λ, j=1 = r + na nn a jj + λ, j=1 = r + na nn [r + λ + = na nn λ j, j=3 λ j ] + λ, o that n j=3 λ j (n 2)a nn, which prove the theorem when A i irreducible In cae A i reducible, conider the matrix C = A + B, where thi time each entry of B i ɛ > (ɛ mall), then C i irreducible and ha eigenvalue r + nɛ, λ, λ 3,, λ n Now applying the reult jut proved we have that max 1 i n (a ii + ɛ) 1 n 2 n j=3 λ j But thi i true for any ɛ arbitrarily mall, o it i true for ɛ = and the theorem i proved 3 j=3

4 Another verion of the inequality in Theorem 1 i the following Theorem 2: Let A = (a ij ) be an n n nonnegative matrix with pectrum σ = {r, a + ib, a ib, λ,, λ n }, where r i the Perron root and n, then b 2 [(n 2) max 1 i n a ii a j= λ j ][n max 1 i n a ii 3a λ j ] Proof: We reaon a before, applying Brauer theorem and then Geršgorin theorem, to arrive at the inequality j= o that (a a nn ) 2 + b 2 (r a nn ) 2, (r + (a nn a jj ) a nn ) 2, j=1 (r + (n 1)a nn λ j ) 2, j=1 (r + (n 1)a nn (r + 2a + λ j )) 2, j= b 2 ((n 1)a nn 2a λ j ) (a a nn ) 2, ((n 2)a nn a j= λ j )(na nn 3a j= j= λ j ) The Nonnegative Invere Eigenvalue Problem Let σ = {λ 1,, λ n } C The nonnegative invere eigenvalue problem i to find neceary and ufficient condition that σ i the et of eigenvalue of an n n nonnegative matrix A (ay) (thi well-known problem i currently unolved except in retricted cae, ee [B-P], [M], [B-H]) By σ we hall mean the complex conjugate of each of the entrie of the et σ The neceary condition σ = σ and k = λ k λ k n = trace(a k ) for k = 1, 2, are eay to ee Johnon [J], Loewy and London [L-L], have hown that for a nonnegative matrix A with pectrum σ = {λ 1,, λ n } and k = λ k λk n = trace(ak ) then n m 1 km m k, for k, m = 1, 2, We hall refer to thee neceary condition henceforth a J-L-L We now olve the nniep for matrice of trace zero Theorem 3: Let σ = {λ 1, λ 2, λ 3, λ } C If 1 =, 2, 3 and 2 2, then there exit a nonnegative matrix with pectrum σ

5 Proof: We remark that the lat inequality among the hypothee of the theorem i the J-L-L neceary condition with n =, k = 2 and m = 2 Let p 1 = λ 1 + λ 2 + λ 3 + λ, p 2 = λ 1 λ 2 + λ 1 λ 3 + λ 1 λ + λ 2 λ 3 + λ 2 λ + λ 3 λ, p 3 = λ 1 λ 2 λ 3 + λ 1 λ 2 λ + λ 1 λ 3 λ + λ 2 λ 3 λ, p = λ 1 λ 2 λ 3 λ, ie p 1, p 2, p 3, p are Newton elementary ymmetric polynomial Newton identitie for ymmetric function [van der W] tate that 1 p 1 =, 2 p p 2 =, 3 p p 2 1 3p 3 =, p p 2 2 p p = We can write thee equation in matrix form with a companion matrix a = p p 3 p 2 p Letting p 1 = 1 = and multiplying both ide on the right by diag( 1, 1 3, 1 2, 1) we get 2 2 = 3 2 ( ) 3 p p 3 p It i eaily verified that = 1 2, 2 3 then multiplying on the left of ( ) with thi matrix we get = 3 2 p 3 p 3 p Finally, performing a imilarity on thi matrix we have = ,

6 which i nonnegative and imilar to the companion matrix with eigenvalue λ 1, λ 2, λ 3, λ, a required Remark: Notice that the analogue of equation ( ) for n n matrice give a factorization of an n n companion matrix, even when the trace i not necearily zero The next theorem give ufficient condition for the exitence of a nonnegative 5 5 matrix of trace zero with eigenvalue λ 1, λ 2, λ 3, λ, λ 5 Theorem : Let σ = {λ 1, λ 2, λ 3, λ, λ 5 } If 1 =, 2, 3, 2 2 and , then there exit a nonnegative 5 5 matrix with pectrum σ Proof: We perform the ame procedure with 5 5 matrice a in Theorem 3 Beginning with the correponding equation ( ) we have = p 5 p p 3 p , }{{}}{{}}{{} C A B o CA = B Then it can be checked that A 1 CA = A 1 B = A in the proof of Theorem 3 we perform a conveniently choen imilarity proving the theorem = , ( )

7 Continuing with our invetigation of olution to the nniep for 5 5 matrice, note that the matrix at ( ) improve on a trace zero companion matrix (in the ene that if the entrie of a companion matrix, for a given σ, are nonnegative then the matrix at ( ) i nonnegative alo, but not converely) which ha the form To illutrate that we have improved on a companion matrix conider σ = {6, 1, 1,, } Concerning the (5,1) and (5,2) entrie of the matrix at ( ), it i worth mentioning that the inequality i not true for all (even trace zero) nonnegative matrice, a can be een from the following nonnegative matrix 1, for which 5 =, whilt 2 = 2 and 3 = 3 Although, for a trace zero 5 5 nonnegative matrix in lower Heenberg form we claim that 2 2, a we now how In fact we how omething lightly tronger than thi Let a a 13 a 1 a 15 a 21 a 23 a 2 a 25 A = a 31 a 32 a 3 a 35 a 1 a 2 a 3 a 5 a 51 a 52 a 53 a 5 Then with the uual notation k = trace(a k ), 1 =, 2 2 = a a 21 +a 13 a 31 +a 1 a 1 +a 15 a 51 +a 23 a 32 +a 2 a 2 +a 25 a 52 +a 3 a 3 +a 35 a 53 +a 5 a 5, and a 23 a 2 a 25 a 13 a 1 a a = 32 a 3 a 35 a + 31 a 3 a 35 8 a 2 a 3 a 5 a 1 a 3 a 5 a 52 a 53 a 5 a 51 a 53 a 5 a a 1 a 15 a a 13 a 15 a a 13 a 1 a + 21 a 2 a 25 a + 21 a 23 a 25 a + 21 a 23 a 2, a 1 a 2 a 5 a 31 a 32 a 35 a 31 a 32 a 3 a 51 a 52 a 5 a 51 a 52 a 53 a 1 a 2 a 3 7

8 = a 32 a 23 (a 5 a 5 + a 15 a 51 + a 1 a 1 ) + a 21 a (a 5 a 5 + a 35 a 53 + a 3 a 3 ) + a 31 a 13 (a 5 a 5 + a 52 a 25 + a 2 a 2 ) + a 2 a 2 a 53 a 35 + a 52 a 25 a 3 a 3 + a 1 a 1 a 53 a 35 + a 51 a 15 a 2 a 2 + a 51 a 15 a 3 a 3 + a 1 a 1 a 52 a 25 x, where x, (grouping ome of the term and letting x include all the negative term) = a 32 a 23 ( 2 2 a a 21 a 13 a 31 a 3 a 3 a 35 a 53 a 23 a 32 a 2 a 2 a 25 a 52 )+ a 21 a ( 2 2 a a 21 a 13 a 31 a 1 a 1 a 15 a 51 a 23 a 32 a 2 a 2 a 25 a 52 )+ a 31 a 13 ( 2 2 a a 21 a 13 a 31 a 1 a 1 a 15 a 51 a 23 a 32 a 3 a 3 a 35 a 53 )+a 2 a 2 a 53 a 35 + a 52 a 25 a 3 a 3 + a 1 a 1 a 35 a 53 + a 51 a 15 a 2 a 2 + a 15 a 51 a 3 a 3 + a 1 a 1 a 52 a 25 x, = a 2 32 a2 23 a2 a2 21 a2 31 a2 13 2a a 21 a 23 a 32 2a 13 a 31 a a 21 2a 31 a 13 a 23 a (a 32a 23 + a a 21 + a 13 a 31 ) y + a 2 a 2 a 53 a 35 + a 52 a 25 a 3 a 3 + a 1 a 1 a 53 a 35 + a 51 a 15 a 2 a 2 + a 51 a 15 a 3 a 3 + a 1 a 1 a 52 a 25 x, where y Collecting appropriate term we can write the above equation a (a 32 a 23 + a 21 a + a 31 a 13 ) (a 32a 23 + a 21 a + a 31 a 13 ) + x + y = a 2 a 2 a 53 a 35 +a 52 a 25 a 3 a 3 +a 1 a 1 a 53 a 35 +a 51 a 15 a 2 a 2 +a 51 a 15 a 3 a 3 +a 1 a 1 a 52 a 25 When a 1 = a 2 = a 15 = a 25 =, or a 15 = a 25 = a 35 =, we have a quadratic with real root and therefore ( 2 2 ) 2 + ( ) (x + y), o 2 2, ie 2 2, proving the claim A i in lower Heenberg form if a 13 = a 1 = a 15 = a 2 = a 25 = a 35 = More Retricted Cae of the NNIEP Notice from the 5 5 companion matrix given earlier that when A ha trace zero then det(a) = , o if det(a), then , thu olving the nniep in a retricted cae, namely trace zero, determinant nonnegative and the matrix A retricted to being in lower Heenberg form (the latter implying that 2 2, while 2 and 3 are alo neceary condition) We conider now the nniep retricted to nonnegative matrice having their diagonal entrie equal It i eay to ee that thi problem reduce to the trace zero cae, ince if A ha diagonal entrie equal, thee diagonal entrie mut be each 1 n and o A 1 n I (where I i the identity matrix) ha trace zero, and having olved the trace zero cae we can add back on the calar matrix 1 n I to obtain a matrix with the eigenvalue of A Looking at thi in more detail, we olve the nniep for matrice where we only conider olving the problem when the diagonal entrie are equal The nniep wa olved by 8

9 Loewy and London in [L-L] for 3 3 matrice It i not difficult to how that retricting the diagonal entrie to be equal in the 2 2 and 3 3 nniep doe not prevent u from olving the general nniep (not counting reducible cae, we will concentrate on the irreducible cae) If a matrix A ha eigenvalue σ = {r, λ, λ 3, λ }, then A 1 I ha trace zero, and a matrix with the eigenvalue of A 1 I i a given in Theorem 3, where 1 = trace(a 1 I) =, 2 = trace([a 1 I]2 ) = trace(a A I) = = which i greater than or equal to zero even in the general nniep Alo, 3 = trace([a 1 I] 3 ) = , mut be greater than or equal to zero for a matrix with equal 8 diagonal entrie, and likewie we mut have 2 2 Thu we have two new neceary condition when the diagonal entrie are equal It i perhap worth noting that = 3(r + λ λ 3 λ )(r λ + λ 3 λ )(r λ λ 3 + λ ), for which, in the cae of four real eigenvalue, we mut have each factor greater than or equal to zero, and thi fact could have been deduced from Theorem 1 (with the λ of Theorem 1 taken a λ, λ 3 and λ ucceively) where here max 1 i a ii = (r+λ+λ 3+λ ) Conidering matrice with diagonal entrie equal doe not prevent u from olving the problem if you only conider the nniep with real eigenvalue (again, excluding reducible cae), ee the olution in [L-L] of thi cae to ee thi (they did not need Suleimanova reult quoted in their proof) Diagonal entrie equal doe however prevent u from olving the general nniep in the remaining cae of σ = {r, λ, a + ib, a ib} (with b ), for which Theorem 1 (or ) implie r + λ 2a To ee that we don t alway have r + λ 2a, conider the matrix with pectrum σ = {6, 1, 3 + i, 3 i} which i nonnegative but r + λ 2a < Acknowledgement Mot of thee reult are taken from my PhD thei while at Univerity College Dublin, Ireland and I will take thi opportunity to thank my advior Thoma J Laffey for hi great patience and encouragement during thoe year Reference 9,

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