A p-adic PERRON-FROBENIUS THEOREM

Transcription

1 A p-adic PERRON-FROBENIUS THEOREM ROBERT COSTA AND PATRICK DYNES Advsor: Clayton Petsche Oregon State Unversty Abstract We prove a result for square matrces over the p-adc numbers akn to the Perron-Frobenus Theorem for square matrces over the real numbers In partcular, we show that f a square n n matrx A has all entres p-adcally close to 1, then ths matrx wll possess a unque maxmal egenvalue 0 such that: a 0 s a p-adc nteger, b 0 has an algebrac multplcty of one, and c there exsts an egenvector assocated to 0 wth all entres p-adcally close to 1 Furthermore, we show that teraton of A/ 0 converges to a projecton operator onto the egenspace of 0 1 Introducton The classcal Perron-Frobenus Theorem gves us nformaton about square matrces wth all entres gven by postve real numbers Theorem 11 Perron-Frobenus, Theorem 1 n Chapter 16 of [3] Let A be a square matrx wth all entres gven by postve real numbers Then there exsts an egenvalue 0 of A of multplcty one such that 0 s a postve real number and < 0 for all other complex egenvalues of A Furthermore, there exsts an egenvector v of A wth egenvalue 0 such that all components of v are postve real numbers A useful applcaton of Theorem 11 s the followng Theorem 12 Theorem 3 n Chapter 16 of [3] Let A be a square n n matrx wth all entres gven by postve real numbers such that, for each column of A, the sum of the entres n that column s 1 Then the maxmal egenvalue 0 of A guaranteed by Theorem 11 s equal to 1, and for any vector x R n wth all entres nonnegatve, the sequence A k x converges to an egenvector of 0 The dynamcal mplcatons of Theorem 12 make t useful for studyng many real-world phenomena Matrces n the form of the matrx A from Theorem 12 are used to model changes n atomc nucle and populatons n ecologcal systems [3, p 241], and Theorem 12 tself forms the bass of Google s search strategy [3, p 242] Our frst result over the p-adc numbers s analogous to Theorem 11 Ths result apples to any p-adc n n matrx A, where: a all entres of A are n 1 + pz p, and b p n For a demonstraton of why these condtons cannot be relaxed, see Example 43 Date: August 13, 2015 Ths work was done durng the Summer 2015 REU program n Mathematcs at Oregon State Unversty Ths work was supported by the Natonal Scence Foundaton Grant DMS

2 2 Robert Costa and Patrck Dynes Theorem 13 Let p be a prme and let n be a postve nteger such that p n Let A be a square n n matrx wth all entres n 1 + pz p ; that s, let 1 + pa pa pa 1n 1 + pa A = pa pa 2n, pa n1 1 + pa n2 1 + pa nn where each a j Z p Then there exsts an egenvalue 0 of A of multplcty one such that 0 Z p, 0 n mod pz p, and p < 0 p for all other egenvalues of A n C p In addton, there exsts an egenvector v of A wth egenvalue 0 such that all components of v are elements of 1 + pz p Our next result states that we can relax the condton of Theorem 13 that p n as long as we strengthen the other condton accordng to the extent to whch p n Specfcally, gven a prme p and postve nteger n, we show that the concluson of Theorem 13 about the exstence of a strctly maxmal egenvalue 0 of A stll holds f we requre that all entres of A be close enough to 1 Theorem 14 Let p be a prme and let n, l be postve ntegers Let A be a square n n matrx wth all entres n 1 + p l Z p ; that s, let 1 + p l a p l a p l a 1n 1 + p A = l a p l a p l a 2n, p l a n1 1 + p l a n2 1 + p l a nn where each a j Z p Suppose l > 2n 1ν p n Then there exsts an egenvalue 0 of A of multplcty one such that 0 Z p, 0 n p p l / n n 1 p, and p < 0 p for all other egenvalues of A n C p Our fnal result analyzes the forward orbts of vectors wth p-adc components under teraton by p-adc square matrces wth a strctly domnant egenvalue of multplcty one We produce a statement for matrces over the p-adc numbers analogous to that of Theorem 12 for matrces over the real numbers Theorem 15 Let A be a square n n matrx wth all entres n Q p Suppose that A has an egenvalue 0 of multplcty one such that p < 0 p for all other egenvalues of A n C p Then teraton of A/ 0 converges to a projecton operator onto the egenspace of 0 2 Background on the p-adc Numbers We begn wth some remarks on the p-adc numbers Q p and ther complete, algebracally closed extenson C p Unless otherwse specfed, we wll take p to be any prme number We refer the reader to [2] for more detals on Q p and C p

3 21 Constructon of Q p A p-adc Perron-Frobenus Theorem 3 Defnton 21 The p-adc valuaton on Z s the functon ν p : Z \ { 0 } Z defned by { max { k N : p k n } f p n ν p n = 3 0 f p n We extend ν p to Q as follows: we defne ν p : Q Z { } by { νp a ν p b f x = a/b Q ν p x = f x = 0 Defnton 22 The p-adc absolute value on Q s the functon p : Q R defned by where we defne p = 0 4 x p = p νpx, 5 We recall that a feld K s complete f every Cauchy sequence n K converges n K It s known that Q s not complete wth respect to any of ts nontrval absolute values [2, Lemma 323] Defnton 23 The feld Q p of p-adc numbers s the completon of Q wth respect to the p-adc absolute value 22 Propertes of Q p There are several useful propertes of Q p that we wll keep n mnd throughout our work The frst of these appears as part of Theorem 3213 n [2] Theorem 24 Fx a prme p and let a, b Q p Then a + b p max { a p, b p } 6 Theorem 25 Proposton 233 n [2] Fx a prme p and let a, b Q p wth a p b p Then a + b p = max { a p, b p } 7 It s a consequence of Theorem 24 that Z p = { x Q p : x p 1 } s a subrng of Q p ; we call ths the rng Z p of p-adc ntegers It follows smlarly that pz p = { x Q p : x p < 1 } s the unque maxmal deal of Z p The quotent feld Z p /pz p s called the resdue feld of Z p ; t s an mmedate consequence of [2, Corollary 336] that Z p /pz p = Z/pZ The followng results allow us to deduce nformaton about the roots of a polynomal fx Z p [X] from the behavor of the reductons of fx and f X modulo pz p f X beng the formal dervatve of fx Theorem 26 Hensel s Lemma, Theorem 341 n [2] Let fx be a polynomal whose coeffcents are n Z p and suppose that there exsts a p-adc nteger α 1 Z p such that and fα 1 0 mod pz p 8 f α 1 0 mod pz p 9 Then there exsts a unque p-adc nteger α Z p such that fα = 0 and α α 1 mod pz p

4 4 Robert Costa and Patrck Dynes Theorem 27 Strong Hensel s Lemma, Lemma 31 and Corollary 1 n [1] Let fx Z p [X] be a polynomal whose coeffcents are n Z p and suppose that there exsts a p-adc nteger α 1 Z p such that fα 1 p < f α 1 2 p 10 Then there exsts a unque p-adc nteger α Z p for whch fα = 0 and α α 1 p fα 1 p f α 1 p 11 We conclude ths subsecton wth a theorem that gves us suffcent condtons for a polynomal n Z p [x] to be rreducble over Q p Theorem 28 Esensten Irreducblty Crteron, Proposton 5311 n [2] Let be a polynomal satsfyng the condtons a n p = 1, a p < 1 for 0 < n, and a 0 p = 1/p Then fx s rreducble over Q p fx = a 0 + a 1 X + + a n X n Z p [x] The Feld C p and Newton Polygons Whle the algebrac closure Q p of Q p s not complete [2, Theorem 574], the completon of Q p s algebracally closed [2, Proposton 578]; we call ths feld C p Defnton 29 Let fx = a 0 + a 1 X + + a n X n C p [X] be a polynomal The Newton polygon of f s the lower convex hull of the set of ponts S = {, ν p a : 1 n } 13 By a slope of the Newton polygon, we wll mean the slope of a lne segment appearng n the polygon Gven a slope of the polygon, by the length of the slope, we wll mean the length of the projecton of the correspondng segment on the x-axs It s clear that snce the Newton polygon s the lower convex hull of a set of ponts n R 2, the slopes form a nondecreasng sequence f we pck the correspondng segments from left to rght Theorem 210 Theorem 647 n [2] Let fx = a 0 + a 1 X + + a n X n C p [X] be a polynomal, and let m 1, m 2,, m r be the slopes of ts Newton polygon n ncreasng order Let 1, 2,, r be the correspondng lengths Then, for each k, 1 k r, fx has exactly k roots countng multplctes of absolute value p m k 3 Background on Lnear Algebra We now revew necessary background materal on lnear algebra We wll take A to be a square n n wth entres from a feld F We refer the reader to [3] and [4] for addtonal detals on lnear algebra

5 A p-adc Perron-Frobenus Theorem 5 31 Calculaton of Determnants A useful formula for computng determnants s provded n [3, Chapter 5, Equaton 16] Ths formula expresses the determnant of A as the sum of sgned products of entres from dstnct rows and columns n A We wll quote ths formula below for convenence Lemma 31 Determnant Formula Let S n denote the symmetrc group on { 1,, n }, and let sgn denote the sgn functon of permutatons n S n Then deta = n sgnσ a,σ 14 σ S n 32 Calculatons Involvng Jordan Normal Forms We brefly revew how to compute powers of matrces gven n Jordan normal form Let A be a matrx n Jordan normal form: =1 J m J J = m2 0, J mj where, for 1 j, J m s the m m Jordan block assocated to the egenvalue of J: J m = Dfferent Jordan blocks may share the same egenvalue, but dfferent egenvalues may not share the same Jordan bock A useful computatonal property of Jordan normal forms, as noted n Secton 55 of [4], s that we can easly compute powers of A: Jm1 n J J n m = n J n m j Thus, n order to compute J n for a postve nteger n, we only have to compute the power J n m of each Jordan block J m By nducton, we wll show that, for any postve nteger n, n n 1 n 1 n n m m 2 2 n n m m n Jm n n n m m 3 3 n n m m 2 2 = n n 1 n n

6 6 Robert Costa and Patrck Dynes If a < b, we say a b = 0, as s standard In the case n = 1, the formula 18 agrees wth the Jordan block J m Suppose nductvely that the formula 18 holds for some nteger n = k 1 That s, We compute J k+1 m : k k 1 k 1 k k m m 2 2 k k m m k n k m m 3 3 k k m m 2 2 Jm k = k k 1 k k k k 1 k 1 k k m m 2 2 k k m m k k k m m 3 3 k k m m Jm k+1 = k k 1 k k k+1 k + k 1 k k k m m k k m m k+1 k k m m k k m m 2 3 = 0 0 k + 21 k 1 k 0 0 k+1 k+1 k+1 k k+1 k+1 m m 2 2 k+1 k+1 m m k+1 k+1 k+1 m m 3 3 k+1 k+1 m m 2 2 = k+1 k+1 k k+1 The fnal step follows by Pascal s dentty for bnomal coeffcents: a + 1 = b + 1 a a + b b for nonnegatve ntegers a and b Hence, the formula 18 holds for n = k + 1 By the Prncple of Mathematcal Inducton, ths formula holds for all postve ntegers n

7 A p-adc Perron-Frobenus Theorem 7 4 Proofs of Man Results 41 Proof of Theorem 13 In ths subsecton, we establsh Theorem 13 Our argument wll make use of the Newton polygon assocated to the characterstc polynomal fx of A Frst we establsh several lemmas Lemma 41 Let p be a prme and let n be a postve nteger Let A be a square n n matrx wth all entres n 1 + pz p ; that s, let Then deta p p 1 n Proof Usng Lemma 31, we calculate: 1 + pa pa pa 1n 1 + pa A = pa pa 2n pa n1 1 + pa n2 1 + pa nn deta = σ S n sgnσ = σ S n sgnσ = σ S n sgnσ = n k=0 where for each I { 1,, n } we defne I {1,,n} I =k n 1 + a,σ p 25 =1 I {1,,n} n k=0 a,σ p 26 I I {1,,n} I =k SI = σ S n sgnσ I a,σ p 27 I p k SI, 28 a,σ 29 Now suppose that 0 k n 2 Let I { 1,, n } wth I = k Snce I n 2, there exsts a transposton ɛ S n that fxes the elements of I Then snce the map σ σɛ s a bjecton from S n to tself, we can wrte SI = σ S n sgnσɛ I a,σɛ 30 = sgnɛ sgnσ a,σ σ S n I 31 = SI, 32

8 8 Robert Costa and Patrck Dynes where n lne 31 we use the fact that a,σɛ = a,σ for all σ S n and for all I The statement SI = SI mples that SI = 0, so lne 28 becomes n deta = p k SI 33 k=n 1 I {1,,n} I =k We see from Theorem 24 that for any subset I of { 1,, n }, SI p 1 Takng the p-adc absolute value of both sdes of equaton 33 and agan applyng Theorem 24, we fnd that n deta p p k max { p p n 1 p, p n p } = p 1 n 34 k=n 1 I {1,,n} I =k Lemma 42 Let the matrx A be as n the statement of Lemma 41 Let the characterstc polynomal fx of A be gven as: fx = X n + C n 1 X n C 1 X + C 0 35 Then C p p 1 n+ for 0 n 1, and f addtonally p n, C n 1 p = 1 Proof Frst we establsh some notatons Gven postve ntegers and j, let δ,j represent the Kronecker delta functon of and j Gven a permutaton σ S n, let Fxσ denote the set of all elements of { 1,, n } that are fxed by σ Gven a subset I of { 1,, n }, let I c represent the set { 1,, n } \ I We apply Lemma 31 to A XI to fnd the characterstc polynomal fx of A: fx = 1 n deta XI 36 = 1 n n sgnσ 1 + pa,σ δ,σ X σ S n 37 = 1 n σ S n sgnσ = 1 n σ S n sgnσ = 1 n I {1,,n} =1 I {1,,n} I Fxσ 1 I X I 1 + pa,σ δ,σ X 38 I c I 1 + pa,σ 1 I X I 39 I c sgnσ 1 + pa,σ 40 I c σ Sn I Fxσ Note that f we let A I denote the n I n I matrx formed by removng the th row and column from A for each I, then lne 40 becomes fx = 1 n 1 I X I deta I 41 I {1,,n}

9 A p-adc Perron-Frobenus Theorem 9 Ths s because, gven I { 1,, n }, applyng Lemma 31 to A I tells us that deta I = sgnσ 1 + pa,σ 42 I c σ SymI c = σ S n I Fxσ sgnσ I c 1 + pa,σ 43 It follows that for 0 n 1, the coeffcent C can be calculated by addng and subtractng the determnants of n n matrces wth entres n 1 + pz p ; Lemma 41 mples that each of these determnants has a p-adc absolute value of at most p 1 n = p 1 n+, and Theorem 24 then mples that C p p 1 n+ 44 Suppose, addtonally, that p n We note by lne 41 that C n 1 p = deta I 45 p I =n 1 = tra p 46 = 1 + a 11 p a nn p p 47 = n + a a nn p p 48 It follows from Theorem 24 that a a nn p 1, and so a a nn p p p 1 But n p = 1 snce p n, so by Theorem 25, lne 48 becomes C n 1 p = max { n p, a a nn p p } = n p = 1 49 Proof of Theorem 13 Let the matrx A be as n the statement of Theorem 13 Suppose that p n Our frst goal s to show the exstence of a strctly maxmal egenvalue 0 of A of multplcty one We reach ths goal by an analyss of the characterstc polynomal fx of A We frst construct a partal Newton polygon for fx Referrng to Lemma 42, we fnd that ν p C n + 1 for 0 n 2, and ν p C n 1 = 0 = ν p 1; see Fgure 1 The slope of the Newton polygon from n 2, ν p C n 2 to n 1, 0 must be at most 0 1 = 1 The fact that the Newton polygon s the lower convex hull of a set of ponts n R 2 mples that the slopes are nonncreasng from rght to left, so we conclude that all slopes to the left of n 1, 0 must be at most 1 Applyng Theorem 210 to ths Newton polygon, we fnd that there are n 1 roots of fx n C p, countng multplctes of absolute value at most p 1, and there s one root 0 wth 0 p = p 0 = 1 50 It follows that 0 s a strctly maxmal egenvalue of A of multplcty one Our second goal s to show that 0 Z p Ths wll requre an applcaton of Theorem 26, Hensel s lemma In partcular, we wll show that 0 n mod pz p Recall we showed n

10 10 Robert Costa and Patrck Dynes n n 1 Remanng Ponts Le n ths Regon n 1 n Fgure 1 Newton Polygon for fx n the proof of Theorem 13 Lemma 42 that f we wrte the characterstc polynomal fx of A as fx = X n + C n 1 X n C 1 X + C 0, 51 then we know that C p p 1 n+ for 0 n 1, and C n 1 p = 1 snce p n Thus, fx Z p [X], so fn and f n are n Z p Note that C n 1 = tra = n a a nn p n mod pz p 52 Reducng fn and f n modulo pz p, we fnd that snce C p p 1 for 0 n 2, whereas fn n n nn n 1 mod pz p 53 0 mod pz p, 54 f n nn n 1 n 2 nn n 2 mod pz p 55 n n n n + n n 1 mod pz p 56 n n 1 0 mod pz p 57 We now apply Theorem 26 to conclude that there exsts a unque p-adc nteger α Z p such that fα = 0 and α n mod pz p Wrtng α = n + x for some x pz p, t follows from Theorem 25 that α p = n p = 1 But we saw n equaton 50 that 0 s the unque root of fx of p-adc absolute value 1, so t follows that 0 = α Z p We now have that 0 n mod pz p 58 What remans s to show that there exsts an egenvector of A wth egenvalue 0 such that all of ts components are elements of 1 + pz p We can solve a lnear system of equatons

11 A p-adc Perron-Frobenus Theorem 11 n Q p to fnd a nonzero egenvector v = v 1,, v n Q n p of 0 If we pck a component v of v such that v p = max { v 1 p,, v n p }, then t follows that s an egenvector of 0 contaned n Z n p Snce u = 1, we see that u = v v 59 Au = 0 u = 0 n mod pz p 60 The matrx A s equvalent to the matrx of all 1 s modulo pz p, so for all j wth 1 j n, calculatng Au j as the scalar product of the jth row of A wth u shows that Au j u u n mod pz p 61 Thus, all of the components of Au j are equvalent to each other modulo pz p, and by equaton 60, they are all equvalent to n modulo pz p But snce Au = 0 u, we also know that Au j = 0 u j nu j mod pz p 62 for all j wth 1 j n Then we have the system of equatons: n nu 1 mod pz p 63 n nu 2 mod pz p 64 n nu n mod pz p 65 Snce n 0 mod pz p, we can dvde each lne above through by n to see that every component u j of u s equvalent to 1 modulo pz p and s hence an element of 1 + pz p Example 43 Here we show that the conclusons of Theorem 13 do not necessarly hold f we relax the condtons that: a all entres of A are n 1 + pz p, and b p n When we appled Theorem 26 n the proof of Theorem 13 to show that the maxmal egenvalue was n Z p, we used the fact that fn 0 mod pz p, but f n 0 mod pz p It stands to reason that f we want the maxmal egenvalue to not be n Z p, t would be worthwhle to consder cases where f n s equvalent to zero modulo pz p We consder the matrx A Mat 2 2 Q 2 defned by [ ] 1 + 2a a A = 12, a a 22 where each a j Z 2 The characterstc polynomal of A s fx = X 2 trax + deta, 67 so f X = 2X tra and f 2 = 4 tra If f 2 = 0, then tra = 4 Then 2 + 2a 11 + a 22 = 4 68 a 11 + a 22 = 1 69

12 12 Robert Costa and Patrck Dynes Let a 11, a 12, a 21, a 22 = 1/3, 0, 0, 2/3, so that each a j Z 2 and a 11 + a 22 = 1 Then equaton 66 becomes 5/3 1 A =, /3 and equaton 67 becomes fx = X 2 4X + 26/9 71 Takng the 2-adc absolute values of the coeffcents of fx n equaton 71, we fnd that 1 2 = 1, 4 2 = 1/4 < 1, and 26/9 2 = 1/2 We conclude by Theorem 28 that fx s rreducble over Q 2 Thus, nether egenvalue of fx s n Q 2, let alone Z 2 If we look at the Newton polygon of fx, we see that ts vertces are 0, 1 and 2, 0, so ts egenvalues have the same 2-adc absolute value, 2 1/2 42 Proof of Theorem 14 In ths subsecton, we generalze Theorem 13 to establsh Theorem 14 Many parts of the argument wll be smlar Lemma 44 Let p be a prme and let n, l be postve ntegers Let A be a square n n matrx wth all entres n 1 + p l Z p ; that s, let 1 + p l a p l a p l a 1n 1 + p A = l a p l a p l a 2n p l a n1 1 + p l a n2 1 + p l a nn Then deta p p l1 n Proof The argument s as n the proof of Lemma 41; the only dfference s that one must replace p wth p l throughout Lemma 45 Let the matrx A be as n the statement of Lemma 44 Let the characterstc polynomal fx of A be gven as: fx = X n + C n 1 X n C 1 X + C 0 73 Then C p p l1 n+ for 0 n 2 and f addtonally l > ν p n, C n 1 p = n p Proof The estmates for C p n the case 0 n 2 follow as n the proof of Lemma 42; the only dfference s that one must replace p wth p l throughout Suppose addtonally that l > ν p n, so n p > p l p p l a a nn p We estmate: C n 1 p = tra p 74 = 1 + p l a p l a nn p 75 = n + p l a a nn p 76 = n p 77

13 A p-adc Perron-Frobenus Theorem 13 Proof of Theorem 14 Let the matrx A be as n the statement of Theorem 14 Suppose n > 1, for otherwse ths dscusson s trval Suppose that l > 2n 1ν p n Our frst goal s to show the exstence of a strctly maxmal egenvalue 0 of A of multplcty one Ths requres a careful analyss of the characterstc polynomal fx of A From Lemma 45, we have that ν p C ln 1 for 0 n 2, and ν p C n 1 = ν p n We note that snce n s a postve nteger larger than 1, the hypotheses mply that l > 2ν p n, so ν p C n 1 < l/2 We construct a partal Newton polygon for fx; see Fgure 2 nl n 1l Remanng Ponts Le n ths Regon 2l l 1 2 n 2 n 1 n Fgure 2 Newton Polygon for fx n the proof of Theorem 14 We note that the lne segment connectng n 1, ν p C n 1 wth n, 0 has the slope 0 ν p C n 1 n n 1 = ν pc n 1 78 > l/2 79 We now consder extendng the lne segment through n 1, ν p C n 1 and n, 0 to the left untl t ntersects the y axs For 0 n 2, we bound the y-coordnate of the pont on the lne segment extenson wth x-coordnate : ν p C n 1 n < l/2n 80 = l/2n 1 + l/2 81 ν p C /2 + l/2 82 ν p C /2 + ν p C /2 83 = ν p C 84

14 14 Robert Costa and Patrck Dynes The calculaton above shows that ths lne segment extenson wll le strctly below the remanng ponts of the Newton polygon We note that the slopes of the segments of the Newton polygon to the left of the pont n 1, ν p C p 1 are bounded above by the slope of the lne segment through n 2, l and n 1, ν p C n 1 We calculate: ν p C n 1 l n 1 n 2 < l/2 l 85 = l/2 86 Applyng Theorem 210 to ths Newton polygon, we fnd that there are n 1 roots of fx n C p, countng multplctes of absolute value strctly less than p l/2 We wll denote these roots by 1, 2,, m Furthermore, there s one root, denoted 0, of absolute value strctly greater than p l/2 Ths root s a strctly maxmal egenvalue of A of multplcty one Now that we have establshed the exstence of a maxmal egenvalue 0, we wll show that 0 s a p-adc nteger Ths wll requre an applcaton of Theorem 27, the strong verson of Hensel s lemma We calculate: fn = n n + C n 1 n n 1 + C n 2 n n C 1 n + C 0 87 = n n n + p l a a nn n n 1 + C n 2 n n C 1 n + C 0 88 = p l a a nn n n 1 + C n 2 n n C 1 n + C 0 89 Snce C p p l for 0 n 2, we have the bound fn p p l We calculate: f n = nn n 1 + n 1C n 1 n n 2 + n 2C n 2 n n C 1 90 = n n n 1n + p l a a nn n n 2 + n 2C n 2 n n C 1 91 = n n + n n 2 n n 1 n + p l a a nn + n 2C n 2 n n C 1 92 = n n 1 + p l n n 2 n n 1 a a nn + n 2C n 2 n n C 1 93 Snce l > n 1ν p n, so that p l < n n 1 p, and snce C p p l for 0 n 2, we have the equalty f n p = n n 1 p Fnally, snce l > 2n 1ν p n, we have the bound: fn p p l 94 < p 2n 1νpn 95 = n 2n 1 p 96 = f n 2 p 97 We now apply Theorem 27 to conclude that there exsts a unque p-adc nteger α so that fα = 0 satsfyng

15 A p-adc Perron-Frobenus Theorem 15 α n p fn p f n p 98 p l n n 1 p 99 < p l p l/2 100 = p l/2 101 We wll show that the none of the nonmaxmal egenvalues are canddates for the root α found by Hensel s lemma Let, 1 m be a nonmaxmal egenvalue of A Snce p < p l/2 and n p > p l/2, we have: n p > p l/2 102 Hence, the nonmaxmal egenvalues are not α Thus 0 = α In partcular, 0 s a p-adc nteger From 99, we obtan 0 n p p l / n n 1 p From the Newton polygon n Fgure 2 and the estmate for C n 1 p from Lemma 45, we calculate: 0 p = C n 1 p 103 = n p Proof of Theorem 15 In ths subsecton, we establsh Theorem 15 Out argument wll make use of the Jordan canoncal form of n n matrces wth a strctly domnatng egenvalue of multplcty one Proof of Theorem 15 Let A be an n n matrx wth an egenvalue 0 of multplcty one such that p < 0 p for all other egenvalues of A We wrte the Jordan canoncal form of A n block form: J A = Q 1 m1 0 Q, J mj where the J m are m m Jordan blocks assocated to the nonmaxmal egenvalues of A We wll denote the nonmaxmal egenvalues, by 1, 2,, j ; note, however, that the egenvalues n ths lst are not necessarly dstnctwe wrte out each matrx J m as:

16 16 Robert Costa and Patrck Dynes We calculate the kth power of these Jordan blocks: J m = k k 1 k 1 k k m m k k k m m 3 3 Jm k = 0 0 k k m 1 k m 2 k m 1 k m k 1 k k For a postve nteger k, the kth power of A can be calculated as k A k = Q 1 0 Jm k 1 0 Q 108 We wll now dvde through by k 0 to obtan: We compute each power J k m / k 0: 0 0 J k m j A/ 0 k = Q 1 0 Jm k 1 / k 0 0 Q Jm k j / k 0 k k k k k m 1 m 1 0 k 0 0 k k m 2 m 2 0 Jm k / k 0 = 0 0 k k k m m

17 A p-adc Perron-Frobenus Theorem 17 Notng that each p < 0 p for 1 j, we see that each block J k m / k 0 vanshes as k tends towards nfnty Thus as as k tends towards nfnty, the sequence A/ 0 k converges to P = Q 1 Q Snce Q and Q 1 nduce a change of bass from the standard bass of Q n p to the bass of generalzed egenvectors of A, and snce the frst column of Q 1 s the egenvector v 0 assocated to the maxmal egenvalue 0, P s the projecton onto the egenspace spanned by v 0 References [1] JWS Cassels Local Felds Cambrdge Unversty Press, frst edton, 1986 [2] Fernando Q Gouvêa p-adc Numbers: An Introducton Sprnger, second edton, 1997 [3] Peter D Lax Lnear Algebra and Its Applcatons Wley, second edton, 2007 [4] Igor R Shafarevch and Alexey O Remzov Lnear Algebra and Geometry Sprnger, frst edton, 2012 Tufts Unversty E-mal address: robert jcosta@tuftsedu Clemson Unversty E-mal address: pdynes@gclemsonedu