Explicit inverses of some tridiagonal matrices - Estudo Geral ...

Linear Algebra and its Applications 325 (2001) 7–21
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Explicit inverses of some tridiagonal matrices
C.M. da Fonseca, J. Petronilho ∗
Depto. de Matematica, Faculdade de Ciencias e Technologia, Univ. Coimbra, Apartado 3008, 3000
Coimbra, Portugal
Received 28 April 1999; accepted 26 June 2000
Submitted by G. Heinig
Abstract
We give explicit inverses of tridiagonal 2-Toeplitz and 3-Toeplitz matrices which generalize
some well-known results concerning the inverse of a tridiagonal Toeplitz matrix. © 2001
Elsevier Science Inc. All rights reserved.
AMS classification:
33C45; 42C05
Keywords: Invertibility of matrices; r-Toeplitz matrices; Orthogonal polynomials
1. Introduction
In recent years the invertibility of nonsingular tridiagonal or block tridiagonal
matrices has been quite investigated in different fields of applied linear algebra (for
historical notes see [8]). Several numerical methods, more or less efficient, have risen
in order to give expressions of the entries of the inverse of this kind of matrices.
Though, explicit inverses are known only in a few cases, in particular when the tridiagonal
matrix is symmetric with constant diagonals and subject to some restrictions
(cf. [3,8,10]). Furthermore, Lewis [5] gave a different way to compute other
explicit inverses of nonsymmetric tridiagonals matrices.
In [1], Gover defines a tridiagonal 2-Toeplitz matrix of order n, as a matrix A =
(a ij ),wherea ij = 0if|i − j| > 1anda ij = a kl if (i, j) ≡ (k, l) mod 2, i.e.,
∗ Corresponding author.
E-mail addresses: cmf@mat.uc.pt (C.M. da Fonseca), josep@mat.uc.pt (J. Petronilho).
0024-3795/01/$ - see front matter 2001 Elsevier Science Inc. All rights reserved.
PII:S0024-3795(00)00289-5

8 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
⎛
⎞
a 1 b 1
c 1 a 2 b 2
c 2 a 1 b 1
A =
c 1 a 2 b 2
. (1)
⎜
.
⎝
c 2 a ..
1
⎟
⎠
. .. . ..
Similarly, a tridiagonal 3-Toeplitz matrix (n × n) is of the form
⎛
⎞
a 1 b 1
c 1 a 2 b 2
c 2 a 3 b 3
c 3 a 1 b 1
B =
c 1 a 2 b 2
. (2)
c 2 a 3 b 3
⎜
.
⎝
c 3 a ..
1
⎟
⎠
. .. . ..
Making use of the theory of orthogonal polynomials, we will give the explicit inverse
of tridiagonal 2-Toeplitz and 3-Toeplitz matrices, based on recent results from
[1,6,7]. As a corollary, we will obtain the explicit inverse of a tridiagonal Toeplitz
matrix, i.e., a tridiagonal matrix with constant diagonals (not necessarily symmetric).
Different proofs of the results by Kamps [3,4] involving the sum of all the entries of
the inverse can be simplified.
Throughout this paper, it is assumed that all the matrices are invertible.
n×n
n×n
2. Inverse of a tridiagonal matrix
Let T be an n × n real nonsingular tridiagonal matrix
⎛
⎞
a 1 b 1 0
c 1 a 2 b 2
T =
. c .. . ..
2 . (3)
⎜
⎝
. .. . ⎟
.. bn−1 ⎠
0 c n−1 a n
Denote T −1 = (α ij ). In [5] Lewis proved the following result.
Lemma 2.1. Let T be the matrix (3) and assume that b l /= 0 for l = 1,...,n− 1.
Then
α ji = γ ij α ij when i

where
C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21 9
γ ij =
j−1
∏
l=i
c l
b l
.
Of course this reduces to α ij = α ji , when T is symmetric.
Using this lemma, one can prove that there exist two finite sequences {u i } and
{v i } (i = 1,...,n− 1) such that
⎛
⎞
u 1 v 1 u 1 v 2 u 1 v 3 ··· u 1 v n
γ 12 u 1 v 2 u 2 v 2 u 2 v 3 ··· u 2 v n
T −1 =
γ 13 u 1 v 3 γ 23 u 2 v 3 u 3 v 3 ··· u 3 v n
. (4)
⎜ . . .
⎝
.
. . . ..
. ⎟
. ⎠
γ 1,n u 1 v n γ 2,n u 2 v n γ 3,n u 3 v n ··· u n v n
Since {u i } and {v i } are only defined up to a multiplicative constant, we make u 1 = 1.
The following result shows how to compute {u i } and {v i }. The determination of these
sequences is very similar to the particular case studied in [8], where the reader may
check for details.
Lemma 2.2. (i) The {v i } (i = 1,...,n− 1) can be computed from
where
v 1 = 1 d 1
,
v k =− b k−1
d k
v k−1 , k = 2,...,n,
d n = a n , d i = a i − b ic i
, i = n − 1,...,1.
d i+1
(ii) The {u i } (i = 1,...,n− 1) can be computed from
u n = 1 ,
δ n v n
where
δ 1 = a 1 ,
u k =− b k
δ k
u k+1 , k = n − 1,...,1,
δ i = a i − b i−1c i−1
δ i−1
, i = 2,...,n− 1.
Notice that for centro-symmetric matrices, i.e., a ij = a n+1−i,n+1−j , we have
v i = u n+1−i .
The following proposition is known in the literature. We give a proof based on the
lecture of [8].
Theorem 2.1. Let T be the n × n tridiagonal matrix defined in (3) and assume that
b l /= 0 for l = 1,...,n− 1.Then

10 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
⎧
(−1) ⎪⎨
i+j d j+1 ···d n
b i ···b j−1 if i j,
δ i ···δ n
(T −1 ) ij =
⎪⎩ (−1) i+j d i+1 ···d n
c j ···c i−1 if i>j
δ j ···δ n
(with the convention that empty product equals 1).
Proof. Let us assume that i j.Then
But
(T −1 ) ij = u i v j = (−1) n−i b i ···b n−1
δ i ···δ n v n
× (−1) j−1 b 1 ···b j−1
d 1 ···d j
.
v n = (−1) n−1 b 1 ···b n−1
.
d 1 ···d n
Therefore
(T −1 ) ij =(−1) n−i b i ···b n−1
δ i ···δ n
× (−1) n−1 d 1 ···d n
b 1 ···b n−1
× (−1) j−1 b 1 ···b j−1
d 1 ···d j
=(−1) i+j b i ···b j−1
d j+1 ···d n
δ i ···δ n
. □
If we set
δ i =
θ i
,
θ i−1
θ 0 = 1, θ 1 = a 1 , (5)
we get the recurrence relation
θ i = a i θ i−1 − b i−1 c i−1 θ i−2 ;
and if we now put
d i =
φ i
, φ n+1 = 1, φ n = a n , (6)
φ i+1
we get the recurrence relation
φ i = a i φ i+1 − b i c i φ i+2 .
As a consequence, we will achieve
⎧
⎨(−1) i+j b i ···b j−1 θ i−1 φ j+1 /θ n if i j,
(T −1 ) ij =
(7)
⎩
(−1) i+j c j ···c i−1 θ j−1 φ i+1 /θ n if i>j.
These are the general relations given by Usmani [9], which we will use throghout
this paper. Notice that
θ n = δ 1 ···δ n = det T.

C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21 11
3. Chebyshev polynomials of the second kind
In the next it is useful to consider the set of polynomials {U n } n0 , such that each
U n is of degree exactly n, satisfying the recurrence relation
U n+1 (x) = 2xU n (x) − U n−1 (x), n 1
with initial conditions
U 0 = 1 and U 1 = 2x.
These polynomials are called Chebyshev polynomials of the second kind and they
have the form
sin(n + 1)θ
U n (x) = , where cos θ = x,
sin θ
when |x| < 1. When |x| > 1, they have the form
sinh(n + 1)θ
U n (x) = , where cosh θ = x.
sinh θ
In particular
U n (±1) = (±1) n (n + 1).
In any case, if |x| /= 1, then
U n (x) = rn+1 + − rn+1 −
,
r + − r −
where
r ± = x ±
√
x 2 − 1
are the two solutions of the quadratic equation r 2 − 2xr + 1 = 0.
The following proposition is a slight extension of Lemma 2.5 of Meurant [8].
Lemma 3.1. Fix a positive integer number n and let a and b be nonzero real numbers
such that U i (a/2b) /= 0 for all i = 1, 2,...,n. Consider the recurrence relation
α 1 = a, α i = a − b2
, i = 2,...,n.
α i−1
Under these conditions, if a =±2b, then
α i =
Otherwise
±b(i + 1)
.
i
α i = b ri+1 −
r+ i − ri −
+ − ri+1
,

12 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
where
r ± = a ± √ a 2 − 4b 2
2b
are the two solutions of the quadratic equation r 2 − (a/b)r + 1 = 0.
Proof. Let us assume that a/=±2b (otherwise it is trivial) and set α i = β i /β i−1 .
Then
β i − aβ i−1 + b 2 β i−2 = 0, β 0 = 1, β 1 = a,
hence β i = b i U i (a/2b) and the conclusion follows.
□
4. Inverse of a tridiagonal 2-Toeplitz matrix
In this section we will determine the explicit inverse of a matrix of type (1). By
Theorem 2.1, this is equivalent to determine δ i and d i , i.e., by (7) and transformations
(5) and (6), θ i and φ i . So, to evaluate θ i ,wehave
θ 0 = 1, θ 1 = a 1 ,
θ 2i = a 2 θ 2i−1 − b 1 c 1 θ 2i−2 ,
θ 2i+1 = a 1 θ 2i − b 2 c 2 θ 2i−1 .
Making the change
z i = (−1) i θ i ,
we get the relations
z 0 = 1, z 1 =−a 1 ,
z 2i =−a 2 z 2i−1 − b 1 c 1 z 2i−2 ,
z 2i+1 =−a 1 z 2i − b 2 c 2 z 2i−1.
According to the results in [6],
z i = Q i (0),
where Q i is a polynomial of degree exactly i, defined by the recurrence relation
Q i+1 (x) = (x − ˜β i )Q i (x) −˜γ i Q i−1 (x), (8)
where { ˜β 2j = a 1 ,
˜β 2j+1 = a 2 ,
and
{
˜γ2j = b 2 c 2 ,
˜γ 2j+1 = b 1 c 1 ,
(9)
and initial conditions Q 0 (x) = 1andQ 1 (x) = x − ˜β 0 .

C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21 13
The solution of the recurrence relation (8) with coefficients (9) is
Q 2i (x) = P i [π 2 (x)] , Q 2i+1 (x) = (x − a 1 )Pi ∗ [π 2 (x)] , (10)
where
π 2 (x) = (x − a 1 )(x − a 2 ), (11)
(√ ) ( )
i x −
Pi ∗ (x) = b1 c 1 − b 2 c 2
b1 b 2 c 1 c 2 Ui
2 √ , (12)
b 1 b 2 c 1 c 2
and
(√ ) [ ( )
i x − b1 c 1 − b 2 c 2
P i (x)= b1 b 2 c 1 c 2 U i
2 √ b 1 b 2 c 1 c
( )] 2
x − b1 c 1 − b 2 c 2
+ βU i−1
2 √ , (13)
b 1 b 2 c 1 c 2
with β 2 = b 2 c 2 /b 1 c 1 . We may conclude that
θ i = (−1) i Q i (0).
Now, to evaluate φ i we put
ψ i = φ n+1−i ,
for i = 1,...,n, and we have to distinguish the cases when the order n of the matrix
(1) is odd and when it is even.
First let us suppose that n is odd. Then we have
ψ 0 = 1, ψ 1 = a 1 ,
ψ 2i = a 2 ψ 2i−1 − b 2 c 2 ψ 2i−2 ,
ψ 2i+1 = a 1 ψ 2i − b 1 c 1 ψ 2i−1.
Following the same steps as in the previous case, we obtain
ψ i = (−1) i Q i (0),
i.e.,
φ i = (−1) i Q n+1−i (0),
where Q i (x) is the same as (10) but with β 2 = b 1 c 1 /b 2 c 2 in (13). Notice that if
b 1 c 1 = b 2 c 2 ,thenθ i = φ n+1−i .
If n is even, we get
ψ 0 = 1, ψ 1 = a 2 ,
ψ 2i = a 1 ψ 2i−1 − b 1 c 1 ψ 2i−2 ,
ψ 2i+1 = a 2 ψ 2i − b 2 c 2 ψ 2i−1 .
Therefore
ψ i = (−1) i Q i (0),

14 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
i.e.,
φ i = (−1) i+1 Q n+1−i (0),
where
Q 2i (x) = P i [π 2 (x)] , Q 2i+1 (x) = (x − a 2 )Pi ∗ [π 2 (x)] ,
with π 2 (x), P i (x) and Pi ∗ (x) the same as in (11), (13) and (12), respectively. Observe
that if a 1 = a 2 , then θ i = φ n+1−i .
We have determined completely the θ i ’s and φ i ’s of (7 ) – thus the inverse of the
matrix – in the case of a tridiagonal 2-Toeplitz matrix (1).
Theorem 4.1. Let A be the tridiagonal matrix (1), with a 1 a 2 /= 0 and b 1 b 2 c 1 c 2 > 0.
Put
π 2 (x):=(x − a 1 )(x − a 2 ), β := √ b 2 c 2 /b 1 c 1
and let {Q i (·; α, γ )} be the sequence of polynomials defined by
(√ ) [ ( )
i π2 (x) − b 1 c 1 − b 2 c 2
Q 2i (x; α, γ )= b1 b 2 c 1 c 2 U i
2 √ b 1 b 2 c 1 c
( )] 2
π2 (x) − b 1 c 1 − b 2 c 2
+γU i−1
2 √ b 1 b 2 c 1 c 2
(√ ) ( )
i π2 (x) − b 1 c 1 − b 2 c 2
Q 2i+1 (x; α, γ ) = (x − α) b1 b 2 c 1 c 2 Ui
2 √ ,
b 1 b 2 c 1 c 2
where α and γ are some parameters. Under these conditions,
⎧
⎪⎨ (−1) i+j b
(A −1 p ⌊(j−i)/2⌋
i
) ij =
⎪⎩
(−1) i+j c ⌊(j−i)/2⌋
p j
b ⌊(j−i+1)/2⌋
q i
θ i−1 φ j+1 /θ n if i j,
c ⌊(j−i+1)/2⌋
q j
θ j−1 φ i+1 /θ n if i>j,
(14)
where p l = (3 − (−1) l )/2, q l = (3 + (−1) l )/2, ⌊z⌋ denotes the greater integer
less or equal to the real number z,
θ i = (−1) i Q i (0; a 1 ,β) , (15)
and
⎧
⎨(−1) i Q n+1−i (0; a 1 , 1/β)
φ i =
⎩
(−1) i+1 Q n+1−i (0; a 2 ,β)
if n is odd,
if n is even.
Remark. As we already noticed, under the conditions of Theorem 4.1, if n is odd
and b 1 c 1 = b 2 c 2 , then θ i = φ n+1−i ;andifn is even and a 1 = a 2 , then also θ i =
φ n+1−i .

C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21 15
As a first application, let us consider a classical problem on the inverse of a tridiagonal
Toeplitz matrix. Suppose that a 1 = a 2 = a, b 1 = b 2 = b and c 1 = c 2 = c in
(1), with a/= 0andbc > 0. Therefore, we have the n × n tridiagonal Toeplitz matrix
⎛
⎞
a b 0
c a b
T =
. c a ..
. (16)
⎜
⎝
. .. . ..
⎟
b ⎠
0 c a
According to (14),
⎧
⎨(−1) i+j b j−i θ i−1 φ j+1 /θ n if i j,
(T −1 ) ij =
⎩
(−1) i+j c i−j θ j−1 φ i+1 /θ n if i>j.
It is well-known and we can easily check using elementary trigonometry that
)
U 2i+1 (x) = 2xU i
(2x 2 − 1 .
Since β = 1and
( a 2 ) (
− 2bc
U i = U i
2bc
from (15) we get
(√ ) ( i a
θ i = bc Ui
( ) a 2
2
2 √ − 1)
=
bc
√ ( )
bc a
a U 2i+1
2 √ ,
bc
)
2 √ ,
bc
and taking into account the above remark the φ i ’s can be computed using the θ i ’s,
giving
(√ ) ( )
n+1−i a
φ i = θ n+1−i = bc Un+1−i
2 √ .
bc
Therefore, the next result comes immediately.
Corollary 4.1. Let T be the matrix in (16) and define d :=a/2 √ bc. The inverse is
given by
⎧
(−1) i+j b j−i U i−1 (d)U n−j (d)
( √bc ) j−i+1
if i j,
U n (d)
⎪⎨
(T −1 ) ij =
(−1)
⎪⎩
i+j c i−j U j−1 (d)U n−i (d)
( √bc ) i−j+1
if i>j.
U n (d)

16 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
This result is well-known. It can be deduced, e.g., using results in the book of
Heinig and Rost [2, p. 28]. The next corollary is Theorem 3.1 of Kamps [3].
Corollary 4.2. Let Σ be the n-square matrix
⎛
⎞
a b 0
b a b
Σ =
. b a ..
. (17)
⎜
⎝
. .. . ⎟ .. b ⎠
0 b a
The inverse is given by
⎧
(−1) ⎪⎨
i+j 1 b
(Σ −1 ) ij =
⎪⎩ (−1) i+j 1 b
U i−1 (a/2b)U n−j (a/2b)
U n (a/2b)
U j−1 (a/2b)U n−i (a/2b)
U n (a/2b)
if i j,
if i>j.
We consider now a second application. In [3,4], the matrices of type (17), with
a>0, b/= 0anda>2|b|, arose as the covariance matrix of one-dependent random
variables Y 1 ,...,Y n , with same expectation. Let us consider the least squares
estimator
ˆµ opt = 1t Σ −1 Y
1 t Σ −1 1 ,
where 1 = (1,...,1) and Y = (Y 1 ,...,Y n ) t , which estimates the parameter µ equal
to the common expectation of the Y i ’s, with variance
V ( ) 1
ˆµ opt =
1 t Σ −1 1 .
According to Kamps, the estimator ˆµ opt is the best unbiased estimator based on Y .
So, the sum of all the entries of the inverse of Σ, 1 t Σ −1 1, has an important role in the
determination of this estimator and therefore in the computation of the variance V
( ˆµ opt ). Kamps used in [3] some sum and product formulas involving different kinds
of Chebyshev polynomials. We will prove the same results in a more concise way.
Corollary 4.3. The sum s i of the ith row (or column) of Σ −1 , for i = 1,...,n is
given by
s i = 1 + b(σ 1i + σ 1,n−i+1 )
,
a + 2b
where σ ij = (Σ −1 ) ij , when i j.
Proof. By Corollary 4.2, for i j,

where
C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21 17
(
σ ij = (−1) i+j 1 r
i
+ − r−) ( )
i r n−j+1
+ − r n−j+1
−
(
) ,
b (r + − r − ) r+ n+1 − rn+1 −
r ± = a ± √ a 2 − 4b 2
.
2b
Thus
∑i−1
n∑
s i = σ ji + σ ij .
j=1
j=1
j=i
By the sum of a geometric progression and the fact that
r + r − = 1,
we have
∑i−1
(
r+ i+1 −
)(
and
σ ji = b (σ 1i + σ ii )
a + 2b
+ 1
a + 2b
n∑
σ ij = b ( )
σ 1,n−i+1 − σ ii
a + 2b
j=i
− 1
a + 2b
(r + − r − )
(
r
i
+ − r i −) ( r n−i
)
r+ n−i+1 − r−
n−i+1
(
)
r+ n+1 − rn+1 −
+ − rn−i −
(
(r + − r − ) r+ n+1 − rn+1 −
)
).
Finally
( )(
r+ i+1 − ri+1 −
= (r + − r − )
r+ n−i+1 − r−
n−i+1
(
r+ n+1 − rn+1 −
) ( )( )
− r+ i − ri − r+ n−i − rn−i −
)
. □
The previous theorem says that s i depends only on the first row of Σ −1 . In fact
(as the following result says) the sum of all entries of Σ −1 depends only on σ 11 and
σ 1n .
Corollary 4.4.
1 t Σ −1 1 = n + 2bs 1
a + 2b .
5. Inverse of a tridiagonal 3-Toeplitz matrix
In an analogous way of the 2-Toeplitz matrices, we may ask about the inverse of a
tridiagonal 3-Toeplitz matrix. To solve this case, we have to compute θ i and φ i in (7).

18 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
For θ i ,wehave
θ 0 = 1, θ 1 = a 1 ,
θ 3i = a 3 θ 3i−1 − b 2 c 2 θ 3i−2 ,
θ 3i+1 = a 1 θ 3i − b 3 c 3 θ 3i−1 ,
θ 3i+2 = a 2 θ 3i+1 − b 1 c 1 θ 3i .
Making the change
z i = (−1) i θ i ,
we get the relations
z 0 = 1, z 1 =−a 1 ,
z 3i =−a 3 z 3i−1 − b 2 c 2 z 3i−2 ,
z 3i+1 =−a 1 z 3i − b 3 c 3 z 3i−1 ,
z 3i+2 =−a 2 z 3i+1 − b 1 c 1 z 3i .
According to the results in [7],
z i = Q i (0),
where Q i is a polynomial of degree exactly i, defined according to
Q i+1 (x) = (x − ˜β i )Q i (x) −˜γ i Q i−1 (x),
where
⎧⎨
⎩
˜β 3j = a 1 ,
˜β 3j+1 = a 2 ,
˜β 3j+2 = a 3 ,
and
⎧
⎨ ˜γ 3j = b 3 c 3 ,
˜γ 3j+1 = b 1 c 1 ,
⎩
˜γ 3j+2 = b 2 c 2 .
The solution of the recurrence relation (8) with coefficients (18) is
(18)
Q 3i (x) = P i [π 3 (x)] + b 3 c 3 (x − a 2 ) P i−1 [π 3 (x)] ,
Q 3i+1 (x) = (x − a 1 )P i [π 3 (x)] + b 1 c 1 b 3 c 3 P i−1 [π 3 (x)] ,
Q 3i+2 (x) = [(x − a 1 )(x − a 2 ) − b 1 c 1 ] P i [π 3 (x)] ,
where
∣ x − a 1 1 1 ∣∣∣∣∣
π 3 (x) =
b 1 c 1 x − a 2 1
∣ b 3 c 3 b 2 c 2 x − a 3
and
P i (x) =
(2 √ ) [ ( )]
i x − b1 c 1 − b 2 c 2 − b 3 c 3
b 1 b 2 b 3 c 1 c 2 c 3 U i
2 √ .
b 1 b 2 b 3 c 1 c 2 c 3

C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21 19
We may conclude that
θ i = (−1) i Q i (0).
In order to compute the φ i ’s, we define
ψ i = φ n+1−i
for i = 1,...,n. In this situation, we have to distinguish three cases. If n ≡ 0
(mod 3), then
ψ 0 = 1, ψ 1 = a 3 ,
ψ 3i = a 1 ψ 3i−1 − b 1 c 1 ψ 3i−2 ,
ψ 3i+1 = a 3 ψ 3i − b 3 c 3 ψ 3i−1 ,
ψ 3i+2 = a 2 ψ 3i+1 − b 2 c 2 ψ 3i .
If n ≡ 1(mod3),then
ψ 0 = 1, ψ 1 = a 1 ,
ψ 3i = a 2 ψ 3i−1 − b 2 c 2 ψ 3i−2 ,
ψ 3i+1 = a 1 ψ 3i − b 1 c 1 ψ 3i−1 ,
ψ 3i+2 = a 3 ψ 3i+1 − b 3 c 3 ψ 3i ,
and if n ≡ 2(mod3),then
ψ 0 = 1, ψ 1 = a 2 ,
ψ 3i = a 3 ψ 3i−1 − b 3 c 3 ψ 3i−2 ,
ψ 3i+1 = a 2 ψ 3i − b 2 c 2 ψ 3i−1 ,
ψ 3i+2 = a 1 ψ 3i+1 − b 1 c 1 ψ 3i .
It is clear that these three recurrence relations can be solved by a similar process
as for the computation of the θ i ’s (mutatis mutandis). Following in this way, we can
state the next proposition.
Theorem 5.1. Let B be the tridiagonal matrix (2), with a 1 a 2 a 3 /= 0 and b 1 b 2 b 3 c 1 c 2
c 3 > 0.Let
( )
a b c
π 3
α β γ ; x x − a 1 1
:=
α x − b 1
∣ γ β x − c∣ ,
( )
a b c
P i
α β γ ; x :=2 √ αβγ
( [ ( )
])
1 a b c
×U i
2 √ π 3
αβγ α β γ ; x − (a + b + c) ,

20 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
( ) ( )
a b c
a b c
Q 3i
α β γ ; x := P i
α β γ ; x
( )
a b c
+γ(x− b) P i−1
α β γ ; x ,
( )
( )
a b c
a b c
Q 3i+1
α β γ ; x := (x − a)P i
α β γ ; x
( )
a b c
+αγ P i−1
α β γ ; x ,
( )
( )
a b c
a b c
Q 3i+2
α β γ ; x := [(x − a)(x − b) − α] P i
α β γ ; x ,
where a, b, c, α, β and γ are some parameters, subject to the restriction αβγ > 0.
Under these conditions,
⎧
⎨(−1) i+j β i β i+1 ···β j θ i−1 φ j+1 /θ n if i j,
(B −1 ) ij =
(19)
⎩
(−1) i+j γ j γ j+1 ···γ i θ j−1 φ i+1 /θ n if i>j,
where
β 3l+s+1 = b s+1 , γ 3l+s+1 = c s+1 (s = 0, 1, 2; l = 0, 1, 2,...),
and the θ i ’s and φ i ’s are explicitly given by
( )
θ i = (−1) i a1 a
Q 2 a 3
i ; 0
(20)
b 1 c 1 b 2 c 2 b 3 c 3
and
⎧
( )
(−1) n+1−i a3 a
Q 2 a 1
n+1−i ; 0 if n ≡ 0, (mod 3),
b 2 c 2 b 1 c 1 b 3 c 3
⎪⎨
( )
φ i = (−1) n+1−i a1 a
Q 3 a 2
n+1−i ; 0
b 3 c 3 b 2 c 2 b 1 c 1
( )
⎪⎩ (−1) n+1−i a2 a
Q 1 a 3
n+1−i ; 0
b 1 c 1 b 3 c 3 b 2 c 2
if n ≡ 1, (mod 3),
if n ≡ 2,(mod 3).
Acknowledgments
This work was supported by CMUC (Centro de Matemática da Universidade de
Coimbra). The authors thank the editor as well as the unknown referees for their
helpful remarks which helped to improve the final version of the paper.

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