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Abstract

The paper is devoted to the asymptotic behavior of the eigenvectors of banded Hermitian Toeplitz matrices as the dimension of the matrices increases to infinity. The main result, which is based on certain assumptions, describes the structure of the eigenvectors in terms of the Laurent polynomial that generates the matrices up to an error term that decays exponentially fast. This result is applicable to both extreme and inner eigenvectors. Mathematics Subject Classification (2000)Primary 47B35–Secondary 15A18–41A25–65F15
On the structure of the eigenvectors of large
Hermitian Toeplitz band matrices
A. ottcher, S. M. Grudsky
1
, and E. A. Maksimenko
The paper is devoted to the asymptotic behavior of the eigenvectors of banded
Hermitian Toeplitz matrices as the dimension of the matrices increases to infinity.
The main result, which is based on certain assumptions, describes the structure of
the eigenvectors in terms of the Laurent polynomial that generates the matrices
up to an error term that decays exponentially fast. This result is applicable to
both extreme and inner eigenvectors.
Keywords: Toeplitz matrix, eigenvector, asymptotic expansions
Mathematics Subject Classification (2000): Primary 47B35; Secondary
15A18, 41A25, 65F15
1 Introduction and main results
Given a function a in L
1
on the complex unit circle T, we denote by a
`
the `th
Fourier coefficient,
a
`
=
1
2π
Z
2π
0
a(e
ix
)e
i`x
dx (` Z),
and by T
n
(a) the n × n Toeplitz matrix (a
jk
)
n
j,k=1
. We assume that a is real-
valued, in which case the matrices T
n
(a) are all Hermitian. Let
λ
(n)
1
λ
(n)
2
. . . λ
(n)
n
be the eigenvalues of T
n
(a) and let
{v
(n)
1
, v
(n)
2
, . . . , v
(n)
n
}
1
This work was partially supported by CONACYT project 80503, Mexico.
1
be an orthonormal basis of eigenvectors such that T
n
(a)v
(n)
j
= λ
(n)
j
v
(n)
j
. The
present paper is dedicated to the asymptotic behavior of the eigenvectors v
(n)
j
as
n .
To get an idea of the kind of results we will establish, consider the function
a(e
ix
) = 2 2 cos x. The range a(T) is the segment [0, 4]. It is well known that
the eigenvalues and eigenvectors of T
n
(a) are given by
λ
(n)
j
= 2 2 cos
πj
n + 1
, x
(n)
j
=
r
2
n + 1
sin
j
n + 1
n
m=1
. (1)
(We denote the eigenvectors in this reference case by x
(n)
j
and reserve the notation
v
(n)
j
for the general case.) Let ϕ be the function
ϕ : [0, 4] [0, π], ϕ(λ) = arccos
2 λ
2
.
We have ϕ(λ
(n)
j
) = πj/(n + 1) and hence, apart from the normalization factor
p
2/(n + 1), x
(n)
j,m
is the value of sin((λ)) at λ = λ
(n)
j
. In other words, an
eigenvector for λ is given by (sin((λ)))
n
m=1
. A speculative question is whether
in the general case we can also find functions
m
such that, at least asymptot-
ically, (Ω
m
(λ))
n
m=1
is an eigenvector for λ. It turns out that this is in general
impossible but that after a slight modification the answer to the question is in
the affirmative. Namely, we will prove that, under certain assumptions, there
are functions
m
, Φ
m
and real-valued functions σ, η such that an eigenvector for
λ = λ
(n)
j
is always of the form
m
(λ) + Φ
m
(λ) + (1)
j+1
e
i(n+1)σ(λ)
e
(λ)
Φ
n+1m
(λ) + error term
n
m=1
. (2)
The error term will be shown to decrease to zero exponentially fast and uniformly
in j and m as n . Moreover, we will show that
m
(λ) is an oscillating
function of m for each fixed λ and that Φ
m
(λ) decays exponentially fast to zero
as m for each λ (which means that Φ
n+1m
(λ) is an exponentially increasing
function of m for each λ). Finally, it will turn out that
n
X
m=1
|Φ
m
(λ)|
2
.
n
X
m=1
|
m
(λ)|
2
= O
1
n
as n , uniformly in λ. Thus, the dominant term in (2) is
m
(λ), while the
terms containing Φ
m
(λ) and Φ
n+1m
(λ) may be viewed as twin babies.
If a is also an even function, a(e
ix
) = a(e
ix
) for all x, then all the matri-
ces T
n
(a) are real and symmetric. In [4], we conjectured that then, again under
additional but reasonable assumptions, the appropriately rotated extreme eigen-
vectors v
(n)
j
are all close to the vectors x
(n)
j
. To be more precise, we conjectured
2
that if n and j (or n j) remains fixed, then there are complex numbers
τ
(n)
j
of modulus 1 such that
τ
(n)
j
v
(n)
j
x
(n)
j
2
= o(1), (3)
where k ·k
2
is the `
2
norm. Several results related to this conjecture were estab-
lished in [3] and [4]. We here prove this conjecture under assumptions that will
be specified in the following paragraph. We will even be able to show that the
o(1) in (3) is O(j/n) if j/n 0 and O(1 j/n) if j/n 1.
Throughout what follows we assume that a is a Laurent polynomial
a(t) =
r
X
k=r
a
k
t
k
(t = e
ix
T)
with r 2, a
r
6= 0, and a
k
= a
k
for all k. The last condition means that a is
real-valued on T. We assume without loss of generality that a(T) = [0, M] with
M > 0 and that a(1) = 0 and a(e
0
) = M for some ϕ
0
(0, 2π). We require that
the function g(x) := a(e
ix
) is strictly increasing on (0, ϕ
0
) and strictly decreasing
on (ϕ
0
, 2π) and that the second derivatives of g at x = 0 and x = ϕ
0
are nonzero.
Finally, we denote by [α, β] [0, M] a segment such that if λ [α, β], then the
2r 2 zeros of the Laurent polynomial a(z) λ that lie in C \ T are pairwise
distinct.
Note that we exclude the case r = 1, because in this case the eigenvalues and
eigenvectors of T
n
(a) are explicitly available. Also notice that if r = 2, which
is the case of pentadiagonal matrices, then for every λ [0, M] the polynomial
a(z) λ has two zeros on T, one zero outside T, and one zero inside T. Thus,
in this situation the last requirement of the previous paragraph is automatically
satisfied for [α, β] = [0, M].
The asymptotic behavior of the extreme eigenvalues and eigenvectors of T
n
(a),
that is, of λ
(n)
j
and v
(n)
j
when j or n j remain fixed, has been studied by several
authors. As for extreme eigenvalues, the pioneering works are [7], [9], [11], [12],
[18], while recent papers on the subject include [3], [6], [8], [10], [13], [14], [15], [19],
[20]. See also the books [1] and [5]. Much less is known about the asymptotics
of the eigenvectors. Part of the results of [4] and [19] may be interpreted as
results on the behavior of the eigenvectors “in the mean” on the one hand and
as insights into what happens if eigenvectors are replaced by pseudomodes on
the other. In [3], we investigated the asymptotics of the extreme eigenvectors of
certain Hermitian (and not necessarily banded) Toeplitz matrices. Our paper [2]
may be considered as a first step to the understanding of the asymptotic behavior
of individual inner eigenvalues of Toeplitz matrices. In the same vein, this paper
intends to understand the nature of individual eigenvectors as part of the whole,
independently of whether they are extreme or inner ones.
3
To state our main results, we need some notation. Let λ [0, M]. Then there
are uniquely defined ϕ
1
(λ) [0, ϕ
0
] and ϕ
2
(λ) [ϕ
0
2π, 0] such that
g(ϕ
1
(λ)) = g(ϕ
2
(λ)) = λ;
recall that g(x) := a(e
ix
). We put
ϕ(λ) =
ϕ
1
(λ) ϕ
2
(λ)
2
, σ(λ) =
ϕ
1
(λ) + ϕ
2
(λ)
2
.
We have
a(z) λ = z
r
a
r
z
2r
+ . . . + (a
0
λ)z
r
+ . . . + a
r
= a
r
z
r
2r
Y
k=1
(z z
k
(λ)),
and our assumptions imply that we can label the zeros z
k
(λ) so that the collection
Z(λ) of the zeros may be written as
{z
1
(λ), . . . , z
r1
(λ), z
r
(λ), z
r+1
(λ), z
r+2
(λ), . . . , z
2r
(λ)}
= {u
1
(λ), . . . , u
r1
(λ), e
1
(λ)
, e
2
(λ)
, 1/u
1
(λ), . . . , 1/u
r1
(λ)} (4)
where |u
ν
(λ)| > 1 for 1 ν r 1 and each u
ν
(λ) depends continuously on
λ [0, M]. Here and in similar places below we write u
k
(λ) := u
k
(λ). We define
δ
0
> 0 by
e
δ
0
= min
λ[0,M]
min
1νr1
|u
ν
(λ)|.
Throughout the following, δ stands for any number in (0, δ
0
). Further, we denote
by h
λ
the function
h
λ
(z) =
r1
Y
ν=1
1
z
u
ν
(λ)
.
The function Θ(λ) = h
λ
(e
1
(λ)
)/h
λ
(e
2
(λ)
) is continuous and nonzero on [0, M]
and we have Θ(0) = Θ(M) = 1. In [2], it was shown that the closed curve
[0, M] C \ {0}, λ 7→ Θ(λ)
has winding number zero. Let θ(λ) be the continuous argument of Θ(λ) for which
θ(0) = θ(M ) = 0.
In [2], we proved that if n is large enough, then the function
f
n
: [0, M] [0, (n + 1)π], f
n
(λ) = (n + 1)ϕ(λ) + θ(λ)
is bijective and increasing and that if λ
(n)
j,
is the unique solution of the equation
f
n
(λ
(n)
j,
) = πj, then the eigenvalues λ
(n)
j
satisfy
|λ
j
λ
(n)
j,
| K e
δn
4
for all j {1, . . . , n}, where K is a finite constant depending only on a. Thus,
we have
(n + 1)ϕ(λ
(n)
j
) + θ(λ
(n)
j
) = πj + O(e
δn
), (5)
uniformly in j {1, . . . , n}.
Now take λ from (α, β). For j {1, . . . , n} and ν {1, . . . , r 1}, we put
A(λ) =
e
(λ)
2i h
λ
(e
1
(λ)
)
, B(λ) =
e
(λ)
2i h
λ
(e
2
(λ)
)
,
D
ν
(λ) =
e
2(λ)
sin ϕ(λ)
(u
ν
(λ) e
1
(λ)
)(u
ν
(λ) e
2
(λ)
)h
0
λ
(u
ν
(λ))
,
F
ν
(λ) =
sin ϕ(λ)
(u
ν
(λ) e
1
(λ)
)(u
ν
(λ) e
2
(λ)
)h
0
λ
(u
ν
(λ))
×
×
|h
λ
(e
1
(λ)
)h
λ
(e
2
(λ)
)|
h
λ
(e
1
(λ)
)h
λ
(e
2
(λ)
)
and define the vector w
(n)
j
(λ) = (w
(n)
j,m
(λ))
n
m=1
by
w
(n)
j,m
(λ) = A(λ)e
imϕ
1
(λ)
B(λ)e
imϕ
2
(λ)
+
r1
X
ν=1
D
ν
(λ)
1
u
ν
(λ)
m
+ F
ν
(λ)
(1)
j+1
e
i(n+1)σ(λ)
u
ν
(λ)
n+1m
.
The assumption that zeros u
ν
(λ) are all simple guarantees that h
0
(u
ν
) 6= 0. We
denote by k · k
2
and k· k
the `
2
and `
norms on C
n
, respectively.
Here are our main results.
Theorem 1.1 As n and if λ
(n)
j
(α, β),
kw
(n)
j
(λ
(n)
j
)k
2
2
=
n
4
1
|h
λ
(e
1
(λ)
)|
2
+
1
|h
λ
(e
2
(λ)
)|
2
λ=λ
(n)
j
+ O(1),
uniformly in j.
Theorem 1.2 Let n and suppose λ
(n)
j
(α, β). Then the eigenvectors v
(n)
j
are of the form
v
(n)
j
= τ
(n)
j
w
(n)
j
(λ
(n)
j
)
kw
(n)
j
(λ
(n)
j
)k
2
+ O
(e
δn
)
!
where τ
(n)
j
T and O
(e
δn
) denotes vectors ξ
(n)
j
C
n
such that kξ
(n)
j
k
Ke
δn
for all j and n with some finite constant K independent of j and n.
5
Note that the previous theorem gives (2) with
m
(λ) = A(λ)e
imϕ
1
(λ)
B(λ)e
imϕ
2
(λ)
, Φ
m
(λ) =
r1
X
ν=1
D
ν
(λ)
u
ν
(λ)
m
,
e
(λ)
=
|h
λ
(e
1
(λ)
)h
λ
(e
2
(λ)
)|
h
λ
(e
1
(λ)
)h
λ
(e
2
(λ)
)
.
Things can be a little simplified for symmetric matrices. Thus, suppose all a
k
are real and a
k
= a
k
for all k. We will show that then {u
1
(λ), . . . , u
r1
(λ)} =
{u
1
(λ), . . . , u
r1
(λ)}. Put
Q
ν
(λ) =
|h
λ
(e
(λ)
)|sin ϕ(λ)
(u
ν
(λ) e
(λ)
)(u
ν
(λ) e
(λ)
)h
0
λ
(u
ν
(λ))
and let y
(n)
j
(λ) = (y
(n)
j,m
(λ))
n
m=1
be given by
y
(n)
j,m
(λ) = sin
(λ) +
θ(λ)
2
r1
X
ν=1
Q
ν
(λ)
1
u
ν
(λ)
m
+
(1)
j+1
u
ν
(λ)
n+1m
. (6)
Theorem 1.3 Let n and suppose λ
(n)
j
(α, β). If a
k
= a
k
for all k, then
ky
(n)
j
(λ
(n)
j
)k
2
2
=
n
2
+ O(1)
uniformly in j, and the eigenvectors v
(n)
j
are of the form
v
(n)
j
= τ
(n)
j
y
(n)
j
(λ
(n)
j
)
ky
(n)
j
(λ
(n)
j
)k
2
+ O
(e
δn
)
!
where τ
(n)
j
T and O
(e
δn
) is as in the previous theorem.
Let J be the n × n matrix with ones on the counterdiagonal and zeros else-
where. Thus, (Jv)
m
= v
n+1m
. A vector v is called symmetric if Jv = v and skew-
symmetric if Jv = v. Trench [17] showed that the eigenvectors v
(n)
1
, v
(n)
3
, . . .
are all symmetric and that the eigenvectors v
(n)
2
, v
(n)
4
, . . . are all skew-symmetric.
From (5) we infer that
sin
(n + 1 m)ϕ(λ
(n)
j
) +
θ(λ
(n)
j
)
2
!
= (1)
j+1
sin
(λ
(n)
j
) +
θ(λ
(n)
j
)
2
!
+ O(e
δn
)
6
and hence (6) implies that
(Jy
(n)
j
(λ
(n)
j
))
m
= (1)
j+1
y
(n)
j,m
(λ
(n)
j
) + O(e
δn
).
Consequently, apart from the term O(e
δn
), the vectors y
(n)
j
(λ
(n)
j
) are symmet-
ric for j = 1, 3, . . . and skew-symmetric for j = 2, 4, . . .. This is in complete
accordance with Trench’s result.
Due to (5), we also have
sin
(λ
(n)
j
) +
θ(λ
(n)
j
)
2
!
= sin

m
n + 1
2
ϕ(λ
(n)
j
)
+ O(e
δn
).
Thus, Theorem 1.3 remains valid with (6) replaced by
y
(n)
j,m
(λ) = sin

m
n + 1
2
ϕ(λ) +
πj
2
r1
X
ν=1
Q
ν
(λ)
1
u
ν
(λ)
m
+
(1)
j+1
u
ν
(λ)
n+1m
. (7)
In this expression, the function θ has disappeared.
Define y
(n)
j
again by (6). The following theorem in conjunction with Theo-
rem 1.3 proves (3).
Theorem 1.4 Let n and suppose λ
(n)
j
(α, β). If a
k
= a
k
for all k, then
y
(n)
j
(λ
(n)
j
)
ky
(n)
j
(λ
(n)
j
)k
2
x
(n)
j
2
= O
j
n
.
The rest of the paper is as follows. We approach eigenvectors by using the
elementary observation that if λ is an eigenvalue of T
n
(a), then every nonzero
column of the adjugate matrix of T
n
(a) λI = T
n
(a λ) is an eigenvector for λ.
In Section 2 we employ “exact” formulas by Trench and Widom for the inverse and
the determinant of a banded Toeplitz matrix to get a representation of the first
column of the adjugate matrix of T
n
(aλ) that will be convenient for asymptotic
analysis. This analysis is carried out in Section 3. On the basis of these results,
Theorems 1.1 and 1.2 are proved in Section 4, while the proofs of Theorems 1.3
and 1.4 are given in Section 4. Section 6 contains numerical results.
2 The first column of the adjugate matrix
The adjugate matrix adj B of an n × n matrix B = (b
jk
)
n
j,k=1
is defined by
(adj B)
jk
= (1)
j+k
det M
kj
7
where M
kj
is the (n 1) ×(n 1) matrix that results from B by deleting the kth
row and the jth column. We have
(A λI) adj (A λI) = (det(A λI))I.
Thus, if λ is an eigenvalue of A, then each nonzero column of adj (A λI) is an
eigenvector. For an invertible matrix B,
adj B = (det B)B
1
. (8)
Formulas for det T
n
(b) and T
1
n
(b) were established by Widom [18] and Trench
[16], respectively. The purpose of this section is to transform Trench’s formula for
the first column of T
1
n
(b) into a form that will be convenient for further analysis.
Theorem 2.1 Let
b(t) =
q
X
k=p
b
k
t
k
= b
p
t
q
p+q
Y
j=1
(t z
j
) (t T)
where p 1, q 1, b
p
6= 0, and z
1
, . . . , z
p+q
are pairwise distinct nonzero complex
numbers. If n > p + q and 1 m n, then the mth entry of the first column of
of adj T
n
(b) is
[adj T
n
(b)]
m,1
=
X
J⊂Z,|J|=p
C
J
W
n
J
X
zJ
S
m,J,z
(9)
where Z = {z
1
, . . . , z
p+q
}, the sum is over all sets J Z of cardinality p, and,
with J := Z \ J,
C
J
=
Y
zJ
z
q
Y
zJ,wJ
1
z w
, W
J
= (1)
p
b
p
Y
zJ
z,
S
m,J,z
=
1
b
p
1
z
m
Y
wJ\{z}
1
z w
.
Proof. It suffices to prove (9) under the assumption that det T
n
(b) 6= 0 because
both sides of (9) are continuous functions of z
1
, . . . , z
p+q
. Thus, let det T
n
(b) 6= 0.
We will employ (8) with B = T
n
(b).
Trench [16] proved that [T
1
n
(b)]
m,1
equals
1
b
p
D
{1,...,p+q}
(0, . . . , q 1, q + n, . . . , q + n + p 2, q + n m)
D
{1,...,p+q}
(0, . . . , q 1, q + n, . . . , q + n + p 1)
(10)
where D
{j
1
,...,j
k
}
(s
1
, . . . , s
k
) denotes the determinant
det
z
s
1
j
1
z
s
2
j
1
. . . z
s
k
j
1
z
s
1
j
2
z
s
2
j
2
. . . z
s
k
j
2
.
.
.
.
.
.
.
.
.
z
s
1
j
k
z
s
2
j
k
. . . z
s
k
j
k
.
8
Note that
D
J
(s
1
+ ξ, . . . , s
k
+ ξ) =
Y
jJ
z
ξ
j
!
D
J
(s
1
, . . . , s
k
),
D
{1,2,...,k}
(0, 1, . . . , k 1) =
Y
j,`J
`>j
(z
`
z
j
).
We first consider the denominator of (10). Put Z = {1, . . . , p + q}. Laplace
expansion along the last p columns gives
D
Z
(0, . . . , q 1, q + n, . . . , q + n + p 1)
=
X
JZ,|J|=p
(1)
inv(J,J)
D
J
(q + n, . . . , q + n + p 1)D
J
(0, . . . , q 1)
=
X
JZ,|J|=p
(1)
inv(J,J)
Y
kJ
z
q+n
k
Y
k,`J
`>k
(z
`
z
k
)
Y
k,`J
`>k
(z
`
z
k
),
where inv(J, J) is the number of inversions in the permutation of length p + q
whose first q elements are the elements of the set J in increasing order and whose
last p elements are the elements of the set J in increasing order. A little thought
reveals that inv(J, J) is just the number of pairs (k, `) with k J, ` J, k < `.
We have
Y
jJ,sJ
(z
j
z
s
) =
Y
`J,kJ
`>k
(z
`
z
k
)
Y
kJ,`J
`>k
(z
k
z
`
)
= (1)
inv(J,J)
Y
`J,kJ
`>k
(z
`
z
k
)
Y
`J,kJ
`>k
(z
`
z
k
) (11)
and hence the denominator is equal to
R
n
X
J⊂Z,|J|=p
C
J
W
n
J
with R
n
:=
(1)
pn
b
n
p
Y
`>k
(z
`
z
k
).
A formula by Widom [18], which can also be found in [1], says that
det T
n
(b) =
X
J⊂Z,|J|=p
C
J
W
n
J
.
Consequently, the denominator of (10) is nothing but R
n
det T
n
(b).
9
Let us now turn to the numerator of (10). This time Laplace expansion along
the last p columns yields
D
Z
(0, . . . , q 1, q + n, . . . , q + n + p 2, q + n m)
=
X
JZ,|J|=p
(1)
inv(J,J)
D
J
(q + n, . . . , q + n + p 1, q + n m)D
J
(0, . . . , q 1)
=
X
JZ,|J|=p
(1)
inv(J,J)
D
J
(0, . . . , q 1)
Y
jJ
z
q+n
j
!
D
J
(0, . . . , p 2, m).
Expanding D
J
(0, . . . , p 2, m) by its last column we get
D
J
(0, . . . , p 2, m) =
X
jJ
(1)
inv(J\{j},j)
z
m
j
D
J\{j}
(0, . . . , p 2) (12)
with inv(J \ {j}, j) being the number of s J \ {j} such that s > j. Thus, (12)
is
X
jJ
(1)
inv(J\{j},j)
z
m
j
Y
k,`J\{j}
`>k
(z
`
z
k
)
=
X
jJ
z
m
j
Y
k,`J
`>k
(z
`
z
k
)
Y
sJ\{j}
1
z
j
z
s
.
This in conjunction with (11) shows that the numerator of (9) equals
b
p
R
n
X
J⊂Z,|J|=p
C
J
W
n
J
X
zJ
S
m,J,z
.
In summary, from (10) we obtain that
[T
1
n
(b)]
m,1
=
1
det T
n
(b)
X
J⊂Z,|J|=p
C
J
W
n
J
X
zJ
S
m,J,z
,
which after multiplication by det T
n
(b) becomes (9).
3 The main terms of the first column
We now apply Theorem 2.1 to
b(t) = a(t) λ = a
r
t
r
2r
Y
k=1
(t z
k
(λ)) (13)
10
where λ (α, β). The set Z = Z(λ) is given by (4). Let
d
0
(λ) = (1)
r
a
r
e
(λ)
r1
Y
k=1
u
k
(λ). (14)
In [2], we showed that d
0
(λ) > 0 for all λ (0, M). The dependence on λ will
henceforth frequently be suppressed in notation. Let
J
1
= {u
1
, . . . , u
r1
, e
1
}, J
2
= {u
1
, . . . , u
r1
, e
2
}
and for ν {1, . . . , r 1}, put
J
0
ν
= {u
1
, . . . , u
r1
, 1/u
ν
}.
Lemma 3.1 If J Z, |J| = r, J / {J
1
, J
2
, J
0
1
, . . . , J
0
r1
}, then
|C
J
W
n
j
S
m,J,z
| K
d
n
0
sin ϕ
e
δn
for all z J, n 1, 1 m n, λ (α, β) with some finite constant K that
does not depend on z, n, m, λ.
Proof. If both e
1
and e
2
belong to J, then
J = {u
ν
1
, . . . , u
ν
k
, e
1
, e
2
, 1/u
s
1
, . . . , 1/u
s
`
}
with k + ` = r 2. Since
min
λ[α,β]
min
j
1
6=j
2
|u
j
1
(λ) u
j
2
(λ)| > 0,
we conclude that |C
J
| K
1
. Here and in the following K
i
denotes a finite
constant that is independent of λ [α, β]. We have k r 2 and thus
|W
J
| = |a
r
|
|u
ν
1
. . . u
ν
k
|
|u
s
1
. . . u
s
`
|
d
0
e
δ
|u
s
1
. . . u
s
`
|
. (15)
If z {u
ν
1
, . . . , u
ν
k
, e
1
, e
2
}, then obviously |S
m,J,z
| K
2
/ sin ϕ and hence
|C
J
W
n
J
S
m,J,z
| K
1
K
2
d
n
0
e
δn
sin ϕ
.
In case z {1/u
s
1
, . . . , 1/u
s
`
}, say z = 1/u
s
1
, we have |S
m,J,z
| K
3
|u
ν
1
|
m
, which
gives
|C
J
W
n
J
S
m,J,z
| K
1
K
3
d
n
0
e
δn
|u
ν
1
|
m
|u
ν
1
|
n
K
1
K
3
d
n
0
e
δn
K
1
K
3
d
n
0
e
δn
sin ϕ
.
11
The only other possibility for J is to be of the type
J = {u
ν
1
, . . . , u
ν
k
, e
1
, 1/u
s
1
, . . . , 1/u
s
`
}
with k + ` r 1, k r 2, ` 1. (The case where e
1
is replaced by
e
2
is completely analogous.) This time, |C
J
| K
4
/ sin ϕ and (15) holds again.
For z {u
ν
1
, . . . , u
ν
k
, e
1
} we have |S
m,J,z
| K
5
and thus get the assertion. If
z = 1/u
s
for some s {s
1
, . . . , s
`
}, say s = s
1
, then |S
m,J,z
| K
6
|u
s
1
|
m
, and the
assertion follows as above, too.
Let
d
1
(λ) =
1
|h
λ
(e
1
(λ)
)h
λ
(e
2
(λ)
)|
r1
Y
k,s=1
1
1
u
k
(λ)u
s
(λ)
1
.
It is easily seen that d
1
(λ) > 0 for all λ [0, M].
Lemma 3.2 If λ = λ
(n)
j
(0, M), then
C
J
1
W
n
J
1
S
m,J
1
,e
1
=
d
1
d
n1
0
sin ϕ
(1)
j
Ae
imϕ
1
+ O(e
δn
)
,
C
J
2
W
n
J
2
S
m,J
2
,e
2
=
d
1
d
n1
0
sin ϕ
(1)
j+1
Be
imϕ
2
+ O(e
δn
)
uniformly in m and λ.
Proof. We abbreviate
Q
r1
k=1
to
Q
k
. Clearly,
W
j
1
= (1)
r
a
r
Y
k
u
k
!
e
1
= (1)
r
a
r
Y
k
u
k
!
e
e
= d
0
e
.
We have
C
J
1
=
(
Q
k
u
r
k
) e
irϕ
1
(e
1
e
2
)
Q
k,s
u
k
1
u
s
Q
k
(u
k
e
2
)
Q
k
e
1
1
u
k
=
e
irϕ
1
e
(e
e
)
Q
k,s
1
1
u
k
u
s
Q
k
1
e
2
u
k
e
i(r1)ϕ
1
Q
k
1
e
1
u
k
=
e
2i sin ϕ
Q
k,s
1
1
u
k
u
s
h(e
2
)h(e
1
)
and because
h(e
2
)h(e
1
) = |h(e
1
)h(e
2
)|e
, (16)
it follows that
C
J
1
=
d
1
e
i(ϕ+θ)
2i sin ϕ
.
12
Furthermore,
S
m,J
1
,e
1
=
1
a
r
1
e
imϕ
1
Q
k
(e
1
u
k
)
=
1
a
r
e
imϕ
1
(1)
r1
(
Q
k
u
k
)
Q
k
(1 e
1
/u
k
)
=
e
im(σ+ϕ)
(1)
r
a
r
(
Q
k
u
k
) h(e
1
)
=
e
im(σ+ϕ)
e
d
0
h(e
1
)
Putting things together we arrive at the formula
C
J
1
W
n
J
1
S
m,J
1
,e
1
=
d
1
d
n1
0
sin ϕ
A e
im(σ+ϕ)
e
i((n+1)ϕ+θ)
.
Obviously, σ + ϕ = ϕ
1
. By virtue of (5),
e
i((n+1)ϕ+θ)
= e
j
(1 + O(e
δn
)) = (1)
j
(1 + O(e
δn
)).
This proves the first of the asserted formulas. Analogously,
W
J
2
= d
0
e
, C
J
2
=
d
1
e
i(ϕ+θ)
2i sin ϕ
, S
m,J
2
,e
2
=
e
im(σϕ)
e
d
0
h(e
2
)
,
which gives the second formula.
Lemma 3.3 If 1 ν r 1 and λ = λ
(n)
j
(α, β), then
C
J
1
W
n
J
1
S
m,J
1
,u
ν
+ C
J
2
W
n
J
2
S
m,J
2
,u
ν
=
d
1
d
n1
0
sin ϕ
(1)
j
D
ν
1
u
m
ν
+ O(e
δn
)
uniformly in m and λ.
Proof. By definition,
S
m,J
1
,u
ν
=
1
a
r
1
u
m
ν
(u
ν
e
1
)
Q
s6=ν
(u
ν
u
s
)
=
u
m
ν
(1)
r1
(
Q
k
u
k
) a
r
(u
ν
e
1
)h
0
(u
ν
))
Since h
0
(z) equals
1
u
1
1
z
u
2
. . .
1
z
u
r1
+ . . . +
1
u
r1
1
z
u
1
. . .
1
z
u
r2
,
we obtain that
h
0
(u
ν
) =
1
u
ν
Y
s6=ν
1
u
ν
u
s
.
13
Thus,
S
m,J
1
,u
ν
=
u
m
ν
(1)
r
a
r
(
Q
k
u
k
) (u
ν
e
1
)h
0
(u
ν
)
=
u
m
ν
e
d
0
(u
ν
e
1
)h
0
(u
ν
)
.
Changing ϕ
1
to ϕ
2
we get
S
m,J
2
,u
ν
=
u
m
ν
e
d
0
(u
ν
e
2
)h
0
(u
ν
)
.
These two expressions along with the expressions for C
J
1
, W
J
1
, C
J
2
, W
J
2
derived
in the proof of Lemma 3.2 show that the sum under consideration is
d
1
d
n1
0
2i sin ϕ
u
m
ν
e
h
0
(u
ν
)
e
i((n+1)ϕ+θ)
u
ν
e
1
e
i((n+1)ϕ+θ)
u
ν
e
2
.
Because of (5), the term in brackets equals
(1)
j
1
u
ν
e
1
1
u
ν
e
2
+ O(e
δn
)
= (1)
j
e
2i sin ϕ
(u
ν
e
1
)(u
ν
e
2
)
+ O(e
δn
).
Lemma 3.4 For 1 ν r 1 and λ (α, β),
C
J
0
ν
W
n
J
0
ν
S
m,J
0
ν
,1/u
ν
=
d
1
d
n1
0
sin ϕ
F
ν
e
i(n+1)σ
u
n+1m
ν
.
Proof. We have C
J
0
ν
= (
Q
k
u
r
k
) /(u
r
ν
P
1
P
2
P
3
) with
P
1
=
1
u
ν
e
1
1
u
ν
e
2
=
(u
ν
e
1
)(u
ν
e
2
)
u
2
ν
e
2
,
P
2
=
Y
k
(u
k
e
1
)
Y
k
(u
k
e
2
) =
Y
k
u
2
k
!
h(e
1
)h(e
2
),
P
3
=
Y
s6=ν
1
u
ν
1
u
s
Y
k
Y
s6=ν
u
k
1
u
s
=
1
u
r2
ν
Y
s6=ν
1
u
ν
u
s
Y
k
u
r2
k
!
Y
k
Y
s6=ν
1
1
u
k
u
s
=
1
u
r3
ν
h
0
(u
ν
)
Y
k
u
r2
k
!
1
d
1
|h(e
1
)h(e
2
)|
1
Q
k
(1 1/(u
k
u
ν
))
.
Thus, C
J
0
ν
equals
d
1
e
2
|h(e
1
)h(e
2
)|
u
ν
(u
ν
e
1
)(u
ν
e
2
)h
0
(u
ν
)h(e
1
)h(e
2
)
Y
k
1
1
u
k
u
ν
.
14
Since W
J
0
ν
= d
0
e
/u
ν
and
S
m,J
0
ν
,1/u
ν
=
1
a
r
u
m
ν
1
Q
k
(1/
u
ν
u
k
)
=
u
m
ν
(1)
r
a
r
(
Q
k
u
k
)
Q
k
1
1
u
k
u
ν
=
u
m
ν
e
d
0
Q
k
1
1
u
k
u
ν
,
we obtain that C
J
0
ν
W
n
J
0
ν
S
m,J
0
ν
,1/u
ν
is equal to
d
1
d
n1
0
u
n+1m
e
e
inσ
|h(e
1
)h(e
2
)|
(u
ν
e
1
)(u
ν
e
2
)h
0
(u
ν
)h(e
1
)h(e
2
)
.
Lemma 3.5 If 1 k r 1 and λ (α, β),
C
J
0
ν
W
n
J
0
ν
S
m,J
0
ν
,u
k
=
d
1
d
n1
0
sin ϕ
O(e
δn
)
uniformly in m and λ.
Proof. This time
C
J
0
ν
W
n
J
0
ν
S
m,J
0
ν
,u
k
=
1
a
r
1
u
m
k
1
(u
k
1/u
ν
)
Q
s6=k
(u
k
u
s
)
=
u
m
k
(1)
r1
a
r
u
k
1
1
u
k
u
ν
Q
s6=k
u
s
Q
s6=k
1
u
k
u
s
=
1
d
0
u
m
k
1
1
1
u
k
u
ν
Q
s6=k
1
u
k
u
s
Expressions for C
J
0
ν
and W
J
0
ν
were given in the proof of Lemma 3.4. It follows
that
C
J
0
ν
W
n
J
0
ν
S
m,J
0
ν
,u
k
= G
ν,k
d
1
d
n1
0
sin ϕ
1
u
n+1
ν
u
m
k
where G
ν,k
equals
e
2
e
inσ
|h(e
1
)h(e
2
)|sin ϕ
(u
ν
e
1
)(u
ν
e
2
)h
0
(u
ν
)h(e
1
)h(e
2
)
Q
s6=k
1
1
u
s
u
ν
Q
s6=k
1
u
k
u
s
.
Since
h
0
(u
ν
) =
1
u
ν
Y
s6=ν
1
u
ν
u
s
,
we see that G
ν,k
remains bounded on [α, β]. Finally,
1
|u
n+1
ν
u
m
k
|
1
|u
ν
|
n
e
δn
.
15
Corollary 3.6 If λ = λ
(n)
j
(α, β), then
[adj T
n
(a λ)]
m,1
= (1)
j
d
1
(λ)d
n1
0
(λ)
sin ϕ(λ)
[w
j,m
(λ) + O(e
δn
)]
uniformly in m and λ.
Proof. This follows from Theorem 2.1 and Lemmas 3.1 to 3.5 along with the fact
that d
1
is bounded and bounded away from zero on [α, β].
4 The asymptotics of the eigenvectors
We now prove Theorem 1.1. There is a finite constant K
1
such that |D
ν
| K
1
and |F
ν
| K
1
for all ν and all λ (α, β). Thus, summing up two finite geometric
series, we get
n
X
m=1
D
ν
1
u
m
ν
+ F
ν
(1)
j+1
e
i(n+1)σ
u
n+1m
ν
2
2K
2
1
1
|u
ν
|
2
1 1/|u
ν
|
2(n+1)
1 1/|u
ν
|
2
K
2
for all ν, n, λ. We further have
n
X
m=1
Ae
imϕ
1
Be
imϕ
2
2
=
n
X
m=1
e
imϕ
1
2h(e
1
)
e
imϕ
2
2h(e
2
)
2
=
n
X
m=1
1
4|h(e
1
)|
2
+
1
4|h(e
1
)|
2
n
X
m=1
e
2imϕ
4h(e
1
)h(e
2
)
+
e
2imϕ
4h(e
1
)h(e
2
)
!
.
The first sum is of the form
P
n
m=1
(γ/4) and therefore equals (n/4)γ. Hence,
because of (16) we are left to prove that
n
X
m=1
e
(λ
(n)
j
)
e
2imϕ(λ
(n)
j
)
K
3
(17)
for all n and j such that λ
(n)
j
(α, β). The sum in (17) is
e
i[(n+1)ϕ(λ
(n)
j
)+θ(λ
(n)
j
)]
sin (λ
(n)
j
)
sin ϕ(λ
(n)
j
)
.
Thus, (17) will follow as soon as we have shown that
sin (λ
(n)
j
)
sin ϕ(λ
(n)
j
)
K
3
16
for all n and j in question. From (5) we infer that
(λ
(n)
j
) = πj ϕ(λ
(n)
j
) θ(λ
(n)
j
) + O(e
δn
),
which implies that
sin (λ
(n)
j
) = (1)
j+1
sin
ϕ(λ
(n)
j
) + θ(λ
(n)
j
)
+ O(e
δn
).
Suppose first that 0 < ϕ(λ
(n)
j
) π/2. Then
sin
ϕ(λ
(n)
j
) + θ(λ
(n)
j
)
sin ϕ(λ
(n)
j
)
π
2
|ϕ(λ
(n)
j
) + θ(λ
(n)
j
)|
|ϕ(λ
(n)
j
)|
π
2
1 +
|θ(λ
(n)
j
)|
|ϕ(λ
(n)
j
)|
!
. (18)
In [2], we proved that |θ/ϕ| is bounded on (0, M). Thus, the right-hand side of
(18) is bounded by some K
3
for all n and j. If π/2 < ϕ(λ
(n)
j
) < π, we may replace
(18) by the upper bound
π
2
1 +
|θ(λ
(n)
j
)|
|π ϕ(λ
(n)
j
)|
!
.
We know again from [2] that |θ/(π ϕ)| is bounded on (0, M). This completes
the proof of Theorem 1.1.
Here is the proof of Theorem 1.2. By virtue of Theorem 1.1, kw
j
(λ
(n)
j
)k
2
> 1
whenever n is sufficiently large. Corollary 3.6 therefore implies that the first
column of adj T
n
(a λ
(n)
j
) is nonzero and thus an eigenvector for λ
(n)
j
for all
n n
0
and all 1 j n such that λ
(n)
j
(α, β). Again by Corollary 3.6, the
mth entry of this column is
d
1
(λ)d
n1
0
(λ)
sin ϕ(λ)
[w
j,m
(λ) + ξ
(n)
j,m
]
λ=λ
(n)
j
where |ξ
(n)
j,m
| Ke
δn
for all n and j under consideration and K does not depend
on m, n, j. It follows that
w
j
(λ
(n)
j
) +
ξ
(n)
j,m
n
m=1
= w
j
(λ
(n)
j
) + O
(e
δn
)
is also an eigenvector for λ
(n)
j
. Consequently,
w
j
(λ
(n)
j
) + O
(e
δn
)
kw
j
(λ
(n)
j
) + O
(e
δn
)k
2
=
w
j
(λ
(n)
j
)
kw
j
(λ
(n)
j
)k
2
+ O
(e
δn
) (19)
is a normalized eigenvector for λ
(n)
j
. From (5) we deduce that all eigenvalues
of T
n
(a) are simple. Thus, v
(n)
j
is a scalar multiple of modulus 1 of (19). This
completes the proof of Theorem 1.2.
17
5 Symmetric matrices
The matrices T
n
(a) are all symmetric if and only if all a
k
are real and a
k
= a
k
for all k. Obviously, this is equivalent to the requirement that the real-valued
function g(x) := a(e
ix
) be even, that is, g(x) = g(x) for all x. Thus, suppose g
is even. In that case
ϕ
0
= π, ϕ
1
(λ) = ϕ
2
(λ) = ϕ(λ), σ(λ) = 0.
Moreover, for t T we have
a
r
t
r
2r
Y
k=1
(t z
k
(λ)) = a(t) λ = a(1/t) λ
= a
r
t
r
2r
Y
k=1
(1/t z
k
(λ)) = a
r
2r
Y
k=1
z
k
(λ)
!
t
r
2r
Y
k=1
(t 1/z
k
(λ)),
which in conjunction with (4) implies that
{u
1
(λ), . . . , u
r1
(λ)} = {u
1
(λ), . . . , u
r1
(λ)}. (20)
The coefficients of the polynomial h
λ
(t) are symmetric functions of
1/u
1
(λ), . . . , 1/u
r1
(λ).
From (20) we therefore see that these coefficients are real. It follows in particular
that h
λ
(e
(λ)
) = h
λ
(e
(λ)
), which gives θ(λ) = 2 arg h
λ
(e
(λ)
) and thus
h
λ
(e
(λ)
) = |h
λ
(e
(λ)
)|e
(λ)/2
, h
λ
(e
(λ)
) = |h
λ
(e
(λ)
)|e
(λ)/2
.
We are now in a position to prove Theorem 1.3. To do so, we use Theorem 1.2.
Consider the vector w
j
(λ
(n)
j
). We now have
Ae
imϕ
1
Be
imϕ
2
=
e
imϕ
2ih(e
)
e
imϕ
2ih(e
)
=
1
2i|h(e
)|
e
imϕ
e
iθ/2
e
imϕ
e
iθ/2
=
1
|h(e
)|
sin
+
θ
2
.
Furthermore,
D
ν
=
sin ϕ
(u
ν
e
)(u
ν
e
)h
0
(u
ν
)
=
Q
ν
|h(e
)|
,
F
ν
=
sin ϕ
(u
ν
e
)(u
ν
e
)h
0
(u
ν
)
|h(e
)h(e
)|
h(e
)h(e
)
=
sin ϕ
(u
ν
e
)(u
ν
e
)h
0
(u
ν
)
.
18
Consequently, from (20) we infer that
r1
X
ν=1
F
ν
u
n+1m
ν
=
r1
X
ν=1
D
ν
u
n+1m
ν
=
r1
X
ν=
1
Q
ν
|h(e
)|u
n+1m
ν
.
In summary, it follows that
w
j
(λ
(n)
j
) =
1
|h(e
(λ
(n)
j
)
)|
y
j
(λ
(n)
j
). (21)
Thus, the representation
v
(n)
j
= τ
(n)
j
"
y
j
(λ
(n)
j
)
ky
j
(λ
(n)
j
)k
2
+ O
(e
δn
)
#
is immediate from Theorem 1.2. Finally, put h
j,n
= |h(e
(λ
(n)
j
)
)| = |h(e
(λ
(n)
j
)
)|.
Theorem 1.1 shows that
kw
j
(λ
(n)
j
)k
2
2
=
n
4
1
|h(e
)|
2
+
1
|h(e
)|
2
λ=λ
(n)
j
+ O(1) =
n
2h
2
j,n
+ O(1),
whence, by (21), ky
j
(λ
(n)
j
)k
2
2
= h
2
j,n
kw
j
(λ
(n)
j
)k
2
2
= n/2 + O(1). The proof of
Theorem 1.3 is complete.
Here is the proof of Theorem 1.4. We first estimate the “small terms” in y
(n)
j
.
Summing up finite geometric series and using the assumption that |u
ν
(λ)| are
separated from 1 we come to
n
X
m=1
r1
X
ν=1
Q
ν
(λ)
1
u
ν
(λ)
m
+
(1)
j+1
u
ν
(λ)
n+1m
2
r1
X
ν=1
4(r 1)|Q
ν
(λ)|
2
1 |u
ν
(λ)|
2
K sin
2
ϕ(λ)
where K is some positive number depending only on a. since ϕ(λ
(n)
j
) = O (j/n),
it follows that
r1
X
ν=1
Q
ν
(λ)
1
u
ν
(λ)
m
+
(1)
j+1
u
ν
(λ)
n+1m
!
n
m=1
2
= O
j
n
. (22)
We next consider the difference between the “main term” of y
(n)
j
and sin
mjπ
n+1
.
Using the elementary estimate
|sin A sin B|
2
= 4 sin
2
A B
2
cos
2
A + B
2
4 sin
2
A B
2
= 2 2 cos(A B),
19
we get
n
X
m=1
sin
(λ
(n)
j
) + θ(λ
(n)
j
)
sin
mjπ
n + 1
2
2n 2
n
X
m=1
cos
m
ϕ(λ
(n)
j
)
πj
n + 1
+ θ(λ
(n)
j
)
.
To simplify the last sum, we use that
n
X
m=1
cos( + ω) =
sin
2
cos
(n+1)ξ
2
+ ω
sin
ξ
2
= n
1 + O(n
2
ξ
2
)
1 + O
(n + 1)ξ
2
+ ω
2
!
.
In our case
ω = θ(λ
(n)
j
) = O
q
λ
(n)
j
= O
j
n
,
ξ = ϕ(λ
(n)
j
)
πj
n + 1
=
θ(λ
(n)
j
)
n + 1
+ O(e
) = O
j
n
2
.
Consequently,
n
X
m=1
sin
(λ
(n)
j
) + θ(λ
(n)
j
)
sin
mjπ
n + 1
2
= O
j
2
n
,
that is,
sin
(λ
(n)
j
) + θ(λ
(n)
j
)
sin
mjπ
n + 1
n
m=1
2
= O
j
n
. (23)
Combining (22) and (23)we obtain that
y
(n)
j
r
n + 1
2
x
(n)
j
2
= O
j
n
+ O
j
n
= O
j
n
, (24)
which implies in particular that
ky
(n)
j
k
2
=
r
n + 1
2
1 + O
j
n

. (25)
Clearly, estimates (24) and (25) yield the asserted estimate. This completes the
proof of Theorem 1.4.
20
6 Numerical results
Given T
n
(a), determine the approximate eigenvalue λ
(n)
j,
from the equation
(n + 1)ϕ(λ
(n)
j,
) + θ(λ
(n)
j,
) = πj.
In [2], we proposed an exponentially fast iteration method for solving this equa-
tion. Let w
(n)
j
(λ) C
n
be as in Section 1 and put
w
(n)
j,
=
w
(n)
j
(λ
(n)
j,
)
kw
(n)
j
(λ
(n)
j,
)k
2
.
We define the distance between the normalized eigenvector v
(n)
j
and the normal-
ized vector w
(n)
j,
by
%(v
(n)
j
, w
(n)
j,
) := min
τT
kτv
(n)
j
w
(n)
j,
k
2
=
q
2 2hv
(n)
j
, w
(n)
j,
i
and put
(n)
= max
1jn
|λ
(n)
j
λ
(n)
j,
|,
(n)
v,w
= max
1jn
%(v
(n)
j
, w
(n)
j,
),
(n)
r
= max
1jn
kT
n
(a)w
(n)
j,
) λ
(n)
j,
w
(n)
j,
k
2
.
The tables following below show these errors for three concrete choices of the
generating function a.
For a(t) = 8 5t 5t
1
+ t
2
+ t
2
we have
n = 10 n = 20 n = 50 n = 100 n = 150
(n)
5.4 · 10
7
1.1 · 10
11
5.2 · 10
25
1.7 · 10
46
9.6 · 10
68
(n)
v,w
2.0 · 10
6
1.1 · 10
10
2.0 · 10
23
1.9 · 10
44
2.0 · 10
65
(n)
r
8.0 · 10
6
2.7 · 10
10
3.4 · 10
23
2.2 · 10
44
1.9 · 10
65
If a(t) = 8 + (4 2i)t + (4 2i)t
1
+ it it
1
then
n = 10 n = 20 n = 50 n = 100 n = 150
(n)
3.8 · 10
8
2.8 · 10
13
2.9 · 10
30
5.9 · 10
58
1.6 · 10
85
(n)
v,w
1.8 · 10
7
4.7 · 10
13
2.0 · 10
29
7.0 · 10
57
2.4 · 10
84
(n)
r
5.4 · 10
7
1.3 · 10
12
2.7 · 10
29
6.7 · 10
57
1.9 · 10
84
In the case where a(t) = 24 + (12 3i)t + (12 + 3i)t
1
+ it
3
it
3
we get
n = 10 n = 20 n = 50 n = 100 n = 150
(n)
6.6 · 10
6
1.2 · 10
10
7.6 · 10
24
1.4 · 10
45
3.3 · 10
67
(n)
v,w
1.9 · 10
6
1.3 · 10
10
2.0 · 10
23
7.2 · 10
45
2.8 · 10
66
(n)
r
2.5 · 10
5
8.6 · 10
10
7.3 · 10
23
1.9 · 10
44
5.9 · 10
66
21
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Albrecht ottcher, Fakult¨at f¨ur Mathematik, TU Chemnitz, 09107 Chemnitz,
Germany
aboettch@mathematik.tu-chemnitz.de
Sergei M. Grudsky, Departamento de Matem´aticas, CINVESTAV del I.P.N.,
Apartado Postal 14-740, 07000 M´exico, D.F., exico
grudsky@math.cinvestav.mx
Egor A. Maksimenko, Departamento de Matem´aticas, CINVESTAV del I.P.N.,
Apartado Postal 14-740, 07000 M´exico, D.F., exico
emaximen@math.cinvestav.mx
23
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Given a symmetric polynomial $P$ in $2n$ variables, there exists a unique symmetric polynomial $Q$ in $n$ variables such that \[ P(x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}) =Q(x_1+x_1^{-1},\ldots,x_n+x_n^{-1}). \] We denote this polynomial $Q$ by $\Phi_n(P)$ and show that $\Phi_n$ is an epimorphism of algebras. We compute $\Phi_n(P)$ for several families of symmetric polynomials $P$: symplectic and orthogonal Schur polynomials, elementary symmetric polynomials, complete homogeneous polynomials, and power sums. Some of these formulas were already found by Elouafi (2014) and Lachaud (2016). The polynomials of the form $\Phi_n(\operatorname{s}_{\lambda/\mu}^{(2n)})$, where $\operatorname{s}_{\lambda/\mu}^{(2n)}$ is a skew Schur polynomial in $2n$ variables, arise naturally in the study of the minors of symmetric banded Toeplitz matrices, when the generating symbol is a palindromic Laurent polynomial, and its roots can be written as $x_1,\ldots,x_n,x^{-1}_1,\ldots,x^{-1}_n$. Trench (1987) and Elouafi (2014) found efficient formulas for the determinants of symmetric banded Toeplitz matrices. We show that these formulas are equivalent to the result of Ciucu and Krattenthaler (2009) about the factorization of the characters of classical groups.
... Note that the asymptotic structure of the eigenvectors in the case of n → ∞ is considered in the papers [28], [29], [30]. ...
Preprint
This paper is devoted to the asymptotic behavior of all eigenvalues of Symmetric (in general non Hermitian) Toeplitz matrices with moderately smooth symbols which trace out a simple loop on the complex plane line as the dimension of the matrices increases to infinity. The main result describes the asymptotic structure of all eigenvalues. The constructed expansion is uniform with respect to the number of eigenvalues. Keywords: Toeplitz matrices, eigenvalues, asymptotic expansions
... However, it mostly provides asymptotic results on the spectra. Notably, some results indicate that the eigenvectors of some banded symmetric Toeplitz matrices become, up to a rotation, close to the sinusoidal, almost equi-spaced eigenvectors observed in equations (3.5) and (3.6) [Böttcher et al., 2010, Ekström et al., 2017. ...
Thesis
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