COMBINATORIAL IDENTITIES DERIVING FROM THE n-th POWER OF A 2 2 MATRIX

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1 COMBINATORIAL IDENTITIES DERIVING FROM THE n-th POWER OF A MATRIX J Mc Laughln 1 Mathematcs Department Trnty College 300 Summt Street, Hartford, CT amesmclaughln@trncolledu Receved:, Accepted:, Publshed: Abstract In ths paper we gve a new formula for the n-th power of a matrx ( a b More precsely, we prove the followng: Let A be an arbtrary matrx, c d T a + d ts trace, D ad bc ts determnant and defne Then, for n 1, y n : A n n/ 0 T n ( D ( yn d y n 1 b y n 1 c y n 1 y n a y n 1 We use ths formula together wth an exstng formula for the n-th power of a matrx, varous matrx denttes, formulae for the n-th power of partcular matrces, etc, to derve varous combnatoral denttes 1 Introducton Throughout the paper, let I denote the -dentty matrx and n an arbtrary postve nteger In [11], Wllams gave the followng formula for the n-th power of matrx A wth egenvalues α and β: ( ( α n A βi + β n A αi, f α β, (11 A n α β β α α n 1 (na (n 1αI, f α β Blatz had gven a smlar, f slghtly more complcated expresson, n [1] 1 mclaugh/

2 For our present purposes (producng combnatoral denttes, t s preferable to express A n drectly n terms of the entres of A, rather than the egenvalues of A If we let T denote the trace of A and D ts determnant, then, wthout loss of generalty, α (T + T 4 D/ and β (T + T 4 D/ A lttle elementary arthmetc gves that (n the case α β, f (1 z n : αn β n α β then (n 1/ m0 m+1 T n m 1 (T 4D m, n 1 (13 A n z n A z n 1 D I Equaton 13 also holds n the case of equal egenvalues (when T 4 D 0, f z n s assumed to have the value on the rght of (1 The ey pont here s that f another closed-form expresson exsts for A n, then equatng the expressons for the entres of A n from ths closed form wth the expressons derved from Equatons 1 and 13 wll produce varous denttes As an llustraton, we consder the followng example Let {F n } n1 denote the Fbonacc sequence, defned by F 0 0, F 1 1 and F +1 F + F 1, for 1 The dentty ( n ( 1 1 Fn+1 F (14 n 1 0 F n F n 1 leads drectly to several of the nown formulae nvolvng the Fbonacc numbers For example, drect substtuton of T D 1 n Equaton 1 gves Formula 91 from Vada s lst [9]: (n 1/ m0 m+1 5 m (15 F n n 1 Many other formulae can be smlarly derved Ths method of dervng combnatoral denttes of course needs two expressons for A n and for a general matrx, these have not been avalable In ths present paper we remedy ths stuaton by presentng a new formula for A n, where the entres are also expressed n terms of the entres n A Man Theorem We prove the followng theorem Theorem 1 Let ( a b (1 A c d

3 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY x (00x, #A11 3 be an arbtrary matrx and let T a + d denote ts trace and D ad bc ts determnant Let n/ ( n ( y n T n ( D Then, for n 1, (3 A n 0 ( yn d y n 1 b y n 1 c y n 1 y n a y n 1 Proof The proof s by nducton Equaton 3 s easly seen to be true for n 1,, so suppose t s true for n 1,, Ths mples that ( a A +1 y + (b c a dy 1 b y c y d y + (b c a dy 1 Thus the result wll follow f t can be shown that y +1 (a + dy + (b c a dy 1 Upon substtutng from Equaton, we have that (a + dy + (b c a dy 1 / ( 0 / 0 ( T +1 ( D + T +1 ( D + If s even, then / ( + 1/, and (a + dy + (b c a dy 1 ( T (+1/ 0 If s odd, then / ( 1/, and (+1/ 1 ( + 1 ( 1/ 0 (+1/ 1 ( 1 T 1 ( D +1 ( T +1 ( D 1 (( + T +1 ( D y +1 ( T +1 ( D 1 ( ( 1/ (( ( (a + dy + (b c a dy 1 T T +1 ( D ( ( + 1/ + T +1 (+1/ ( D (+1/ ( + 1/ 1 (+1/ ( + 1 T +1 ( D y +1 Ths completes the proof 0

4 4 Remar: Schwerdtfeger ([7], pages outlnes a method due to Jacobsthal [3] for fndng the n power of an arbtrary matrx A whch, after a lttle manpulaton, s essentally equvalent to the method descrbed n Theorem 1 We were not aware of Jacobsthal s result before our own dscovery The am now becomes to use combnatons of the formulae at (13 and (3, together wth varous devces such wrtng A n (A n, wrtng A B + C or A B C where B and C commute, etc, to produce combnatoral denttes One can also consder partcular matrces A whose n-th power has a smple form, and then use the formulae at (13 and (3 to derve combnatoral denttes, As an mmedate consequence of (13 and (3, we have the followng corollary to Theorem 1 Corollary 1 For 1 (n 1/, (4 ( (n 1/ n ( ( 1 n (n 1/ n (, Proof For the frst dentty, we equate the (1, entres at (13 and (3 and then let T x and D x y Ths gves that (n 1/ 0 1 (x n 1 (y x (n 1/ 0 ( n x n 1 y + 1 We now expand the left sde and equate coeffcents of equal powers of y/x For the second dentty, we smlarly equate the (1, entres at (13 and (3, expand (T 4D m by the bnomal theorem and compare coeffcents of le powers of ( D/T Remar: The lterature on combnatoral denttes s qute extensve and all of the denttes we descrbe n ths paper may already exst elsewhere However, we beleve that at least some of the methods used to generate/prove them are new

5 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY x (00x, #A Some Formulae for the Fbonacc Numbers The dentty ([10], pp 18-0 (31 F n (n 1/ 0 ( n 1 follows from (14, ( and (3, upon settng T D b 1 The Lucas sequence {L n } n1 s defned by L 1 1, L 3 and L n+1 L n + L n 1, for n The dentty A n (A n mples ( ( Fn +1 F n Fn+1 F n F n F n 1 F n F n 1 The facts that L n F n+1 + F n 1 and F n+1 F n 1 F n ( 1 n together wth Theorem 1 now mply the followng dentty from [4] (page 5: ( 1/ ( 1 (3 F n F n L 1 n ( 1 (n+1 0 Let ı : ( ( 1 and defne B : 1 + ı ı Then B ı 1, leadng to the 1 0 dentty: (33 F 1 1 ( 1 m ( + 1 m ( 1 m, ı 1 m m0 a varant of Ram s formula labelled FeP at [6] Many smlar denttes for the Fbonacc and Lucas sequences can also be easly derved from Theorem 1 4 A Bnomal Expanson from Wllams Formula Proposton 1 Let n N and 1 t n, t ntegral Then (41 ( n t n 1 n 1 m m0 0 ( 1 m n 1 m 1 m m ( ( n 1 m m t 1 ( f + e 1 Proof Let A The egenvalues of A are clearly f + e and f Wth the 0 f notaton of Theorem 1, T (e+f and D f(f +e If we now equate the (1, entry from the left sde of Equaton 11 wth the (1, entry from the left sde of Equaton 3,

6 6 we get, after a lttle manpulaton, that (f + e n f n + e (n 1/ m0 Next, replace e by e/ to get that (4 (f + e n f n + e (n 1/ m0 1 m m 1 m m ((e + f n 1 m ( f(e + f m (e + f n 1 m ( f(e + f m Fnally, set f 1, expand both sdes of Equaton 4 and compare coeffcents of le powers of e to get the result As before, let We wrte 5 Commutatng Matrces I ( a b A c d (51 A (ma + w I + ((1 ma w I, notng that the matrces on the rght commute n ( n (5 A n (ma + w I n ((1 ma w I 0 and we mght hope to use the formulae at (11 and (1 to derve new denttes We gve two examples Proposton Let n N and, r ntegers wth 0 r n Then ( n n ( ( ( n n, r n, (53 ( 1 n ( 1 r 0 0, r n Proof Expand the rght sde of Equaton 5 and re-ndex to get that A n n w n r A r (1 m r ( 1 r r0 n 0 ( m ( 1 1 m n 0 ( ( 1 r

7 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY x (00x, #A11 7 If we compare coeffcents of w on both sdes, t s clear that n ( m n ( ( ( n n (m 1 n, r n, ( 1 ( 1 1 m r 0 0 0, r n Equvalently, n n m (1 m n ( 1 n 0 0 ( ( 1 r 1, r n, 0, r n Fnally, equatng coeffcents of powers of m on both sdes of the last equalty gve the result We note that the case for r n follows snce the left sde then must equal (m + (1 m n Proposton 3 Let n N and t an nteger such that 0 t n For each nteger [0, (n 1/ ], ( 1/ 1 r n ( ( ( ( n 1 r 1 r r (54 r 1 r 0 ( ( n 1 r ( 1 r+t++n+, t 0, r + n + t 0, t 0 Proof We set m 0 n Equatons 51and 5, so that ( n a b n ( a w w c d n c 0 b d w If we use Theorem 1 and compare values n the (1, poston, we deduce that (55 (n 1/ 0 n 1 1 w n ( 1/ (a + d n 1 (bc ad r0 ( 1 r r (a + d w 1 r (bc (a w(d w r For ease of notaton we replace a + d by T and ad bc by D If we then expand the rght sde and collect powers of w, we get that (n 1/ 0 1 n 1 ( 1/ r0 T n 1 ( D 1 r 0 r m m0 p0 ( ( 1 r 1 r r ( r m ( m p ( 1 +r m+p T 1 r +m p D r m w n ++m+p

8 8 If we solve t n + + m + p for p, we have that (n 1/ 0 1 T n 1 ( D n 1 ( 1/ 1 r n r ( ( ( ( n 1 r 1 r r w t r m t0 1 r0 0 m0 ( m ( 1 r+t++n T m+n t r 1 D r m m n + t Next, cancel a factor of T n 1 from both sdes, replace T by 1/y, D by x/y and m by r to get that (n 1/ 0 1 n 1 (wy t t0 n 1 t0 x n 1 ( 1/ r0 (wy t (n 1/ 0 ( n 1 ( 1 r 0 r 0 ( 1 r r r r + n + t ( 1/ r 1 r 0 r r + n + t ( 1 r ( 1 r+t++n+ x ( ( 1 r 1 r r ( 1 r+t++n+ x ( r r ( r Fnally, compare coeffcents of x (wy t to get the result Corollary For every complex number w dfferent from 0, 1/, φ (the golden rato or 1/φ, (56 F n n 1 ( 1/ r0 ( 1 r r w n (1 w 1 r (1 + w w r Proof Let a b c 1, d 0 n Equaton 55 and once agan use the dentty at (31 Ths dentty contans several of the nown denttes for the Fbonacc sequence as specal cases For example, lettng w φ gves Bnet s Formula (after summng usng the Bnomal Theorem; w 0 gves the formula at (31; w 1/ gves (15; w 1

9 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY x (00x, #A11 9 gves the dentty F n n ( n ( 1 1 F, 1 whch s a specal case of Formula 46 from Vada s lst ([9], page Commutatng Matrces II ( a b Let A wth T a + d and D ad bc and I the dentty matrx as c d before Let g be an arbtrary real or complex number such that g + T g + D 0 It s easy to chec that (61 A 1 (A + g I(g A + D I, g + T g + D and that A + g I and g A + D I commute If both sdes of Equaton 61 are now rased to the n-th power, the rght sde expanded va the Bnomal Theorem and powers of A collected, we get the followng proposton: Proposton 4 Let A be an arbtrary matrx wth trace T and determnant D 0 Let g be a complex number such that g + T g + D 0, g 0 and let n be a postve nteger Then ( (6 A n g D g + T g + D n n r0 r 0 ( n r ( D g ( g r A r D Corollary 3 Let n be a postve nteger and let m be an nteger wth 0 m n Then for n w n, (63 n 1 ( ( ( n 1 n + w ( 1 w + m w 0 m w ( ( ( n n + n + w m 1 ( 1 + w n + + w m w n+m+1 Remar: Consderaton of the bnomal coeffcents on ether sde of (63 shows that ths dentty s non-trval only for m/ (n 1/ w m, but we prefer to state the lmts on w as we have done for neatness reasons

10 10 Proof If Equaton 6 s multpled by (g +T g+d n, both sdes expanded and coeffcents of le powers of g compared, we get that for each nteger m [0, n], (64 A n n 0 ( ( n D n T m m n r0 1 + ( 1 r+m+n ( ( n n r m+n r+m n D (3n r m/ A r If we next assume that (1, entry of A s non-zero, apply Theorem 1 to the varous powers of A on each sde of Equaton 64 and compare the (1, entres on each sde of (64, we get that (65 n r0 (n 1/ ( 1 r+m+n T n 1 ( D n 0 ( ( n n r m+n r+m n D (3n r m/ ( ( n D n T m m r 1 0 ( r 1 Equvalently, after re-ndexng and changng the order of summaton, (66 n v0 0 n ( ( ( ( v n 1 n + n v DT ( 1 + n v m n + v n v+m n ( ( n n v + n v + n m v0 v 3n+m+1 ( v + 3n m 1 T r 1 ( D ( 1 ( D T v The result follows from comparng coeffcents of equal powers of D/T and replacng v by n w Corollary 4 Let n be a postve nteger and let m {0, 1, n} Then (67 n 0 (68 (n + 1 ( ( ( n m m n 0 n 0 ( ( n m m n r0, r+m+n even ( ( n n 3 m m n r0, r+m+n even r m+n r m+n ( n r+m n, ( n r+m n (r + 1,

11 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY x (00x, #A11 11 and (610 F n+ n 0 ( ( n ( 1 m+ m n r0, r+m+n even n r0, r+m+n even r m+n r m+n ( n r+m n (3n r m/ ( r+1 1, ( n r+m n ( 1 ( n+r m/ F r+ Proof Equatons (67 to (610 follow from comparng (1, 1 entres at (64, usng, respectvely, the followng denttes: I n I, ( n ( 1 n + 1 n, 1 0 n n + 1 ( n ( 3 1 n+1 1 n 1 0 n+1 + n, + ( n ( 1 ( n Fn+ F n F n F n Corollary 5 If g s a complex number dfferent from 0, φ or 1/φ, then ( n n g r ( ( n n (611 F n ( 1 r+ g r F g r + g 1 r r0 0 ( 1 1 Proof Let A and compare the (1, entres on each sde of Equaton 6, usng 1 0 (14 7 Mscellaneous denttes derved from partcular matrces The denttes n Corollary 4 derved from the fact that the n-th power of each of the varous matrces mentoned n the proof has a smple, elegant form It s of course easy to construct such( matrces whose n-th power has a smlar nce form by a b constructng matrces wth a predetermned par of egenvalues α and β and c d

12 1 usng the formula of Wllams at (11 matrx to produce denttes of the form (n 1/ 0 1 As an example, the matrx followng dentty for n 1: (71 n (a + d n 1 ( (ad bc ( (n 1/ 0 Theorem 1 can then be appled to any such α n β n, f α β, α β nα n 1, f α β has both egenvalues equal to 1, leadng to the 1 n 1 ( 1 We gve one addtonal example Proposton 5 Let n be a postve nteger, and s an nteger wth 0 s n 1 Then ( ( s n 1 (7 ( 1 0 s 1 Proof We consder n n/ 0 0 ( n 0 and use Theorem 1 to get that 0 ( + n ( (n 1/ 0 1 Upon expandng powers of + and re-ndexng, we get that (73 n 1 n n + s n s s0 { s ( + n 1 ( ( 1 n 1 s Fnally, we note that ( ( ( ( n n n 1 n 1 s s ( s ( n 1 s 1 and compare coeffcents of s n s on both sdes of (73 to get the result } ( 1 8 concludng remars Theorem 1 can be appled n a number of other ways The dentty ( ( n cos nθ sn nθ cos θ sn θ sn nθ cos nθ sn θ cos θ

13 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY x (00x, #A11 13 wll gve alternatve (to those derved from De Movre s Theorem expressons for cos nθ and sn nθ If (x n, y n denotes the n-th largest par of postve ntegers satsfyng the Pell equaton x my 1, the dentty ( ( n xn m y n x1 m y 1 y n x n y 1 x 1 wll gve alternatve expressons for x n and y n to those derved from the formula x n + myn (x 1 + my 1 n Smlarly, one can use Theorem 1 to fnd varous denttes for the Brahmagupta polynomals [8] x n and y n defned by ( ( n xn y n x1 y 1 t y n x n t y 1 x 1 and the Morgan-Voyce polynomals [5] B n and b n defned by ( ( Bn B n 1 x + 1 B n 1 B n 1 0 and b n B n B n 1 It may be possble to extend some of the methods used n ths paper to develop new combnatoral denttes For example, t may be possble to explot the equaton A (m 1 A + w 1 I + (m A + w I + ((1 m 1 m A (w 1 + w I or, more generally, the equaton (( p A (m A + w I n p m A 1 p w I to fnd new combnatoral denttes n a way smlar to the use of Equaton 51 n the secton Commutng Matrces I We hope to explore some of these deas n a later paper 1 References [1] P J Blatz, On the Arbtrary Power of an Arbtrary ( -Matrx (n Bref Versons The Amercan Mathematcal Monthly, Vol 75, No 1 (Jan, 1968, pp [] H W Gould, A hstory of the Fbonacc Q-matrx and a hgher-dmensonal problem Fbonacc Quart 19 (1981, no 3, [3] E Jacobsthal, Fbonaccsche Polynome und Kresthelungsglechungen Stzungsberschte der berlner Math Gesellschaft, 17, (1919 0, [4] R C Johnson, Matrx methods for Fbonacc and related sequences mathsduracu/ dma0rc/ped/fbpdf, (August 003 [5] A M Morgan-Voyce, Ladder Networ Analyss Usng Fbonacc Numbers IRE Trans Crcut Th CT-6, 31-3, Sep 1959 [6] Raesh Ram, Fbonacc number formulae, (September 003 [7] Hans Schwerdtfeger, Geometry of complex numbers Mathematcal Expostons, No 13 Unversty of Toronto Press, Toronto 196 x+186 pp

14 14 [8] E R Suryanarayan, The Brahmagupta Polynomals Fb Quart 34, 30-39, 1996 [9] S Vada, Fbonacc & Lucas numbers, and the golden secton Theory and applcatons Ells Horwood Seres: Mathematcs and ts Applcatons Ells Horwood Ltd, Chchester; Halsted Press [John Wley & Sons, Inc], New Yor, pp [10] N N Vorob ev, Fbonacc numbers Translated from the Russan by Halna Moss; translaton edtor Ian N Sneddon Blasdell Publshng Co (a dvson of Random House, New Yor-London 1961 v+66 pp [11] Kenneth S Wllams, The nth Power of a Matrx (n Notes Mathematcs Magazne, Vol 65, No 5 (Dec, 199, p 336