Perron Vectors of an Irreducble Nonnegatve Interval Matrx Jr Rohn August 4 2005 Abstract As s well known an rreducble nonnegatve matrx possesses a unquely determned Perron vector. As the man result of ths paper we gve a descrpton of the set of Perron vectors of all the matrces contaned n an rreducble nonnegatve nterval matrx A. Ths result s then appled to show that there exsts a subset A of A parameterzed by n parameters (nstead of n 2 ones n the descrpton of A) such that for each A A there exsts a matrx A A havng the same spectral radus and the same Perron vector as A. Key words. Nonnegatve matrx rreducble matrx nterval matrx Perron vector. AMS Subect Classfcatons. 15A18 15A48 65G40. 1 Irreducble nterval matrces In ths paper we consder only square n n matrces. Such a matrx A s called nonnegatve f all ts entres are nonnegatve. A nonnegatve matrx A R n n s sad to be reducble f there exsts a permutaton matrx P such that ( ) B C P T AP = 0 D where B and D are square matrces (.e. at least of sze 1 1) and t s called rreducble f t s not reducble. The basc egenvalue propertes of rreducble nonnegatve matrces are summed up n the Perron-Frobenus theorem (see Horn and Johnson [3] p. 508). We formulate here only a porton of t relevant to the scope of ths paper; ϱ(a) denotes the spectral radus of A e = (1 1... 1) T R n and x > 0 means that all entres of x are postve. Insttute of Computer Scence Czech Academy of Scences Pod vodárenskou věží 2 182 07 Prague Czech Republc (rohn@cs.cas.cz). 1
Theorem 1. For each rreducble nonnegatve matrx A there exsts a unque vector x satsfyng and no egenvalue λ ϱ(a) has a postve egenvector. Ax = ϱ(a)x (1) e T x = 1 (2) x > 0 (3) The postve egenvector determned unquely by (1) (3) s called the Perron vector of A; we shall denote t by x(a). Gven A A R n n wth A A the set A = [A A] = { A A A A } s called an nterval matrx wth the bounds A and A (see e.g. Neumaer [4] for basc facts concernng nterval matrces). A s sad to be nonnegatve f A 0 whch s the same as to say that all matrces n A are nonnegatve. A nonnegatve nterval matrx A s called rreducble f each A A s rreducble. It turns out that checkng rreducblty of A = [A A] reduces to checkng ths property for A only. The followng proposton s a consequence of a more general result (Berman and Plemmons [1] Corollary 1.3.21) but we nclude an elementary proof of t for the sake of completeness. Proposton 2. A nonnegatve nterval matrx [A A] s rreducble f and only f A s rreducble. Proof. If each A [A A] s rreducble then so s A. Conversely assume that A s rreducble and that some A [A A] s reducble so that there exsts a permutaton matrx P such that ( ) B C P T AP = 0 D where 0 s of sze at least 1 1. Then from 0 A A t follows ( ) ( ) ( 0 0 P T B1 C AP = 1 B C P T AP = 0 0 E 1 D 1 0 D whch mples that E 1 = 0 hence A s reducble. Ths contradcton shows that each A [A A] s rreducble and the proof s complete. In ths paper we are nterested n descrpton of the set of Perron vectors of all the matrces contaned n a gven rreducble nonnegatve nterval matrx A. As far as we know ths topc has not been studed yet. ) 2
2 Perron vectors of an nterval matrx The set of spectral rad of all the matrces contaned n an rreducble nonnegatve nterval matrx A = [A A] s easy to descrbe: { ϱ(a) A A } = [ϱ(a) ϱ(a)] because the spectral radus s a contnuous functon of A (Horn and Johnson [3] p. 313) hence the real functon ψ(t) = ϱ(a + t(a A)) s contnuous n [0 1] so that t attans all the ntermedate values between the endpont values ϱ(a) and ϱ(a) and no spectral radus can exceed ths nterval because 0 A A A mples that ϱ(a) ϱ(a) ϱ(a) (Horn and Johnson [3] p. 491). The followng man result of ths paper presents a descrpton of the set { x(a) A A } of the Perron vectors of all matrces contaned n a gven rreducble nonnegatve nterval matrx A. Theorem 3. Let A = [A A] be an rreducble nonnegatve nterval matrx. Then a vector x R n s the Perron vector of some matrx A A f and only f t satsfes Axx T xx T A T (4) e T x = 1 (5) x > 0. (6) Proof. Let x be the Perron vector of some matrx A [A A] so that (1) (3) hold. Then from A A A n vew of postvty of x we obtan hence for each = 1... n we have Ax Ax = ϱ(a)x Ax ϱ(a) (Ax) and thus also (Axx T ) = (Ax) = (xx T A T ) whch proves (4); (5) and (6) are gven by (2) (3). Conversely let x satsfy (4) (6). Then for each we have = (Axx T ) (xx T A T ) = (Ax) hence whch mples that (Ax) mn 3 (Ax).
Let us choose any λ satsfyng λ mn (Ax). Then from the frst nequalty t follows that Ax λx whereas the second one gves λx Ax together Ax λx Ax. (7) For each = 1... n defne a real functon of one real varable t by ϕ (t) = ((A + t(a A))x λx). Then ϕ (0) = (Ax λx) 0 and ϕ (1) = (Ax λx) 0 by (7) hence by contnuty of ϕ there exsts a t [0 1] such that ϕ (t ) = 0. Now put A = A + dag (t 1... t n )(A A) (where dag (t 1... t n ) denotes the dagonal matrx wth dagonal entres t 1... t n ) then A [A A] because t [0 1] for each and we have (Ax λx) = ϕ (t ) = 0 for each hence Ax = λx. Snce e T x = 1 and x > 0 by (5) (6) Theorem 1 gves that λ = ϱ(a) and x = x(a) hence s the Perron vector of A whch proves the second mplcaton. The nequalty (4) could also be wrtten n a more symmetrc form Axx T (Axx T ) T but we prefer the form (4) whch as we have seen arses naturally n the proof. The constructon gven n the second part of the proof s worth summarzng as a separate asserton. Theorem 4. Let x satsfy (4) (6). Then and for each λ wth mn λ mn (Ax) (8) (Ax) (9) there holds λ = ϱ(a) and x = x(a) where the matrx A [A A] s gven by wth t = A = A + dag (t 1... t n )(A A) (10) { (λx Ax) /((A A)x) f ((A A)x) > 0 1 f ((A A)x) = 0 ( = 1... n). (11) 4
Proof. As everythng else has been stated n the proof of Theorem 3 t remans to explan the formula (11) for t only. Ths value s a soluton of the equaton ϕ (t ) = 0.e. t satsfes t ((A A)x) = (λx Ax). (12) If ((A A)x) > 0 then ths equaton has the unque soluton t = (λx Ax) ((A A)x). If ((A A)x) = 0 then snce we know from the proof of Theorem 3 that equaton (12) has a soluton t must be (λx Ax) = 0 hence the equaton s satsfed for any t R thus also for our choce t = 1. 3 The subset A In accordance wth the constructon made n (10) denote A = { A + T (A A) 0 T I } so that A s a subset of A. Let us compare t wth the descrpton of A whch can also be wrtten as A = { A + T (A A) 0 T ee T }. We can see that the descrpton of A nvolves n parameters t [0 1] ( = 1... n) whereas that of A contans n 2 parameters t [0 1] ( = 1... n). Nevertheless the followng consequence of Theorem 4 shows that all the spectral rad and Perron vectors of A are attaned over ts subset A. Theorem 5. Let A be an rreducble nonnegatve nterval matrx. Then for each A A there exsts an A A such that ϱ(a) = ϱ(a ) and x(a) = x(a ). Proof. Let A A. Then x = x(a) satsfes (4) (6) by Theorem 3 and there holds ϱ(a) = (Ax) k x k for each k so that from A A A t follows (Ax) k x k ϱ(a) (Ax) k x k for each k hence λ = ϱ(a) satsfes (9) and a drect applcaton of Theorem 4 gves that ϱ(a) = ϱ(a ) and x(a) = x(a ) where A s gven by (10) (11) and thus belongs to A. Fnally we note that egenvectors of nterval matrces were examned from another pont of vew by Hartfel [2]. 5
Acknowledgment The author thanks two anonymous referees for ther comments. Ths work was supported by the Czech Republc Natonal Research Program Informaton Socety proect 1ET400300415. References [1] A. Berman and R. J. Plemmons Nonnegatve Matrces n the Mathematcal Scences SIAM Phladelpha 1994. [2] D. J. Hartfel Egenvector sets for ntervals of matrces Lnear Algebra and Its Applcatons 262 (1997) pp. 1 10. [3] R. A. Horn and C. R. Johnson Matrx Analyss Cambrdge Unversty Press Cambrdge 1985. [4] A. Neumaer Interval Methods for Systems of Equatons Cambrdge Unversty Press Cambrdge 1990. 6