# Extremal Inverse Eigenvalue Problem for a Special Kind of Matrices

Authors: Zhibing Liu; Yeying Xu; Kanmin Wang; Chengfeng Xu
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We consider the following inverse eigenvalue problem: to construct a special kind of matrix (real symmetric doubly arrow matrix) from the minimal and maximal eigenvalues of all its leading principal submatrices. The necessary and sufficient condition for the solvability of the problem is derived. Our results are constructive and they generate algorithmic procedures to construct such matrices.


---

## Body

## 1. Introduction

Peng et al. in [1] solved two inverse eigenvalue problems for symmetric arrow matrices and, in the other article [2], a correction, for one of the problems stated in [1], has been presented as well. In recent paper [3], Nazari and Beiranvand introduced an algorithm to construct symmetric quasi- antibidiagonal matrices that having its given eigenvalues. Pickmann et al. in [4] introduced an algorithm for inverse eigenvalue problem on symmetric tridiagonal matrices. In this paper we introduced symmetric doubly arrow matrix as follows:
(1)
A
=
(
a
1
b
1
⋯
b
s
-
1
0
⋯
0
b
1
a
2
⋯
0
0
⋯
0
⋮
⋮
⋱
0
0
⋯
0
b
s
-
1
0
⋯
a
s
b
s
⋯
b
n
-
1
0
0
⋯
b
s
a
s
+
1
⋯
0
⋮
⋮
⋯
⋮
⋮
⋱
⋮
0
0
0
b
n
-
1
0
⋯
a
n
)
,
a
j
,
b
j
∈
ℝ
,
where b
j
≥
0,  1
≤
s
≤
n. If s
=
1 or s
=
n; then the matrix A of the form (1) is a symmetric arrow matrix as follows:
(2)
B
=
(
a
1
b
1
⋯
b
n
-
1
b
1
a
2
⋯
0
⋮
⋮
⋱
⋮
b
n
-
1
0
⋯
a
n
)
,
a
j
,
b
j
∈
ℝ
.This family of matrices appears in certain symmetric inverse eigenvalue and inverse Sturm-Liouville problems [5, 6], which arise in many applications [7–12], including modern control theory and vibration analysis [7, 8]. In this paper, we construct matrix A of the form (1), from a special kind of spectral information, which only recently is being considered. Since this type of matrix structure generalizes the well-known arrow matrices, we think that it will also become of interest in applications.We will denoteI
j as the identity matrix of order j; A
j as the j
×
j leading principal submatrix of A; P
j
(
λ
) as the characteristic polynomial of A
j; and λ
1
(
j
)
≤
λ
2
(
j
)
≤
⋯
≤
λ
j
(
j
) as the eigenvalues of A
j.We want to solve the following problem.Problem 1.
Given the2
n
-
1 real numbers λ
1
(
j
) and λ
j
(
j
), j
=
1,2
,
…
,
n, find an n
×
n matrix A of the form (1) such that λ
1
(
j
) and λ
j
(
j
) are, respectively, the minimal and maximal eigenvalues of A
j,  j
=
1,2
,
…
,
n.Our work is motivated by the results in [2]. There, the authors solved this kind of inverse eigenvalue problem for symmetric arrow matrix B of the form (2).Theorem 2 (see [2]).
Let the real numbersλ
1
(
j
) and λ
j
(
j
), j
=
1,2
,
…
,
n, be given. Then there exists an n
×
n matrix B of the form (2), such that λ
1
(
j
) and λ
j
(
j
) are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix B
j, j
=
1,2
,
…
,
n, if and only if
(3)
λ
1
(
n
)
≤
⋯
≤
λ
1
(
3
)
≤
λ
1
(
2
)
≤
λ
1
(
1
)
≤
λ
2
(
2
)
≤
λ
3
(
3
)
≤
⋯
≤
λ
n
(
n
)
.Theorem 3 (see [2]).
Let the real numbersλ
1
(
j
) and λ
j
(
j
),  j
=
1,2
,
…
,
n, be given. Then there exists a unique n
×
n matrix B of the form (2), with a
j
∈
ℝ and b
j
>
0, such that λ
1
(
j
) and λ
j
(
j
) are, respectively, the minimal and the maximal eigenvalues of its leading principal submatrix B
j,  j
=
1,2
,
…
,
n, if and only if
(4)
λ
1
(
n
)
<
⋯
<
λ
1
(
3
)
<
λ
1
(
2
)
<
λ
1
(
1
)
<
λ
2
(
2
)
<
λ
3
(
3
)
<
⋯
<
λ
n
(
n
)
.In this paper, we will show that Theorems2 and 3 are also right for symmetric doubly arrow matrix A in (1) by a similar method.The paper is organized as follows. In Section2, we discuss some properties of A. In Section 3, we solve Problem 1 by giving a necessary and sufficient condition for the existence of the matrix A in (1)  and also solve the case in which the matrix A, in Problem 1, is required to have all its entries b
i positive. Finally, In Section 4 we show some examples to illustrate the results.


## 2. Properties of the MatrixA

Lemma 4.
LetA be a matrix of the form (1). Then the sequence of characteristic polynomials {
P
j
(
λ
)
}
j
=
1
n satisfies the recurrence relation:
(5)
p
1
(
λ
)
=
(
λ
-
a
1
)
,
p
j
(
λ
)
=
(
λ
-
a
j
)
p
j
-
1
(
λ
)
-
b
j
-
1
2
∏
i
=
2
j
-
1
‍
(
λ
-
a
i
)
,
222222222222222222
j
=
2,3
,
…
,
s
,
p
j
(
λ
)
=
(
λ
-
a
j
)
p
j
-
1
(
λ
)
-
b
j
-
1
2
∏
i
=
s
+
1
j
-
1
‍
(
λ
-
a
i
)
p
s
-
1
(
λ
)
,
22222222222222222222222
j
=
s
+
1
,
…
,
n
.Proof.
It is easy to verify by expanding the determinant.Lemma 5 (see [2]).
Letp
(
λ
) be a monic polynomial of degree n with all real zeroes. If λ
1 and λ
n are, respectively, the minimal and maximal zeroes of p
(
λ
), then(1)
ifμ
<
λ
1, we have that (
-
1
)
n
p
(
μ
)
>
0;
(2)
ifμ
>
λ
n, we have that p
(
μ
)
>
0.Observe that, from the Cauchy interlacing property, the minimal and the maximal eigenvalue,λ
1
(
j
) and λ
j
(
j
), respectively, of each leading principal submatrix A
j,  j
=
1,2
,
…
,
n, of the matrix A in (1) satisfy the relations
(6)
λ
1
(
n
)
≤
⋯
≤
λ
1
(
3
)
≤
λ
1
(
2
)
≤
λ
1
(
1
)
≤
λ
2
(
2
)
≤
λ
3
(
3
)
≤
⋯
≤
λ
n
(
n
)
,
(7)
λ
1
(
j
)
≤
a
i
≤
λ
j
(
j
)
,
i
=
1,2
,
…
,
j
;
j
=
1,2
,
…
,
n
.Lemma 6.
Let{
P
j
(
λ
)
}
j
=
1
n be the polynomials defined in (5), whose minimal and maximal zeroes, λ
1
(
j
) and λ
j
(
j
),  j
=
1,2
,
…
,
n, respectively, satisfy the relations (6) and (7), and
(8)
h
j
=
{
p
j
-
1
(
λ
1
(
j
)
)
∏
i
=
2
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
-
p
j
-
1
(
λ
j
(
j
)
)
∏
i
=
2
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
j
=
2,3
,
…
,
s
,
p
j
-
1
(
λ
1
(
j
)
)
p
s
-
1
(
λ
j
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
-
p
j
-
1
(
λ
j
(
j
)
)
p
s
-
1
(
λ
1
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
,
j
=
s
+
1
,
…
,
n
.
Then
(9)
h
~
j
=
(
-
1
)
j
-
1
h
j
≥
0
,
j
=
2,3
,
…
,
n
.Proof.
From Lemma5, we have
(10)
(
-
1
)
j
-
1
p
j
-
1
(
λ
1
(
j
)
)
≥
0
,
p
j
-
1
(
λ
j
(
j
)
)
≥
0
,
222222222222222222222222
j
=
2,3
,
…
,
n
,
(
-
1
)
s
-
1
p
s
-
1
(
λ
1
(
j
)
)
≥
0
,
p
s
-
1
(
λ
j
(
j
)
)
≥
0
,
2222222222222222222222
j
=
s
+
1
,
…
,
n
.
Moreover, from (7)
(11)
∏
i
=
2
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
≥
0
,
(
-
1
)
j
∏
i
=
2
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
≥
0
,
2222222222222222222222222222
j
=
2,3
,
…
,
s
,
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
≥
0
,
(
-
1
)
j
-
s
+
1
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
≥
0
,
2222222222222222222222222222222
j
=
s
+
1
,
…
,
n
.
Clearly h
~
j
≥
0 from (10) and (11).Lemma 7 (see [2]).
LetA be a matrix of the form (2) with b
i
≠
0
,
i
=
1,2
,
…
,
n
-
1. Let λ
1
(
j
) and λ
j
(
j
), respectively, be the minimal and maximal eigenvalues of the leading principal submatrix A
j,  j
=
1,2
,
…
,
n, of A. Then
(12)
λ
1
(
j
)
<
⋯
<
λ
1
(
3
)
<
λ
1
(
2
)
<
λ
1
(
1
)
<
λ
2
(
2
)
<
λ
3
(
3
)
<
⋯
<
λ
j
(
j
)
,
λ
1
(
j
)
<
a
i
<
λ
j
(
j
)
,
i
=
2,3
,
…
,
j
for each j
=
2,3
,
…
,
n.


## 3. Solution of Problem1

The following theorem solves Problem1. In particular, the theorem shows that condition (6) is necessary and sufficient for the existence of the matrix A in (1).Theorem 8.
Let the real numbersλ
1
(
j
) and λ
j
(
j
),  j
=
1,2
,
…
,
n, be given. Then there exists an n
×
n matrix A of the form (1), such that λ
1
(
j
) and λ
j
(
j
) are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix A
j,  j
=
1,2
,
…
,
n, if and only if
(13)
λ
1
(
n
)
≤
⋯
≤
λ
1
(
3
)
≤
λ
1
(
2
)
≤
λ
1
(
1
)
≤
λ
2
(
2
)
≤
λ
3
(
3
)
≤
⋯
≤
λ
n
(
n
)
.Proof.
Letλ
1
(
j
) and λ
j
(
j
),  j
=
1,2
,
…
,
n, satisfy (13). Observe that
(14)
A
1
=
(
a
1
)
=
(
λ
1
(
1
)
)
and p
1
(
λ
)
=
λ
-
a
1. From Theorem 2, there exists A
j,  j
=
2
,
…
,
s with λ
1
(
j
) and λ
j
(
j
) as its minimal and maximal eigenvalues, respectively. To show the existence of A
j,  j
=
s
+
1
,
…
,
n with λ
1
(
j
) and λ
j
(
j
) as its minimal and maximal eigenvalues, respectively, is equivalent to showing that the system of equations
(15)
p
j
(
λ
1
(
j
)
)
=
(
λ
1
(
j
)
-
a
j
)
p
j
-
1
(
λ
1
(
j
)
)
-
b
j
-
1
2
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
p
s
-
1
(
λ
1
(
j
)
)
=
0
,
p
j
(
λ
j
(
j
)
)
=
(
λ
j
(
j
)
-
a
j
)
p
j
-
1
(
λ
j
(
j
)
)
-
b
j
-
1
2
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
p
s
-
1
(
λ
j
(
j
)
)
=
0
has real solution a
j and b
j
-
1,  j
=
s
+
1
,
…
,
n. If the determinant
(16)
h
j
=
p
j
-
1
(
λ
1
(
j
)
)
p
s
-
1
(
λ
j
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
-
p
j
-
1
(
λ
j
(
j
)
)
p
s
-
1
(
λ
1
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
of the coefficient matrix of the system (15) is nonzero, then the system has unique solutions a
j and b
j
-
1
2,  j
=
s
+
1
,
…
,
n. In this case, from Lemma 6 we have h
~
j
>
0. By solving the system (15) we obtain
(17)
a
j
=
(
∏
i
=
s
+
1
j
-
1
‍
λ
1
(
j
)
p
j
-
1
(
λ
1
(
j
)
)
p
s
-
1
(
λ
j
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
-
λ
j
(
j
)
p
j
-
1
(
λ
j
(
j
)
)
p
s
-
1
(
λ
1
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
)
(
h
j
)
-
1
,
b
j
-
1
2
=
(
λ
j
(
j
)
-
λ
1
(
j
)
)
p
j
-
1
(
λ
1
(
j
)
)
p
j
-
1
(
λ
j
(
j
)
)
h
j
.
Since
(18)
(
-
1
)
j
-
1
(
λ
j
(
j
)
-
λ
1
(
j
)
)
p
j
-
1
(
λ
1
(
j
)
)
p
j
-
1
(
λ
j
(
j
)
)
≥
0
;
then b
j
-
1 is a real number and therefore, there exists A with the spectral properties required.
Now we will show that, ifh
j
=
0, the system (15) still has a solution. We do this by induction by showing that the rank of the coefficients matrix is equal to the rank of the augmented matrix.
Letj
=
s
+
1. If h
s
+
1
=
0, then
(19)
h
~
s
+
1
=
(
-
1
)
s
h
s
+
1
=
(
-
1
)
s
(
p
s
(
λ
1
(
s
+
1
)
)
p
s
-
1
(
λ
s
+
1
(
s
+
1
)
)
-
p
s
(
λ
s
+
1
(
s
+
1
)
)
p
s
-
1
(
λ
1
(
s
+
1
)
)
)
=
0
,
which, from Lemma 5, is equivalent to
(20)
p
s
(
λ
1
(
s
+
1
)
)
p
s
-
1
(
λ
s
+
1
(
s
+
1
)
)
=
0
,
p
s
(
λ
s
+
1
(
s
+
1
)
)
p
s
-
1
(
λ
1
(
s
+
1
)
)
=
0
.
In this case the augmented matrix is
(21)
(
p
s
(
λ
1
(
s
+
1
)
)
p
s
-
1
(
λ
1
(
s
+
1
)
)
λ
1
(
s
+
1
)
p
s
(
λ
1
(
s
+
1
)
)
p
s
(
λ
s
+
1
(
s
+
1
)
)
p
s
-
1
(
λ
s
+
1
(
s
+
1
)
)
λ
s
+
1
(
s
+
1
)
p
s
(
λ
s
+
1
(
s
+
1
)
)
)
,
and the ranks of both matrices, the coefficient matrix and the augmented matrix, are equal. Hence A
s
+
1 exists.
Now we considerj
≥
s
+
2. If h
j
=
0, then
(22)
h
~
j
=
(
-
1
)
j
-
1
h
j
=
(
-
1
)
j
-
1
(
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
p
j
-
1
(
λ
1
(
j
)
)
p
s
-
1
(
λ
j
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
-
p
j
-
1
(
λ
j
(
j
)
)
p
s
-
1
(
λ
1
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
)
=
0
.
From Lemma 5
(23)
p
j
-
1
(
λ
1
(
j
)
)
=
0
⋁
p
s
-
1
(
λ
j
(
j
)
)
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
=
0
,
p
j
-
1
(
λ
j
(
j
)
)
=
0
⋁
p
s
-
1
(
λ
1
(
j
)
)
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
=
0
.
Then h
j
=
0 leads us to the following cases:(i)
λ
1
(
j
)
=
λ
1
(
j
-
1
)
∧
λ
j
-
1
(
j
-
1
)
=
λ
j
(
j
),
(ii)
λ
1
(
j
)
=
λ
1
(
j
-
1
)
∧
p
s
-
1
(
λ
1
(
j
)
)
∏
i
=
s
+
1
j
-
1
(
λ
1
(
j
)
-
a
i
)
=
0,
(iii)
p
s
-
1
(
λ
j
(
j
)
)
∏
i
=
s
+
1
j
-
1
(
λ
j
(
j
)
-
a
i
)
=
0
∧
λ
j
-
1
(
j
-
1
)
=
λ
j
(
j
),
(iv)
p
s
-
1
(
λ
j
(
j
)
)
∏
i
=
s
+
1
j
-
1
(
λ
j
(
j
)
-
a
i
)
=
0
∧
p
s
-
1
(
λ
1
(
j
)
)
∏
i
=
s
+
1
j
-
1
(
λ
1
(
j
)
-
a
i
)
=
0,and the augmented matrix is
(24)
(
p
j
-
1
(
λ
1
(
j
)
)
p
s
-
1
(
λ
1
(
j
)
)
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
λ
1
(
j
)
p
j
-
1
(
λ
1
(
j
)
)
p
j
-
1
(
λ
j
(
j
)
)
p
s
-
1
(
λ
j
(
j
)
)
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
λ
j
(
j
)
p
j
-
1
(
λ
j
(
j
)
)
)
.
By replacing conditions (i)–(iii) in (24), it is clear that the coefficients matrix and the augmented matrix have the same rank. From condition (iv), the system of (15) becomes
(25)
p
j
-
1
(
λ
1
(
j
)
)
a
j
=
λ
1
(
j
)
p
j
-
1
(
λ
1
(
j
)
)
,
p
j
-
1
(
λ
j
(
j
)
)
a
j
=
λ
j
(
j
)
p
j
-
1
(
λ
j
(
j
)
)
.
If p
j
-
1
(
λ
1
(
j
)
)
≠
0 and p
j
-
1
(
λ
j
(
j
)
)
≠
0, then a
j
=
λ
1
(
j
)
=
λ
j
(
j
) and from (13)
(26)
λ
1
(
j
)
=
λ
1
(
j
-
1
)
=
⋯
=
λ
1
(
1
)
=
⋯
=
λ
j
-
1
(
j
-
1
)
=
λ
j
(
j
)
.
Thus, p
j
-
1
(
λ
1
(
j
)
)
=
p
j
-
1
(
λ
j
(
j
)
)
=
0, which is a contradiction. Hence, under condition (iv) p
j
-
1
(
λ
1
(
j
)
)
=
0 or p
j
-
1
(
λ
j
(
j
)
)
=
0 and therefore the coefficients matrix and the augmented matrix have also the same rank. By taking b
j
-
1
2
≥
0, there exists a j
×
j matrix A
j with the required spectral properties. The necessity comes from the Cauchy interlacing property.We have seen in the proof of Theorem8 that if the determinant h
j of the coefficients matrix of the system (15) is nonzero, then the Problem 1 has a unique solution except for the sign of the b
i entries.Now we solve the Problem1 in the case that the b
i entries are required to be positive. We need the following lemma.Lemma 9.
LetA be a matrix of the form (1) with b
i
≠
0
,
i
=
1,2
,
…
,
n
-
1. Let λ
1
(
j
) and λ
j
(
j
), respectively, be the minimal and maximal eigenvalues of the leading principal submatrix A
j,  j
=
1,2
,
…
,
n, of A. Then
(27)
λ
1
(
j
)
<
⋯
<
λ
1
(
3
)
<
λ
1
(
2
)
<
λ
1
(
1
)
<
λ
2
(
2
)
<
λ
3
(
3
)
<
⋯
<
λ
j
(
j
)
,
λ
1
(
j
)
<
a
i
<
λ
j
(
j
)
,
i
=
2,3
,
…
,
j
for each j
=
2,3
,
…
,
n.Proof.
From Lemma7, (27) hold for j
≤
s. For j
=
s
+
1, we have from (5)
(28)
p
s
+
1
(
λ
)
=
(
λ
-
a
s
+
1
)
p
s
-
b
s
2
p
s
-
1
(
λ
)
.
As b
s
≠
0, then from Lemma 7, we have p
s
+
1
(
λ
1
(
s
)
)
≠
0, p
s
+
1
(
λ
s
(
s
)
)
≠
0, and
(29)
λ
1
(
s
+
1
)
<
λ
1
(
s
)
<
⋯
<
λ
1
(
2
)
<
λ
1
(
1
)
<
λ
2
(
2
)
<
⋯
<
λ
s
(
s
)
<
λ
s
+
1
(
s
+
1
)
.
If λ
1
(
s
+
1
)
=
a
s
+
1 or λ
s
+
1
(
s
+
1
)
=
a
s
+
1, then
(30)
0
=
p
s
+
1
(
a
s
+
1
)
=
-
b
s
2
p
s
-
1
(
a
s
+
1
)
contradicts b
s
≠
0 or (29) and from (7) we have
(31)
λ
1
(
s
+
1
)
<
a
s
+
1
<
λ
s
+
1
(
s
+
1
)
.
Let j
=
s
+
2. Then from (5)
(32)
p
s
+
2
(
λ
1
(
s
+
1
)
)
=
-
b
s
+
1
2
(
λ
1
(
s
+
1
)
-
a
s
+
1
)
p
s
-
1
(
λ
1
(
s
+
1
)
)
≠
0
.
In the same way p
s
+
2
(
λ
s
+
1
(
s
+
1
)
)
≠
0. Hence λ
1
(
s
+
1
) and λ
s
+
1
(
s
+
1
) are not zeroes of p
s
+
2
(
λ
) and from (6)
(33)
λ
1
(
s
+
2
)
<
λ
1
(
s
+
1
)
<
λ
1
(
s
)
<
⋯
<
λ
1
(
2
)
<
λ
1
(
1
)
<
λ
2
(
2
)
<
⋯
<
λ
s
(
s
)
<
λ
s
+
1
(
s
+
1
)
<
λ
s
+
2
(
s
+
2
)
.
Now suppose that λ
1
(
s
+
2
)
=
a
s
+
2. Then
(34)
0
=
p
s
+
2
(
a
s
+
2
)
=
-
b
s
+
1
2
(
a
s
+
2
-
a
s
+
1
)
p
s
-
1
(
a
s
+
2
)
=
-
b
s
+
1
2
(
λ
1
s
+
2
-
a
s
+
1
)
p
s
-
1
(
a
s
+
2
)
contradicts the inequalities (31) and (33). The same  occurs if we assume that λ
s
+
2
(
s
+
2
)
=
a
s
+
2. Then from (7) and Lemma 7 we have
(35)
λ
1
(
s
+
2
)
<
a
i
<
λ
s
+
2
(
s
+
2
)
,
i
=
2,3
,
…
,
s
+
2
.
Now, suppose that (27) hold for s
+
3
≤
j
≤
n
-
1 and consider
(36)
p
j
+
1
(
λ
)
=
(
λ
-
a
j
+
1
)
p
j
(
λ
)
-
b
j
2
∏
i
=
s
+
1
j
‍
(
λ
-
a
i
)
p
s
-
1
(
λ
)
.
Since b
j
≠
0 and λ
1
(
j
)
<
a
i
<
λ
j
(
j
)
,
i
=
2,3
,
…
,
j, then ∏
i
=
s
+
1
j
(
λ
1
j
-
a
i
)
≠
0 and ∏
i
=
s
+
1
j
(
λ
j
j
-
a
i
)
≠
0. Hence neither λ
1
j nor λ
j
j are zeroes of p
j
+
1
(
λ
). Then from (6) we have
(37)
λ
1
(
j
+
1
)
<
λ
1
(
j
)
<
⋯
<
λ
1
(
2
)
<
λ
1
(
1
)
<
λ
2
(
2
)
<
⋯
<
λ
j
(
j
)
<
λ
j
+
1
(
j
+
1
)
.
Finally, if λ
1
(
j
+
1
)
=
a
j
+
1, then
(38)
0
=
p
j
+
1
(
a
j
+
1
)
=
-
b
j
2
∏
i
=
s
+
1
j
‍
(
a
j
+
1
-
a
i
)
p
s
-
1
(
a
j
+
1
)
=
-
b
j
2
∏
i
=
s
+
1
j
‍
(
λ
1
j
+
1
-
a
i
)
p
s
-
1
(
λ
1
j
+
1
)
contradicts (33). Then
(39)
λ
1
(
j
+
1
)
<
a
i
<
λ
j
+
1
(
j
+
1
)
,
i
=
2,3
,
…
,
j
+
1
.The following theorem solves Problem1 with b
j
>
0.Theorem 10.
Let the real numbersλ
1
(
j
) and λ
j
(
j
),  j
=
1,2
,
…
,
n, be given. Then there exists a unique n
×
n matrix A of the form (1), with a
j
∈
ℝ and b
j
>
0, such that λ
1
(
j
) and λ
j
(
j
) are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix A
j, j
=
1,2
,
…
,
n, if and only if
(40)
λ
1
(
n
)
<
⋯
<
λ
1
(
3
)
<
λ
1
(
2
)
<
λ
1
(
1
)
<
λ
2
(
2
)
<
λ
3
(
3
)
<
⋯
<
λ
n
(
n
)
.Proof.
The proof is quite similar to the proof of Theorem8: Let λ
1
(
j
) and λ
j
(
j
), j
=
2
,
…
,
n, satisfy (40). From Theorem 3, there exists A
j,  j
=
2
,
…
,
s with λ
1
(
j
) and λ
j
(
j
) as its minimal and maximal eigenvalues, respectively. To show the existence of A
j,  j
=
s
+
1
,
…
,
n with the required spectral properties, is equivalent to showing that the system of (15) has real solutions a
j and b
j
-
1, with b
j
-
1
>
0
,
j
=
s
+
1
,
s
+
2
,
…
,
n. To do this it is enough to show that the determinant of the coefficients matrix
(41)
h
j
=
p
j
-
1
(
λ
1
(
j
)
)
p
s
-
1
(
λ
j
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
-
p
j
-
1
(
λ
j
(
j
)
)
p
s
-
1
(
λ
1
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)is nonzero.
From Lemmas6 and 9 it follows that h
~
j
=
(
-
1
)
j
-
1
h
j
>
0. Hence h
j
≠
0 and the system (15) has real and unique solutions:
(42)
a
j
=
(
∏
i
=
s
+
1
j
-
1
‍
λ
1
(
j
)
p
j
-
1
(
λ
1
(
j
)
)
p
s
-
1
(
λ
j
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
j
(
j
)
-
a
i
)
-
λ
j
(
j
)
p
j
-
1
(
λ
j
(
j
)
)
p
s
-
1
(
λ
1
(
j
)
)
×
∏
i
=
s
+
1
j
-
1
‍
(
λ
1
(
j
)
-
a
i
)
)
(
h
j
)
-
1
,
b
j
-
1
2
=
(
λ
j
(
j
)
-
λ
1
(
j
)
)
p
j
-
1
(
λ
1
(
j
)
)
p
j
-
1
(
λ
j
(
j
)
)
h
j
,
where
(43)
(
-
1
)
j
-
1
(
λ
j
(
j
)
-
λ
1
(
j
)
)
p
j
-
1
(
λ
1
(
j
)
)
p
j
-
1
(
λ
j
(
j
)
)
>
0
.
Then it is clear that b
j
-
1
2
>
0. Therefore, the b
j
-
1 can be chosen positive and then there exists a unique matrix A
j with the required spectral properties. The necessity of the result comes from Lemma 9.


## 4. Examples

Now we give an algorithm to construct the solutionA of Problem 1.Algorithm.
(1)
Input a positive integers and real numbers λ
1
(
j
) and λ
j
(
j
),  j
=
1,2
,
…
,
n;
(2)
leta
1
=
λ
1
(
1
).  p
1
(
λ
)
=
λ
-
a
1;
(3)
forj
=
2
,
…
,
n, calculate p
j
(
λ
) according to (5);
(4)
computea
j and b
j
-
1 according to (17).Example 1.
The following numbers [2]
(44)
λ
1
(
5
)
λ
1
(
4
)
λ
1
(
3
)
λ
1
(
2
)
λ
1
(
1
)
λ
2
(
2
)
-
11.2369
-
11.1921
-
10.9106
-
8.7760
-
6.0043
-
2.6295
λ
3
(
3
)
λ
4
(
4
)
λ
5
(
5
)
1.8532
8.4266
10.4020
satisfy the necessary and sufficient condition (40) of Theorem 10. Then the doubly arrow matrix with b
i
>
0 and s
=
3 is
(45)
A
=
(
-
6.0043
3.0584
5.2453
3.0584
-
5.4011
5.2453
-
2.3357
4.4747
1.8880
4.4747
6.1414
1.8880
9.8361
)
.
From the above 5
×
5 doubly arrow matrix A, we recompute the spectrum σ
(
A
j
) of its submatrix A
j by MATLAB 7.0, j
=
1,2
,
…
,
5, and get(46)
σ
(
A
1
)
λ
1
(
1
)
¯
=
-
6.0043
,
σ
(
A
2
)
λ
1
(
2
)
¯
=
-
8.7759
λ
2
(
2
)
¯
=
-
2.6295
,
σ
(
A
3
)
λ
1
(
3
)
¯
=
-
10.9106
λ
2
(
3
)
¯
=
-
4.6837
λ
3
(
3
)
¯
=
1.8532
,
σ
(
A
4
)
λ
1
(
4
)
¯
=
-
11.1921
λ
2
(
4
)
¯
=
-
5.0836
λ
3
(
4
)
¯
=
0.2494
λ
4
(
4
)
¯
=
8.4266
,
σ
(
A
5
)
λ
1
(
5
)
¯
=
-
11.2369
λ
2
(
5
)
¯
=
-
5.1390
λ
3
(
5
)
¯
=
0.1384
λ
4
(
5
)
¯
=
8.0719
λ
5
(
5
)
¯
=
10.4020
.Example 2.
We modify the previous example, that some given eigenvalues  become equal to [2]
(47)
λ
1
(
5
)
λ
1
(
4
)
λ
1
(
3
)
λ
1
(
2
)
λ
1
(
1
)
λ
2
(
2
)
-
11.2369
-
10.9106
-
10.9106
-
8.7760
-
6.0043
-
6.0043
λ
3
(
3
)
λ
4
(
4
)
λ
5
(
5
)
1.8532
8.4266
10.4020
.
These numbers satisfy the necessary and sufficient condition (13) of Theorem 8. One solution of Problem 1 with s
=
3 is the matrix(48)
A
=
(
-
6.0043
0
6.2090
0
-
8.7760
6.2090
-
3.0531
0
4.0562
0
8.4266
4.0562
8.9205
)
.
Recomputing the spectrum of A, we have(49)
σ
(
A
1
)
λ
1
(
1
)
¯
=
-
6.0043
,
σ
(
A
2
)
λ
1
(
2
)
¯
=
-
8.7760
λ
2
(
2
)
¯
=
-
6.0043
,
σ
(
A
3
)
λ
1
(
3
)
¯
=
-
10.9106
λ
2
(
3
)
¯
=
-
8.7760
λ
3
(
3
)
¯
=
1.8532
,
σ
(
A
4
)
λ
1
(
4
)
¯
=
-
10.9106
λ
2
(
4
)
¯
=
-
8.7760
λ
3
(
4
)
¯
=
1.8532
λ
4
(
4
)
¯
=
8.4266
,
σ
(
A
5
)
λ
1
(
5
)
¯
=
-
11.2369
λ
2
(
5
)
¯
=
-
8.7760
λ
3
(
5
)
¯
=
0.6980
λ
4
(
5
)
¯
=
8.4266
λ
5
(
5
)
¯
=
10.4020
.



---

*Source: 513513-2014-02-05.xml*