A FORMULA FOR COMPUTING INTEGER POWERS FOR ONE TYPE OF TRIDIAGONAL MATRIX

Transcription

1 Hacettepe Journal of Mathematcs and Statstcs Volume FORMUL FOR COMPUTING INTEGER POWERS FOR ONE TYPE OF TRIDIGONL MTRIX H Kıyak I Gürses F Yılmaz and D Bozkurt Receved :08 :009 : ccepted 5 :0 :0 bstract In ths paper we derve the general expresson of the r th power r N for one type of trdagonal matrx Keywords: Trdagonal matrx Matrx power Egenvalues Chebyshev polynomal 000 MS Classfcaton: C C E Introducton In the present paper we derve a general expresson for the rth power for one type of trdagonal matrx r N and N denotes the set of natural numbers In [4] Rmas derved a general expresson for the lth power for one type of symmetrc trdagonal matrces of even order General expressons for the rth power of a matrx are obtaned by usng the equalty r = PJ r P [] J s the Jordan form of and P the transformng matrx We need the egenvalues and egenvectors of the matrx to compute the matrces J and P The egenvalues of are the roots x nk = cos kπ k = n n + of the nth degree Chebyshev polynomal U nx = of the second knd snn + arccosx x snarccosx Selcuk Unversty Scence Faculty Department of Mathematcs 450 Campus Konya Turkey E-mal: H Kıyak kyak8@hotmalcom I G Gürses rematca@hotmalcom F Yılmaz fylmaz@selcukedutr D Bozkurt dbozkurt@selcukedutr Correspondng uthor

2 35 K Kıyak I G Gürses F Yılmaz D Bozkurt Man results Let be the followng n-square trdagonal matrx = 0 0 n n 0 The egenvalues of are the roots of the characterstc equaton λe = 0 and also the roots of U nx the nth degree Chebyshev polynomal of the second knd [3] In the present paper we nvestgate nteger powers of ths matrx Intally we wll scrutnze the nteger powers for even orders Let us wrte α D nα = λe = α α and α 3 nα = α α α = λ R By and 3 we obtan the followng two results 4 D nα = nα 5 nα = α n α n α α = α α = α and 0α = We obtan 4 by the nducton prncple For n = D α = α = α

3 Computng Integer Powers 353 Let D k α = k α for n = k We have to show that D k+ α = k+ α for n = k + Let α D k+ α = α k+ k+ If we expand the determnant accordng to frst row then we have α 0 α 7 D k+ α = α + α α and f we expand the frst determnant n 7 accordng to last row and the second determnant accordng to frst column then we obtan α D k+ α = α α α + α 0 α α α }} D k α= k α } } k α

4 354 K Kıyak I G Gürses F Yılmaz D Bozkurt Then α D k+ α = α α k α k α α } } k α = α[α k α k α] k α }} k+ α = α k+ α k α = k+ α By solvng the dfference equaton 5 we obtan [] α 8 nα = U n It s known that the roots of the polynomal U nx are [3] 9 x nk = cos kπ k = n n + whch are ncluded n the nterval [ ] Takng and 7 nto account we fnd the egenvalues of the matrx to be λ k = cos kπ k = n n + Snce the multplcty of each of the egenvalues λ k s and applyng the relaton λ k = λ n k+ k = n we wrte down the Jordan form of the matrx as J = dag λ n λ n λ n λ n + λ n + λ n λ n λ n Let us solve the homogeneous lnear equatons λ E x = 0 λ s the th egenvalue of the matrx n By elementary row operatons the coeffcent matrx of the system s λ λ λ λ λ D n λ

5 Computng Integer Powers 355 Snce D n λ = 0 rankλ E = n Then x + λ x x 3 = 0 x λ x 3 x 4 = 0 x n λ x n x n = 0 x n + λ x n = 0 Now we wll scrutnze the solutons of the above system of lnear equatons accordng to n and k λ Let x n = = U 0 and for any n and k defne n k or mod 4 a = n k 0 or 3 mod 4 n or mod 4 e = n 0 or 3 mod 4 We consder the followng two cases: Case Let n be an even number If k s an odd number then λ x n = λ = U x n = λ λ = U x k+ = aλ n k + cλ n k n k x k+ = aλ n k + bλ n k 3 + dλ n k 5 x k = aλ n k + a bλ n k λ λ = au n k λ + = au n k + c dλ n k n k + λ λ = au n k x = eλ n en λ n n λ = eun λ b = an k c = an k 3 and d s the sum of the coeffcents of the terms λ n k 5 n x k 3 and λ n k n x k

6 35 K Kıyak I G Gürses F Yılmaz D Bozkurt If k s an even number then λ x n = λ = U x n = λ λ = U x k+3 = aλ n k n k x k+ = aλ n k + cλ n k 4 x k+ = aλ n k + bλ n k 3 λ = au n k 3 λ + = au n k λ + dλ n k n k + + n λ = eun λ λ λ = au n k x k = aλ n k + b aλ n k + c + dλ n k 4 λ + = au n k x = eλ n en λ n 3 b = an k c = an k 3 and d s the dfference of the coeffcents of the terms λ n k 5 x k+ Case Let n be an odd number If k s an odd number x n = λ = U λ x n = λ λ = U x k+3 = aλ n k n k x k+ = aλ n k + cλ n k 4 x k+ = aλ n k + bλ n k 3 λ = au n k 3 λ + = au n k λ + dλ n k n k + = eu n λ n x k+3 and λ n k λ λ = au n k x k = aλ n k + b aλ n k + c + dλ n k 4 λ + = au n k x = eλ n en λ n 3 b = an k c = an k 3 n and d s the dfference of the coeffcents of the terms λ n k 5 x k+ n x k+3 and λ n k n

7 Computng Integer Powers 357 If k s an even number then λ x n = λ = U x n = λ λ = U x k+ = aλ n k + cλ n k n k x k+ = aλ n k + bλ n k 3 + dλ n k 5 x k = aλ n k + a bλ n k λ λ = au n k λ + = au n k + c dλ n k n k + λ λ = au n k x = eλ n en λ n 3 b = an k c = an k 3 + = eu n λ and d s the sum of the coeffcents of the terms λ n k 5 Takng P = x λ nto account P = [ P P P n ] n x k 3 and λ n k n x k Usng ths expresson we can wrte down the transformng matrx P as eu n λ eu n λ eu n λn au n k λ au n k λ au n k λn P = U 3 λ U 3 λ U 3 λn U λ U λ U λn U λ U λ U λn U 0 λ U 0 λ U 0 λn Denotng the th row of P by R = n from we have 3 R = [ t U n λ t U n λ t U n λn ] t = n or mod 4 n 0 or 3 mod 4

8 358 K Kıyak I G Gürses F Yılmaz D Bozkurt Denotng jth column of the nverse matrx P by ρ j we obtan m j n+ n j m j U λ n+ n j 4 ρ j = m 3 j U λ3 n+ n j m n j U λn n+ n j m j = n j or mod 4 n j 0 or 3 mod 4 Takng expresson 4 nto account we obtan the matrx P as follows: e n+ n n+ 3 n+ P e U λ n+ n U λ n+ 3 U λ n+ = e n U λn n+ n n U λn n n+ 3 U λn n+ n+ n+ n n+ n+ 0 U λ U λ n+ 0 U λn n n+ U λn 0 Usng the equalty r = PJ r P we derve r } j = PJ r P } j = RJr ρ j = tmj n + n k= λ r k4 λ λk ku n λ r m j n+ n j λ r m j U λ n+ n j = R λ r 3m 3 j U λ3 n+ n j λ r nm n j U λn n+ n j U n j λk λ are the egenvalues of the matrx U k x s the kth degree Chebyshev polynomal of the second knd j = n and we equate ths matrx wth ts unque element as usual

9 Computng Integer Powers 359 Now we consder the case when the order n s odd Workng as before we obtan α 5 D nα = λe = α α and α nα = α α for odd orders α = λ R lso the equaltes and 3 are vald for odd orders Snce the multplcty of all the egenvalues λ k s and applyng the relaton λ k = λ n k+ k = n and λ = 0 we can wrte down the Jordan form of the matrx as: 7 J = dag λ n λ n λ n λ n+3 0 λ n+3 λ n λ n λ n Denotng jth column of the nverse matrx P by ρ j we obtan 8 ρ j = λ m + j n+ λ m j λ m + j n+ λ m +3 j n+ +n n+ n j U λ n j U λ3 n j U λn n j m j = n j or mod 4 n j 0 or 3 mod 4

10 30 K Kıyak I G Gürses F Yılmaz D Bozkurt Takng the expresson 8 nto account we obtan the matrx P as follows: P e = λ e + n+ λ + n+ λ e +n n+ n U λ n λ + n+ λ + n+ U λn λ n +n n+ 3 U λ 3 λ + n+ λ + n+ U λ U λn λ +n 3 n+ U λn Usng the equalty r = PJ r P we derve r } j = PJ r P } j = RJr ρ j = r t m j n + n k= λ + n+ λ + n+ U λ λ +n n+ = R λ λ r nm j λ r kλ λk +kun λ + n+ λ + n+ U λn λ +n n+ λ λ r m + j n+ λ λ r m + j n+ λ λ r 3m +3 j n+ +n n+ U n j λk n j U λ n j U λ3 n j U λn n j 0 U λ 0 U λn 0 λ are the egenvalues of the matrx U k x s the kth degree Chebyshev polynomal of the second knd j = n and we equate the matrx wth ts unque element 3 Numercal examples Takng nto account the derved expressons for n even one can see the effectveness of the formula For n = 4 we obtan; J = dag λ 4 λ 3 λ 3 λ 4 = dag a b b a a = cos π 5 b = cos π 5

11 Computng Integer Powers 3 If r s even r} = a r 4 a a 3 a + b r 4 b b 3 b r} = r} = 0 r} = r } 3 3 = a r+ 4 a a 3 a + b r+ 4 b b 3 b r} = r} = r} = a r 4 a a + b r 4 b b r} = r} = r} = r} = 4 4 a r 4 a a + b r 4 b b r} = 33 a r+ 4 a + b r+ 4 b r} = r} = r} = 44 a r 4 a + b r 4 b For example for r = 4; = If r s odd r} = 0 r} = r} = [ a r 4 a a 3 aa + b r 4 b b 3 bb ] r} = r} = r} = r} = 4 4 a r 4 a a 3 a + b r 4 b b 3 b r} = 0 r} = r} = 3 3 a r+ 4 a a + b r+ 4 b b r} = r} = r} = 0 33 r} = r} = a r+ 4 a + b r+ 4 b r} = 0 44 For example for r = =

12 3 K Kıyak I G Gürses F Yılmaz D Bozkurt We wll now see the effectveness of the formula for odd orders For example f n = 5 we obtan If r s even J = dag λ 5 λ 4 0 λ 4 λ 5 = dag c d 0 d c c = 3 d = r } = c r d + c d r = 3 r + 3 r } = r} = 0 r } 3 = r} 3 = 3 cr d = 3 r r } 4 = r} 4 = 0 r } = r} = 5 5 c r d c d r = 3 r 3 r } = cr+ d + c d r = 3 r+ + 3 r } 3 = r} 3 = 0 r } = r} = 4 4 c d r c r+ d = 3 3 r+ r } = r} = r } 33 = 3 ar b = 3 cr = c r = 3 r r } 34 = r} 43 = 0 r } 35 = r} 53 = 3 cr d = c r = 3 r r } = 44 c r+ d + c d r = 3 r+ + 3 } r = r} = } r = 55 c r d + c d r = 3 r + 3 For example for r = 4 we have = If r s odd r } = 0 r } = r} = c r+ d + c d r = 3 r+ + 3 r } = r} = r } = r} = 4 4 c r+ d c d r = 3 r+ 3 } r = r} = r } = 0

13 Computng Integer Powers 33 r } 3 = r} 3 = 3 cr+ d = 3 r r } 4 = r} 4 = 0 r } = r} = 5 5 c d r c r+ d = 3 3 r+ } r = 0 33 r } 34 = r} 43 = 3 cr+ d = c r = 3 r r } 35 = r} 53 = 0 r } 44 = 0 } r = r} = c r+ d + c d r = 3 r+ + 3 r } 55 = 0 For example for r = 3 we have: = cknowledgement We are grateful to Jonas Rmas for comments and contrbutons References [] garwal R P Dfference Equatons and Inequltes Marcel Dekker New York 99 [] Horn R and Johnson Ch Matrx nalyss Cambrdge Unversty Press Cambrdge 985 [3] Mason J C and Handscomb D C Chebyshev Polynomals CRC Press Washngton 003 [4] Rmas J On computng of arbtrary postve nteger powers for one type of symmetrc trdagonal matrces of even order-i ppled Mathematcs and Computaton [5] Rmas J On computng of arbtrary postve nteger powers for one type of symmetrc trdagonal matrces of odd order-i ppled Mathematcs and Computaton