Ranks of quotients, remainders and p-adic digits of matrices

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1 axv: v2 [math.nt] 31 Jan 2014 Ranks of quotents, emandes and p-adc dgts of matces Mustafa Elshekh Andy Novocn Mak Gesbecht Abstact Fo a pme p and a matx A Z n n, wte A as A = p(a quo p)+ (A em p) whee the emande and quotent opeatons ae appled element-wse. Wte the p-adc expanson of A as A = A [0] + pa [1] + p 2 A [2] + whee each A [] Z n n has entes between [0, p 1]. Uppe bounds ae poven fo the Z-anks of A em p, and A quo p. Also, uppe bounds ae poven fo the Z/pZ-ank of A [] fo all 0 when p = 2, and a conjectue s pesented fo odd pmes. Keywods: Matx ank, Intege matx, Remande and quotent, p-adc expanson. AMS classfcaton: 15A03, 15B33, 15B36, 11C20. Outlne Ths pape pesents two elated esults on ntege matces afte applyng element-wse dvson wth emande. Fst, let A be an n n ntege matx wth ank ove Z and ank 0 ove Z/pZ. If n > p 0 then Theoem 1 n Secton 1 shows that ank(a em p) (p 0 1)(p + 1)/(2(p 1)) and ank(a quo p) + (p 0 1)(p + 1)/(2(p 1)). The second esult s concened wth the Z/pZ-anks of p-adc dgts of an ntege matx. Let U, S, V Z n n such that U, V have entes fom {0, 1}, det U det V 0 (mod 2), S = dag(1,..., 1, 0,..., 0), be the ank of S ove Cheton School of Compute Scence, Unvesty of Wateloo, Wateloo, Ontao, Canada (melshek@uwateloo.ca, andy@novocn.com, mwg@uwateloo.ca). Suppoted by the Natual Scences and Engneeng Reseach Councl (NSERC) of Canada. 1

2 Ranks of quotents, emandes and p-adc dgts of matces 2 Z/2Z, and n 2. If M = USV Z n n, then Theoem 16 n Secton 2 shows that ank of M [] ove Z/2Z s ( ) 2 fo all 1. A conjectue s pesented n Secton 2.3 fo the same setup, but fo p an odd pme. A esult on ntege ank of Latn squaes s also obtaned. Let A be the ntege matx of ank one fomed by the oute poduct between the vecto (1, 2,..., p 1) and ts tanspose. Then A em p s a Latn squae on the symbols {1,..., p 1}. It s shown n Coollay 10 n Secton 1.3 that the ntege ank of ths Latn squae s (p + 1)/2. 1 Quotent and Remande Matces Fo any ntege n and any pme p, let n em p and n quo p denote the (nonnegatve) emande and quotent n the Eucldean dvson n = qp + whee 0 < p. The opeatos em p and quo p ae natually extended to vectos and matces usng element-wse applcaton. Thoughout, we utlze the noton of Smth nomal fom of an ntege matx. Fo any matx A Z n n of ank, thee exst unmodula matces U, V Z n n and a unque n n ntege matx S = dag(s 1, s 2,..., s n ) such that A = USV. Futhemoe, s s +1 fo all 1 n and s = 0 fo all < n. S s called the Smth nomal fom of A. Fo a dscusson on exstence and unqueness of Smth nomal fom, we efe to the eade to the textbook by Newman [3]. We use two notons of anks. The ntege ank of A Z n n s denoted by ank(a). The ank of the mage of A n the fnte feld Z/pZ s denoted by ank p (A). Altenatvely, f = ank(a) and the Smth fom of A s S = dag(s 1,..., s, 0,..., 0), then ank p (A) = 0 s the maxmal ndex such that p s. Fnally, we use the notaton A,j fo the jth column of A Z n n and a,j fo the enty (, j) of A. 1.1 Rank Theoem The followng theoem s the man esult of Secton 1. Theoem 1. Let A be an n n matx ove Z, = ank(a), 0 = ank p (A), and assume n > p 0. Then () ank(a em p) (p 0 1)(p + 1)/(2(p 1)). () ank(a quo p) + (p 0 1)(p + 1)/(2(p 1)).

3 Ranks of quotents, emandes and p-adc dgts of matces 3 Poof. We wll pove pat () n Lemma 2. Fo pat (), we have A = (A em p) + p(a quo p), o p(a quo p) = A (A em p). Fo matces X = Y + Z, ank s sub-addtve and ank(x) ank(y ) + ank(z). Scalng a matx by p o 1 does not change ts ank. So ank(a quo p) ank(a) + ank(a em p) = + ank(a em p). Lemma 2. ank(a em p) (p 0 1)(p + 1)/(2(p 1)). Poof. Let A = USV be the Smth nomal fom of A, wth S = S + ps q whee S q = S quo p and S = S em p. Then A em p = USV em p = (US V + pus q V ) em p = US V em p. (1) If 0 = ank p (A) then S = dag(σ 1,..., σ 0, 0,..., 0) whee σ [1, p 1] fo all 1 0. The jth column of A em p s ( 0 ) ( 0 ) A,j em p = σ l v l,j U,l em p = c l,j U,l em p, (2) l=1 whee c l,j [0, p 1]. If we only consde the non-zeo coeffcents c l,j, then the ght-hand sde of (2) s an -tem sum (c l1,ju,l c l,ju,l ) em p, whee 1 0 and 1 l 1 < l 2 <... < l 0. The coeffcents c lk,j ae elements n [1, p 1] whch ae unts modulo p. In patcula, we can facto c l1,j fom the sum, and e-wte (2) as: A,j em p = (c l1,j(u,l1 + α l2,ju l2,j α l,ju,l )) em p, (3) whee α lk,j [1, p 1] fo all k. Fx some, j and some non-zeo assgnment of α l2,j,..., α l,j n (3) and let û = U,l1 + α l2,ju l2,j... + α l,ju,l. Then (3) becomes A,j em p = (c l1,jû) em p. Thee ae p 1 possble values fo c l1,j and hence the possble values of A,j em p ae: {û em p, (2û) em p, ((p 1)û) em p}. (4) We ae nteested n gettng an uppe bound on the ank of ths set of vectos. Fst note that (xy) em p = (x em p)(y em p) em p. So (û) em p = ((û em p)) em p fo [1, p 1]. Hence the maxmal ank one can acheve fom (4) occus when (up to pemutaton) û em p = (0, 1, 2,..., p 1,...). The est of the entes ae duplcates fom the same ange [0, p 1] by the l=1

4 Ranks of quotents, emandes and p-adc dgts of matces 4 pgeonhole pncple. Now apply Lemma 3 to conclude that the vectos n (4) have ank at most (p + 1)/2. Thus fo each, j and non-zeo assgnment of α l2,j,..., α l,j, thee ae at most (p + 1)/2 lnealy ndependent columns of A em p. We now count the maxmal possble numbe of dstnct A,j s. Thee ae ( 0 ) possble ways to select dffeent columns fom the fst 0 columns of U. Fo each choce, thee ae 1 coeffcents: α l2,j,..., α l,j, and (p 1) 1 possble ways to assgn the non-zeo values fom [1, p 1]. Each choce gves a set of vectos as n (4) whose ank s at most (p + 1)/2. Summng ove all [1, 0 ], the maxmal possble ank fom the span of columns n (2) s 0 ( 0 =1 usng the bnomal theoem. ) (p ) 1 p Remande of Rank-1 Matces In ths secton we pove the followng auxlay esult. = p 0 1 p + 1 p 1 2, (5) Lemma 3. Let p be any odd pme, n p. Let u Z n be any non-zeo vecto whee the entes of u em p nclude {1, 2,..., p 1}. Then the set of vectos {u em p, (2u) em p,..., ((p 1)u) em p} s lnealy dependent and has ank (p + 1)/2. Fst we pove ths esult fo n = p 1. A genealzaton follows. Let u = (1, 2,..., p 1) Z (p 1) and M Z (p 1) (p 1) be the ank-1 matx M = uu T and R = M em p. Lemma 4. ank(r) = (p + 1)/2. Poof. Lemma 5 shows that (p + 1)/2 s an uppe bound on the ank and Lemma 7 shows that (p + 1)/2 s a lowe bound. Lemma 5. ank(r) (p + 1)/2. Poof. Let 1 j (p 1)/2 and 1 p 1. Wte j = qp + whee 0 < p. Also, j < p = p p j, whch mples 0. Then (p j) = ( q 1)+(p ) whee 0 < (p ) < p. So j em p+(p j) em p = + (p ) = p. But R,j = j em p, so fo all 1 (p 1)/2 we have R, = (p, p,..., p) T R,p. Thus thee ae (p 1)/2 lnealy dependent columns, and no moe than (p + 1)/2 lnealy ndependent columns.

5 Ranks of quotents, emandes and p-adc dgts of matces 5 To pove that (p + 1)/2 s also a lowe bound on the ank, t suffces (usng Lemma 5) to consde the matx B of sze (p 1) p+1 whch s 2 fomed by the fst (p 1)/2 columns of R and the column B,(p+1)/2 = R,(p+1)/2 +R,(p 1)/2 = (p,..., p) T. The matx B has the followng stuctue: p p p 1 p B = 3 6 em p 3 p 1 em p p (p 1) (p 1) em p 2(p 1) em p 2 em p p 2 Lemma 6. Ethe the ght kenel of B s empty, o the fst (p 1)/2 columns of B ae lnealy dependent. Poof. If the ght kenel of B s not empty, then thee exsts (p+1)/2 nteges c 1,..., c (p+1)/2 not dentcally zeo, such that c 1 B,1 + c 2 B, c (p+1)/2 B,(p+1)/2 = 0. (6) Apply ths lnea combnaton smultaneously to the fst two ows of B to get c 1 + 2c c (p 1)/2 (p 1)/2 = c (p+1)/2 p (7) 2c 1 + 4c c (p 1)/2 (p 1) = c (p+1)/2 p (8) But (7) mples ethe a contadcton n (8): the ght kenel of B s empty, o c (p+1)/2 = 0 and the fst (p 1)/2 columns of B ae lnealy dependent. Lemma 7. (p + 1)/2 ank(r). Poof. Usng Lemma 6, povng a lowe bound on the ank of R can be educed to showng that the fst (p 1)/2 columns of B ae lnealy ndependent. We use nducton. Consde the sequence of matces B (k) fomed by the fst k columns of B, whee 2 k (p 1)/2. The base case of nducton, B (2), has ank 2 whch s staghtfowad to vefy. Fo the nductve case, we assume B (k 1) has ank k 1, and use Lemma 9 to deduce that B (k) has ank k. The followng lemma s needed befoe povng Lemma 9. Lemma 8. Fo all j 1, (3j em p) 3j = pq fo some ntege q 0.

6 Ranks of quotents, emandes and p-adc dgts of matces 6 Poof. Wte 3j as 3j = qp + whee = 3j em p and q = 3j quo p. Then 3j = qp. Lemma 9. Let B (k) be the (p 1) k ntege matx n poof of Lemma 7. Ethe B (k) has column ank k, o B (k 1) s ank defcent. Poof. If the ght kenel of B (k) was not empty, then thee exsts nteges c 1,..., c k not dentcally zeo, such that 1 2 k 2 4 2k c em p (3k) em p. = 0.. (9)... c k 0 We then pefom the followng ow opeatons on the left-hand sde of (9): eplace (ow 3) by (ow 3) 3 (ow 1), then dvde ow 3 by p. Fom Lemma 8, we have that ow 3 s now [ q ], fo some q (n fact, q = (3k) quo p). We then pefom the followng column opeatons: let l denote the column ndex whee the fst 1 appeas n ow 3 (l s guaanteed to be geate than o equal 1 snce fo all p > 3, k (p 1)/2, we have 3k > p.) Pvot on enty l n ow 3 and elmnate all entes of ow 3 wth ndces between l + 1 and k 1. Subtact q 1 multples of column l fom column k. Then pvot on enty k of ow 3 and subtact column k fom column l. Effectvely, ths sequence of opeatons tansfoms ow 3 nto: [ ]. The ght-hand sde of (9) s zeo, and hence not effected by the afoementoned elementay opeatons. Fnally, the tansfomed ow 3 mples ethe that c k s zeo, o the exstence of c 1,..., c k s contadctoy. Ths poves the statement of the lemma. We ae now eady to genealze Lemma 4 and pove Lemma 3. Poof of Lemma 3. Fo the column vecto u Z n 1, consde the matx R Z n n = uu T em p, whch s analogous to the matx R of Lemma 4.

7 Ranks of quotents, emandes and p-adc dgts of matces 7 The mage of u em p has entes fom the nteval [0, p 1]. If n > p then, by the pgeonhole pncple, the vecto u em p wll contan duplcate (and zeo) entes, whch coespond to duplcate and zeo ows n R. So up to ow/column pemutatons, R contans R as a submatx, and the exta ows/columns ae duplcate and/o zeo. Hence ank( R) = ank(r). 1.3 A Note on Ranks of Latn Squaes It s woth notng that Lemma 4 also mples a esult on the anks of Latn squaes of cetan odes. As befoe, let p be an odd pme, and let R be the (p 1) (p 1) ntege matx whose (, j)th enty s j em p. We show that R s a Latn squae as follows. R s the Cayley multplcaton table of the fnte feld Z/pZ, excludng the element 0. Snce Z/pZ s an ntegal doman, we have j em p j em p wheneve j j (whee, j, j [1, p 1]). So evey ow/column of R has the esdues {1,..., p 1} appeang only once, and R s a Latn squae of ode p 1. R has ank 1 ove Z/pZ and non-tval ank ove Z by Lemma 4 as stated n the followng coollay. Coollay 10. Let p be any odd pme, and let R be any Latn squae of ode p 1 on the symbols {1,..., p 1}. Then the ntege ank of R, taken as a (p 1) (p 1) ntege matx, s (p + 1)/2. 2 p-adc Matces We now swtch the focus to anks of p-adc matces. Ranks n ths secton ae ove the fnte feld wth p elements, wth esdue classes {0, 1,..., p 1}. Fo any pme p and any matx M Z n n wth entes m,j < β, the p-adc expanson of M s M = M [0] + pm [1] p s M [s] whee the entes of each matx M [] ae between [0, p 1], and s log p β. We call M [] the th p-adc matx dgt of M. We extend the supescpt [] notaton to vectos and nteges n the obvous way. We pesent esults concenng the 2-adc matx dgts. Fo odd pmes, we only pesent a conjectue. It s an open queston to study the combnatoal stuctue of the column space of the p-adc matx dgts fo pmes othe than 2. The two anks, ove Z and ove Z/pZ, ae equal unless p s an elementay dvso of the matx.

8 Ranks of quotents, emandes and p-adc dgts of matces , Fgue 1: An example of A (left) and M = AA T (ght), whee = 4. The ows of A ae pattoned by the numbe of non-zeo entes n each ow. The coespondng blocks n the symmetc matx M ae shown wth bodes. The column pattons of M ae m 0, m 1, m 2, m 3, m 4. And ank p (M [0] ) = ank p (m [0] 1 ) = 4, ank p (M [1] ) = ank p (m [1] 2 ) = 6, ank p (M [2] ) = ank p (m [2] 4 ) = Bnay code matces Fx ( p = 2. The goal of ths secton s to show that fo all 1, ank p (M [] ) = ) 2 whee M = AA T fo some specally constucted A, whch we call bnay code matx. We wll genealze the constucton of M n a subsequent secton. Fo now, A s constucted as follows. Stat wth the 2 matx whose, j enty s the jth bt n the bnay expanson of. Then apply ow pemutatons to A such that the fst ( 0) ows have have exactly 0 non-zeo entes, followed by ( 1) ows whch have exactly 1 non-zeo entes, followed by ( 2) ows whch have exactly 2 non-zeo entes and so on. See Fgue 2.1 fo an example whee = 4.

9 Ranks of quotents, emandes and p-adc dgts of matces 9 The lth column of M s gven by: M,l = a 1,l A, a,l A, = j J l A,j, (10) whee J l {1, 2,..., } and the second equalty holds because a,l {0, 1}. We call J l the summng ndex set of M,l. Let m k denote the 2 ( ) k submatx of M, whch ncludes all columns of the fom: M,l = j J l A,j whee J l {1, 2,..., } and J l = k. Then the columns of M can be pattoned nto: M = [ m 0 m 1 m 2... m 2 m m ]. (11) The next lemma shows that [ ] M [] = m [] m []... m [] (12) Lemma 11. If k < 2, then m [] k = 0 fo all 1. Poof. Columns of m k ae gven by j J A,j whee J = k. The entes of A ae ethe 0 o 1. So the lagest enty n m k s = k. The esult follows by appealng to the bnay expanson of k. We expect ank p (m [] ) ( ) [] 2 2 snce m s a matx of dmenson 2 ( ) 2 2. The next lemma shows that the ank s, n fact, equal to ths uppe bound. Lemma 12. ank p (m [] = ( fo all 1. Poof. Let c 1,..., c ( ) be the column ndces of m 2 n M. Let S(m 2) be the ( 2 ) ( 2 ) 2 submatx of m2 fomed by the ows c 1,..., c ( 2), and S(A) be the ( ) 2 submatx of A fomed by the ows c1,..., c ( 2). Rows of S(A) have exactly 2 non-zeo entes because of the constucton of A. If we teat A and M as block matces then S(m 2 ) = S(A)S(A) T s the 2 th dagonal block of M (See Fgue 2.1). The entes n ow ρ of S(m 2 ) ae gven by lnea combnatons of the entes n ow ρ of S(A). The summng ndex sets J j, whee J j = 2, ae exactly the locatons of the non-zeo entes of ows of S(A), whch ae all dffeent by constucton. Hence thee s only one enty n ow ρ of S(m 2 ) whose summng set matches the locatons of the non-zeo entes n ow ρ of

10 Ranks of quotents, emandes and p-adc dgts of matces 10 S(A). The value of ths enty s = 2. The othe entes have values less than 2. Now appeal to the bnay expanson of 2 to get that S(m [] s an dentty (sub)matx of m [] 2 whose sze s ( (. Theefoe, m [] 2 has ank (. Next we wll pove that ank p (M [] ) = ( ) 2 by showng that all the columns of m [], 2 +1 m[],..., 2 +2 m[] 2 ae lnealy dependent on those of m[]. 2 Lemma 13. Consde any column m n m 2 +z, whee z 1. Then m [] s a lnea combnaton of columns of m [] 2. Poof. Let J be the summng ndex set of m, whee J = 2 + z. Let I be the set of all subsets of J of sze 2, so I = ( ) 2 +z 2. Fo evey I I, thee s a unque coespondng column c I n m 2 whose summng set s I. We wll show that m [] can be obtaned by addng up c I s. In othe wods, m [] I I c [] I (mod 2). (13) Let A J denote the submatx of A fomed by the columns ndexed by J. Fo any ow ρ of A J, let 2 + k ρ be the numbe of 1 s n that ow, whee 2 k ρ z. Fst, f k ρ < 0, then the coespondng sum of 1 s at ths ow s less than 2. By Lemma 11, we have the coespondng entes n both m [] 2 and m [] ae zeos and (13) tvally holds. On the othe hand, f 0 k 2 +z ρ z, then the ρth enty of the ght-hand sde of (13) s ( 2 +k ρ ) 2 (mod 2) snce I = ( 2 +k ρ. (Recall that the numbe of non-zeo entes n ow ρ s 2 +k ρ athe than 2 +z.) The ρth enty of the left-hand sde of (13) s (2 + k ρ ) quo 2. The (2 + k ρ ) tem coesponds to addng (2 + k ρ ) nonzeo entes, and the quo 2 opeaton coesponds to the th bt of the bnay expanson of m. By Lemma 15 (below), we have (2 + k ρ ) quo 2 ( 2 +k ρ ) 2 (mod 2), and (13) holds. The poof of the next (auxlay) lemma uses a theoem due to Kumme [2]. Ths s tue n the example of Fgue 2.1 wthout any eodeng, because we constucted the ow blocks of A such that the bnay expanson of comes afte the bnay expanson of j wheneve > j. Wthout such odeng, the dentty block asseton holds up to ow and column pemutatons.

11 Ranks of quotents, emandes and p-adc dgts of matces 11 Fact 14 (Kumme s Theoem). The exact powe of p dvdng ( ) a+b a s equal to the numbe of caes when pefomng the addton of (a + b) wtten n base p. A coollay of Kumme s theoem s that ( ) a+b a s odd (esp. even) f addng (a + b) wtten n bnay expanson geneates no (esp. some) caes. Lemma 15. (2 + k) quo 2 ( 2 +k (mod 2). Poof. We wll show that (2 + k) quo 2 and ( ) 2 +k 2 have the same paty. Wte k = Q2 + R fo a quotent Q 0 and a emande 0 R < 2. Thee ae two cases fo Q. If Q s even, then the th bt of k s 0 and hence no caes ae geneated when addng k and 2 n base 2. So by Kumme s Theoem, ( ) ( 2 +k 2 s odd and 2 ) +k 2 1 (mod 2). If Q s odd, then the th bt of k s 1 and the numbe of caes geneated when addng 2 + k n base 2 s at least 1. So by Kumme s theoem ( ) ( 2 +k 2 s even and 2 ) +k 2 0 (mod 2). We have shown that ( 2 +k and Q have opposte pates. Now, substtute k = Q2 + R to get (2 + k) quo 2 = Q + 1. Hence, modulo 2, (2 + k) quo 2 also have an opposte paty to that of Q. Ths concludes ou poof. 2.2 Non-symmetc Matces So fa we have shown that ank p (M [] ) = ank p (m [] = (, whee M = AA T fo some specally constucted A. We now put the esults togethe nto a moe geneal theoem. Theoem 16. Assume U, S, V Z n n, such that U, V have entes fom {0, 1}, det U det V 0 (mod 2), S = dag(1,..., 1, 0,..., 0), ank p (S) =, and n 2. If M = USV Z n n, then ank p (M [] ) = ( fo all 1. Poof. Snce S = SS, we have M = USV = USSV = LR, whee L = US Z n, and R = SV Z n. Let A Z 2 be the bnay code matx of the dgts {0,..., 2 1}. Consde the matces L = A, R = A T and M = L R. If we stat wth L (esp. R) and augment t wth the appopate (n 2 ) addtonal ows (esp. columns), and apply the appopate ow and column pemutatons, then we could tansfom L nto L (esp. R nto R), and n effect, tansfom M nto M. Ou goal s to show that the ank aguments of the pevous lemmas hold unde the afoementoned opeatons..e. the coeffcent of 2 n the bnay expanson of k.

12 Ranks of quotents, emandes and p-adc dgts of matces 12 We fst note that ow and column pemutatons peseve anks. Also, by a smple enumeaton agument ove the bnay tuples of sze, and by the gven fact that n 2, we can conclude that any addtonal ows (esp. columns) augmented to L (esp. R) wll be lnealy dependent. In fact, any such ows (esp. columns) wll be duplcates of exstng ows (esp. columns). Now, consde addng exta columns to R. The esultng exta columns n M ae duplcates of exstng columns and hence the anks n Lemma 12 ae not affected. Fnally, addng exta ows to L does not change the cadnalty of the summng ndex sets n (10). The est of the esults ae staghtfowad to vefy. 2.3 Odd Pmes Fo p = 2, the non-zeo pattens of the bnay code matx A concdes wth the summng ndces n (10). Ths s not tue fo odd pmes, whee the lnea combnatons can have coeffcents othe than 0 and 1. Thus t s an open queston to devse constucton a smla to bnay code matces, whch exposes the combnatoal stuctue of the column space of M = AA T. Howeve, we pesent the followng conjectue towads undestandng the p- adc anks fo odd pmes. Conjectue 17. Assume p = 2k + 1 s an odd pme, U, S, V Z n n such that U, V have entes fom [0, p 1], det U det V 0 (mod p), S s a 0, 1 dagonal matx and ank p (S) =. Let M = USV = M [0] + M [1] p + whee M [] (Z/pZ) n n. It s conjectued that ank p (M [1] ) k =0 ( ) ( ) + 2k 1 2 (14) 2k Futhemoe, n the genec case whee the entes of U, V ae unfomly chosen at andom fom [0, p 1], and n s abtaly lage, the anks ae equal to the stated bound. Ths conjectue fst appeaed n [1]. It shows that a poduct of matces wth small entes and small ank can stll have vey lage ank, but not full, p-adc expanson. In othe wods, the caes fom the poduct U SV wll mpact many dgts n the expanded poduct.

13 Ranks of quotents, emandes and p-adc dgts of matces 13 Acknowledgment The authos would lke to thank Andew Anold, Kevn Hae, Davd McKnnon, Jason Peasgood, and B. Davd Saundes. Refeences [1] M. Elshekh, M. Gesbecht, A. Novocn, and B. D. Saundes. Fast computaton of Smth foms of spase matces ove local ngs. In Poceedngs of the 37th Intenatonal Symposum on Symbolc and Algebac Computaton, ISSAC 12, pages , New Yok, NY, USA, ACM. [2] E. E. Kumme. Übe de Egänzungssätze zu den allgemenen Recpoctätsgesetzen. Jounal f u de ene und angewandte Mathematk, 44:93 146, [3] M. Newman. Integal Matces. Academc Pess, New Yok, NY, USA, 1972.