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when, (2m+, matrices, where, eigenvalues, tridiagonal, given, eigenvectors, even,, formulas, that, cos2, odd,, sin(n−, Kouachi, 1)θk, An(σ), When, conditions, with, entries, Eigenvalues, have, 2)θk, characteristic, matrix, using, formula, u(k)j, same

S. Kouachi (Khenchela and Annaba)

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 =

a1 b2 c2 0 . . . 0

0 a2 b3

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

aσ1 b2 cσ2 0 . . . 0

0 aσ2 b3

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

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

kπ n , k = 1, . . . , n,

and

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)

1−jaσ1 . . . aσj−1u

(k)

× 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,

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 + 2d1d2 cos θk, k = 1, . . . ,m− 1,

b−

d21 + d

2 + 2d1d2 cos θk, k = m, . . . , 2m− 2,

b+

, k = n− 1,

b+

, 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

θk − β

d1 d2 sin

n− j − 1

when j is odd,

d1d2

d1d2 sin

n− j

θk + (d22 − β(b− λk)) sin

n− j

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

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:

4 + 16(36) cos2

4 + 16(36) cos2

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

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 ∆

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∆

n−1 − d2∆0n−2, n = 3, 4, . . . ,when n = 2m+ 1,(6.a.1)

∆0n = Y2∆

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∆

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

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

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

, k = 1, . . . ,m,

(b1 + b2) +

(b1 − b2)2 + 16d2 cos2 θk

, 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

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

a2 Y3

0 . . . 0 an−2 Yn−1

where E1n−1 and E

n−1 are (n− 1)× (n− 1) matrices of the form (1),

Ein−1 =

Yi ci 0 . . . 0

ai Yi+1

. . . . . . . . . cn+i−3

. . . . . . 0 an+i−3 Yn+i−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 +

d2

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

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

(b1 + b2)−

(b1 − b2)2 + 16d2 cos2 θk

(b1 + b2) +

(b1 − b2)2 + 16d2 cos2 θk

, k = m, . . . , 2m− 2,

(b1 + b2−α−β)−

(b1 + b2)2 +(α+β)2−2(b1− b2)(α−β)

, k = n− 1,

(b1 + b2 − α− β) +

(b1 + b2)2 + (α+ β)2 − 2(b1 − b2)(α− β)

, k = n,

Eigenvalues and eigenvectors of tridiagonal matrices 115

where

θk =

kπ 2m

(k −m+ 1)π

2m , k = m, . . . , 2m− 2.

Proof. By (4.1), the characteristic determinant (8.b.1) becomes

∆n = d2m−2

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

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

a′1 b2 c

0 a′2 b3

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

(k)

1 + cσ1u

(k)

aσ1u

(k)

(k)

2 u

(k)

2 + cσ2u

(k)

aσn−1u

(k)

(k)

n )u

(k)

n = 0,

116 S. Kouachi

where

(k)

i = Y

(k)

i = bi − λk(4.2)

b1 − λk if i is odd,

b2 − λk if i is even,

given by formulas (4), and θk, k = 1, . . . , n, are solutions of

(11.a) [ξ(k)1 − (α+ β)] sin (2m+ 2)θk +

d2

(k)

sin 2mθk = 0

when n = 2m+ 1, and

(11.b) d2 sin (2m+ 2)θk − (αξ

(k)

(k)

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

(k)

aσ2 ξ

(k)

(k)

n

u (k)

u (k)

u (k)

n

−aσ1u

(k)

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) =

(k)

j (σ)

(k)

n−1

, j, k = 1, . . . , n,

where

Eigenvalues and eigenvectors of tridiagonal matrices 117

(k)

j (σ)

(k)

2 cσ2 0 . . . −aσ1u

(k)

aσ2 ξ

(k)

. . . . . . cσj−2

0 0 aσj−2 ξ

(k)

. . . aσj−1 0 cσj

... aσj+1

. . . . . . cσn−1

(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

(k)

j (σ) =

(k)

j−1(σ) 0

0 S(k)n−j(σ)

where

(k)

j−1(σ) =

−aσ1u

(k)

(k)

0 aσ2

. . . . . . ξ(k)j−1

is a supertriangular (j−1)×(j−1) matrix with diagonal (−aσ1u

(k)

aσj−1) and

(k)

n−j(σ) =

(k)

j+1 cσj+1 0 · · · 0

aσj+1

(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)

(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 =

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 =

(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)

(k)

n−j

(k)

n−1

, j = 2, . . . , n, k = 1, . . . , n.

Finally,

(18.a) u(k)j (σ) =

µj(σ)u

(k)

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)

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,

(k)

(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)

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

(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

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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