On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices
On formulae for the determinant of symmetric pentadiagonal Toeplitz matrices
Mohamed Elouafi
Mathematics Subject Classification
⎜ b a ⎜ ⎜⎜ c b Pm = Pm (
We show that the characteristic polynomial of a symmetric pentadiagonal Toeplitz matrix is the product of two polynomials given explicitly in terms of the Chebyshev polynomials. We consider here the problem of finding the determinant of the m × m symmetric pentadiagonal Toeplitz matrix b

1 Introduction
⎛ a
0
. . .
. . .
. . .
b
c
· · · 0 ⎞
. . . ... ⎟
. . ⎟⎟
. 0 ⎟⎟ .
b c ⎟⎟
⎟
a b ⎠⎟
b a
This class of matrices arises naturally in many applications, such as signal processing, trigonometric
moment problems, integral equations and elliptic partial differential equations with boundary conditions [
9
].
Computing the determinant of the matrix Pm have intrigued the researchers for decades. If c = 0, then Pm is
reduced to a tridiagonal matrix and there exists a closed form of det (Pm ) from which the eigenvalues of the
matrix are explicitly given. It is becoming a challenge to find similar formulae for the general case and so
far, little is known about the eigenvlaues of Pm [
1,2,5,8
]. In [
5,7
], det (Pm ) is explicitly computed using the
kernel of the Chebyshev polynomials {Tn} , {Un} , {Vn} and {Wn} [
11
] and, as a consequence, the eigenvalues
of the matrix Pm are localized by means of explicitly given rational functions. The formulae are simplified to
give det(Pm ) as polynomials of the parameters a, b, c [
6
].
In the new formula presented here, det(Pm ) is given as the product of two polynomials given in a standard
form. Here is our main result:
Theorem 1.1 We have
and
where
det(P2n+1) = 2
γn,k cn+1−k bk Tk
det(P2n) =
μn,k cn−k bk Vk
n+1
k=0
n
k=0
,
γn,k = (−1)k n + 1 + k , μn,k = (−1)k n + 1 + k .
n + 1 − k n − k
2 Proof of the main result
Since det(Pm (a, b, c)) = cm det Pm ac , bc , 1 , then we can assume for simplicity that c = 1. We denote by
ζ j , ζ1j , j = 1, 2, the roots of the polynomial g(x) = x4 + bx3 + ax2 + bx + 1 assumed pairwise distinct and
different of ±1.
Recall that the Chebyshev polynomials {Tn}, {Un}, {Vn} and {Wn} are orthogonal polynomials over (−1, 1)
with respect to the weight √11−x2 , √1 − x2, 11+−xx and 11−+xx , respectively, and we have for ζ ∈ C∗
⎪⎧⎨ Tn 21 ζ + ζ
1
⎪⎩ Vn 21 ζ + ζ
1
= 21 ζ n + ζ −n , Un 21 ζ + ζ
1
=
ζ n+1/2+ζ −n−1/2
ζ 1/2+ζ −1/2 ,
Wn 21 ζ + ζ
1
= ζ n+ζ1−−ζζ−−1n−1 ,
=
ζ n+1/2−ζ −n−1/2
ζ 1/2−ζ −1/2 .
We shall use the following formula for det(Pm):
Lemma 2.1 For J ⊂ {1, 2}, let IJ (k) =
1 i f k ∈ J
−1 i f k ∈/ J , and
(1)
(2)
(3)
ωJ =
ζkIJ (k), γJ =
(ζkIJ (k) − ζ jIJ ( j)).
2
k=1
1
We have
1 1 .
where d = ζ2 + ζ2 − ζ1 − ζ1
Proof See [4].
det (Pm) = d2 2
k=1(ζk − ζk−1)
J
Let us put α = ζ1ζ2 , β = ζ1ζ2−1 and u = 21 α + α−1 , v = 21 β + β−1 . We have by the Vieta’
formulae:
1≤ j<k≤2
m+1
(−1)J γJ ωJ 2
×
J
2
b a
= −
Proof Using the notations from Lemma 2.1, we obtain
det(Pm) =
and
Remark that and hence
Similarly, we obtain that
αm2+1+1
α− 2
m+1
−
α 1
−
= α1/2
m+2
α 2 α− 2
−
m+2
−
1
α−1/2
1/2
= Um+1 α
2
+
α−1/2
= Um+1
1 u
+
2
(F1)
By the same method, we get
J
γJ ωnJ+1 =
(ζ2−1 − ζ1)(β − 1)
β
,
A straightforward computation (using the Maple software for example) shows that
and this completes the proof of the Lemma.
tRheemsaerckon2d.3orWdeerheaqvueautio+nvx +2−2 =a2 a+2+1 1xa+nd (b2u +2=1)0(.v T+h1is)g=ivesb2 fo2r. eTxhaemn,pule+ 1 and v + 1 are the zeros of
J
2
C =
and
The term Um2 +1 1+2u2(u−−Uvm)2 +1 1+2v is a symmetric polynomial of u + 1 and v + 1 and, consequently,
it can be expressed in terms of the elementary symmetric polynomials u + 1 + v + 1 = a2 + 1 and
(u + 1) (v + 1) = b2 2 . For this, we distinguish two cases:
Case 1: m = 2n + 1. Using the following expression of U2n+2 (x )[
3
]:
U2n+2 (x ) =
(−1)k
n+1
k=0
2n + 2 − k
k
(2x )2n+2−2k
n+1
k=0
= (−1)n+1
γn,k (2x )2k , γn,k = (−1)k
n + 1 + k ,
n + 1 − k
U2n+2
1 +2 v = (−1)n+1 n+1γn,k2k (1 + u)k + (1 + v)k ,
On the other hand, we have for x, y :
we obtain
and for k ≥ 1
and
gives
and for k ≥ 1:
and thus
Applying those formulae for x = √1 + u and y = √1 + v where
x2k + y2k = (xy)k x k x −k
y + y
= 2(xy)k Tk 2y + 2yx ,
x
x2k − y2k = (xy)k x k x −k
y − y
= (xy)k xy − xy Uk−1 2y + 2x
x y
= x2 − y2 (xy)k−1 Uk−1 2y + 2yx .
x
xy = b2, x2 − y2 = u − v,
2xy + 2yx = x22+xyy2 = 2 + u + v = a2+b2,
b
(1 + u)k + (1 + v)k = 2 b2 k Tk a2+b2 ,
Case 2: m = 2n. We have [
3
]:
(1 + u)k − (1 + v)k = (u − v) b2 k−1 Uk−1 a2+b2
k=0
n
U2n+1 (x) = (−1)k 2n +k1 − k (2x)2n+1−2k
n
= (−1)n μn,k (2x)2k+1 , μn,k = (−1)k n +n −1 +k k ,
k=0
U2n+1
As for the odd case, we have for x , y :
x 2k+1 + y2k+1 = (x y)k+1/2
x 2k+1 − y2k+1 = (x y)k+1/2
= (x y)k+1/2
= (x y)k+1/2
and
and
This implies
x
y
x
y
x
y
x
y
√
√
n
,
,
(1 + u)k+1/2 + (1 + v)k+1/2 =
1 + u +
1 + v
(1 + u)k+1/2 − (1 + v)k+1/2 =
1 + u −
1 + v
which completes the proof of Theorem 1.1.
3 Numerical computation of det (Pm)
In this section, we shall derive from the formulae (1) and (2) an efficient algorithm for computing det (Pm ) .
We are lead to evaluate sums of the form
N
k=0
SN =
αk Pk (x )
where x = a +2b2c , and { Pr } are polynomials that satisfy the threeterm recurrence
Such sums can be computed efficiently through the following method described in [
11
]:
Equation (4) may be written in matrix notation as M p = q, where M is the (N + 1) × (N + 1) matrix
Pr (x ) − 2x Pr −1(x ) + Pr −2(x ) = 0.
(4)
0
.
.
.
0
0
be the row vector such that
It follows that
yT = (y0, y1, . . . yN )
yT M = uT = (α0, α1, . . . αN ) .
yk = 2x yk+1 − yk+2 + αk , for k = N , . . . , 0.
SN = uT p = yT Mp = yT q =y0 P0 (x) + (P1(x) − 2x P0(x)) y1.
Thus, yk are computed by putting yN+1 = yN+2 = 0 and performing the threeterm recurrence
Let
and
and
where yn+1 = yn+2 = 0 and
yk = 2x yk+1 − yk+2 + μn,k cn−k bk , for k = n, . . . , 0.
For Pk = Tk and Pk = b1 Uk−1, with U−1 = 0, respectively, we obtain
where yn+2 = yn+3 = 0 and
For Pk = Vk and Pk = Wk , respectively, we obtain
yk = 2x yk+1 − yk+2 + γn,k cn+1−k bk , for k = n + 1, . . . , 0.
n+1
k=0
n+1
k=1
n
k=0
n
k=0
γn,k cn+1−k bk Tk (x) = y0 − x y1
1
γn,k cn+1−k bk−1Uk−1 (x) = b y1,
μn,k cn−k bk Vk (x) = y0 − y1
μn,k cn−k bk Wk (x) = y0 + y1,
Here is the implementation of the algorithm in Maple (To accelerate the algorithm, the terms γn,k cn+1−k bk
and μn,k cn−k bk are computed recursively at the same time as yk . Implementation details are omitted):
## Computing det(P_2n+1)
##
detP1:=proc(n,a,b,c)
local i,j,r,s,x,k,t,z;
i := 0;
j := 0;
r := (1)ˆ(n+1)*bˆ(n+1);
x:=(a+2*c)/b;
t:=2*n;
z:=c/b;
for k from 0 to n+1 do
s:=i;
i:=r+x*ij;
## i:=simplify(r+x*ij); if the purpose
## is to compute the characteristic
## polynomial with variable a
j:=s;
r:=r*z*((t+2 )*(t+1 ))/((t+k+2 )*(k+1 ));
t:=t2;
od;
return 2*j*(i(j*x/2)/b; ## return simplify(2*j*(i(j*x/2)/b);
## if the purpose is to compute the
## characteristic polynomial with
## variable a
end;
## Computing det(P_2n)
##
detP2:=proc(n,a,b,c)
local i,j,r,s,x,k;
i := 0;
j := 0;
r := (1)ˆn*bˆn;
x:=(a+2*c)/b;
t:=2*n;
z:=c/b;
for k from 0 to n do
s:=i;
i:=r+x*ij;
## i:=simplify(r+x*ij); if the purpose
## is to compute the characteristic
## polynomial with variable a
j:=s;
r:=r*z*((t+1 )*t))/(t+k+1 )*(k+1 ));
t:=t2;
od;
return iˆ(2)jˆ(2); ## return simplify(iˆ(2)jˆ(2));
## if the purpose is to compute
## the characteristic polynomial
## with variable a
end;
One can easily check that the complexity of the algorithm is about 7N , where N is the size of the
matrix. Thus, the algorithm is the fastest among many other recently proposed (we exclude those based on
the roots of certain polynomials which are approximative) [
10
]. Moreover, subject to minor modifications as
explained in Algorithm 1, the algorithm is suitable for computing the characteristic polynomial of a symmetric
pentadiagonal Toeplitz matrix using computer algebra systems such as MAPLE, MATHEMATICA, MATLAB
and MACSYMA.
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