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Explicit expression for powers of tridiagonal 2-Toeplitz matrix of odd order
Linear Algebra and its Applications 436 (2012) 3493–3506
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Explicit expression for powers of tridiagonal 2-Toeplitz matrix of odd order Jonas Rimas Department of Applied Mathematics, Faculty of Fundamental Sciences, Kaunas University of Technology, Kaunas 51368, Lithuania
ARTICLE INFO
ABSTRACT
Article history: Received 6 September 2011 Accepted 19 December 2011 Available online 13 January 2012
In this paper, we give the eigenvalue decomposition for odd order tridiagonal 2-Toeplitz matrix and derive the explicit expression for integer powers of such matrix. © 2011 Elsevier Inc. All rights reserved.
Submitted by R.A. Brualdi AMS classification: 15A18 15B05 Keywords: Toeplitz matrices Eigenvalues Eigenvectors Chebyshev polynomials
1. Introduction In [1] Gover has defined a tridiagonal 2-Toeplitz matrix of order n as a matrix Bn
aij
= 0 if |i − j| > 1, aij = akl if (i, j) ≡ (k, l) mod 2, i.e. ⎡
Bn
= tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) =
a1 b1 ⎢ ⎢c a b ⎢ 1 2 2 ⎢ ⎢ c2 a1 b1 ⎢ ⎢ ⎢ c1 a2 b2 ⎢ ⎢ ⎢ ⎢ c2 a1 ⎣
⎤
..
. .. .. . .
E-mail address: [email protected] 0024-3795/$ - see front matter © 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2011.12.025
=[aij ] where
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
(1.1)
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and has solved the eigenvalue problem of this matrix. The tridiagonal 2-Toeplitz matrices were analyzed by other authors, as well. The explicit expressions for eigenvalues and eigenvectors of tridiagonal 2-Toeplitz matrix of order n were found in [2]. In [3] the explicit inverse of the tridiagonal 2-Toeplitz matrix of order n was got with use of the theory of orthogonal polynomials. Making use of the results obtained in [1], we give the explicit expression for powers of a tridiagonal 2-Toeplitz matrix of odd order which is some generalization of the expressions of powers of tridiagonal matrices, obtained in [4,5]. The paper is organized as follows: In Section 2 we present expressions of the eigenvalues and eigenvectors of odd order matrix Bn given in (1.1). In Section 3 we introduce the eigenvector matrix T and prove Theorem 3.1 that provides an expression of the inverse matrix T −1 and Theorem 3.2 that gives a general expression for powers of Bn . In Section 4 the numerical example is presented. The paper finishes with a conclusion.
2. Eigenvalues and eigenvectors of tridiagonal 2-Toeplitz matrix of odd order In [1, Theorem 2.3 and Theorem 3.2] Gover proved the following results. Theorem 2.1 [1]. The eigenvalues of the tridiagonal 2-Toeplitz matrix of order n are a1 and the solutions of the quadratic equation
(a1 − λ)(a2 − λ) − b1 c1 − b2 c2 −
√
b1 c1 b2 c2
= 2m + 1 given in (1.1)
αk = 0, k = 1, 2, . . . , m,
(2.1)
where
αk = 2 cos
kπ m+1
.
(2.2)
Theorem 2.2 [1]. The eigenvector of B2m+1 in (1.1) associated with the eigenvalue λk which is a solution of (2.1) is given by 1 1 x = q0 (αk ), − (a1 − λk )p0 (αk ), sq1 (αk ), − s(a1 − λk )p1 (αk ), b1 b1 T 1 (2.3) . . . , − sm−1 (a1 − λk )pm−1 (αk ), sm qm (αk ) b1 and the eigenvector associated with the eigenvalue a1 is given by
T c1 c1 2 c1 m , 0, − , 0, . . . , − , x = 1, 0, − b2 b2 b2 where
s
=
c1 c2 b1 b2
,
(2.4)
(2.5)
pi (y), qi (y) are polynomials, satisfying the three term recurrence formula Fi (y)
= yFi−1 (y) − Fi−2 (y)
(2.6)
with initial conditions p0 (y) = 1, p1 (y) b2 c2
β=
b1 c1
.
= y, q0 (y) = 1, q1 (y) = y + β,
(2.7) (2.8)
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Here and in what follows the superscript T means the operation of transposition. Based on Theorem 2.1 we state following proposition. Theorem 2.3. Consider ai ∈ R, bi > 0, ci > 0 (i = 1, 2). Let Bn = tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) π be tridiagonal 2-Toeplitz matrix (1.1) of order n = 2m + 1 (m ∈ N ) and αk = 2 cos mk+ for every 1 1 k n. Then all eigenvalues of Bn are real, have multiplicity 1 and are given by: ⎧ √ ⎪ a1 − a2 2 ⎪ a1 +a2 − ⎪ + b1 c1 + b2 c2 + b1 c1 b2 c2 αk , if 1 k m, ⎪ 2 2 ⎪ ⎨ λk = a1 , if k = m + 1, (2.9) ⎪ ⎪ 2 ⎪ √ ⎪ a1 +a2 ⎪ a − a 1 2 ⎩ + + b1 c1 + b2 c2 + b1 c1 b2 c2 αk , if m + 2 k n. 2 2 Proof. Rewrite (2.1) in the form
λ2 − (a1 + a2 ) λ + a1 a2 − b1 c1 − b2 c2 − Solving this quadratic equation, we get a1 + a2 a1 − a 2 2
λk1,k2 =
∓
2
2
√
b1 c1 b2 c2
+ b1 c1 + b2 c2 +
αk = 0. √
b1 c1 b2 c2
αk .
(2.10)
αk = 2 cos mk+π1 follows, that for any k = 1, m, m + 2, n αk assumes different values and √ √ |αk | < 2. If bi > 0 and ci > 0 (i = 1, 2), we have b1 c1 + b2 c2 + b1 c1 b2 c2 αk = ( b1 c1 − √ √ 2 b2 c2 ) + (2 + αk ) b1 c1 b2 c2 > 0 (k = 1, m, m + 2, n). So, with any ai ∈ R (i = 1, 2) and for every k = 1, m, m + 2, n λk1,k2 in (2.10) are different and real. Using the equality From
2(n − k + 1)π 2kπ (n − k + 1)π 2kπ = 2 cos = 2 cos 2π − = 2 cos m+1 n+1 n+1 n+1 kπ = 2 cos = αk , m+1
αn−k+1 = 2 cos
which holds when n we get (2.9).
= 2m + 1 (m ∈ N ), and taking into account that a1 is an eigenvalue of B2m+1 ,
Let Ui (y) be the ith degree Chebyshev polynomial of the second kind with i Ui (y)
=
sin((i + 1) arccos y) sin(arccos y)
∈N
{−1, 0} [6]:
, −1 y 1.
Chebyshev polynomial of the second kind satisfies the three term recurrence formula Ui (y)
= 2yUi−1 (y) − Ui−2 (y)
(U−1 (y) = 0, U0 (y) = 1, U1 (y) = 2y)
(2.11)
and the equality
Ui
y 2
=
y 1 . . . 1 y , . .. . . . 1 1 y
where determinant have order i (i
(2.12)
= 1, 2, 3, . . .) [6].
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On the base of Theorem 2.2 we prove following statement. Theorem 2.4. Let ai ∈ R, bi > 0, ci > 0 (i = 1, 2). Then the eigenvector Tk of B2m+1 in (1.1) associated with the eigenvalue λk in (2.9) is given by ⎡ ⎤ α α s0 U0 2k + β U−1 2k ⎢ ⎥ ⎢ ⎥ αk 0 λk −a1 ⎢ ⎥ U s 0 b 2 ⎢ ⎥ 1 ⎢ ⎥ αk αk 1 ⎢ ⎥ + β U s U 1 0 ⎢ ⎥ 2 2 ⎢ ⎥ λ −a α ⎢ ⎥ s1 kb 1 U1 2k ⎢ ⎥ 1 ⎥ , k = 1, m, m + 2, n, Tk = ⎢ ⎢ ⎥ . .. ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ αk αk ⎢ sm−1 U + β Um−2 2 ⎥ ⎢ ⎥ m−1 2 ⎢ ⎥ ⎢ ⎥ αk m−1 λk −a1 U s ⎢ ⎥ m − 1 2 ⎣ b1 ⎦ αk αk m s Um 2 + β Um−1 2 ⎡ 0 ⎤
−
Tk
=
c1
b2 ⎢ ⎢ ⎢ 0 ⎢ 1 ⎢ ⎢ − c1 ⎢ b2 ⎢ ⎢ 0 ⎢ ⎢ 2 ⎢ ⎢ − bc1 2 ⎢ ⎢ .. ⎢ ⎢ . ⎢ ⎢ ⎢ 0 ⎣
−
c1 m b2
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
k
= m + 1;
(2.13)
here β is defined by (2.8). Proof. From (2.6), (2.7) follows y 1 . 1 y .. pi (y) = , qi (y) .. .. . . 1 1 y
=
y+β 1 . 1 y . . , .. .. . . 1 1 y
(2.14)
where both determinants have order i (i = 1, 2, 3, . . .). Taking into account properties of determinants, we get qi (y)
= pi (y) + β pi−1 (y), i = 1, 2, 3, . . .
Using (2.6), (2.7), (2.11), (2.12), (2.14) and (2.15), we write y pi (y) = Ui 2 , y y qi (y) = Ui 2 + β Ui−1 2 ; here i = 0, 1, 2, 3, . . . From (2.3), (2.4) and (2.16) we get (2.13). So, the theorem is proved.
(2.15)
(2.16)
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3. The main result and its applications The powers of odd order matrix Bn (n similarity transformation [7]
= 2m + 1, m ∈ N ) given in (1.1) can be found applying the
(Bn )l = TJ l T −1 , where l T
(3.1)
∈ N, = [T1 , T2 , . . . , Tn ]
(3.2)
is the transforming matrix (the eigenvector matrix), Tk (k corresponding to eigenvalue λk (k = 1, n) in (2.9), J
= 1, n) is the eigenvector (see (2.13)) of Bn
= diag(λ1 , λ2 , . . . , λn )
is the Jordan’s form of the matrix Bn . Concerning the inverse matrix T −1 we prove following theorem. Theorem 3.1. Let ai is given by T −1 where n
∈ R, bi > 0, ci > 0 (i = 1, 2). Then the inverse matrix T −1 of the matrix T in (3.2)
= [H1 , H2 , . . . , Hn ]T , = 2m + 1 (m ∈ N ), ⎡
Hk
=
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
tk s0
=
U0
αk
+ β U−1
αk
⎤T
⎥ ⎥ ⎥ ⎥ c1 2 ⎥ ⎥ tk αk αk ⎥ U1 2 + β U 0 2 s1 ⎥ ⎥ tk λk −a1 αk ⎥ U ⎥ 1 1 c1 2 s ⎥ ⎥ .. ⎥ ⎥ . ⎥ ⎥ tk αk αk ⎥ U + β U m − 1 m − 2 ⎥ m − 1 2 2 s ⎥ ⎥ tk λk −a1 αk ⎥ U m − 1 m − 1 ⎥ c1 2 s ⎦ tk αk αk U + β U m m−1 sm 2 2 tk s0
b 0 t − c1 2 ⎢ k ⎢ ⎢ 0 ⎢ ⎢ 1 ⎢ ⎢ tk − b1 c2 ⎢ ⎢ ⎢ 0 ⎢ ⎢ 2 ⎢ ⎢ tk − b1 ⎢ c2 ⎢ ⎢ .. ⎢ . ⎢ ⎢ ⎢ ⎢ 0 ⎢ ⎣ b m tk − c 1 ⎡
Hk
(3.3)
2
2
λk −a1
U0
αk
2
, k = 1, m, m + 2, n,
⎤T ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
, k = m + 1,
(3.4)
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J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
tk
=
⎧ 4−αk2 1 ⎪ , ⎪ ⎪ ⎨ n+1 1+β αk +β 2 + (λk −a1 )2 b1 c1 1 ⎪ ⎪ 1− β 2 ⎪ , if k = m + 1, ⎩ 1 1−
if k
= 1, m, m + 2, n, (3.5)
β n+1
αk , s, β and λk are defined by (2.2), (2.5), (2.8) and (2.9), respectively. Proof. The subscript n (n = 2m + 1, m sake of simplicity. From (3.1), when l = 1, we have HB
∈ N ) in Bn while performing the proof we shall omit for the
= J H;
(3.6)
here H
= T −1 .
The matrix H can be presented as follows: T H = H1 H2 · · · Hn ; here Hk = Hk1 Hk2 · · · Hkn (k = 1, n) is row vector (the kth row of the matrix H ) and left eigenvector of the matrix B corresponding to eigenvalue λk (k = 1, n) [7]. Since the left eigenvector is defined with accuracy of constant factor, the row vector Hk can be expressed as Hk
= tk hk (k = 1, n),
(3.7)
where tk is a nonzero constant factor and hk is some left eigenvector of the matrix B corresponding to eigenvalue λk . Write down (3.6) in the expanded form: ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ λ 0 ··· 0 1 H1 ⎢ ⎥ ⎢ H1 ⎥ ⎥ ⎢ ⎢ . ⎥ ⎥ ⎢ . ⎢ ⎥ ⎢H ⎥ ⎢ H2 ⎥ 2⎥ ⎢ 0 λ2 . . .. ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎢ . ⎥. ⎢ . ⎥B=⎢ (3.8) ⎢ ⎥ . ⎢ . ⎥ ⎢ . ⎥ . . ⎢ ⎥ . . . ⎢ . ⎥ ⎢ . ⎥ . . ⎢ ⎥ . 0 ⎦ ⎦ ⎣ ⎣ ⎦ ⎣ Hn Hn 0 · · · 0 λn From (3.8) follows: Hk (B − λk I )
= 0 (k = 1, n),
(3.9)
where the right hand side is the zero row vector, I is the identity matrix. Applying operation of transposition to (3.9), we get (B − λk I )T HkT = 0T (k
(B −
λk I )T hTk
=0
T
= 1, n) and (see (3,7))
(k = 1, n).
(3.10) we present in the expanded form: ⎡ a − λk c1 ⎢ 1 ⎢ b a − λk c2 ⎢ 1 2 ⎢ ⎢ b2 a1 − λk c1 ⎢ ⎢ .. .. ⎢ ⎢ . . b1 ⎢ ⎢ . .. a − λ ⎢ c2 2 k ⎣ b2 a1 − λ k
(3.10) ⎤⎡ ⎥ ⎢ hk1 ⎥⎢ ⎥ ⎢ hk2 ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ .. ⎥⎢ . ⎥⎢ ⎥⎢ ⎥⎣ ⎦ hkn
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
⎡
⎤ 0
=
⎢ ⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢.⎥ ⎢.⎥ ⎢.⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 0
(k = 1, n).
(3.11)
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From (3.1) we similarly can obtain the relation (B − λk I )Tk = 0 (k = 1, n), where the right hand side is the zero column vector. The expanded form of the latter expression is following: ⎤⎡ ⎡ ⎤ ⎡ ⎤ a1 − λ k b1 0 ⎥ ⎢ T1k ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎢ ⎢ ⎥ ⎥ ⎢ T2k ⎥ ⎢ c1 a2 − λ k b2 ⎥ ⎢0⎥ ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ c a − λ b ⎥ ⎢ ⎥ 2 1 k 1 ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎥ ⎢ . ⎥ (k = 1, n). = (3.12) . . ⎥ ⎢ .. ⎥ ⎢ .. .. ⎢.⎥ ⎥⎢ . ⎥ ⎢ c ⎢.⎥ 1 ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ ⎥ ⎢ ⎥ .. ⎥⎢ ⎢ ⎥ ⎢ ⎥ . a2 − λ k ⎥⎣ ⎢ b2 ⎦ ⎣ ⎦ ⎦ ⎣ T 0 nk c2 a1 − λ k Comparing (3.11) and (3.12) we conclude, that expressions of the elements of left eigenvector hk = hk1 hk2 · · · hkn (k = 1, n) can be obtained from the expressions of the corresponding elements T of eigenvector Tk = T1k T2k · · · Tnk (k = 1, n) by interchanging bi and ci (i = 1, 2) in these expressions. Taking this into account and using the relations (3.7) and (2.13), we get (3.4). The constant factor tk (k = 1, n) we shall find by demanding TH = I would be that the equality fulfilled (I is the identity matrix). From this equality follows ni=1 Hki Tik = ni=1 tk hki Tik = 1 (k = 1, n) and tk = n 1h T (k = 1, n). The latter expression can be written as ⎧ ⎨
tk
=⎩
i=1
1 , S1 +S2 1 , if S3
ki ik
if k k
= 1, m, m + 2, n,
(3.13)
= m + 1;
here n+1
=
S1
2 2 αk αk + β Ui−2 , S2 Ui−1 2 2 i=1 n+1
=
S3
2 b1 c1 i−1 b2 c2
i=1
n−1
=
2 (λk − a1 )2
b1 c1
i=1
Ui2−1
αk 2
,
n+1
=
2
i=1
1
β 2i−2
(β is defined by (2.8)). Using formula of sum of geometric progression, we find S3
=
1 − β n1+1 1 − β12
.
(3.14)
Now, we shall find the sums S1 and S2 . These sums can be expressed by S1
= S11 + 2β S12 + β 2 S13
S2
=
(3.15)
and
(λk − a1 )2 b1 c1
S21 ,
(3.16)
where n+1
S11
=
2
i=1
Ui2−1
αk 2
n+1
, S12 =
2
i=1
Ui−1
αk 2
Ui−2
αk 2
n+1
, S13 =
2
i=1
Ui2−2
αk 2
,
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=
S21
2
i=1
Ui2−1
αk
2
.
Since Um (y)
sin(m + 1) arccos y
=
sin arccos y
sin(m + 1)γ
=
(γ = arccos y),
sin γ
we write n+1
=
S11
n+1
2
1
sin (iγ ), 2
sin2 γ i=1
S12
2
1
=
sin2 γ i=1
n+1
=
S13
n−1
2
1
sin(iγ ) sin((i − 1)γ ),
2
sin
sin2 γ i=1
((i − 1)γ ) =
2
1
sin2 γ i=1
sin2 (iγ ),
(3.17)
n−1
=
S21
2
1
sin2 (iγ )
sin2 γ i=1
= S13 .
(3.18)
We shall find sums in (3.17), applying relations [8] N k=1 N k=1
sin2 (kx) sin(kx)
N
=
sin
=
1 sin(Nx)
−
2
2
(N +1)x 2
sin
sin x sin
cos((N
+ 1)x),
(3.19)
Nx 2
(3.20)
x 2
and the equality Uν
αk
=
2
sin
(ν + 1) arccos
αk
sin
(ν + 1) mk+π1 π sin mk+ 1
2
=
α sin arccos 2k
=
sin
(ν + 1) arccos cos mk+π1
sin arccos cos
⎧ ⎪ 1, if ν ⎪ ⎪ ⎨ −1, if ⎪ ⎪ ⎪ ⎩ 0, if ν
=
kπ m+1
= n + 1 = 2m + 2, ν = n − 1 = 2m, =
n−1 2
= m.
Using (3.19), we write ⎛ S11
= =
1 sin2 γ 1 2
sin
⎝
n+1 4 n+1
γ
4
− −
1 sin
2
sin γ
2 1
n+1 γ 2
U n−1 2
αk 2
cos
T n+3 2
n+3
αk
⎞
γ ⎠
2 2
;
here Ti (y)
= cos(i arccos(y)) (−1 y 1)
is the Chebyshev polynomial of ith degree of the first kind. Using relation [6] U n−1 (x)T n+3 (x) 2
2
=
1 2
(Un+1 (x) − 1)
(3.21)
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and (3.21) we get S11 =
1
n+1
1 − cos2 γ
1
−
4
4
αk
Un+1
2
1
+
=
4
1 4 − αk2
n + 2 − Un+1
αk
2
=
n+1
.
4 − αk2 (3.22)
Similarly, we obtain ⎛
=
S13
=
1 sin2 γ 1
⎝
n−1 4 n−1
sin2 γ
4
−
1 sin
2
⎞ γ ⎠
2
αk
T n+1
2
2
n+1
αk
U n−3
cos
sin γ
2 1
−
n−1 γ 2
2
2
.
Taking into account relation [6] U n−3 (x)T n+1 (x) 2
2
=
1 2
(Un−1 (x) − 1)
and (3.21), we have S13
=
1
n−1
1 − cos2 γ
4
−
1 4
αk
Un−1
2
+
1 4
=
1 4 − αk2
αk n − Un−1 2
=
n+1
.
4 − αk2 (3.23)
Now, we shall find S12 . After simple trigonometric transformations, we get n+1
S12
=
2
1
sin2 γ i=1
sin(iγ ) (sin(iγ ) cos γ
n+1
=
2 cos γ
2
sin
γ
i=1
− cos(iγ ) sin γ )
n+1
sin (iγ ) −
2
1
2
2 sin γ i=1
sin(2iγ )
= cos γ S11 −
1 2 sin γ
σ,
where (see (3.20))
n+1
σ =
2
i=1
sin(2iγ )
=
sin
n+3 γ 2
sin
n+1 γ 2
sin γ
.
Taking into account (3.21), we have
S12
= cos γ
n+1
−
1
sin
2 sin γ 4 − αk2 αk n+1 = cos arccos 2 4 − αk2
n+3 γ 2
−
sin
n+1 γ 2
sin γ 1 αk αk U n+1 U n−1 2 2 2 2 2
=
Using (3.15), (3.16), (3.22), (3.23), (3.24) and (3.18), we obtain ! n+1 (λk − a1 )2 2 1 + βαk + β + . S1 + S2 = b1 c1 4 − αk2
αk n + 1 . 2 4 − αk2
(3.24)
(3.25)
From (3.13), (3.14) and (3.25) follows (3.5). This completes the proof. The following result provides a general expression for the elements of the powers of odd order matrix (1.1) in terms of the Chebyshev polynomials of the second kind.
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J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
Theorem 3.2. Consider aν ∈ R, bν > 0, cν > 0 (ν = 1, 2). Let Bn = tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) π be tridiagonal 2-Toeplitz matrix (1.1) of order n = 2m + 1 (m ∈ N ) and αk = 2 cos mk+ for every 1 1 k n. Then ⎧ i−j n ⎪ αk αk αk αk l ⎪ 2 λ t s + β U + β U U U ⎪ i − 1 i − 3 j − 1 j − 3 k ⎪ k=1 (k=m+1) k 2 2 2 2 ⎪ 2 2 2 2 ⎪ ⎪ i−1 j−1 ⎪ ⎪ 2 2 c b ⎪ l ⎪ − c21 , i, j= 1, 3, 5, . . . , n, ⎪ +λm+1 tm+1 − b12 ⎪ ⎪ i−j 2 n ⎪ (λ αk α l ⎪ k −a1 ) ⎪ k=1 (k=m+1) λk tk s 2 U U j−2 2k , i, j = 2, 4, 6, . . . , n − 1, i−2 ⎪ b ⎨ 1 c1 2 2 2 i − j + 1 αk αk λk −a1 α l Bnl = n 2 λ t s + β U U j−2 2k , U i − 1 i − 3 ⎪ k=1 (k=m+1) k k ij 2 2 c 1 ⎪ 2 2 2 ⎪ ⎪ ⎪ ⎪ i = 1, 3, 5, . . . , n; j = 2, 4, 6, . . . , n − 1, ⎪ ⎪ ⎪ ⎪ i−j−1 ⎪ α α ⎪ n ⎪ λlk tk s 2 λkb−1 a1 U i−2 α2k , U j−1 2k + β U j−3 2k ⎪ k = 1 ( k = m + 1 ) ⎪ 2 ⎪ 2 2 ⎪ ⎪ ⎩ i = 2, 4, 6, . . . , n − 1; j = 1, 3, 5, . . . , n (3.26) for all l
∈ N and 1 i, j n, where s, β, λk , and tk are defined by (2.5), (2.8), (2.9) and (3.5), respectively.
Proof. The transforming matrix T and its inverse T −1 (see (3.2), (2.13) and (3.3), (3.4)) we present as follows: T T = L1 L2 · · · Ln , T −1 = τ1 τ2 · · · τn ; here row vectors Li (i ⎡ s
Li
=
Li
=
i−1 2
= 1, n) and column vectors τj (j = 1, n) are given by
U i−1
α1
+ β U i−3
α1
⎤T
⎢ ⎥ 2 2 2 2 ⎢ ⎥ i−1 ⎢ ⎥ α2 α2 2 U i−1 2 + β U i−3 2 ⎢ s ⎥ ⎢ ⎥ 2 2 ⎢ ⎥ .. ⎢ ⎥ . ⎢ ⎥ ⎢ ⎥ i−1 ⎢ ⎥ α α ⎢ s 2 U i−1 2m + β U i−3 2m ⎥ ⎢ ⎥ , i = 1, 3, 5, . . . , n, 2 2 i−1 ⎢ ⎥ ⎢ ⎥ 2 c1 − ⎢ ⎥ b 2 ⎢ i−1 ⎥ ⎢ 2 ⎥ αm+2 αm+2 + β U i−3 U i−1 ⎢s ⎥ 2 2 ⎢ ⎥ 2 2 ⎢ ⎥ . ⎢ ⎥ . ⎢ ⎥ . ⎣ ⎦ i−1 α α n n s 2 U i−1 2 + β U i−3 2 2 2 ⎤T ⎡ i−2 λ1 −a1 α1 s 2 U i−2 b1 2 2 ⎢ ⎥ ⎢ s i−22 λ2 −a1 U i−2 α2 ⎥ ⎢ ⎥ b1 2 2 ⎢ ⎥ .. ⎢ ⎥ ⎢ ⎥ . ⎢ ⎥ i−2 ⎢ ⎥ λ − a α m 1 m ⎢ s 2 ⎥ U i−2 2 ⎢ ⎥ , i = 2, 4, 6, . . . , n − 1, b1 2 ⎢ ⎥ ⎢ ⎥ 0 ⎢ i−2 ⎥ ⎢ ⎥ λ − a1 α m + 2 m + 2 ⎢s 2 ⎥ U i−2 b1 2 ⎢ ⎥ 2 ⎢ ⎥ . ⎢ ⎥ .. ⎣ ⎦ i−2 λ − a α n 1 n s 2 U i−2 2 b 1
2
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
⎡
τj =
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡
τj =
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
t1 j−1 s 2 t2
s
U j−1 2
j−1 s 2
2
U j−1
j−1 s 2
2
αm
+ β U j−3
2
−
αm+2 2
2
tn j−1 s 2
+ β U j−3
.. .
2
U j−1
2
α2
tm+1
tm+2
+ β U j−3
2
tm
α1
2
U j−1 2
t1 j−2 s 2
λ1 −a1
t2
λ2 −a1
b1 c2
j−1
+ β U j−3 2
.. .
αn 2
2
2
+ β U j−3
α1
2
2
1
2
α2
⎤
⎥ ⎥ ⎥ ⎥ ⎥ 2 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ αm ⎥ ⎥ 2 ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ αm+2 ⎥ ⎥ 2 ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ α
2
α2
j
= 1, 3, 5, . . . , n,
n
2
2
U j−2
α1
⎤
⎥ ⎥ ⎥ ⎥ U j − 2 j−2 ⎥ c1 2 2 ⎥ s 2 ⎥ .. ⎥ ⎥ . ⎥ ⎥ ⎥ tm λm −a1 αm U j−2 2 ⎥ j−2 c1 ⎥ 2 s 2 ⎥, ⎥ ⎥ 0 ⎥ ⎥ tm+2 λm+2 −a1 αm+2 ⎥ U j−2 ⎥ j−2 c1 2 ⎥ 2 s 2 ⎥ ⎥ .. ⎥ . ⎥ ⎥ ⎦ tn λn −a1 αn U j−2 2 j−2 c c1
j
= 2, 4, 6, . . . , n − 1.
2
= TJ l T −1 and J l = diag λl1 , λl2 , . . . , λln , we can write ⎡
U j−1
j−1 2
s
Since Bl
Bnl
ij
= Li J l τj =
λl1
t1
j−1 s 2
⎢ ⎢ ⎢ ⎢ λl2 jt−2 1 ⎢ ⎢ s 2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ λlm tj−m1 ⎢ ⎢ s 2 Li ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ l tm+2 ⎢λ ⎢ m+2 j−1 ⎢ s 2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ l tn
λn
s
j−1 2
U j−1
α1
+ β U j−3
α1
⎤
⎥ ⎥ ⎥ ⎥ U j−1 2 + β U j−3 2 ⎥ 2 2 ⎥ ⎥ ⎥ .. ⎥ . ⎥ ⎥ ⎥ αm αm ⎥ U j−1 2 + β U j−3 2 ⎥ 2 2 ⎥ ⎥ j−1 ⎥ 2 b1 l ⎥ λm+1 tm+1 − c2 ⎥ ⎥ ⎥ αm+2 αm+2 ⎥ + β U j−3 U j−1 ⎥ 2 2 2 2 ⎥ ⎥ .. ⎥ ⎥ . ⎥ ⎥ ⎦ αn αn U j−1 2 + β U j−3 2
2
2
2
α2
2
2
2
α2
3503
3504
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506 n
=
k=1(k =m+1)
λlk tk s
+ λlm+1 tm+1
−
⎡
Bnl
ij
= Li J l τj =
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ Li ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
n
=
k=1 (k =m+1)
i, j
Bnl
ij
αk αk αk αk + β U i−3 + β U j−3 U i−1 U j−1 2 2 2 2 2 2 2 2
i−j 2
i−1
c1
2
b2
λl1
s
λl2
t1 j−2 2 t2
s
λlm
j−2 2
tm s
j−2 2
λlm+2 tmj−+22 s
λln λlk tk s
s
2
tn j−2 2
i−j 2
−
b1 c2
λ1 −a1
= Li J l τj =
k=1 (k =m+1)
; i, j = 1, 3, 5, . . . , n,
U j−2
α1
(3.27)
⎤
⎥ ⎥ ⎥ ⎥ c1 2 ⎥ 2 ⎥ .. ⎥ ⎥ . ⎥ ⎥ λm −a1 αm ⎥ U j−2 ⎥ c1 2 2 ⎥ ⎥ ⎥ 0 ⎥ ⎥ λm+2 −a1 αm+2 ⎥ U j−2 ⎥ c1 2 2 ⎥ ⎥ .. ⎥ ⎥ . ⎥ ⎦ λn −a1 αn U j−2 2 c c1
λ2 −a1
2
U j−2
1
(λk − a1 ) b1 c1
λlk tk s
i−j+1 2
2
α2
2
2
= 2, 4, 6, . . . , n − 1, n
j−21
U i−2 2
αk
U j−2
2
2
αk
2
;
αk αk + β U i−3 U i−1 2 2 2 2
λ k − a1 c1
U j−2
(3.28)
αk 2
2
;
= 1, 3, 5, . . . , n; j = 2, 4, 6, . . . , n − 1, (3.29) n i−j−1 λk − a1 α α α k k k = Li J l τj = λlk tk s 2 U i−2 + β U j−3 ; U j−1 i
Bnl
ij
k=1 (k =m+1)
i
c1
2
2
2
2
2
2
= 2, 4, 6, . . . , n − 1; j = 1, 3, 5, . . . , n.
(3.30)
From (3.27) to (3.30) follows (3.26). Thus we complete the proof. The eigenvalue decomposition for the tridiagonal 2-Toeplitz matrix of odd order, presented in the paper, lets us calculate not only powers (as shown in the Theorem 3.2) but also any functions defined on spectrum of this matrix. Proposition 3.1. Consider aν ∈ R, bν > 0, cν > 0 (ν = 1, 2). Let Bn = tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) be tridiagonal 2-Toeplitz matrix (1.1) of order n = 2m + 1 (m ∈ N ) and λk (1 k n) be spectrum of Bn given by (2.9). If f (x) is any function defined on spectrum of Bn , then f (Bn )
= T diag(f (λ1 ), . . . , f (λn )) T −1 ;
(3.31)
here T and T −1 are the eigenvector matrix and its inverse, defined by (3.2) and (3.3), respectively. Proof. See, e.g. [9]. 4. Example We can find the arbitrary positive integer powers of the tridiagonal 2-Toeplitz matrix Bn (n 2m + 1, m ∈ N ) (see (1.1)), taking into account derived expressions.
=
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
For example, if n
λ1 = λ3 =
a1
= 3, we would have: J = diag(λ1 , λ2 , λ3 ),
+ a2 2
a1
+ a2 2
3505
− +
a1
− a2
2
2 a1
− a2
2
2
α1 = α3 = 0, t1 =
+ b1 c1 + b2 c2 , λ2 = a1 , + b1 c1 + b2 c2 ,
1 (λ −a
)2
, t2 =
1 β2 , t3 = , 2 (λ −a )2 2 1+ β 1 + β + 3b c 1 1 1
1 + β 2 + 1b c 1 1 1 0 ⎤ ⎡ α1 α1 0 s U0 2 + β U−1 2 − bc12 s0 U0 α23 + β U−1 α23 ⎥ ⎢ ⎥ ⎢ λ −a α λ −a α ⎥ T =⎢ s0 1b 1 U0 21 0 s0 3b 1 U0 23 1 1 ⎣ 1 ⎦ α1 α1 c1 α3 α3 1 1 s U1 2 + β U 0 2 − b2 s U1 2 + β U 0 2 ⎡ ⎤ 1 1 1 ⎢ ⎥ ⎢ λ1 −a1 ⎥ λ − a1 3 ⎥, =⎢ 0 b1 b1 ⎣ ⎦ c1 s(α1 + β) − b s(α3 + β) 2 ⎤ ⎡ t1 α1 α t1 λ1 −a1 α t α α U0 21 s10 U1 21 + β U0 21 U0 2 + β U−1 21 0 0 b s s 1 ⎢ ⎥ 0 1 ⎢ ⎥ b b T −1 = ⎢ ⎥ t − 1 0 t2 − c 1 ⎣ 2 c2 2 ⎦ α3 α3 t3 λ3 −a1 α3 t3 α3 α3 t3 U0 2 U0 2 + β U−1 2 U1 2 + β U 0 2 b1 s0 s0 s0 ⎡ ⎤ α +β λ −a t t 1 1 t1 1 s ⎢ 1 1 c1 ⎥ ⎢ ⎥ = ⎢ t2 0 −t2 bc21 ⎥ ⎣ ⎦ α +β λ −a t3 t3 3c 1 t3 3 s 1
and ⎡
(B3 )l = TJ l T −1 =
a a a ⎢ 11 12 13 ⎢ ⎢ a21 a22 a23 ⎣ a31 a32 a33
⎤ ⎥ ⎥ ⎥; ⎦
here a11 (l)
= λl1 t1 + λl2 t2 + λl3 t3 , a12 (l) = λl1 t1 λ1c−1 a1 + λl3 t3 λ3c−1 a1 ,
a13 (l)
= λl1 t1 α1 +β − λl2 t2 s1β + λl3 t3 α3 +β , s s
a21 (l)
= λl1 t1 λ1b−1 a1 + λl3 t3 λ3b−1 a1 , a22 (l) = λl1 t1 (λ1b−1 ca11 ) + λl3 t3 (λ3b−1 ca11 ) ,
a23 (l)
= λl1 t1 λ1b−1 a1 α1 +β + λl3 t3 λ3b−1 a1 α3 +β , s s
a31 (l)
= λl1 t1 s(α1 + β) − λl2 t2 βs + λl3 t3 s(α3 + β),
2
a32 (l)
= λl1 t1 s(α1 + β) λ1c−1 a1 + λl3 t3 s(α3 + β) λ3c−1 a1 ,
a33 (l)
= λl1 t1 (α1 + β)2 + λl2 t2 β12 + λl3 t3 (α3 + β)2 ,
s and β are defined by (2.5) and (2.8), respectively, l
∈ N.
2
3506
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
5. Conclusion We have shown that the integer powers of tridiagonal 2-Toeplitz matrix of odd order can be expressed in terms of Chebyshev polynomials of the second kind. When the condition of non singularity of the matrix is satisfied (all eigenvalues are non zero), derived expression can be applied for computing negative integer powers (including inverse matrix). References [1] M.J.C. Gover, The eigenproblem of a tridiagonal 2-Toeplitz matrix, Linear Algebra Appl. 197 (1994) 63–78. [2] R. Alvarez-Nodarse, J. Petronilho, N.R. Quintero, On some k-Toeplitz matrices: algebraic and analytical aspects. Applications, J. Comput. Appl. Math. 184 (2005) 518–537. [3] C.M. da Fonseca, J. Petronilho, Explicit inverses of some tridiagonal matrices, Linear Algebra Appl. 325 (2001) 7–21. [4] J. Rimas, On computing of arbitrary integer powers for one type of symmetric tridiagonal matrices of odd order-I, Appl. Math. Comput. 171 (2005) 1214–1217. [5] J. Rimas, On computing of arbitrary integer powers for one type of symmetric tridiagonal matrices of odd order-II, Appl. Math. Comput. 174 (2006) 676–683. [6] J.C. Mason, D.C. Handscomb, Chebyshev Polynomials, Chapman & Hall/CRC, New York, 2003. [7] R. Horn, C. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1995. [8] A. Jeffrey, D. Zwillinger, I.M. Ryzhik, Table of Integrals, Series and Products, Academic Press, 2007. [9] P. Lankaster, Theory of Matrices, Academic Press, 1969.
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Explicit expression for powers of tridiagonal 2-Toeplitz matrix of odd order Jonas Rimas Department of Applied Mathematics, Faculty of Fundamental Sciences, Kaunas University of Technology, Kaunas 51368, Lithuania
ARTICLE INFO
ABSTRACT
Article history: Received 6 September 2011 Accepted 19 December 2011 Available online 13 January 2012
In this paper, we give the eigenvalue decomposition for odd order tridiagonal 2-Toeplitz matrix and derive the explicit expression for integer powers of such matrix. © 2011 Elsevier Inc. All rights reserved.
Submitted by R.A. Brualdi AMS classification: 15A18 15B05 Keywords: Toeplitz matrices Eigenvalues Eigenvectors Chebyshev polynomials
1. Introduction In [1] Gover has defined a tridiagonal 2-Toeplitz matrix of order n as a matrix Bn
aij
= 0 if |i − j| > 1, aij = akl if (i, j) ≡ (k, l) mod 2, i.e. ⎡
Bn
= tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) =
a1 b1 ⎢ ⎢c a b ⎢ 1 2 2 ⎢ ⎢ c2 a1 b1 ⎢ ⎢ ⎢ c1 a2 b2 ⎢ ⎢ ⎢ ⎢ c2 a1 ⎣
⎤
..
. .. .. . .
E-mail address: [email protected] 0024-3795/$ - see front matter © 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2011.12.025
=[aij ] where
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
(1.1)
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J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
and has solved the eigenvalue problem of this matrix. The tridiagonal 2-Toeplitz matrices were analyzed by other authors, as well. The explicit expressions for eigenvalues and eigenvectors of tridiagonal 2-Toeplitz matrix of order n were found in [2]. In [3] the explicit inverse of the tridiagonal 2-Toeplitz matrix of order n was got with use of the theory of orthogonal polynomials. Making use of the results obtained in [1], we give the explicit expression for powers of a tridiagonal 2-Toeplitz matrix of odd order which is some generalization of the expressions of powers of tridiagonal matrices, obtained in [4,5]. The paper is organized as follows: In Section 2 we present expressions of the eigenvalues and eigenvectors of odd order matrix Bn given in (1.1). In Section 3 we introduce the eigenvector matrix T and prove Theorem 3.1 that provides an expression of the inverse matrix T −1 and Theorem 3.2 that gives a general expression for powers of Bn . In Section 4 the numerical example is presented. The paper finishes with a conclusion.
2. Eigenvalues and eigenvectors of tridiagonal 2-Toeplitz matrix of odd order In [1, Theorem 2.3 and Theorem 3.2] Gover proved the following results. Theorem 2.1 [1]. The eigenvalues of the tridiagonal 2-Toeplitz matrix of order n are a1 and the solutions of the quadratic equation
(a1 − λ)(a2 − λ) − b1 c1 − b2 c2 −
√
b1 c1 b2 c2
= 2m + 1 given in (1.1)
αk = 0, k = 1, 2, . . . , m,
(2.1)
where
αk = 2 cos
kπ m+1
.
(2.2)
Theorem 2.2 [1]. The eigenvector of B2m+1 in (1.1) associated with the eigenvalue λk which is a solution of (2.1) is given by 1 1 x = q0 (αk ), − (a1 − λk )p0 (αk ), sq1 (αk ), − s(a1 − λk )p1 (αk ), b1 b1 T 1 (2.3) . . . , − sm−1 (a1 − λk )pm−1 (αk ), sm qm (αk ) b1 and the eigenvector associated with the eigenvalue a1 is given by
T c1 c1 2 c1 m , 0, − , 0, . . . , − , x = 1, 0, − b2 b2 b2 where
s
=
c1 c2 b1 b2
,
(2.4)
(2.5)
pi (y), qi (y) are polynomials, satisfying the three term recurrence formula Fi (y)
= yFi−1 (y) − Fi−2 (y)
(2.6)
with initial conditions p0 (y) = 1, p1 (y) b2 c2
β=
b1 c1
.
= y, q0 (y) = 1, q1 (y) = y + β,
(2.7) (2.8)
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
3495
Here and in what follows the superscript T means the operation of transposition. Based on Theorem 2.1 we state following proposition. Theorem 2.3. Consider ai ∈ R, bi > 0, ci > 0 (i = 1, 2). Let Bn = tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) π be tridiagonal 2-Toeplitz matrix (1.1) of order n = 2m + 1 (m ∈ N ) and αk = 2 cos mk+ for every 1 1 k n. Then all eigenvalues of Bn are real, have multiplicity 1 and are given by: ⎧ √ ⎪ a1 − a2 2 ⎪ a1 +a2 − ⎪ + b1 c1 + b2 c2 + b1 c1 b2 c2 αk , if 1 k m, ⎪ 2 2 ⎪ ⎨ λk = a1 , if k = m + 1, (2.9) ⎪ ⎪ 2 ⎪ √ ⎪ a1 +a2 ⎪ a − a 1 2 ⎩ + + b1 c1 + b2 c2 + b1 c1 b2 c2 αk , if m + 2 k n. 2 2 Proof. Rewrite (2.1) in the form
λ2 − (a1 + a2 ) λ + a1 a2 − b1 c1 − b2 c2 − Solving this quadratic equation, we get a1 + a2 a1 − a 2 2
λk1,k2 =
∓
2
2
√
b1 c1 b2 c2
+ b1 c1 + b2 c2 +
αk = 0. √
b1 c1 b2 c2
αk .
(2.10)
αk = 2 cos mk+π1 follows, that for any k = 1, m, m + 2, n αk assumes different values and √ √ |αk | < 2. If bi > 0 and ci > 0 (i = 1, 2), we have b1 c1 + b2 c2 + b1 c1 b2 c2 αk = ( b1 c1 − √ √ 2 b2 c2 ) + (2 + αk ) b1 c1 b2 c2 > 0 (k = 1, m, m + 2, n). So, with any ai ∈ R (i = 1, 2) and for every k = 1, m, m + 2, n λk1,k2 in (2.10) are different and real. Using the equality From
2(n − k + 1)π 2kπ (n − k + 1)π 2kπ = 2 cos = 2 cos 2π − = 2 cos m+1 n+1 n+1 n+1 kπ = 2 cos = αk , m+1
αn−k+1 = 2 cos
which holds when n we get (2.9).
= 2m + 1 (m ∈ N ), and taking into account that a1 is an eigenvalue of B2m+1 ,
Let Ui (y) be the ith degree Chebyshev polynomial of the second kind with i Ui (y)
=
sin((i + 1) arccos y) sin(arccos y)
∈N
{−1, 0} [6]:
, −1 y 1.
Chebyshev polynomial of the second kind satisfies the three term recurrence formula Ui (y)
= 2yUi−1 (y) − Ui−2 (y)
(U−1 (y) = 0, U0 (y) = 1, U1 (y) = 2y)
(2.11)
and the equality
Ui
y 2
=
y 1 . . . 1 y , . .. . . . 1 1 y
where determinant have order i (i
(2.12)
= 1, 2, 3, . . .) [6].
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J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
On the base of Theorem 2.2 we prove following statement. Theorem 2.4. Let ai ∈ R, bi > 0, ci > 0 (i = 1, 2). Then the eigenvector Tk of B2m+1 in (1.1) associated with the eigenvalue λk in (2.9) is given by ⎡ ⎤ α α s0 U0 2k + β U−1 2k ⎢ ⎥ ⎢ ⎥ αk 0 λk −a1 ⎢ ⎥ U s 0 b 2 ⎢ ⎥ 1 ⎢ ⎥ αk αk 1 ⎢ ⎥ + β U s U 1 0 ⎢ ⎥ 2 2 ⎢ ⎥ λ −a α ⎢ ⎥ s1 kb 1 U1 2k ⎢ ⎥ 1 ⎥ , k = 1, m, m + 2, n, Tk = ⎢ ⎢ ⎥ . .. ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ αk αk ⎢ sm−1 U + β Um−2 2 ⎥ ⎢ ⎥ m−1 2 ⎢ ⎥ ⎢ ⎥ αk m−1 λk −a1 U s ⎢ ⎥ m − 1 2 ⎣ b1 ⎦ αk αk m s Um 2 + β Um−1 2 ⎡ 0 ⎤
−
Tk
=
c1
b2 ⎢ ⎢ ⎢ 0 ⎢ 1 ⎢ ⎢ − c1 ⎢ b2 ⎢ ⎢ 0 ⎢ ⎢ 2 ⎢ ⎢ − bc1 2 ⎢ ⎢ .. ⎢ ⎢ . ⎢ ⎢ ⎢ 0 ⎣
−
c1 m b2
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
k
= m + 1;
(2.13)
here β is defined by (2.8). Proof. From (2.6), (2.7) follows y 1 . 1 y .. pi (y) = , qi (y) .. .. . . 1 1 y
=
y+β 1 . 1 y . . , .. .. . . 1 1 y
(2.14)
where both determinants have order i (i = 1, 2, 3, . . .). Taking into account properties of determinants, we get qi (y)
= pi (y) + β pi−1 (y), i = 1, 2, 3, . . .
Using (2.6), (2.7), (2.11), (2.12), (2.14) and (2.15), we write y pi (y) = Ui 2 , y y qi (y) = Ui 2 + β Ui−1 2 ; here i = 0, 1, 2, 3, . . . From (2.3), (2.4) and (2.16) we get (2.13). So, the theorem is proved.
(2.15)
(2.16)
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
3497
3. The main result and its applications The powers of odd order matrix Bn (n similarity transformation [7]
= 2m + 1, m ∈ N ) given in (1.1) can be found applying the
(Bn )l = TJ l T −1 , where l T
(3.1)
∈ N, = [T1 , T2 , . . . , Tn ]
(3.2)
is the transforming matrix (the eigenvector matrix), Tk (k corresponding to eigenvalue λk (k = 1, n) in (2.9), J
= 1, n) is the eigenvector (see (2.13)) of Bn
= diag(λ1 , λ2 , . . . , λn )
is the Jordan’s form of the matrix Bn . Concerning the inverse matrix T −1 we prove following theorem. Theorem 3.1. Let ai is given by T −1 where n
∈ R, bi > 0, ci > 0 (i = 1, 2). Then the inverse matrix T −1 of the matrix T in (3.2)
= [H1 , H2 , . . . , Hn ]T , = 2m + 1 (m ∈ N ), ⎡
Hk
=
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
tk s0
=
U0
αk
+ β U−1
αk
⎤T
⎥ ⎥ ⎥ ⎥ c1 2 ⎥ ⎥ tk αk αk ⎥ U1 2 + β U 0 2 s1 ⎥ ⎥ tk λk −a1 αk ⎥ U ⎥ 1 1 c1 2 s ⎥ ⎥ .. ⎥ ⎥ . ⎥ ⎥ tk αk αk ⎥ U + β U m − 1 m − 2 ⎥ m − 1 2 2 s ⎥ ⎥ tk λk −a1 αk ⎥ U m − 1 m − 1 ⎥ c1 2 s ⎦ tk αk αk U + β U m m−1 sm 2 2 tk s0
b 0 t − c1 2 ⎢ k ⎢ ⎢ 0 ⎢ ⎢ 1 ⎢ ⎢ tk − b1 c2 ⎢ ⎢ ⎢ 0 ⎢ ⎢ 2 ⎢ ⎢ tk − b1 ⎢ c2 ⎢ ⎢ .. ⎢ . ⎢ ⎢ ⎢ ⎢ 0 ⎢ ⎣ b m tk − c 1 ⎡
Hk
(3.3)
2
2
λk −a1
U0
αk
2
, k = 1, m, m + 2, n,
⎤T ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
, k = m + 1,
(3.4)
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J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
tk
=
⎧ 4−αk2 1 ⎪ , ⎪ ⎪ ⎨ n+1 1+β αk +β 2 + (λk −a1 )2 b1 c1 1 ⎪ ⎪ 1− β 2 ⎪ , if k = m + 1, ⎩ 1 1−
if k
= 1, m, m + 2, n, (3.5)
β n+1
αk , s, β and λk are defined by (2.2), (2.5), (2.8) and (2.9), respectively. Proof. The subscript n (n = 2m + 1, m sake of simplicity. From (3.1), when l = 1, we have HB
∈ N ) in Bn while performing the proof we shall omit for the
= J H;
(3.6)
here H
= T −1 .
The matrix H can be presented as follows: T H = H1 H2 · · · Hn ; here Hk = Hk1 Hk2 · · · Hkn (k = 1, n) is row vector (the kth row of the matrix H ) and left eigenvector of the matrix B corresponding to eigenvalue λk (k = 1, n) [7]. Since the left eigenvector is defined with accuracy of constant factor, the row vector Hk can be expressed as Hk
= tk hk (k = 1, n),
(3.7)
where tk is a nonzero constant factor and hk is some left eigenvector of the matrix B corresponding to eigenvalue λk . Write down (3.6) in the expanded form: ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ λ 0 ··· 0 1 H1 ⎢ ⎥ ⎢ H1 ⎥ ⎥ ⎢ ⎢ . ⎥ ⎥ ⎢ . ⎢ ⎥ ⎢H ⎥ ⎢ H2 ⎥ 2⎥ ⎢ 0 λ2 . . .. ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎢ . ⎥. ⎢ . ⎥B=⎢ (3.8) ⎢ ⎥ . ⎢ . ⎥ ⎢ . ⎥ . . ⎢ ⎥ . . . ⎢ . ⎥ ⎢ . ⎥ . . ⎢ ⎥ . 0 ⎦ ⎦ ⎣ ⎣ ⎦ ⎣ Hn Hn 0 · · · 0 λn From (3.8) follows: Hk (B − λk I )
= 0 (k = 1, n),
(3.9)
where the right hand side is the zero row vector, I is the identity matrix. Applying operation of transposition to (3.9), we get (B − λk I )T HkT = 0T (k
(B −
λk I )T hTk
=0
T
= 1, n) and (see (3,7))
(k = 1, n).
(3.10) we present in the expanded form: ⎡ a − λk c1 ⎢ 1 ⎢ b a − λk c2 ⎢ 1 2 ⎢ ⎢ b2 a1 − λk c1 ⎢ ⎢ .. .. ⎢ ⎢ . . b1 ⎢ ⎢ . .. a − λ ⎢ c2 2 k ⎣ b2 a1 − λ k
(3.10) ⎤⎡ ⎥ ⎢ hk1 ⎥⎢ ⎥ ⎢ hk2 ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ .. ⎥⎢ . ⎥⎢ ⎥⎢ ⎥⎣ ⎦ hkn
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
⎡
⎤ 0
=
⎢ ⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢.⎥ ⎢.⎥ ⎢.⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 0
(k = 1, n).
(3.11)
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
3499
From (3.1) we similarly can obtain the relation (B − λk I )Tk = 0 (k = 1, n), where the right hand side is the zero column vector. The expanded form of the latter expression is following: ⎤⎡ ⎡ ⎤ ⎡ ⎤ a1 − λ k b1 0 ⎥ ⎢ T1k ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎢ ⎢ ⎥ ⎥ ⎢ T2k ⎥ ⎢ c1 a2 − λ k b2 ⎥ ⎢0⎥ ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ c a − λ b ⎥ ⎢ ⎥ 2 1 k 1 ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎥ ⎢ . ⎥ (k = 1, n). = (3.12) . . ⎥ ⎢ .. ⎥ ⎢ .. .. ⎢.⎥ ⎥⎢ . ⎥ ⎢ c ⎢.⎥ 1 ⎥⎢ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ ⎥ ⎢ ⎥ .. ⎥⎢ ⎢ ⎥ ⎢ ⎥ . a2 − λ k ⎥⎣ ⎢ b2 ⎦ ⎣ ⎦ ⎦ ⎣ T 0 nk c2 a1 − λ k Comparing (3.11) and (3.12) we conclude, that expressions of the elements of left eigenvector hk = hk1 hk2 · · · hkn (k = 1, n) can be obtained from the expressions of the corresponding elements T of eigenvector Tk = T1k T2k · · · Tnk (k = 1, n) by interchanging bi and ci (i = 1, 2) in these expressions. Taking this into account and using the relations (3.7) and (2.13), we get (3.4). The constant factor tk (k = 1, n) we shall find by demanding TH = I would be that the equality fulfilled (I is the identity matrix). From this equality follows ni=1 Hki Tik = ni=1 tk hki Tik = 1 (k = 1, n) and tk = n 1h T (k = 1, n). The latter expression can be written as ⎧ ⎨
tk
=⎩
i=1
1 , S1 +S2 1 , if S3
ki ik
if k k
= 1, m, m + 2, n,
(3.13)
= m + 1;
here n+1
=
S1
2 2 αk αk + β Ui−2 , S2 Ui−1 2 2 i=1 n+1
=
S3
2 b1 c1 i−1 b2 c2
i=1
n−1
=
2 (λk − a1 )2
b1 c1
i=1
Ui2−1
αk 2
,
n+1
=
2
i=1
1
β 2i−2
(β is defined by (2.8)). Using formula of sum of geometric progression, we find S3
=
1 − β n1+1 1 − β12
.
(3.14)
Now, we shall find the sums S1 and S2 . These sums can be expressed by S1
= S11 + 2β S12 + β 2 S13
S2
=
(3.15)
and
(λk − a1 )2 b1 c1
S21 ,
(3.16)
where n+1
S11
=
2
i=1
Ui2−1
αk 2
n+1
, S12 =
2
i=1
Ui−1
αk 2
Ui−2
αk 2
n+1
, S13 =
2
i=1
Ui2−2
αk 2
,
3500
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506 n−1
=
S21
2
i=1
Ui2−1
αk
2
.
Since Um (y)
sin(m + 1) arccos y
=
sin arccos y
sin(m + 1)γ
=
(γ = arccos y),
sin γ
we write n+1
=
S11
n+1
2
1
sin (iγ ), 2
sin2 γ i=1
S12
2
1
=
sin2 γ i=1
n+1
=
S13
n−1
2
1
sin(iγ ) sin((i − 1)γ ),
2
sin
sin2 γ i=1
((i − 1)γ ) =
2
1
sin2 γ i=1
sin2 (iγ ),
(3.17)
n−1
=
S21
2
1
sin2 (iγ )
sin2 γ i=1
= S13 .
(3.18)
We shall find sums in (3.17), applying relations [8] N k=1 N k=1
sin2 (kx) sin(kx)
N
=
sin
=
1 sin(Nx)
−
2
2
(N +1)x 2
sin
sin x sin
cos((N
+ 1)x),
(3.19)
Nx 2
(3.20)
x 2
and the equality Uν
αk
=
2
sin
(ν + 1) arccos
αk
sin
(ν + 1) mk+π1 π sin mk+ 1
2
=
α sin arccos 2k
=
sin
(ν + 1) arccos cos mk+π1
sin arccos cos
⎧ ⎪ 1, if ν ⎪ ⎪ ⎨ −1, if ⎪ ⎪ ⎪ ⎩ 0, if ν
=
kπ m+1
= n + 1 = 2m + 2, ν = n − 1 = 2m, =
n−1 2
= m.
Using (3.19), we write ⎛ S11
= =
1 sin2 γ 1 2
sin
⎝
n+1 4 n+1
γ
4
− −
1 sin
2
sin γ
2 1
n+1 γ 2
U n−1 2
αk 2
cos
T n+3 2
n+3
αk
⎞
γ ⎠
2 2
;
here Ti (y)
= cos(i arccos(y)) (−1 y 1)
is the Chebyshev polynomial of ith degree of the first kind. Using relation [6] U n−1 (x)T n+3 (x) 2
2
=
1 2
(Un+1 (x) − 1)
(3.21)
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
3501
and (3.21) we get S11 =
1
n+1
1 − cos2 γ
1
−
4
4
αk
Un+1
2
1
+
=
4
1 4 − αk2
n + 2 − Un+1
αk
2
=
n+1
.
4 − αk2 (3.22)
Similarly, we obtain ⎛
=
S13
=
1 sin2 γ 1
⎝
n−1 4 n−1
sin2 γ
4
−
1 sin
2
⎞ γ ⎠
2
αk
T n+1
2
2
n+1
αk
U n−3
cos
sin γ
2 1
−
n−1 γ 2
2
2
.
Taking into account relation [6] U n−3 (x)T n+1 (x) 2
2
=
1 2
(Un−1 (x) − 1)
and (3.21), we have S13
=
1
n−1
1 − cos2 γ
4
−
1 4
αk
Un−1
2
+
1 4
=
1 4 − αk2
αk n − Un−1 2
=
n+1
.
4 − αk2 (3.23)
Now, we shall find S12 . After simple trigonometric transformations, we get n+1
S12
=
2
1
sin2 γ i=1
sin(iγ ) (sin(iγ ) cos γ
n+1
=
2 cos γ
2
sin
γ
i=1
− cos(iγ ) sin γ )
n+1
sin (iγ ) −
2
1
2
2 sin γ i=1
sin(2iγ )
= cos γ S11 −
1 2 sin γ
σ,
where (see (3.20))
n+1
σ =
2
i=1
sin(2iγ )
=
sin
n+3 γ 2
sin
n+1 γ 2
sin γ
.
Taking into account (3.21), we have
S12
= cos γ
n+1
−
1
sin
2 sin γ 4 − αk2 αk n+1 = cos arccos 2 4 − αk2
n+3 γ 2
−
sin
n+1 γ 2
sin γ 1 αk αk U n+1 U n−1 2 2 2 2 2
=
Using (3.15), (3.16), (3.22), (3.23), (3.24) and (3.18), we obtain ! n+1 (λk − a1 )2 2 1 + βαk + β + . S1 + S2 = b1 c1 4 − αk2
αk n + 1 . 2 4 − αk2
(3.24)
(3.25)
From (3.13), (3.14) and (3.25) follows (3.5). This completes the proof. The following result provides a general expression for the elements of the powers of odd order matrix (1.1) in terms of the Chebyshev polynomials of the second kind.
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J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
Theorem 3.2. Consider aν ∈ R, bν > 0, cν > 0 (ν = 1, 2). Let Bn = tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) π be tridiagonal 2-Toeplitz matrix (1.1) of order n = 2m + 1 (m ∈ N ) and αk = 2 cos mk+ for every 1 1 k n. Then ⎧ i−j n ⎪ αk αk αk αk l ⎪ 2 λ t s + β U + β U U U ⎪ i − 1 i − 3 j − 1 j − 3 k ⎪ k=1 (k=m+1) k 2 2 2 2 ⎪ 2 2 2 2 ⎪ ⎪ i−1 j−1 ⎪ ⎪ 2 2 c b ⎪ l ⎪ − c21 , i, j= 1, 3, 5, . . . , n, ⎪ +λm+1 tm+1 − b12 ⎪ ⎪ i−j 2 n ⎪ (λ αk α l ⎪ k −a1 ) ⎪ k=1 (k=m+1) λk tk s 2 U U j−2 2k , i, j = 2, 4, 6, . . . , n − 1, i−2 ⎪ b ⎨ 1 c1 2 2 2 i − j + 1 αk αk λk −a1 α l Bnl = n 2 λ t s + β U U j−2 2k , U i − 1 i − 3 ⎪ k=1 (k=m+1) k k ij 2 2 c 1 ⎪ 2 2 2 ⎪ ⎪ ⎪ ⎪ i = 1, 3, 5, . . . , n; j = 2, 4, 6, . . . , n − 1, ⎪ ⎪ ⎪ ⎪ i−j−1 ⎪ α α ⎪ n ⎪ λlk tk s 2 λkb−1 a1 U i−2 α2k , U j−1 2k + β U j−3 2k ⎪ k = 1 ( k = m + 1 ) ⎪ 2 ⎪ 2 2 ⎪ ⎪ ⎩ i = 2, 4, 6, . . . , n − 1; j = 1, 3, 5, . . . , n (3.26) for all l
∈ N and 1 i, j n, where s, β, λk , and tk are defined by (2.5), (2.8), (2.9) and (3.5), respectively.
Proof. The transforming matrix T and its inverse T −1 (see (3.2), (2.13) and (3.3), (3.4)) we present as follows: T T = L1 L2 · · · Ln , T −1 = τ1 τ2 · · · τn ; here row vectors Li (i ⎡ s
Li
=
Li
=
i−1 2
= 1, n) and column vectors τj (j = 1, n) are given by
U i−1
α1
+ β U i−3
α1
⎤T
⎢ ⎥ 2 2 2 2 ⎢ ⎥ i−1 ⎢ ⎥ α2 α2 2 U i−1 2 + β U i−3 2 ⎢ s ⎥ ⎢ ⎥ 2 2 ⎢ ⎥ .. ⎢ ⎥ . ⎢ ⎥ ⎢ ⎥ i−1 ⎢ ⎥ α α ⎢ s 2 U i−1 2m + β U i−3 2m ⎥ ⎢ ⎥ , i = 1, 3, 5, . . . , n, 2 2 i−1 ⎢ ⎥ ⎢ ⎥ 2 c1 − ⎢ ⎥ b 2 ⎢ i−1 ⎥ ⎢ 2 ⎥ αm+2 αm+2 + β U i−3 U i−1 ⎢s ⎥ 2 2 ⎢ ⎥ 2 2 ⎢ ⎥ . ⎢ ⎥ . ⎢ ⎥ . ⎣ ⎦ i−1 α α n n s 2 U i−1 2 + β U i−3 2 2 2 ⎤T ⎡ i−2 λ1 −a1 α1 s 2 U i−2 b1 2 2 ⎢ ⎥ ⎢ s i−22 λ2 −a1 U i−2 α2 ⎥ ⎢ ⎥ b1 2 2 ⎢ ⎥ .. ⎢ ⎥ ⎢ ⎥ . ⎢ ⎥ i−2 ⎢ ⎥ λ − a α m 1 m ⎢ s 2 ⎥ U i−2 2 ⎢ ⎥ , i = 2, 4, 6, . . . , n − 1, b1 2 ⎢ ⎥ ⎢ ⎥ 0 ⎢ i−2 ⎥ ⎢ ⎥ λ − a1 α m + 2 m + 2 ⎢s 2 ⎥ U i−2 b1 2 ⎢ ⎥ 2 ⎢ ⎥ . ⎢ ⎥ .. ⎣ ⎦ i−2 λ − a α n 1 n s 2 U i−2 2 b 1
2
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
⎡
τj =
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡
τj =
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
t1 j−1 s 2 t2
s
U j−1 2
j−1 s 2
2
U j−1
j−1 s 2
2
αm
+ β U j−3
2
−
αm+2 2
2
tn j−1 s 2
+ β U j−3
.. .
2
U j−1
2
α2
tm+1
tm+2
+ β U j−3
2
tm
α1
2
U j−1 2
t1 j−2 s 2
λ1 −a1
t2
λ2 −a1
b1 c2
j−1
+ β U j−3 2
.. .
αn 2
2
2
+ β U j−3
α1
2
2
1
2
α2
⎤
⎥ ⎥ ⎥ ⎥ ⎥ 2 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ αm ⎥ ⎥ 2 ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ αm+2 ⎥ ⎥ 2 ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ α
2
α2
j
= 1, 3, 5, . . . , n,
n
2
2
U j−2
α1
⎤
⎥ ⎥ ⎥ ⎥ U j − 2 j−2 ⎥ c1 2 2 ⎥ s 2 ⎥ .. ⎥ ⎥ . ⎥ ⎥ ⎥ tm λm −a1 αm U j−2 2 ⎥ j−2 c1 ⎥ 2 s 2 ⎥, ⎥ ⎥ 0 ⎥ ⎥ tm+2 λm+2 −a1 αm+2 ⎥ U j−2 ⎥ j−2 c1 2 ⎥ 2 s 2 ⎥ ⎥ .. ⎥ . ⎥ ⎥ ⎦ tn λn −a1 αn U j−2 2 j−2 c c1
j
= 2, 4, 6, . . . , n − 1.
2
= TJ l T −1 and J l = diag λl1 , λl2 , . . . , λln , we can write ⎡
U j−1
j−1 2
s
Since Bl
Bnl
ij
= Li J l τj =
λl1
t1
j−1 s 2
⎢ ⎢ ⎢ ⎢ λl2 jt−2 1 ⎢ ⎢ s 2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ λlm tj−m1 ⎢ ⎢ s 2 Li ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ l tm+2 ⎢λ ⎢ m+2 j−1 ⎢ s 2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ l tn
λn
s
j−1 2
U j−1
α1
+ β U j−3
α1
⎤
⎥ ⎥ ⎥ ⎥ U j−1 2 + β U j−3 2 ⎥ 2 2 ⎥ ⎥ ⎥ .. ⎥ . ⎥ ⎥ ⎥ αm αm ⎥ U j−1 2 + β U j−3 2 ⎥ 2 2 ⎥ ⎥ j−1 ⎥ 2 b1 l ⎥ λm+1 tm+1 − c2 ⎥ ⎥ ⎥ αm+2 αm+2 ⎥ + β U j−3 U j−1 ⎥ 2 2 2 2 ⎥ ⎥ .. ⎥ ⎥ . ⎥ ⎥ ⎦ αn αn U j−1 2 + β U j−3 2
2
2
2
α2
2
2
2
α2
3503
3504
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506 n
=
k=1(k =m+1)
λlk tk s
+ λlm+1 tm+1
−
⎡
Bnl
ij
= Li J l τj =
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ Li ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
n
=
k=1 (k =m+1)
i, j
Bnl
ij
αk αk αk αk + β U i−3 + β U j−3 U i−1 U j−1 2 2 2 2 2 2 2 2
i−j 2
i−1
c1
2
b2
λl1
s
λl2
t1 j−2 2 t2
s
λlm
j−2 2
tm s
j−2 2
λlm+2 tmj−+22 s
λln λlk tk s
s
2
tn j−2 2
i−j 2
−
b1 c2
λ1 −a1
= Li J l τj =
k=1 (k =m+1)
; i, j = 1, 3, 5, . . . , n,
U j−2
α1
(3.27)
⎤
⎥ ⎥ ⎥ ⎥ c1 2 ⎥ 2 ⎥ .. ⎥ ⎥ . ⎥ ⎥ λm −a1 αm ⎥ U j−2 ⎥ c1 2 2 ⎥ ⎥ ⎥ 0 ⎥ ⎥ λm+2 −a1 αm+2 ⎥ U j−2 ⎥ c1 2 2 ⎥ ⎥ .. ⎥ ⎥ . ⎥ ⎦ λn −a1 αn U j−2 2 c c1
λ2 −a1
2
U j−2
1
(λk − a1 ) b1 c1
λlk tk s
i−j+1 2
2
α2
2
2
= 2, 4, 6, . . . , n − 1, n
j−21
U i−2 2
αk
U j−2
2
2
αk
2
;
αk αk + β U i−3 U i−1 2 2 2 2
λ k − a1 c1
U j−2
(3.28)
αk 2
2
;
= 1, 3, 5, . . . , n; j = 2, 4, 6, . . . , n − 1, (3.29) n i−j−1 λk − a1 α α α k k k = Li J l τj = λlk tk s 2 U i−2 + β U j−3 ; U j−1 i
Bnl
ij
k=1 (k =m+1)
i
c1
2
2
2
2
2
2
= 2, 4, 6, . . . , n − 1; j = 1, 3, 5, . . . , n.
(3.30)
From (3.27) to (3.30) follows (3.26). Thus we complete the proof. The eigenvalue decomposition for the tridiagonal 2-Toeplitz matrix of odd order, presented in the paper, lets us calculate not only powers (as shown in the Theorem 3.2) but also any functions defined on spectrum of this matrix. Proposition 3.1. Consider aν ∈ R, bν > 0, cν > 0 (ν = 1, 2). Let Bn = tridiagn (c1 , a1 , b1 ; c2 , a2 , b2 ) be tridiagonal 2-Toeplitz matrix (1.1) of order n = 2m + 1 (m ∈ N ) and λk (1 k n) be spectrum of Bn given by (2.9). If f (x) is any function defined on spectrum of Bn , then f (Bn )
= T diag(f (λ1 ), . . . , f (λn )) T −1 ;
(3.31)
here T and T −1 are the eigenvector matrix and its inverse, defined by (3.2) and (3.3), respectively. Proof. See, e.g. [9]. 4. Example We can find the arbitrary positive integer powers of the tridiagonal 2-Toeplitz matrix Bn (n 2m + 1, m ∈ N ) (see (1.1)), taking into account derived expressions.
=
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
For example, if n
λ1 = λ3 =
a1
= 3, we would have: J = diag(λ1 , λ2 , λ3 ),
+ a2 2
a1
+ a2 2
3505
− +
a1
− a2
2
2 a1
− a2
2
2
α1 = α3 = 0, t1 =
+ b1 c1 + b2 c2 , λ2 = a1 , + b1 c1 + b2 c2 ,
1 (λ −a
)2
, t2 =
1 β2 , t3 = , 2 (λ −a )2 2 1+ β 1 + β + 3b c 1 1 1
1 + β 2 + 1b c 1 1 1 0 ⎤ ⎡ α1 α1 0 s U0 2 + β U−1 2 − bc12 s0 U0 α23 + β U−1 α23 ⎥ ⎢ ⎥ ⎢ λ −a α λ −a α ⎥ T =⎢ s0 1b 1 U0 21 0 s0 3b 1 U0 23 1 1 ⎣ 1 ⎦ α1 α1 c1 α3 α3 1 1 s U1 2 + β U 0 2 − b2 s U1 2 + β U 0 2 ⎡ ⎤ 1 1 1 ⎢ ⎥ ⎢ λ1 −a1 ⎥ λ − a1 3 ⎥, =⎢ 0 b1 b1 ⎣ ⎦ c1 s(α1 + β) − b s(α3 + β) 2 ⎤ ⎡ t1 α1 α t1 λ1 −a1 α t α α U0 21 s10 U1 21 + β U0 21 U0 2 + β U−1 21 0 0 b s s 1 ⎢ ⎥ 0 1 ⎢ ⎥ b b T −1 = ⎢ ⎥ t − 1 0 t2 − c 1 ⎣ 2 c2 2 ⎦ α3 α3 t3 λ3 −a1 α3 t3 α3 α3 t3 U0 2 U0 2 + β U−1 2 U1 2 + β U 0 2 b1 s0 s0 s0 ⎡ ⎤ α +β λ −a t t 1 1 t1 1 s ⎢ 1 1 c1 ⎥ ⎢ ⎥ = ⎢ t2 0 −t2 bc21 ⎥ ⎣ ⎦ α +β λ −a t3 t3 3c 1 t3 3 s 1
and ⎡
(B3 )l = TJ l T −1 =
a a a ⎢ 11 12 13 ⎢ ⎢ a21 a22 a23 ⎣ a31 a32 a33
⎤ ⎥ ⎥ ⎥; ⎦
here a11 (l)
= λl1 t1 + λl2 t2 + λl3 t3 , a12 (l) = λl1 t1 λ1c−1 a1 + λl3 t3 λ3c−1 a1 ,
a13 (l)
= λl1 t1 α1 +β − λl2 t2 s1β + λl3 t3 α3 +β , s s
a21 (l)
= λl1 t1 λ1b−1 a1 + λl3 t3 λ3b−1 a1 , a22 (l) = λl1 t1 (λ1b−1 ca11 ) + λl3 t3 (λ3b−1 ca11 ) ,
a23 (l)
= λl1 t1 λ1b−1 a1 α1 +β + λl3 t3 λ3b−1 a1 α3 +β , s s
a31 (l)
= λl1 t1 s(α1 + β) − λl2 t2 βs + λl3 t3 s(α3 + β),
2
a32 (l)
= λl1 t1 s(α1 + β) λ1c−1 a1 + λl3 t3 s(α3 + β) λ3c−1 a1 ,
a33 (l)
= λl1 t1 (α1 + β)2 + λl2 t2 β12 + λl3 t3 (α3 + β)2 ,
s and β are defined by (2.5) and (2.8), respectively, l
∈ N.
2
3506
J. Rimas / Linear Algebra and its Applications 436 (2012) 3493–3506
5. Conclusion We have shown that the integer powers of tridiagonal 2-Toeplitz matrix of odd order can be expressed in terms of Chebyshev polynomials of the second kind. When the condition of non singularity of the matrix is satisfied (all eigenvalues are non zero), derived expression can be applied for computing negative integer powers (including inverse matrix). References [1] M.J.C. Gover, The eigenproblem of a tridiagonal 2-Toeplitz matrix, Linear Algebra Appl. 197 (1994) 63–78. [2] R. Alvarez-Nodarse, J. Petronilho, N.R. Quintero, On some k-Toeplitz matrices: algebraic and analytical aspects. Applications, J. Comput. Appl. Math. 184 (2005) 518–537. [3] C.M. da Fonseca, J. Petronilho, Explicit inverses of some tridiagonal matrices, Linear Algebra Appl. 325 (2001) 7–21. [4] J. Rimas, On computing of arbitrary integer powers for one type of symmetric tridiagonal matrices of odd order-I, Appl. Math. Comput. 171 (2005) 1214–1217. [5] J. Rimas, On computing of arbitrary integer powers for one type of symmetric tridiagonal matrices of odd order-II, Appl. Math. Comput. 174 (2006) 676–683. [6] J.C. Mason, D.C. Handscomb, Chebyshev Polynomials, Chapman & Hall/CRC, New York, 2003. [7] R. Horn, C. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1995. [8] A. Jeffrey, D. Zwillinger, I.M. Ryzhik, Table of Integrals, Series and Products, Academic Press, 2007. [9] P. Lankaster, Theory of Matrices, Academic Press, 1969.