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APPLICATIONES MATHEMATICAE
35,1 (2008), pp. 107–120
S. Kouachi (Khenchela and Annaba)
EIGENVALUES AND EIGENVECTORS OF SOME
TRIDIAGONAL MATRICES WITH
NON-CONSTANT DIAGONAL ENTRIES
Abstract. We give explicit expressions for the eigenvalues and eigenvectors of some tridiagonal matrices with non-constant diagonal entries. Our
techniques are based on the theory of recurrent sequences.
1. Introduction. We consider tridiagonal matrices of the form
(1) An =

−α+ b1 c1 0 0 . . . 0
a1 b2 c2 0 . . . 0
0 a2 b3
. . . . . .
...
0 0
. . . . . . . . . 0
...
...
. . . . . . . . . cn−1
0 . . . . . . 0 an−1 −β + bn

,
where aj and cj , j = 1, . . . , n − 1, and α and β are complex numbers. We
suppose that
(2) ajcj = d2, j = 1, . . . , n− 1,
and
(3) bj =
{
b1 if j is odd,
b2 if j is even,
j = 1, . . . , n− 1,
where d, b1 and b2 are complex numbers with d 6= 0.
If σ is a mapping (not necessarily a permutation) from the set of the
integers from 1 to n − 1 into the set N∗ of integers different from zero, we
2000 Mathematics Subject Classification: Primary 15A18.
Key words and phrases: eigenvectors, tridiagonal matrices.
The author was supported by the National Agency of Development and Scientific
Research (ANDRU) under grant 01/06/163/99.
[107] c© Instytut Matematyczny PAN, 2008
108 S. Kouachi
denote by An(σ) the n× n matrix
(1.1) An(σ) =

−α+ b1 cσ1 0 0 . . . 0
aσ1 b2 cσ2 0 . . . 0
0 aσ2 b3
. . . . . .
...
0 0
. . . . . . . . . 0
...
...
. . . . . . . . . cσn−1
0 . . . . . . 0 aσn−1 −β + bn

and by ∆n(σ) = |An(σ) − λIn| its characteristic polynomial. If σ = id,
where id is the identity, then An(id) and its characteristic polynomial ∆n(id)
coincide with An and ∆n respectively. Our aim is to establish the eigenvalues
and the corresponding eigenvectors of the matrices An(σ).
We mention that the matrices An(σ) are of circulant type when α = β =
a1 = a2 = · · · = 0 and all the entries on the subdiagonal are equal. They
are of Toeplitz type when α = β = 0 and all the entries on the subdiagonal
are equal and those on the superdiagonal are also equal (see U. Grenander
and G. Szegö [4]).
When a1 = a2 = · · · = c1 = c2 = · · · = 1, b1 = b2 = −2 and α = β = 0,
the eigenvalues of An have been found by J. F. Elliott [1] and R. T. Gregory
and D. Carney [2] to be
λk = −2 + 2 cos
kπ n+ 1
, k = 1, . . . , n.
When a1 = a2 = · · · = c1 = c2 = · · · = 1, b1 = b2 = −2 and α = 1 and
β = 0 or β = 1, the eigenvalues have been reported to be
λk = −2 + 2 cos
kπ n , k = 1, . . . , n,
and
λk = −2 + 2 cos
2kπ
2n+ 1
, k = 1, . . . , n,
respectively, without proof.
W. C. Yueh [7] has generalized the results of J. F. Elliott [1] and
R. T. Gregory and D. Carney [2] to the case when a1 = a2 = · · · = a,
c1 = c2 = · · · = c, b1 = b2 = b and α = 0, β =

ac or α = 0, β = −

ac or α = −β =

ac or α = β =

ac or α = β = −

ac. He has calculated, in
this case, the eigenvalues and their corresponding eigenvectors
λk = b+ 2

ac cos θk, k = 1, . . . , n,
where θk = 2kπ/(2n+ 1), (2k − 1)π/(2n+ 1), (2k − 1)π/(2n), kπ/n and
(k − 1)π/n, k = 1, . . . , n, respectively.
Eigenvalues and eigenvectors of tridiagonal matrices 109
In [6], we have generalized the results of [7] to more general matrices
of the form (1.1) for any complex constants satisfying condition (2) but
with b1 = b2. We have proved that the eigenvalues remain the same as in
the case when the ai’s and the ci’s are equal but the components of the
eigenvector u(k)(σ) associated to the eigenvalue λk, which we denote by
u (k)
j (σ), j = 1, . . . , n, are of the form
u (k)
j (σ) = (−d)
1−jaσ1 . . . aσj−1u
(k)
1
× d sin (n− j + 1)θk − β sin (n− j)θk
d sinnθk − β sin (n− 1)θk
, j = 1, . . . , n,
where θk is given by the formula
d2 sin (n+ 1)θk − d(α+ β) sinnθk + αβ sin (n− 1)θk = 0, k = 1, . . . , n.
Recently [4], we generalized the above results to tridiagonal matrices (1.1)
satisfying
ajcj =
{
d21 if j is odd,
d22 if j is even,
j = 1, 2, . . . ,
where d1 and d2 are complex numbers, but we always required the diagonal
entries to be equal. We have given explicit eigenvalues for many cases, for
example when n is even and αβ = d22, we have
λk =

b+
d21 + d
2
2 + 2d1d2 cos θk, k = 1, . . . ,m− 1,
b−
d21 + d
2
2 + 2d1d2 cos θk, k = m, . . . , 2m− 2,
b+ (α+ β) +

(α− β)2 + 4d21
2
, k = n− 1,
b+ (α+ β)−

(α− β)2 + 4d21
2
, k = n.
The corresponding eigenvectors u(k)(σ) = (u(k)1 (σ), . . . , u
(k)
n (σ))t, k = 1, . . . ,
n− 2, where t is the transposition symbol, are given by
u (k)
j (σ)
= %j(σ)

(b− λk − β) sin
(
n− j − 1
2
)
θk − β
d1 d2 sin
(
n− j − 1
2
)
θk
when j is odd,
1√
d1d2
[
d1d2 sin
(
n− j
2
+ 1
)
θk + (d22 − β(b− λk)) sin
n− j
2
θk
]
when j is even,
where
%j(σ) = (−

d1d2)n−jaσ1 . . . aσj−1 , j = 1, . . . , n,
and
110 S. Kouachi
θk =
{
2kπ/n, k = 1, . . . ,m− 1,
2(k −m+ 1)π/n, k = m, . . . , 2m− 2.
The eigenvectors u(n−1)(σ) and u(n)(σ) associated respectively with the
eigenvalues λn−1 and λn are given by another formula. The matrices studied
by J. F. Elliott [1], R. T. Gregory and D. Carney [2] and W. C. Yueh [7] are
special cases of those considered by S. Kouachi [6] which are, in their turn,
special cases of those considered in this paper since we allow unequal entries
on the principal diagonal. All the conditions imposed in the above papers
are very restrictive and the techniques used are complicated and are not (in
general) applicable to tridiagonal matrices considered in this paper, even for
small n. For example our techniques are applicable to the 6× 6 matrices
A6 =

5− 4

3 6 0 0 0 0
6 3 18 0 0 0
0 2 5 −4 0 0
0 0 −9 3 5 + i

11 0
0 0 0 5− i

11 5 −18i
0 0 0 0 2i 3− 3

3

,
A′6 =

5− 4

3 9 0 0 0 0
4 3 36 0 0 0
0 1 5 12i 0 0
0 0 −3i 3 4 + 2i

5 0
0 0 0 4− 2i

5 5 −6
0 0 0 0 −6 3− 3

3

and guarantee that they possess the same eigenvalues and in addition they
give their exact expressions (formulas (9.b) below) since conditions (2) and
(3) are satisfied:
λ1, λ4 = 4±
1
2

4 + 16(36) cos2
(
π
6
)
= 4±

109,
λ2, λ5 = 4±
1
2

4 + 16(36) cos2
(
π
3
)
= 4±

37,
λ3, λ6 = 4−
(
2 +
3
2
)√
3
± 1
2

(5− 3)2 + (4

3 + 3

3)2 − 2(5− 3)(4

3− 3

3)
= 4−
(
2 +
3
2
)√
3± 1
2

151− 4

3,
Eigenvalues and eigenvectors of tridiagonal matrices 111
whereas the previous techniques are restricted to the case when the entries
on each diagonal are equal and the direct calculation only gives the characteristic polynomial
P (λ) = λ6 + λ5(7

3− 24) + λ4(93− 139

3)
+ λ3(82

3 + 1072) + λ2(7734

3− 6093)
+ λ(−10 953

3− 1944)− 52 731

3 + 29 295
whose roots are difficult to calculate.
2. The eigenvalues. When α = β = 0, the matrix An(σ) and its
characteristic polynomial will be denoted respectively by A0n(σ) and ∆
0
n(σ),
and in the general case they will be denoted by An and ∆n. We put
(4) Y1Y2 = 4d2 cos2 θ,
where
(4.1) Y1 = b1 − λ and Y2 = b2 − λ.
The main result of this paper is
Theorem 1. The eigenvalues of the matrices A0n(σ) are independent of
the entries (ai, ci, i = 1, . . . , n − 1) and of the mapping σ provided that
conditions (2) and (3) are satisfied and their characteristic determinants
are given by
∆0n(σ) = d
2mY1 sin (2m+ 2)θ
sin 2θ
when n = 2m+ 1,(5.a)
∆0n(σ) = d
2m sin (2m+ 1)θ
sin θ
when n = 2m.(5.b)
Proof. Since the right hand sides of formulas (5.a) and (5.b) are independent of σ, to prove that so also is ∆0n, it suffices to prove these formulas
for σ = id. Then, by expanding ∆0n in terms of its last column and using
(2), (3) and (4.1), we get
∆0n = Y1∆
0
n−1 − d2∆0n−2, n = 3, 4, . . . ,when n = 2m+ 1,(6.a.1)
∆0n = Y2∆
0
n−1 − d2∆0n−2, n = 3, 4, . . . ,when n = 2m.(6.b.1)
Let us begin by proving (5.b). By writing the expressions of ∆0n for n =
2m+2, 2m+1 and 2m, multiplying∆02m+1 and∆
0
2m by Y2 and d
2 respectively
and adding the three resulting equations, we get
(6.b.2) ∆02m+2 = (Y1Y2 − 2d2)∆02m − d4∆02m−2.
We now prove (5.b) by induction on m. For m = 0 and m = 1 the formula is
satisfied. Suppose that it is satisfied for all integers < m. Then from (6.b.2)
and using (4), we get
112 S. Kouachi
∆02m+2 = d
2m+2
[
(4 cos2 θ − 2) sin (2m+ 1)θ − sin (2m− 1)θ
sin θ
]
= d2m+2
[
2 cos 2θ sin (2m+ 1)θ − sin (2m− 1)θ
sin θ
]
.
Using the well known trigonometric formula
(∗) 2 sin η cos ζ = sin(η + ζ) + sin (η − ζ)
for η = (2m+ 1)θ and ζ = 2θ, we deduce (5.b) for n = 2m+ 2.
When n = 2m+ 1, applying (6.b.1) for n = 2m+ 2, we get
(6.a.2) ∆02m+1 =
∆02m+2 + d
2∆02m
Y2
.
By direct application of (5.b) twice, for n = 2m + 2 and n = 2m, to the
right hand side of the last expression, we get
∆02m+1 = d
2m+2 sin (2m+ 3)θ + sin (2m+ 1)θ
Y2 sin θ
.
Using again the trigonometric formula (∗) for η = (2m + 3)θ and ζ =
(2m+ 1)θ, we deduce that
∆02m+1 = 2d
2m+2 sin (2m+ 2)θ cos θ
Y2 sin θ
.
By (4), the last expression becomes
∆02m+1 = d
2mY1 sin (2m+ 2)θ
2 sin θ cos θ
,
which gives (5.a) by applying (∗) for η = ζ = θ.
Theorem 2. When conditions (2) and (3) are satisfied , the eigenvalues
of A0n(σ) are given by
(7) λk =

(b1 + b2)−

(b1 − b2)2 + 16d2 cos2 θk
2
, k = 1, . . . ,m,
(b1 + b2) +

(b1 − b2)2 + 16d2 cos2 θk
2
, k = m+ 1, . . . , 2m,
b1 k = n,
where
θk =

kπ 2m+ 2
, k = 1, . . . ,m,
(k −m)π
2m+ 2
, k = m+ 1, . . . , 2m,
when n = 2m+ 1,(7.a)
θk =

kπ 2m+ 1
, k = 1, . . . ,m,
(k −m)π
2m+ 1
, k = m+ 1, . . . , 2m,
when n = 2n.(7.b)
Eigenvalues and eigenvectors of tridiagonal matrices 113
Proof. In formulas (7), the symbol √ denotes any concrete branch of the
square root. When n = 2m + 1, the eigenvalues are trivial consequences of
formula (5.a) by putting Y1 sin (2m+ 2)θ = 0, which gives, by using (4) and
(4.1),
λ2 − (b1 + b2)λ+ b1b2 − 4d2 cos2 θk = 0, k = 1, . . . ,m,
where θk = kπ/(2m+ 2), k = 1, . . . ,m and λn = b1.
When n = 2m, the same reasoning with sin (2m+ 1)θ = 0 yields (7).
If we suppose that α 6= 0 or β 6= 0, then following the same reasoning
as in S. Kouachi [4] and [5] by expanding ∆n in terms of the first and
last columns and using the linearity of determinants with respect to their
columns, we get
∆n = ∆0n − α|E2n−1| − β|E1n−1|+ αβ
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
Y2 c2 0 . . . 0
a2 Y3
. . . . . .
...
0
. . . . . . . . . 0
...
. . . . . . . . . cn−2
0 . . . 0 an−2 Yn−1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
,
where E1n−1 and E
2
n−1 are (n− 1)× (n− 1) matrices of the form (1),
Ein−1 =

Yi ci 0 . . . 0
ai Yi+1
. . . . . .
...
0
. . . . . . . . . 0
...
. . . . . . . . . cn+i−3
. . . . . . 0 an+i−3 Yn+i−2

, i = 1, 2.
Since all the entries ai on the subdiagonal and ci on the superdiagonal
satisfy conditions (2) and (3), using formulas (5.a) and (5.b) and taking
into account the order of the entries ai, ci and Yi, we deduce
∆n = d2m
Y1 sin (2m+ 2)θ
sin 2θ
− (α+ β)d2m sin (2m+ 1)θ
sin θ
+ αβd2m−2
Y2 sin 2mθ
sin 2θ
= d2m
Y1 sin (2m+ 2)θ − 2(α+ β) cos θ sin (2m+ 1)θ + αβd2 Y2 sin 2mθ
sin 2θ
.
Applying the trigonometric formula (∗) for η = (2m+ 1)θ and ζ = θ, we get
(8.a) ∆n = d2m
[Y1 − (α+ β)] sin(2m+ 2)θ +
[αβ
d2 Y2 − (α+ β)
]
sin 2mθ
sin 2θ
114 S. Kouachi
when n = 2m+ 1, and
∆n = d2m
sin (2m+ 1)θ
sin θ
− αd2m−2 Y2 sin 2mθ
sin 2θ
− βd2m−2 Y1 sin 2mθ
sin 2θ
+ αβd2m−2
sin (2m− 1)θ
sin θ
= d2m−2
2d2 sin(2m+1)θ cos θ−(αY2+βY1) sin 2mθ+2αβ sin(2m−1)θ cos θ
sin 2θ
when n = 2m. Applying (∗) twice, first for η = (2m + 1)θ and ζ = θ and
then for = (2m− 1)θ and ζ = θ, we get
(8.b) ∆n =
d2m−2
d2 sin (2m+ 2)θ−(αY2+βY1−αβ−d2) sin 2mθ+αβ sin (2m−2)θ
sin 2θ
.
If αβ = d2, then by application of (∗) for η = 2mθ and ζ = 2θ, we get
∆n = d2m−2
(2d2 cos 2θ − αY2 − βY1 + 2d2) sin 2mθ
sin 2θ
.
Using (4) and the formula cos 2θ = 2 cos2 θ − 1, we deduce
(8.b.1) ∆n = d2m−2
(Y1Y2 − αY2 − βY1) sin 2mθ
sin 2θ
.
Before proceeding further, let us deduce from formulas (8) a proposition for
the matrix Bn(σ) which is obtained from An(σ) by changing α and β to b1
and b2 respectively.
Proposition 1. When n is even, the eigenvalues of Bn(σ) are the same
as those of An(σ).
We have
Theorem 3. When n = 2m and αβ = d2, if conditions (2) and (3) are
satisfied , then the eigenvalues of all the matrices An(σ) are given by
(9.b) λk =
(b1 + b2)−

(b1 − b2)2 + 16d2 cos2 θk
2
, k = 1, . . . ,m− 1,
(b1 + b2) +

(b1 − b2)2 + 16d2 cos2 θk
2
, k = m, . . . , 2m− 2,
(b1 + b2−α−β)−

(b1 + b2)2 +(α+β)2−2(b1− b2)(α−β)
2
, k = n− 1,
(b1 + b2 − α− β) +

(b1 + b2)2 + (α+ β)2 − 2(b1 − b2)(α− β)
2
, k = n,
Eigenvalues and eigenvectors of tridiagonal matrices 115
where
θk =

kπ 2m , k = 1, . . . ,m− 1,
(k −m+ 1)π
2m , k = m, . . . , 2m− 2.
Proof. By (4.1), the characteristic determinant (8.b.1) becomes
∆n = d2m−2
[λ2 − (b1 + b2 − α− β)λ− (αb2 + βb1 − b1b2)] sin 2mθ
sin 2θ
.
Then the eigenvalues λk, k = 1, . . . , 2m − 2, are trivial by putting 2mθ =
kπ, k = 1, . . . ,m− 1, and using (4). This gives
λ2 − (b1 + b2)λ+ b1b2 − 4d2 cos2 θk = 0, k = 1, . . . ,m− 1.
The eigenvalues λn−1 and λn are deduced by solving the equation
λ2 − (b1 + b2 − α− β)λ− (αb2 + βb1 − b1b2) = 0.
Let us see what formulas (8) say and what they do not say. They say
that if a′i, c

i, i = 1, . . . , n− 1, are other constants satisfying conditions (2)
and (3) and
(1.2) A′n =

−α+ b1 c′1 0 0 . . . 0
a′1 b2 c

2 0 . . . 0
0 a′2 b3
. . . . . .
...
0 0
. . . . . . . . . 0
...
...
. . . . . . . . . c′n−1
0 . . . . . . 0 a′n−1 −β + bn

,
then the matrices An, A′n and An(σ) possess the same characteristic polynomial and hence the same eigenvalues. Therefore we have this immediate
consequence of formulas (4):
Corollary 1. The matrices An, A′n and An(σ) are all similar provided
that all the entries satisfy conditions (2) and (3).
3. The eigenvectors. The components of the eigenvector u(k)(σ), k =
1, . . . , n, associated to the eigenvalue λk, k = 1, . . . , n, which we denote by
u (k)
j , j = 1, . . . , n, are solutions of the linear system of equations
(10)

(−α+ ξ(k)1 )u
(k)
1 + cσ1u
(k)
2 = 0,
aσ1u
(k)
1 + ξ
(k)
2 u
(k)
2 + cσ2u
(k)
3 = 0,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
aσn−1u
(k)
n−1 + (−β + ξ
(k)
n )u
(k)
n = 0,
116 S. Kouachi
where
ξ
(k)
i = Y
(k)
i = bi − λk(4.2)
=
{
b1 − λk if i is odd,
b2 − λk if i is even,
i = 1, . . . , n− 1, k = 1, . . . , n,
given by formulas (4), and θk, k = 1, . . . , n, are solutions of
(11.a) [ξ(k)1 − (α+ β)] sin (2m+ 2)θk +
[
αβ
d2 ξ
(k)
2 − (α+ β)
]
sin 2mθk = 0
when n = 2m+ 1, and
(11.b) d2 sin (2m+ 2)θk − (αξ
(k)
2 + βξ
(k)
1 − αβ − d
2) sin 2mθk
+ αβ sin (2m− 2)θk = 0
when n = 2m.
Since these n equations are linearly dependent, by eliminating the first
equation we obtain a system of n− 1 equations in n− 1 unknowns, written
in matrix form as
(12)

ξ
(k)
2 cσ2 0 . . . 0
aσ2 ξ
(k)
3
. . . . . .
...
0
. . . . . . . . . 0
...
. . . . . . . . . cσn−1
0 . . . 0 aσn−1 −β + ξ
(k)
n 

u (k)
2
u (k)
3
...
...
u (k)
n 
=

−aσ1u
(k)
1
0
...
...
0

.
The determinant of this system is given by formulas (8) for α = 0 and n
replaced by n− 1 and equals
(13.a) ∆(k)n−1 = d
2m−2 d
2 sin (2m+ 2)θk − (βξ
(k)
2 − d2) sin 2mθk
sin 2θk
when n = 2m+ 1, and
(13.b) ∆(k)n−1 = d
2m−2 (ξ
(k)
2 − β) sin 2mθk − β sin (2m− 2)θk
sin 2θk
when n = 2m, and for all k = 1, . . . , n.
(14) u(k)j (σ) =
Γ
(k)
j (σ)

(k)
n−1
, j, k = 1, . . . , n,
where
Eigenvalues and eigenvectors of tridiagonal matrices 117
Γ
(k)
j (σ)
=
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
ξ
(k)
2 cσ2 0 . . . −aσ1u
(k)
1 0 . . . 0
aσ2 ξ
(k)
3
. . . . . . 0 0 . . .
...
0
. . . . . . cσj−2
...
... . . .
...
0 0 aσj−2 ξ
(k)
j−1 0 0 . . .
...
...
...
. . . aσj−1 0 cσj
. . .
...
...
...
. . . . . . 0 ξ(k)j+1
. . . 0
...
... . . .
. . .
... aσj+1
. . . . . . cσn−1
0 . . . . . . . . . 0 0 aσn−1 −β + ξ
(k)
n ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
,
j = 2, . . . , n, k = 1, . . . , n. By permuting the first j − 2 columns with the
(j − 1)th one and using the properties of determinants, we get
(15) u(k)j (σ) = (−1)
j−2 Λ
(k)
j (σ)
∆n−1
, j = 2, . . . , n,
where Λ(k)j (σ) is the determinant of the matrix
C
(k)
j (σ) =
(
T
(k)
j−1(σ) 0
0 S(k)n−j(σ)
)
,
where
T
(k)
j−1(σ) =

−aσ1u
(k)
1 ξ
(k)
2 cσ2 0 · · · 0
0 aσ2
. . . . . . . . .
...
...
. . . . . . . . . . . . 0
... 0
. . . . . . . . . cσj−2
... 0 0
. . . . . . ξ(k)j−1
0 · · · · · · · · · 0 aσj−1

is a supertriangular (j−1)×(j−1) matrix with diagonal (−aσ1u
(k)
1 , aσ2 , . . . ,
aσj−1) and
S
(k)
n−j(σ) =

ξ
(k)
j+1 cσj+1 0 · · · 0
aσj+1
. . . . . . . . .
...
0
. . . . . . . . . 0
...
. . . . . . . . . cσn−1
0 · · · 0 aσn−1 −β + ξ
(k)
n 
,
118 S. Kouachi
is an (n − j) × (n − j) tridiagonal matrix of the form (1.1) and satisfying
conditions (2) and (3). Thus
|C(k)j (σ)| = |T
(k)
j−1(σ)| |S
(k)
n−j(σ)|(16)
= −aσ1 . . . aσj−1u
(k)
1 ∆
(k)
n−j , j = 2, . . . , n and k = 1, . . . , n,
where ∆(k)n−j(σ) is given by formulas (8) for α = 0 and n replaced by n− j:
(17.a) ∆(k)n−j =
 dn−j−2
d2 sin (n−j+2)θk−(βξ
(k)
2 −d2) sin (n−j)θk
sin 2θk
for j odd,
dn−j−1

(k)
1 −β] sin (n−j+1)θk−β sin(n−j−1)θk
sin 2θk
for j even,
when n is odd, and
(17.b) ∆(k)n−j =
 dn−j−1

(k)
2 −β] sin (n−j+1)θk−β sin (n−j−1)θk
sin 2θk
for j odd,
dn−j−2
d2 sin (n−j+2)θk−(βξ
(k)
1 −d2) sin (n−j)θk
sin 2θk
for j even,
when n is even, for all j = 2, . . . , n and k = 1, . . . , n. Using formulas (13)–
(17), we get
(18) u(k)j (σ) = (−1)
j−1aσ1 . . . aσj−1u
(k)
1

(k)
n−j

(k)
n−1
, j = 2, . . . , n, k = 1, . . . , n.
Finally,
(18.a) u(k)j (σ) =
µj(σ)u
(k)
1

d1−j
d2 sin (n−j+2)θk−(βξ
(k)
2 −d2) sin (n−j)θk
d2 sin (n+1)θk−(βξ
(k)
2 −d2) sin (n−1)θk
for j odd,
d2−j

(k)
1 −β) sin(n−j+1)θk−β sin(n−j−1)θk
d2 sin (n+1)θk−(βξ
(k)
2 −d2) sin (n−1)θk
for j even,
when n is odd, and
(18.b) u(k)j (σ) =
µj(σ)u
(k)
1

d1−j

(k)
2 −β) sin(n−j+1)θk−β sin(n−j−1)θk

(k)
2 −β) sinnθk−β sin(n−2)θk
for j odd,
d−j
d2 sin(n−j+2)θk−(βξ
(k)
1 −d2) sin(n−j)θk

(k)
2 −β) sinnθk−β sin(n−2)θk
for j even,
for all j = 2, . . . , n and k = 1, . . . , n, when n is even, where
(†) µj(σ) = (−1)1−jaσ1 . . . aσj−1 , j = 2, . . . , n,
ξ
(k)
1 , ξ
(k)
2 and θk, k = 1, . . . , n, are given by formulas (4) and (11). If we
choose
Eigenvalues and eigenvectors of tridiagonal matrices 119
u (k)
1 (σ) = (−d)
n−1
{
d2 sin(n+ 1)θk − (βξ
(k)
2 − d2) sin(n− 1)θk for n odd,
(ξ(k)2 − β) sinnθk − β sin(n− 2)θk for n even,
and put
(‡) %j(σ) = (−d)n−jaσ1 . . . aσj−1 , j = 2, . . . , n,
we get
Theorem 4. The eigenvectors u(k)(σ) = (u(k)1 (σ), . . . , u
(k)
n (σ))t, k =
1, . . . , n of the matrices An(σ) are given by
(19.a) u(k)j (σ) = %j(σ)

d2 sin(n− j + 2)θk − (βξ
(k)
2 − d2) sin(n− j)θk
for j odd ,
(ξ(k)1 − β)d sin(n− j + 1)θk − βd sin(n− j − 1)θk
for j even,
when n is odd , and
(19.b) u(k)j (σ) = %j(σ)

(ξ(k)2 − β) sin(n− j + 1)θk − β sin(n− j − 1)θk
for j odd ,
d sin(n− j + 2)θk − (βξ
(k)
1 /d− d) sin(n− j)θk
for j even,
when n is even. Here %j(σ), j = 2, . . . , n, is given by (‡) and θk, ξ
(k)
1 and
ξ
(k)
2 , k = 1, . . . , n, are given by formulas (4) and (11).
Acknowledgements. The author thanks the Superior Education Ministry of Algeria and the National Agency of Development and Scientific Research (ANDRU) for their financial support. He also thanks the anonymous
referees for their suggestions.
References
[1] J. F. Elliott, The characteristic roots of certain real symmetric matrices, master
thesis, Univ. of Tennessee, 1953.
[2] R. T. Gregory and D. Carney, A Collection of Matrices for Testing Computational
Algorithms, Wiley-Interscience, 1969.
[3] U. Grenander and G. Szegö, Toeplitz Forms and Their Applications, Univ. of California Press, Berkeley and Los Angeles, 1958.
[4] S. Kouachi, Eigenvalues and eigenvectors of tridiagonal matrices, Electronic J. Linear
Algebra 15 (2006), 115–133.
[5] —, Explicit eigenvalues of tridiagonal matrices, in preparation.
[6] —, Eigenvalues and eigenvectors of several tridiagonal matrices, submitted to Bull.
Belgian Math. Soc.
120 S. Kouachi
[7] W. C. Yueh, Eigenvalues of several tridiagonal matrices, Appl. Math. E-Notes 5
(2005), 66–74.
Centre Universitaire de Khenchela
40100 Algeria
E-mail: kouachi@hotmail.com
Current address:
Laboratoire de Mathématiques
Université d’Annaba
23200 Algeria
Received on 25.1.2008 (1912)

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