Riordan arrays and the LDU decomposition of symmetric Toeplitz plus Hankel matrices

0.13 MB·English

by  Paul Barry

Riordan arrays and the LDU decomposition
of symmetric Toeplitz plus Hankel matrices
Paul Barry
School of Science
Aoife Hennessy
1
1 Department of Computing, Mathematics and Physics
0
2 Waterford Institute of Technology, Cork Road, Waterford, Ireland

n [email protected]
a
J

3
1 Abstract

] We examine a result of Basor and Ehrhardt concerning Hankel and Toeplitz plus
O
Hankel matrices, within the context of the Riordan group of lower-triangular matrices.
C
Thisallows ustodeterminetheLDU decompositionofcertainsymmetricToeplitzplus
.
h
Hankel matrices. We also determine the generating functions and Hankel transforms
t
a of associated sequences.
m
[
Keywords: Toeplitz-plus-Hankel,Riordanarray,LDUdecomposition. 15A30,15A15,40C05

1
v
5 1 Introduction
0
6
2 In [1] Basor and Ehrhardt studied the transformation
.
1
0 n 1
1 − n 1
b = − (a +a ), (1)
1 n k 1−n+2k 2−n+2k
: k=0(cid:18) (cid:19)
v X
i
X definedforsequences a inthecontextofrelatingthedeterminantsofcertainToeplitz
{ n}∞n=
r plus Hankel matrices to the−d∞eterminants of related Hankel matrices.
a
In this note, we shall study an equivalent transformation, which we will construct with
the aid of Riordan arrays [9]. We call this the B-transform. We shall then use our results to
examine the LDU decomposition of the resulting Toeplitz plus Hankel matrices.
In the next section, we shall detail the notations that will be used in this note, and
give a basic introduction to the relevant theory of Riordan arrays. We shall follow this
with a section which defines the B-transform, studies some of its properties, and shows its
equivalence the Basor and Ehrhardt transform. In particular, we derive an expression for
the generating function of the image sequence, which for instance allows us to determine
the Hankel transform of image sequence in many cases. A final section then looks at the
LDU decomposition of the related Toeplitz plus Hankel matrices, with examples involving
Riordan arrays.
1
2 Notation and basic Riordan array theory
Although many of our results will be valid for sequences a with values in C, we shall in
n
the sequel assume that the sequences we deal with are integer sequences, a Z. For an
n
integer sequence an, that is, an element of ZN, the power series f(x) = ∞k=0a∈kxk is called
the ordinary generating function or g.f. of the sequence. a is thus the coefficient of xn in
n
P
this series. We denote this by a = [xn]f(x). For instance, F = [xn] x is the n-th
n n 1 x x2
Fibonacci number, while C = [xn]1 √1 4x is the n-th Catalan number. We−u−se the notation
n − 2x−
0n = [xn]1 for the sequence 1,0,0,0,... Thus 0n = [n = 0] = δ = 0 . Here, we have used
n,0 n
the Iverson bracket notation [3], defined by [ ] = 1 if the proposition is true, and [ ] = 0
P (cid:0) P(cid:1) P
if is false.
P
For a power series f(x) = ∞n=0anxn with f(0) = 0 we define the reversion or composi-
¯ ¯
tional inverse of f to be the power series f(x) such that f(f(x)) = x. We sometimes write
¯ P
f = Revf.
The Hankel transform [6] of a sequence a is the sequence h = a n . If the sequence
n n | i+j|i,j=0
a has a g.f. that has a continued fraction expansion of the form
n
a
0
,
β x2
1
1 α x
− 0 − β x2
2
1 α x
− 1 − β x2
3
1 α x
2
− − 1
−···
then the Hankel transform of a is given by [5]
n
n
h = an+1βnβn 1 β2 β = an+1 βn+1 k. (2)
n 0 1 2− ··· n 1 n 0 k −
−
k=1
Y
The LDU decomposition of Hankel matrices has been studied in [7, 8].
Some of the lower-triangular matrices that we shall meet will be coefficient arrays of
families of orthogonal polynomials. General references for orthogonal polynomials include
[2, 4, 11].
At will denote the transpose of the matrix A, and we will on occasion use A B to denote
·
theproduct of thematrices AB, where thismakes reading thetext easier. This also conforms
with the use of “ ” for the product in the Riordan group (see below).
·
The Riordan group [9, 10], is a set of infinite lower-triangular integer matrices, where
each matrix is defined by a pair of generating functions g(x) = 1 + g x + g x2 + and
1 2
···
f(x) = f x + f x2 + where f = 0 [10]. We assume in addition that f = 1 in what
1 2 1 1
··· 6
follows. The associated matrix is the matrix whose i-th column is generated by g(x)f(x)i
(the first column being indexed by 0). The matrix corresponding to the pair g,f is denoted
by (g,f) or (g,f). The group law is then given by
R
(g,f) (h,l) = (g,f)(h,l) = (g(h f),l f).
· ◦ ◦
The identity for this law is I = (1,x) and the inverse of (g,f) is (g,f) 1 = (1/(g f¯),f¯)
−
¯ ◦
where f is the compositional inverse of f.
2
A Riordan array of the form (g(x),x), where g(x) is the generating function of the se-
quence a , is called the sequence array of the sequence a . Its (n,k)-th term is a . Such
n n n k
−
arrays are also called Appell arrays as they form the elements of the Appell subgroup.
If M is the matrix (g,f), and a = (a ,a ,...) is an integer sequence with ordinary generat-
0 1 ′
ing function (x), then the sequence Ma has [9] ordinary generating function g(x) (f(x)).
A A
This result is often called “the Fundamental Theorem of Riordan arrays”. The (infinite)
matrix (g,f) can thus be considered to act on the ring of integer sequences ZN by multi-
plication, where a sequence is regarded as a (infinite) column vector. We can extend this
action to the ring of power series Z[[x]] by
(g,f) : (x) (g,f) (x) = g(x) (f(x)).
A 7→ ·A A
Example 1. The so-called binomial matrix B is the element ( 1 , x ) of the Riordan
1 x 1 x
group. It has general element n , and hence as an array coincide−s wit−h Pascal’s triangle.
k
More generally, Bm is the element ( 1 , x ) of the Riordan group, with general term
(cid:0) (cid:1) 1 mx 1 mx
n mn k. It is easy to show that the in−verse−B m of Bm is given by ( 1 , x ).
k − − 1+mx 1+mx
(cid:0)Fo(cid:1)r a sequence a ,a ,a ,... with g.f. g(x), the “aeration” of the sequence is the sequence
0 1 2
a ,0,a ,0,a ,... with interpolated zeros. Its g.f. is g(x2). We note that since c(x) = 1 √1 4x
0 1 2 − 2x−
has the well-known continued fraction expansion
1
c(x) = ,
x
1
− x
1
− 1 ...
−
c(x2) has the continued fraction expansion
1
c(x2) = . (3)
x2
1
− x2
1
− 1 ...
−
The aeration of a (lower-triangular) matrix M with general term m is the matrix whose
i,j
general term is given by
1+( 1)i j
mr − − ,
i+j,i−j 2
2 2
where mr is the i,j-th element of the reversal of M:
i,j
mr = m .
i,j i,i j
−
In the case of a Riordan array (or indeed any lower triangular array), the row sums of the
aeration are equal to the diagonal sums of the reversal of the original matrix.
3
Example 2. The Riordan array (c(x2),xc(x2)) is the aeration of the Riordan array
(c(x),xc(x)) = (1 x,x(1 x)) 1.
−
− −
Here
1 √1 4x
c(x) = − −
2x
is the g.f. of the Catalan numbers. Indeed, the reversal of (c(x),xc(x)) is the matrix with
general element
n+k n k +1
[k n+1] − ,
≤ k n+1
(cid:18) (cid:19)
which begins
1 0 0 0 0 0 ...
1 1 0 0 0 0 ...
 
1 2 2 0 0 0 ...
 1 3 5 5 0 0 ... .
 
 1 4 9 14 14 0 ... 
 
 1 5 14 28 42 42 ... 
 
 ... ... ... ... ... ... ... 
 
 
Then (c(x2),xc(x2)) has general element
n+1 k +11+( 1)n k
−
− ,
n k n+1 2
−
(cid:18) 2 (cid:19)
and begins
1 0 0 0 0 0 ...
0 1 0 0 0 0 ...
 
1 0 1 0 0 0 ...
 0 2 0 1 0 0 ... .
 
 2 0 3 0 1 0 ... 
 
 0 5 0 4 0 1 ... 
 
 .. .. .. .. .. .. .. 
 . . . . . . . 
 
 
We have
1
1 x −
(c(x2),xc(x2)) = , .
1+x2 1+x2
(cid:18) (cid:19)
B
3 The -transform
We let
1
1 x x −
L = − , ,
1+x2 1+x2
(cid:18) (cid:19)
4
where we recall that the Riordan matrix (g,f) is the lower triangular matrix whose k-th
column has generating function g(x)f(x)k, for suitable g and f. Then L has (n,k)-th term
n
,
n k
−
(cid:18)⌊ 2 ⌋(cid:19)
and L 1 = 1 x , x is the coefficient array of the generalized Chebyshev polynomials
− 1+−x2 1+x2
defined by
(cid:0) (cid:1)
P (x) = xP (x) P (x), P (x) = 1, P (x) = x 1.
n n 1 n 2 0 1
− − − −
The matrix L begins
1 0 0 0 0 0 ...
1 1 0 0 0 0 ...
 
2 1 1 0 0 0 ...
L =  3 3 1 1 0 0 ... .
 
 6 4 4 1 1 0 ... 
 
 10 10 5 5 1 1 ... 
 
 ... ... ... ... ... ... ... 
 
We now define B to be the matrix 
B = L (1+x,x)t. (4)
·
This matrix begins
1 1 0 0 0 0 ...
1 2 1 0 0 0 ...
 
2 3 2 1 0 0 ...
B =  3 6 4 2 1 0 ... .
 
 6 10 8 5 2 1 ... 
 
 10 20 15 10 6 2 ... 
 
 .. .. .. .. .. .. .. 
 . . . . . . . 
 
Since the matrix (1+x,x)t is given by 
1 1 0 0 0 0 ...
0 1 1 0 0 0 ...
 
0 0 1 1 0 0 ...
(1+x,x)t =  0 0 0 1 1 0 ... 
 
 0 0 0 0 1 1 ... 
 
 0 0 0 0 0 1 ... 
 
 ... ... ... ... ... ... ... 
 
 
1 0 0 0 0 0 ... 0 1 0 0 0 0 ...
0 1 0 0 0 0 ... 0 0 1 0 0 0 ...
   
0 0 1 0 0 0 ... 0 0 0 1 0 0 ...
=  0 0 0 1 0 0 ... + 0 0 0 0 1 0 ... ,
   
 0 0 0 0 1 0 ...   0 0 0 0 0 1 ... 
   
 0 0 0 0 0 1 ...   0 0 0 0 0 0 ... 
   
 ... ... ... ... ... ... ...   ... ... ... ... ... ... ... 
   
   
5
we see that the (n,k)-th element of B is given by
n n n
b = + 0k.
n,k n k n k+1 − n ·
− −
(cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊2⌋(cid:19)
Now let a be a sequence. We define the B transform of a to be the sequence b
n n 0 n n n 0
{ } ≥ { } ≥
given by
n+1
b = b a (5)
n n,k k
k=0
X
where B = (b ) .
n,k n,k 0
≥
Example 3. The B-transform of the Fibonacci numbers is given by
n+1 n+1
n n n
b = b F = ( + 0k)F .
n n,k k n k n k+1 − n · k
− −
k=0 k=0 (cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊2⌋(cid:19)
X X
This sequence starts
1,3,7,17,39,91,207,475,1075,2445,5515,....
It has the interesting property that its Hankel transform is ( 2)n.
−
Proposition 4. We have
1
B = 1 , x − T, (6)
1+x2 1+x2 ·
(cid:18) (cid:19)
where T is the matrix
1
T = ,x (1+x,x)t. (7)
1 x ·
(cid:18) − (cid:19)
Proof. We have
1
B = L (1+x,x)t = L 1 ,x − 1 ,x (1+x,x)t.
· · 1 x · 1 x ·
(cid:18) − (cid:19) (cid:18) − (cid:19)
Now
1 1 1
1 − 1 x x − 1 −
L ,x = − , ,x
· 1 x 1+x2 1+x2 · 1 x
(cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) − (cid:19)
1
1 1 x x −
= ,x − ,
1 x · 1+x2 1+x2
(cid:18)(cid:18) − (cid:19) (cid:18) (cid:19)(cid:19)
1
1 x −
= , .
1+x2 1+x2
(cid:18) (cid:19)
6
We recall that the matrix 1+1x2, 1+xx2 −1 = (c(x2),xc(x2)), where c(x) = 1−√21x−4x is the g.f.
of the Catalan numbers, has general element
(cid:0) (cid:1)
n+1 k +11+( 1)n k
−
− .
n k n+1 2
−
(cid:18) 2 (cid:19)
In addition, 1 ,x (1+x,x)t is the matrix T given by
1 x ·
−
(cid:0) (cid:1)
1 1 0 0 0 0 ...
1 2 1 0 0 0 ...
 
1 2 2 1 0 0 ...
T =  1 2 2 2 1 0 ... .
 
 1 2 2 2 2 1 ... 
 
 1 2 2 2 2 2 ... 
 
 .. .. .. .. .. .. .. 
 . . . . . . . 
 
 
Now note that
1 0 0 0 0 0 ... 0 1 0 0 0 0 ...
1 1 0 0 0 0 ... 0 1 1 0 0 0 ...
   
1 1 1 0 0 0 ... 0 1 1 1 0 0 ...
T =  1 1 1 1 0 0 ... + 0 1 1 1 1 0 ... ,
   
 1 1 1 1 1 0 ...   0 1 1 1 1 1 ... 
   
 1 1 1 1 1 1 ...   0 1 1 1 1 1 ... 
   
 ... ... ... ... ... ... ...   ... ... ... ... ... ... ... 
   
   
and hence the action of T on a sequence a is to return the sequence with n-th term equal
n
to
n n+1 n
a + a = 2 a +a a .
k k k n+1 0
−
k=0 k=1 k=0
X X X
Thus we have
Proposition 5. We have
n n+1 k +11+( 1)n k k k+1
−
b = − ( a + a ),
n n k n+1 2 j j
−
k=0(cid:18) 2 (cid:19) j=0 j=1
X X X
or equivalently,
n n+1 k +11+( 1)n k k
−
b = − (2 a +a a ).
n n k n+1 2 j k+1 − 0
−
k=0(cid:18) 2 (cid:19) j=0
X X
In the following, we will be interested in determining the g.f. of the image of a . We
n
have the following result.
7
Proposition 6. Let f(x) be the g.f. of a . Then b = n+1b a has g.f. given by
n n k=0 n,k k
1 P
1 x − (1+x)f(x) a0
, − . (8)
1+x2 1+x2 · x(1 x)
(cid:18) (cid:19) (cid:18) − (cid:19)
Equivalently, the g.f. of b is given by
n
1
1 x x − (1+x)f(x) a0
− , − . (9)
1+x2 1+x2 · x
(cid:18) (cid:19) (cid:18) (cid:19)
Proof. The result follows from the fact that the generating function of n a + n+1a
k=0 k k=1 k
is given by
P P
1 1 f(x) a (1+x)f(x) a
0 0
f(x)+ − = − .
1 x 1 x x x(1 x)
− − (cid:18) (cid:19) −
Corollary 7. The g.f. of b = n b a is given by
n k=0 n,k k
P(1+xc(x2))f(xc(x2)) a
0
− .
x(1 xc(x2))
(cid:18) − (cid:19)
Proof. This follows from the fundamental theorem of Riordan arrays since
1
1 x −
, = (c(x2),xc(x2)).
1+x2 1+x2
(cid:18) (cid:19)
Example 8. The g.f. of the B-transform of the Fibonacci numbers F is given by
n
1+xc(x2) xc(x2)
.
x(1 xc(x2))1 xc(x2) x2c(x2)2
− − −
This follows since
x
F = [xn]
n
1 x x2
− −
and F = 0. This g.f. may be simplified to
0
1 4x2 +x√1 4x2 1 4x2 +x√1 4x2
− − = − − .
1 2x 5x2 +10x3 (1 2x)(1 5x2)
− − − −
By solving the equation
1
u = ,
x2
1 3x+
− 1+x x2c(x2)
−
8
we see that this g.f. may be expressed (using Eq. (3)) as the continued fraction
1
,
2x2
1 3x+
− x2
1+x
− x2
1
− x2
1
− 1 ...
−
which shows that the the Hankel transform of the B-transform of the Fibonacci numbers is
( 2)n.
−
We have defined the B matrix using the Riordan array 1 , x . This matrix is
1+x2 1+x2
associated with the Chebyshev polynomials of the second kind U (x) (it is the coefficient
n
(cid:0) (cid:1)
array of U (x/2)). The matrix 1 x2, x is related to the Chebyshev polynomials of the
n 1−+x2 1+x2
first kind Tn. We have (cid:16) (cid:17)
Proposition 9. We have
1 x2 x −1
B = − , (1+x,x) (1+x,x)t. (10)
1+x2 1+x2 · ·
(cid:18) (cid:19)
Proof. We have
B = L (1+x,x)t = L (1+x,x) 1 (1+x,x) (1+x,x)t
−
· · · ·
1
= L ,x (1+x,x) (1+x,x)t
· 1+x · ·
(cid:18) (cid:19)
1 x x 1
= − , ,x (1+x,x) (1+x,x)t
1+x2 1+x2 · 1+x · ·
(cid:18) (cid:19) (cid:18) (cid:19)
1 x2 x −1
= − , (1+x,x) (1+x,x)t.
1+x2 1+x2 · ·
(cid:18) (cid:19)
We can decompose (1+x,x) (1+x,x)t as the sum of two matrices:
·
1 0 0 0 0 0 ... 0 1 0 0 0 0 ...
1 1 0 0 0 0 ... 0 1 1 0 0 0 ...
   
0 1 1 0 0 0 ... 0 0 1 1 0 0 ...
 0 0 1 1 0 0 ... + 0 0 0 1 1 0 ... ,
   
 0 0 0 1 1 0 ...   0 0 0 0 1 1 ... 
   
 0 0 0 0 1 1 ...   0 0 0 0 0 1 ... 
   
 .. .. .. .. .. .. ..   .. .. .. .. .. .. .. 
 . . . . . . .   . . . . . . . 
   
   
9
which is the sum of (1+x,x) and a shifted version of (1+x,x). To obtain B we multiply
1
by 1 x2, x − . This gives us, once again
1+−x2 1+x2
(cid:16) (cid:17)
1 0 0 0 0 0 ... 0 1 0 0 0 0 ...
1 1 0 0 0 0 ... 0 1 1 0 0 0 ...
   
2 1 1 0 0 0 ... 0 2 1 1 0 0 ...
B =  3 3 1 1 0 0 ... + 0 3 3 1 1 0 ... ,
   
 6 4 4 1 1 0 ...   0 6 4 4 1 1 ... 
   
 10 10 5 5 1 1 ...   0 10 10 5 5 1 ... 
   
 ... ... ... ... ... ... ...   ... ... ... ... ... ... ... 
   
   
where the first member of the sum is the Riordan array
1 x2 x −1 1 x2 x −1
− , (1+x,x) = − , = L.
1+x2 1+x2 · (1+x)(1+x2) 1+x2
(cid:18) (cid:19) (cid:18) (cid:19)
Theorem 10. Let b = n+1b a where b is the (n,k)-th element of B. Then
n k=0 n,k k n,k
P n
n
b = (a +a ),
n n 2k n 2k+1
k − −
k=0(cid:18) (cid:19)
X
where we have extended a to negative n by setting a = a .
n n n
−
Proof. We have seen that B has general term
n n n
+ 0k .
n k n k+1 − · n
− −
(cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊2⌋(cid:19)
Thus the B transform of a is given by
n
n+1
n n n
+ 0k a
n k n k+1 − · n k
− −
k=0(cid:18)(cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊2⌋(cid:19)(cid:19)
X
which can also be written as
n+1
n n
(1 0k)+ a
n k − n k+1 k
− −
k=0(cid:18)(cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊ 2 ⌋(cid:19)(cid:19)
X
since 0k n = 0k n . We can also write this as
n n−k
· ⌊2⌋ · ⌊ 2 ⌋
(cid:0) (cid:1) (cid:0) (cid:1)
n+1
n n
b = (1 0n k+1)+ a . (11)
n k−1 − − k n−k+1
k=0(cid:18)(cid:18)⌊ 2 ⌋(cid:19) (cid:18)⌊2⌋(cid:19)(cid:19)
X
10