Powers of tridiagonal matrices with constant diagonals

Applied Mathematics and Computation 206 (2008) 885–891

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Powers of tridiagonal matrices with constant diagonals Jesús Gutiérrez-Gutiérrez CEIT and Tecnun (University of Navarra), Manuel de Lardizábal 15, E-20018 San Sebastián, Spain

a r t i c l e

i n f o

a b s t r a c t In this paper, we derive a general expression for the entries of the qth power ðq 2 NÞ of the n  n complex tridiagonal matrix tridiagn ða1 ; a0 ; a1 Þ for all n 2 N, in terms of the Chebyshev polynomials of the second kind. Ó 2008 Elsevier Inc. All rights reserved.

Keywords: Tridiagonal matrices Eigenvalues Eigenvectors Jordan’s form Chebyshev polynomials

1. Introduction In the present paper we study the entries of positive integer powers of an n  n complex tridiagonal Toeplitz (constant diagonals) matrix

0

a0

B Ba B 1 B B B0 An ¼ tridiagn ða1 ; a0 ; a1 Þ ¼ B B . B .. B B . B . @ . 0

a1

0

a0

a1

 .. .

a1 .. .

a0 .. . .. . 

a1 .. . .. . 0



 ..

.

..

.

..

.

a1

0 .. . .. .

1

C C C C C C C; C 0 C C C C a1 A a0

where a1 a1 –0. The paper is organized as follows. In Section 2, we present an eigenvalue decomposition of the matrix An ¼ tridiagn ða1 ; a0 ; a1 Þ. In Section 3, based on this decomposition we introduce Theorem 2 that provides a general expression for the entries of Aqn for all q; n 2 N, in terms of the Chebyshev polynomials of the second kind. This novel expression is an extension of the one obtained in [1] for the powers of the Hermitian tridiagonal matrix tridiagn ða1 ; a0 ; a1 Þ with n 2 N. From Theorem 2 we will also deduce the expression given by Rimas for the entries of the powers of the real skew-symmetric tridiagonal matrix tridiagn ð1; 0; 1Þ with n 2 N (see [2 or 3] for the even case and [4 or 5] for the odd case). The paper finishes with a numerical example. 2. An eigenvalue decomposition of An It is known (see e.g. [6]) that V n diagða1 ; . . . ; an ÞV 1 n is an eigenvalue decomposition of An ¼ tridiagn ða1 ; a0 ; a1 Þ, where the entries of the eigenvector matrix V n are

E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2008.10.005

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J. Gutiérrez-Gutiérrez / Applied Mathematics and Computation 206 (2008) 885–891

½V n j;k ¼

rffiffiffiffiffiffiffiffij1 a1 jkp sin ; a1 nþ1

1 6 j;

k6n

and the eigenvalues are given by

aj ¼ a0 þ 2a1

rffiffiffiffiffiffiffiffi a1 jp cos ; a1 nþ1

1 6 j 6 n:

ð1Þ

The following result presents the eigenvalue decomposition of An that we will really use in Section 3. Theorem 1. Let a1 ; a0 ; a1 2 C; a1 a1 –0 and n 2 N. Then W n Dn W 1 n is an eigenvalue decomposition of An ¼ tridiagn ða1 ; a0 ; a1 Þ, where the entries of the eigenvector matrix W n are

½W n j;k ¼

rffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffij 2 a1 jkp sin ; nþ1 a1 nþ1

1 6 j;

k 6 n;

and Dn ¼ diagða1 ; . . . ; an Þ, with aj given by (1). Moreover, W n can be written as

W n ¼ Kn  n ; where Kn is the diagonal matrix

rffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffin ! a1 a1 ; ;...; a1 a1

Kn ¼ diag

and  n is the unitary matrix

½ n j;k ¼

rffiffiffiffiffiffiffiffiffiffiffiffi 2 jkp sin ; nþ1 nþ1

1 6 j;

k 6 n:

Proof. We will split the proof into three steps: Step 1. In this step we will prove the following equality that will be used in the other two steps: n X

cos

h¼1

mhp ¼ nþ1



1 if m is even; 0

ð2Þ

if m is odd;

for every n 2 N; 1 6 m 6 2n þ 1. P P We first consider the even case. If x–1 is an n þ 1th root of unity then nh¼0 xh ¼ 0 and consequently nh¼1 xh ¼ 1. Obmpi mpi mpi serve that enþ1 is an n þ 1th root of unity. Moreover, since 0 < m < 2ðn þ 1Þ then enþ1 –1. Therefore, by taking x ¼ enþ1 , we deduce that " #

1 ¼ Re

n X

xh ¼

n X

h¼1

Re½xh  ¼

h¼1

n X

cos

h¼1

mhp ; nþ1

where Re denotes real part. The odd case is a direct consequence of the fact that

cos

mhp mðn þ 1  hÞp ¼  cos ; nþ1 nþ1

1 6 h 6 n:

Step 2. In the second step we will show that  n is unitary. Obviously, we only have to prove that  n  n ¼ In , where  denotes conjugate transpose (i.e.,  n ¼  n > ) and In is the n  n identity matrix. To show that ½ n  n j;k is the Kronecker delta dj;k , we first consider the case j ¼ k. We have

½ n  n j;j

n X ¼ ½ n j;h ½ n j;h ¼ h¼1

!  n n  n X 2 X 1 X 2jhp 1 2jhp 2 jhp : sin 1  cos cos n ¼ ¼ n þ 1 h¼1 nþ1 n þ 1 n þ 1 h¼1 nþ1 nþ1 h¼1

Consequently, by using (2) we obtain that ½ n  n j;j ¼ 1. We now consider the case j–k. We have

½ n  n j;k ¼

n X ½ n j;h ½ n k;h ¼ h¼1

 n n  2 X jhp khp 1 X ðj  kÞhp ðj þ kÞhp sin cos sin ¼  cos : n þ 1 h¼1 nþ1 n þ 1 n þ 1 h¼1 nþ1 nþ1

From (2) n X h¼1

cos

n ðj þ kÞhp X ðj  kÞhp cos ¼ ¼ nþ1 nþ1 h¼1

and therefore ½ n  n j;k ¼ 0.



1 if j þ k is even; 0

if j þ k is odd;

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Step 3. Obviously,

½Kn  n j;k ¼

n X ½Kn j;h ½ n h;k ¼ ½Kn j;j ½ n j;k ¼ h¼1

rffiffiffiffiffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffij 2 a1 jkp sin ¼ ½W n j;k : nþ1 a1 nþ1

To finish the proof we need to show that An ¼ W n Dn W 1 n . We have

rffiffiffiffiffiffiffiffij X n a1 ½ n Dn  n j;h ½K1 n h;k a1 h¼1 rffiffiffiffiffiffiffiffij rffiffiffiffiffiffiffiffijk rffiffiffiffiffiffiffiffijk X n a1 a1 a1  ¼ ½ n Dn  n j;k ½K1  ¼ ½ D   ¼ ½ n j;h ½Dn  n h;k n n n j;k n k;k a1 a1 a1 h¼1 rffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffijk X rffiffiffiffiffiffiffiffijk X  n n  a1 a1 a1 hp ¼ ½ n j;h ah ½ n k;h ¼ a0 þ 2a1 cos ½ n j;h ½ n k;h : a1 a1 a1 nþ1 h¼1 h¼1

 1  1 ½W n Dn W 1 n j;k ¼ ½Kn  n Dn  n Kn j;k ¼ ½Kn j;j ½ n Dn  n Kn j;k ¼

Therefore, ½W n Dn W 1 n j;k can be written as

½W n Dn W 1 n j;k ¼

rffiffiffiffiffiffiffiffijk a1 ðL1 þ L2 Þ; a1

ð3Þ

where

L1 ¼ a0

n X ½ n j;h ½ n k;h ¼ a0 ½ n  n j;k ¼ a0 ½In j;k ¼ a0 dj;k

ð4Þ

h¼1

and

rffiffiffiffiffiffiffiffi X rffiffiffiffiffiffiffiffi X a1 n hp 4a1 a1 n hp jhp khp cos cos ½ n j;h ½ n k;h ¼ sin sin a1 h¼1 nþ1 n þ 1 a1 h¼1 nþ1 nþ1 nþ1 rffiffiffiffiffiffiffiffi X   2a1 a1 n hp ðj  kÞhp ðj þ kÞhp ¼ cos cos  cos n þ 1 a1 h¼1 nþ1 nþ1 nþ1 rffiffiffiffiffiffiffiffi"X   X # n  a1 a1 n ðj  k þ 1Þhp ðj  k  1Þhp ðj þ k þ 1Þhp ðj þ k  1Þhp  : cos cos ¼ þ cos þ cos n þ 1 a1 h¼1 nþ1 nþ1 nþ1 nþ1 h¼1

L2 ¼ 2a1

If j  k þ 1–0 and j  k  1–0 then by using (2) we obtain that n X

cos

h¼1

n n n ðj þ k þ 1Þhp X ðj þ k  1Þhp X ðj  k þ 1Þhp X ðj  k  1Þhp cos cos cos ¼ ¼ ¼ ¼ nþ1 n þ 1 n þ 1 nþ1 h¼1 h¼1 h¼1

and therefore L2 ¼ 0. If j  k þ 1 ¼ 0 then from (2) we have n X

cos

h¼1

n n ðj þ k þ 1Þhp X ðj þ k  1Þhp X ðj  k  1Þhp cos cos ¼ ¼ ¼ 1 nþ1 n þ 1 nþ1 h¼1 h¼1

and consequently

L2 ¼

a1 nþ1

! # rffiffiffiffiffiffiffiffi" X rffiffiffiffiffiffiffiffi n a1 a1 1  1  ð1  1Þ ¼ a1 : a1 h¼1 a1

If j  k  1 ¼ 0 then by using (2) we obtain that n X

cos

h¼1

n n ðj þ k þ 1Þhp X ðj þ k  1Þhp X ðj  k þ 1Þhp cos cos ¼ ¼ ¼ 1; nþ1 n þ 1 nþ1 h¼1 h¼1

and therefore

a1 L2 ¼ nþ1

! # rffiffiffiffiffiffiffiffi" rffiffiffiffiffiffiffiffi n X a1 a1 1  ð1  1Þ ¼ a1 1 þ : a1 a 1 h¼1



1 if j þ k þ 1 is even 0

if j þ k þ 1 is odd

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Summarizing

( L2 ¼

a1

qffiffiffiffiffiffi a1 a1

0

if j  k 2 f1; 1g;

ð5Þ

in other case:

By substituting expressions (4) and (5) into (3) we have

½W n Dn W 1 n j;k

that is, ½An j;k ¼

8 a1 > > > a1 > > : 0

½W n Dn W 1 n j;k .

if j  k ¼ 1; if j  k ¼ 0; if j  k ¼ 1; in other case; h

Observe that

rffiffiffiffiffiffiffiffi2    a1     ¼ 1 ()  a1  ¼ 1 () ja1 j ¼ ja1 j: W n W n ¼ In () Kn Kn ¼ In ()  a  a1  1 Therefore, the eigenvector matrix W n of An is unitary if and only if ja1 j ¼ ja1 j. Thus, W n is unitary, e.g., when An is symmetric, skew-symmetric, Hermitian or skew-Hermitian.

3. General expression for the entries of Aqn Let U m ðxÞ be the mth degree Chebyshev polynomial of the second kind, with m 2 N [ f0g:

U m ðxÞ ¼

sin½ðm þ 1Þ arccos x ; sin arccos x

1 6 x 6 1:

ð6Þ

The following result provides a general expression for the entries of Aqn for all q; n 2 N in terms of the Chebyshev polynomials of the second kind. qffiffiffiffiffiffi hp Theorem 2. Consider a1 ; a0 ; a1 2 C; a1 a1 –0 and n 2 N. Let An ¼ tridiagn ða1 ; a0 ; a1 Þ, b ¼ aa11 and kh ¼ 2 cos nþ1 for every 1 6 h 6 n. Then

½Aqn j;k

"   bn2c X bjk knhþ1 ¼ ð4  k2nhþ1 ÞU j1 2ð1 þ ð1Þnþ1 Þaq0 U j1 ð0ÞU k1 ð0Þ þ 2n þ 2 2 h¼1    knhþ1  U k1 ½ða0 þ a1 bknhþ1 Þq þ ð1Þjþk ða0  a1 bknhþ1 Þq  2

for all q 2 N and 1 6 j; k 6 n, where bxc denotes the largest integer less than or equal to x. Proof. By using Theorem 1 we obtain that q q 1 q  1 q  1 jk ½Aqn j;k ¼ ½ðW n Dn W 1 ½ n Dqn  n j;k n Þ j;k ¼ ½W n Dn W n j;k ¼ ½Kn  n Dn  n Kn j;k ¼ ½Kn j;j ½ n Dn  n j;k ½Kn k;k ¼ b   q n n n X X 2bjk X hp jhp khp ¼ bjk ½ n j;h ½Dqn  n h;k ¼ bjk ½ n j;h aqh ½ n k;h ¼ a0 þ 2a1 b cos sin sin : n þ 1 n þ 1 n þ 1 n þ1 h¼1 h¼1 h¼1

Observe that from expression (6) we have

  p sin mh hp nþ1 ¼ U m1 cos ; h p nþ1 sin nþ1

m ¼ j; k;

1 6 h 6 n:

Consequently,

½Aqn j;k ¼

q     n  2bjk X hp hp hp 2 hp U k1 cos : a0 þ 2a1 b cos sin U j1 cos n þ 1 h¼1 nþ1 nþ1 nþ1 nþ1

p hp Since cos nþ1 ¼  cos ðnþ1hÞ , 1 6 h 6 n, then nþ1

knhþ1 ¼ 2 cos

hp ¼ kh ; nþ1

1 6 h 6 n:

ð7Þ

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Therefore,

½Aqn j;k ¼

    n bjk X knhþ1 knhþ1 ða0 þ a1 bknhþ1 Þq ð4  k2nhþ1 ÞU j1 U k1 : 2n þ 2 h¼1 2 2

By using (7), ½Aqn j;k can be written as

½Aqn j;k

¼

8 <

bjk ½S 2nþ2 1

:

bjk ½S 2nþ2 1

þ S2  þ S2 þ

if n is even; 4aq0 U j1 ð0ÞU k1 ð0Þ

if n is odd;

where

    knhþ1 knhþ1 U k1 ; 2 2 h¼1     b2nc X knhþ1 knhþ1 S2 ¼ ða0  a1 bknhþ1 Þq ð4  k2nhþ1 ÞU j1  U k1  : 2 2 h¼1 n

S1 ¼

b2c X

ða0 þ a1 bknhþ1 Þq ð4  k2nhþ1 ÞU j1

ð8Þ

i bjk h S1 þ S2 þ 2ð1 þ ð1Þnþ1 Þaq0 U j1 ð0ÞU k1 ð0Þ : 2n þ 2

ð9Þ

Thus,

½Aqn j;k ¼ Since

U m ðxÞ ¼ ð1Þm U m ðxÞ;

m 2 N [ f0g

ð10Þ

(i.e., the mth degree Chebyshev polynomial of the second kind is an even or odd function, when m is even or odd, respectively), then n

S2 ¼ ð1Þjþk

b 2c X

ða0  a1 bknhþ1 Þq ð4  k2nhþ1 ÞU j1

h¼1

    knhþ1 knhþ1 U k1 : 2 2

ð11Þ

The theorem follows by substituting expressions (8) and (11) into (9). h Corollary 3. Consider a1 ; a0 ; a1 2 C; a1 a1 –0 and n 2 N. Let An ¼ tridiagn ða1 ; a0 ; a1 Þ and b ¼

qffiffiffiffiffiffi

a1 . a1

Then

½Aqn k;j ¼ b2k2j ½Aqn j;k for all q 2 N and 1 6 j; k 6 n. Proof. From the previous theorem we have

½Aqn k;j ½Aqn j;k

¼

bkj bjk

¼ b2k2j : 

From Theorem 2 we can now directly obtain the expression given in [1] for the powers of the Hermitian tridiagonal matrix tridiagn ða1 ; a0 ; a1 Þ. hp for every 1 6 h 6 n. Then Theorem 4. Consider a0 2 R; a1 –0 and n 2 N. Let An ¼ tridiagn ða1 ; a0 ; a1 Þ and kh ¼ 2 cos nþ1

½Aqn j;k

 jk " bn2c X 1 a1 nþ1 q ¼ 2ð1 þ ð1Þ Þa0 U j1 ð0ÞU k1 ð0Þ þ ð4  k2nhþ1 Þ 2n þ 2 ja1 j h¼1      knhþ1 knhþ1 U k1 ½ða0 þ ja1 jknhþ1 Þq þ ð1Þjþk ða0  ja1 jknhþ1 Þq   U j1 2 2

for all q 2 N and 1 6 j; k 6 n. Finally, it should be mentioned that from Theorem 2 we can also obtain the expression given by Rimas for the entries of the powers of the real skew-symmetric tridiagonal matrix tridiagn ð1; 0; 1Þ (see [2 or 3] for the even case and [4 or 5] for the odd case).

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hp Theorem 5. Suppose An ¼ tridiagn ð1; 0; 1Þ with n 2 N. Let kh ¼ 2 cos nþ1 for every 1 6 h 6 n. Then

    b 2c  ð1Þjþk ðq þ k  jÞp X kh kh U k1 kqh 4  k2h U j1 cos nþ1 2 2 2 h¼1 n

½Aqn j;k ¼

for all q 2 N and 1 6 j; k 6 n. Proof. From Theorem 2, by taking b ¼ i and by using expressions (7) and (10), we have

    b 2c 1 þ ð1Þqþjþk X kh kh U k1 : kqh ð4  k2h ÞU j1 2n þ 2 2 2 h¼1 n

½Aqn j;k ¼ ðiÞqþjk

To conclude the proof we need to show that

ðiÞqþjk

1 þ ð1Þqþjþk ðq þ k  jÞp ¼ ð1Þjþk cos : 2 2

p The case when q þ k  j is odd (or equivalently, q þ j þ k is odd) is clear, since 1 þ ð1Þqþjþk ¼ cos ðqþkjÞ ¼ 0. On the other 2 hand, if q þ k  j is even then

ðiÞqþjk

qþjk qþkj qþkj 1 þ ð1Þqþjþk 2 qþjk ¼ ðiÞqþjk ¼ ð1Þqþjk ði Þ 2 ¼ ð1Þ 2 ¼ ð1Þjk ð1Þ 2 ¼ ð1Þjþk ð1Þ 2 2 ðq þ k  jÞp ¼ ð1Þjþk cos :  2

4. Numerical example Let An ¼ tridiagn ða1 ; a0 ; a1 Þ, with a1 ; a0 ; a1 2 C; a1 a1 –0 and n 2 N. Observe that since An is a persymmetric matrix, Aqn is also persymmetric, with q 2 N. From Theorem 2, we can find any positive integer power of An with n 2 N. As an example, we consider here the cases n ¼ 3 and n ¼ 4. If n ¼ 3 then Aq3 ; q 2 N, has the form

0

1

c1 c4 c6 B C Aq3 ¼ @ c2 c5 c4 A c3 c2 c1 and by using Theorem 2 and Corollary 3 we obtain that

 1 q 4a0 þ a2 ðða0 þ a1 baÞq þ ða0  a1 baÞq Þ ; 8 1 c2 ¼ ba3 ðða0 þ a1 baÞq  ða0  a1 baÞq Þ; 8  1

c3 ¼ b2 4aq0 þ a2 ðða0 þ a1 baÞq þ ða0  a1 baÞq Þ ; 8 1 c4 ¼ 2 c2 ; b

c1 ¼

1 8 1

c5 ¼ a4 ðða0 þ a1 baÞq þ ða0  a1 baÞq Þ; c6 ¼ with a ¼

b4

c3 ;

qffiffiffiffiffiffi pffiffiffi 2 and b ¼ aa11 .

If n ¼ 4 then Aq4 ; q 2 N, has the form

0

c1 Bc B 2 q A4 ¼ B @ c3 c4

c5 c6 c7 c3

1

c8 c10 c9 c8 C C C; c6 c5 A c2 c1

J. Gutiérrez-Gutiérrez / Applied Mathematics and Computation 206 (2008) 885–891

891

and from Theorem 2 and Corollary 3 we deduce that

i 1 h 2 ð4  a2 Þðða0 þ a1 baÞq þ ða0  a1 baÞq Þ þ ð4  b Þðða0 þ a1 bbÞq þ ða0  a1 bbÞq Þ ; 10 i 1 h c2 ¼ b að4  a2 Þðða0 þ a1 baÞq  ða0  a1 baÞq Þ þ bð4  b2 Þðða0 þ a1 bbÞq  ða0  a1 bbÞq Þ ; 10 i 1 h c3 ¼ b2 að4  a2 Þðða0 þ a1 baÞq þ ða0  a1 baÞq Þ  bð4  b2 Þðða0 þ a1 bbÞq þ ða0  a1 bbÞq Þ ; 10 i 1 h c4 ¼ b3 ð4  a2 Þðða0 þ a1 baÞq  ða0  a1 baÞq Þ  ð4  b2 Þðða0 þ a1 bbÞq  ða0  a1 bbÞq Þ ; 10 1 c5 ¼ 2 c2 ; b i 1 h 2 2 2 c6 ¼ a ð4  a2 Þðða0 þ a1 baÞq þ ða0  a1 baÞq Þ þ b ð4  b Þðða0 þ a1 bbÞq þ ða0  a1 bbÞq Þ ; 10 i 1 h c7 ¼ b a2 ð4  a2 Þðða0 þ a1 baÞq  ða0  a1 baÞq Þ  b2 ð4  b2 Þðða0 þ a1 bbÞq  ða0  a1 bbÞq Þ ; 10 1 c8 ¼ 4 c3 ; b 1 c9 ¼ 2 c7 ; b 1 c10 ¼ 6 c4 ; b

c1 ¼

with a ¼ 2 cos p5 ; b ¼ 2 cos 25p and b ¼

qffiffiffiffiffiffi

a1 . a1

A different expression for the entries of the powers of A3 and A4 can be found in [7]. Acknowledgments I would like to thank Dr. F.J. Williams for encouraging the writing of this paper and for his helpful comments and criticism. This work was partially supported by the Spanish Ministry of Education and Science, and by the European Regional Development Fund through the MultiMIMO Project TEC2007-68020-C04-03 and the program CONSOLIDER-INGENIO 2010 CSD2008-00010 COMONSENS. References [1] J. Gutiérrez-Gutiérrez, Positive integer powers of certain tridiagonal matrices, Appl. Math. Comput. 202 (2008) 133–140. [2] J. Rimas, On computing of arbitrary positive integer powers for one type of even order skew-symmetric tridiagonal matrices with eigenvalues imaginary axis – I, Appl. Math. Comput. 174 (2006) 997–1000. [3] J. Rimas, On computing of arbitrary positive integer powers for one type of even order skew-symmetric tridiagonal matrices with eigenvalues imaginary axis – II, Appl. Math. Comput. 181 (2006) 1120–1125. [4] J. Rimas, On computing of arbitrary positive integer powers for one type of odd order skew-symmetric tridiagonal matrices with eigenvalues imaginary axis – I, Appl. Math. Comput. 183 (2006) 1378–1380. [5] J. Rimas, On computing of arbitrary positive integer powers for one type of odd order skew-symmetric tridiagonal matrices with eigenvalues imaginary axis – II, Appl. Math. Comput. 186 (2007) 872–878. [6] W.F. Trench, On the eigenvalue problem for Toeplitz band matrices, Linear Algebra Appl. 64 (1985) 199–214. [7] R. Witula, D. Slota, Some phenomenon of the powers of certain tridiagonal and asymmetric matrices, Appl. Math. Comput. 202 (2008) 348–359.

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