Journal of Applied Mathematics

Volume 2014, Article ID 513513, 7 pages

http://dx.doi.org/10.1155/2014/513513

## Extremal Inverse Eigenvalue Problem for a Special Kind of Matrices

^{1}School of Sciences, Jiujiang University, Jiujiang 332005, China^{2}Faculty of Library, Jiujiang University, Jiujiang 332005, China

Received 11 June 2013; Accepted 13 December 2013; Published 5 February 2014

Academic Editor: Hui-Shen Shen

Copyright © 2014 Zhibing Liu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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.

#### 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: where , . If or ; then the matrix of the form (1) is a symmetric arrow matrix as follows:

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 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 denote as the identity matrix of order ; as the leading principal submatrix of ; as the characteristic polynomial of ; and as the eigenvalues of .

We want to solve the following problem.

*Problem 1. *Given the real numbers and , , find an matrix of the form (1) such that and are, respectively, the minimal and maximal eigenvalues of , .

Our work is motivated by the results in [2]. There, the authors solved this kind of inverse eigenvalue problem for symmetric arrow matrix of the form (2).

Theorem 2 (see [2]). *Let the real numbers and , , be given. Then there exists an matrix of the form (2), such that and are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix , , if and only if
*

*Theorem 3 (see [2]). Let the real numbers and , , be given. Then there exists a unique matrix of the form (2), with and , such that and are, respectively, the minimal and the maximal eigenvalues of its leading principal submatrix , , if and only if
*

*In this paper, we will show that Theorems 2 and 3 are also right for symmetric doubly arrow matrix in (1) by a similar method.*

*The paper is organized as follows. In Section 2, we discuss some properties of . In Section 3, we solve Problem 1 by giving a necessary and sufficient condition for the existence of the matrix in (1) and also solve the case in which the matrix , in Problem 1, is required to have all its entries positive. Finally, In Section 4 we show some examples to illustrate the results.*

*2. Properties of the Matrix *

*2. Properties of the Matrix*

*Lemma 4. Let be a matrix of the form (1). Then the sequence of characteristic polynomials satisfies the recurrence relation:
*

*Proof. *It is easy to verify by expanding the determinant.

*Lemma 5 (see [2]). Let be a monic polynomial of degree with all real zeroes. If and are, respectively, the minimal and maximal zeroes of , then(1)if , we have that ;(2)if , we have that .*

*Observe that, from the Cauchy interlacing property, the minimal and the maximal eigenvalue, and , respectively, of each leading principal submatrix , , of the matrix in (1) satisfy the relations
*

*Lemma 6. Let be the polynomials defined in (5), whose minimal and maximal zeroes, and , , respectively, satisfy the relations (6) and (7), and
Then
*

*Proof. *From Lemma 5, we have
Moreover, from (7)
Clearly from (10) and (11).

*Lemma 7 (see [2]). Let be a matrix of the form (2) with . Let and , respectively, be the minimal and maximal eigenvalues of the leading principal submatrix , , of . Then
for each .*

*3. Solution of Problem 1*

*3. Solution of Problem 1*

*The following theorem solves Problem 1. In particular, the theorem shows that condition (6) is necessary and sufficient for the existence of the matrix in (1).*

*Theorem 8. Let the real numbers and , , be given. Then there exists an matrix of the form (1), such that and are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix , , if and only if
*

*Proof. *Let and , , satisfy (13). Observe that
and . From Theorem 2, there exists , with and as its minimal and maximal eigenvalues, respectively. To show the existence of , with and as its minimal and maximal eigenvalues, respectively, is equivalent to showing that the system of equations
has real solution and , . If the determinant
of the coefficient matrix of the system (15) is nonzero, then the system has unique solutions and , . In this case, from Lemma 6 we have . By solving the system (15) we obtain
Since
then is a real number and therefore, there exists with the spectral properties required.

Now we will show that, if , 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.

Let . If , then
which, from Lemma 5, is equivalent to
In this case the augmented matrix is
and the ranks of both matrices, the coefficient matrix and the augmented matrix, are equal. Hence exists.

Now we consider . If , then
From Lemma 5
Then leads us to the following cases:(i),
(ii),
(iii),
(iv),and the augmented matrix isBy 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
If and , then and from (13)
Thus, , which is a contradiction. Hence, under condition (iv) or and therefore the coefficients matrix and the augmented matrix have also the same rank. By taking , there exists a matrix with the required spectral properties. The necessity comes from the Cauchy interlacing property.

*We have seen in the proof of Theorem 8 that if the determinant of the coefficients matrix of the system (15) is nonzero, then the Problem 1 has a unique solution except for the sign of the entries.*

*Now we solve the Problem 1 in the case that the entries are required to be positive. We need the following lemma.*

*Lemma 9. Let be a matrix of the form (1) with . Let and , respectively, be the minimal and maximal eigenvalues of the leading principal submatrix , , of . Then
for each .*

*Proof. *From Lemma 7, (27) hold for . For , we have from (5)
As , then from Lemma 7, we have , , and
If or , then
contradicts or (29) and from (7) we have
Let . Then from (5)
In the same way . Hence and are not zeroes of and from (6)
Now suppose that . Then
contradicts the inequalities (31) and (33). The same occurs if we assume that . Then from (7) and Lemma 7 we have
Now, suppose that (27) hold for and consider
Since and , then and . Hence neither nor are zeroes of . Then from (6) we have
Finally, if , then
contradicts (33). Then

*The following theorem solves Problem 1 with . *

*Theorem 10. Let the real numbers and , , be given. Then there exists a unique matrix of the form (1), with and , such that and are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix , , if and only if
*

*Proof. *The proof is quite similar to the proof of Theorem 8: Let and , , satisfy (40). From Theorem 3, there exists , with and as its minimal and maximal eigenvalues, respectively. To show the existence of , with the required spectral properties, is equivalent to showing that the system of (15) has real solutions and , with . To do this it is enough to show that the determinant of the coefficients matrix
is nonzero.

From Lemmas 6 and 9 it follows that . Hence and the system (15) has real and unique solutions:
where
Then it is clear that . Therefore, the can be chosen positive and then there exists a unique matrix with the required spectral properties. The necessity of the result comes from Lemma 9.

*4. Examples*

*4. Examples**Now we give an algorithm to construct the solution of Problem 1.*

*Algorithm.*(1)Input a positive integer and real numbers and , ;(2)let . ;(3)for , calculate according to (5);(4)compute and according to (17).

*Example 1. *The following numbers [2]satisfy the necessary and sufficient condition (40) of Theorem 10. Then the doubly arrow matrix with and is
From the above doubly arrow matrix , we recompute the spectrum of its submatrix by MATLAB 7.0, , and get

*Example 2. *We modify the previous example, that some given eigenvalues become equal to [2]These numbers satisfy the necessary and sufficient condition (13) of Theorem 8. One solution of Problem 1 with is the matrix
Recomputing the spectrum of , we have

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments*

*This work is supported by the Natural Science Foundation of Jiangxi, China (nos. 20114BAB201015, 20122BAB201013, and 20132BAB201056), and Scientific and Technological Project of Jiangxi Education Office, China (no. KJLD13093).*

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