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:
(1)A=(a1b1⋯bs-10⋯0b1a2⋯00⋯0⋮⋮⋱00⋯0bs-10⋯asbs⋯bn-100⋯bsas+1⋯0⋮⋮⋯⋮⋮⋱⋮000bn-10⋯an),aj,bj∈ℝ,
where bj≥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=(a1b1⋯bn-1b1a2⋯0⋮⋮⋱⋮bn-10⋯an),aj,bj∈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 denote Ij as the identity matrix of order j; Aj as the j×j leading principal submatrix of A; Pj(λ) as the characteristic polynomial of Aj; and λ1(j)≤λ2(j)≤⋯≤λj(j) as the eigenvalues of Aj.

We want to solve the following problem.

Problem 1.

Given the 2n-1 real numbers λ1(j) and λj(j), j=1,2,…,n, find an n×n matrix A of the form (1) such that λ1(j) and λj(j) are, respectively, the minimal and maximal eigenvalues of Aj, 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 [<xref ref-type="bibr" rid="B11">2</xref>]).

Let the real numbers λ1(j) and λj(j), j=1,2,…,n, be given. Then there exists an n×n matrix B of the form (2), such that λ1(j) and λj(j) are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix Bj, j=1,2,…,n, if and only if
(3)λ1(n)≤⋯≤λ1(3)≤λ1(2)≤λ1(1)≤λ2(2)≤λ3(3)≤⋯≤λn(n).

Theorem 3 (see [<xref ref-type="bibr" rid="B11">2</xref>]).

Let the real numbers λ1(j) and λj(j), j=1,2,…,n, be given. Then there exists a unique n×n matrix B of the form (2), with aj∈ℝ and bj>0, such that λ1(j) and λj(j) are, respectively, the minimal and the maximal eigenvalues of its leading principal submatrix Bj, j=1,2,…,n, if and only if
(4)λ1(n)<⋯<λ1(3)<λ1(2)<λ1(1)<λ2(2)<λ3(3)<⋯<λn(n).

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

The paper is organized as follows. In Section 2, 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 bi positive. Finally, In Section 4 we show some examples to illustrate the results.

2. Properties of the Matrix <inline-formula>
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M56">
<mml:mrow>
<mml:mi>A</mml:mi></mml:mrow>
</mml:math></inline-formula>
Lemma 4.

Let A be a matrix of the form (1). Then the sequence of characteristic polynomials {Pj(λ)}j=1n satisfies the recurrence relation:
(5)p1(λ)=(λ-a1),pj(λ)=(λ-aj)pj-1(λ)-bj-12∏i=2j-1(λ-ai),222222222222222222j=2,3,…,s,pj(λ)=(λ-aj)pj-1(λ)-bj-12∏i=s+1j-1(λ-ai)ps-1(λ),22222222222222222222222j=s+1,…,n.

Proof.

It is easy to verify by expanding the determinant.

Lemma 5 (see [<xref ref-type="bibr" rid="B11">2</xref>]).

Let p(λ) be a monic polynomial of degree n with all real zeroes. If λ1 and λn are, respectively, the minimal and maximal zeroes of p(λ), then

if μ<λ1, we have that (-1)np(μ)>0;

if μ>λn, we have that p(μ)>0.

Observe that, from the Cauchy interlacing property, the minimal and the maximal eigenvalue, λ1(j) and λj(j), respectively, of each leading principal submatrix Aj, j=1,2,…,n, of the matrix A in (1) satisfy the relations
(6)λ1(n)≤⋯≤λ1(3)≤λ1(2)≤λ1(1)≤λ2(2)≤λ3(3)≤⋯≤λn(n),(7)λ1(j)≤ai≤λj(j),i=1,2,…,j;j=1,2,…,n.

Lemma 6.

Let {Pj(λ)}j=1n be the polynomials defined in (5), whose minimal and maximal zeroes, λ1(j) and λj(j), j=1,2,…,n, respectively, satisfy the relations (6) and (7), and
(8)hj={pj-1(λ1(j))∏i=2j-1(λj(j)-ai)-pj-1(λj(j))∏i=2j-1(λ1(j)-ai),mmmmmmmmmmmmmmmmmmj=2,3,…,s,pj-1(λ1(j))ps-1(λj(j))×∏i=s+1j-1(λj(j)-ai)-pj-1(λj(j))ps-1(λ1(j))×∏i=s+1j-1(λ1(j)-ai),j=s+1,…,n.
Then
(9)h~j=(-1)j-1hj≥0,j=2,3,…,n.

Proof.

From Lemma 5, we have
(10)(-1)j-1pj-1(λ1(j))≥0,pj-1(λj(j))≥0,222222222222222222222222j=2,3,…,n,(-1)s-1ps-1(λ1(j))≥0,ps-1(λj(j))≥0,2222222222222222222222j=s+1,…,n.
Moreover, from (7)
(11)∏i=2j-1(λj(j)-ai)≥0,(-1)j∏i=2j-1(λ1(j)-ai)≥0,2222222222222222222222222222j=2,3,…,s,∏i=s+1j-1(λj(j)-ai)≥0,(-1)j-s+1∏i=s+1j-1(λ1(j)-ai)≥0,2222222222222222222222222222222j=s+1,…,n.
Clearly h~j≥0 from (10) and (11).

Lemma 7 (see [<xref ref-type="bibr" rid="B11">2</xref>]).

Let A be a matrix of the form (2) with bi≠0,i=1,2,…,n-1. Let λ1(j) and λj(j), respectively, be the minimal and maximal eigenvalues of the leading principal submatrix Aj, j=1,2,…,n, of A. Then
(12)λ1(j)<⋯<λ1(3)<λ1(2)<λ1(1)<λ2(2)<λ3(3)<⋯<λj(j),λ1(j)<ai<λj(j),i=2,3,…,j
for each j=2,3,…,n.

3. Solution of Problem <xref ref-type="statement" rid="problem1">1</xref>

The following theorem solves Problem 1. 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 numbers λ1(j) and λj(j), j=1,2,…,n, be given. Then there exists an n×n matrix A of the form (1), such that λ1(j) and λj(j) are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix Aj, j=1,2,…,n, if and only if
(13)λ1(n)≤⋯≤λ1(3)≤λ1(2)≤λ1(1)≤λ2(2)≤λ3(3)≤⋯≤λn(n).

Proof.

Let λ1(j) and λj(j), j=1,2,…,n, satisfy (13). Observe that
(14)A1=(a1)=(λ1(1))
and p1(λ)=λ-a1. From Theorem 2, there exists Aj, j=2,…,s with λ1(j) and λj(j) as its minimal and maximal eigenvalues, respectively. To show the existence of Aj, j=s+1,…,n with λ1(j) and λj(j) as its minimal and maximal eigenvalues, respectively, is equivalent to showing that the system of equations
(15)pj(λ1(j))=(λ1(j)-aj)pj-1(λ1(j))-bj-12∏i=s+1j-1(λ1(j)-ai)ps-1(λ1(j))=0,pj(λj(j))=(λj(j)-aj)pj-1(λj(j))-bj-12∏i=s+1j-1(λj(j)-ai)ps-1(λj(j))=0
has real solution aj and bj-1, j=s+1,…,n. If the determinant
(16)hj=pj-1(λ1(j))ps-1(λj(j))×∏i=s+1j-1(λj(j)-ai)-pj-1(λj(j))ps-1(λ1(j))×∏i=s+1j-1(λ1(j)-ai)
of the coefficient matrix of the system (15) is nonzero, then the system has unique solutions aj and bj-12, j=s+1,…,n. In this case, from Lemma 6 we have h~j>0. By solving the system (15) we obtain
(17)aj=(∏i=s+1j-1λ1(j)pj-1(λ1(j))ps-1(λj(j))×∏i=s+1j-1(λj(j)-ai)-λj(j)pj-1(λj(j))ps-1(λ1(j))×∏i=s+1j-1(λ1(j)-ai))(hj)-1,bj-12=(λj(j)-λ1(j))pj-1(λ1(j))pj-1(λj(j))hj.
Since
(18)(-1)j-1(λj(j)-λ1(j))pj-1(λ1(j))pj-1(λj(j))≥0;
then bj-1 is a real number and therefore, there exists A with the spectral properties required.

Now we will show that, if hj=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.

Let j=s+1. If hs+1=0, then
(19)h~s+1=(-1)shs+1=(-1)s(ps(λ1(s+1))ps-1(λs+1(s+1))-ps(λs+1(s+1))ps-1(λ1(s+1)))=0,
which, from Lemma 5, is equivalent to
(20)ps(λ1(s+1))ps-1(λs+1(s+1))=0,ps(λs+1(s+1))ps-1(λ1(s+1))=0.
In this case the augmented matrix is
(21)(ps(λ1(s+1))ps-1(λ1(s+1))λ1(s+1)ps(λ1(s+1))ps(λs+1(s+1))ps-1(λs+1(s+1))λs+1(s+1)ps(λs+1(s+1))),
and the ranks of both matrices, the coefficient matrix and the augmented matrix, are equal. Hence As+1 exists.

Now we consider j≥s+2. If hj=0, then
(22)h~j=(-1)j-1hj=(-1)j-1(∏i=s+1j-1(λ1(j)-ai)pj-1(λ1(j))ps-1(λj(j))×∏i=s+1j-1(λj(j)-ai)-pj-1(λj(j))ps-1(λ1(j))×∏i=s+1j-1(λ1(j)-ai))=0.
From Lemma 5(23)pj-1(λ1(j))=0⋁ps-1(λj(j))∏i=s+1j-1(λj(j)-ai)=0,pj-1(λj(j))=0⋁ps-1(λ1(j))∏i=s+1j-1(λ1(j)-ai)=0.
Then hj=0 leads us to the following cases:

and the augmented matrix is
(24)(pj-1(λ1(j))ps-1(λ1(j))∏i=s+1j-1(λ1(j)-ai)λ1(j)pj-1(λ1(j))pj-1(λj(j))ps-1(λj(j))∏i=s+1j-1(λj(j)-ai)λj(j)pj-1(λj(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)pj-1(λ1(j))aj=λ1(j)pj-1(λ1(j)),pj-1(λj(j))aj=λj(j)pj-1(λj(j)).
If pj-1(λ1(j))≠0 and pj-1(λj(j))≠0, then aj=λ1(j)=λj(j) and from (13)
(26)λ1(j)=λ1(j-1)=⋯=λ1(1)=⋯=λj-1(j-1)=λj(j).
Thus, pj-1(λ1(j))=pj-1(λj(j))=0, which is a contradiction. Hence, under condition (iv) pj-1(λ1(j))=0 or pj-1(λj(j))=0 and therefore the coefficients matrix and the augmented matrix have also the same rank. By taking bj-12≥0, there exists a j×j matrix Aj 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 hj of the coefficients matrix of the system (15) is nonzero, then the Problem 1 has a unique solution except for the sign of the bi entries.

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

Lemma 9.

Let A be a matrix of the form (1) with bi≠0,i=1,2,…,n-1. Let λ1(j) and λj(j), respectively, be the minimal and maximal eigenvalues of the leading principal submatrix Aj, j=1,2,…,n, of A. Then
(27)λ1(j)<⋯<λ1(3)<λ1(2)<λ1(1)<λ2(2)<λ3(3)<⋯<λj(j),λ1(j)<ai<λj(j),i=2,3,…,j
for each j=2,3,…,n.

Proof.

From Lemma 7, (27) hold for j≤s. For j=s+1, we have from (5)
(28)ps+1(λ)=(λ-as+1)ps-bs2ps-1(λ).
As bs≠0, then from Lemma 7, we have ps+1(λ1(s))≠0, ps+1(λs(s))≠0, and
(29)λ1(s+1)<λ1(s)<⋯<λ1(2)<λ1(1)<λ2(2)<⋯<λs(s)<λs+1(s+1).
If λ1(s+1)=as+1 or λs+1(s+1)=as+1, then
(30)0=ps+1(as+1)=-bs2ps-1(as+1)
contradicts bs≠0 or (29) and from (7) we have
(31)λ1(s+1)<as+1<λs+1(s+1).
Let j=s+2. Then from (5)
(32)ps+2(λ1(s+1))=-bs+12(λ1(s+1)-as+1)ps-1(λ1(s+1))≠0.
In the same way ps+2(λs+1(s+1))≠0. Hence λ1(s+1) and λs+1(s+1) are not zeroes of ps+2(λ) and from (6)
(33)λ1(s+2)<λ1(s+1)<λ1(s)<⋯<λ1(2)<λ1(1)<λ2(2)<⋯<λs(s)<λs+1(s+1)<λs+2(s+2).
Now suppose that λ1(s+2)=as+2. Then
(34)0=ps+2(as+2)=-bs+12(as+2-as+1)ps-1(as+2)=-bs+12(λ1s+2-as+1)ps-1(as+2)
contradicts the inequalities (31) and (33). The same occurs if we assume that λs+2(s+2)=as+2. Then from (7) and Lemma 7 we have
(35)λ1(s+2)<ai<λs+2(s+2),i=2,3,…,s+2.
Now, suppose that (27) hold for s+3≤j≤n-1 and consider
(36)pj+1(λ)=(λ-aj+1)pj(λ)-bj2∏i=s+1j(λ-ai)ps-1(λ).
Since bj≠0 and λ1(j)<ai<λj(j),i=2,3,…,j, then ∏i=s+1j(λ1j-ai)≠0 and ∏i=s+1j(λjj-ai)≠0. Hence neither λ1j nor λjj are zeroes of pj+1(λ). Then from (6) we have
(37)λ1(j+1)<λ1(j)<⋯<λ1(2)<λ1(1)<λ2(2)<⋯<λj(j)<λj+1(j+1).
Finally, if λ1(j+1)=aj+1, then
(38)0=pj+1(aj+1)=-bj2∏i=s+1j(aj+1-ai)ps-1(aj+1)=-bj2∏i=s+1j(λ1j+1-ai)ps-1(λ1j+1)
contradicts (33). Then
(39)λ1(j+1)<ai<λj+1(j+1),i=2,3,…,j+1.

The following theorem solves Problem 1 with bj>0.

Theorem 10.

Let the real numbers λ1(j) and λj(j), j=1,2,…,n, be given. Then there exists a unique n×n matrix A of the form (1), with aj∈ℝ and bj>0, such that λ1(j) and λj(j) are, respectively, the minimal and maximal eigenvalues of its leading principal submatrix Aj, j=1,2,…,n, if and only if
(40)λ1(n)<⋯<λ1(3)<λ1(2)<λ1(1)<λ2(2)<λ3(3)<⋯<λn(n).

Proof.

The proof is quite similar to the proof of Theorem 8: Let λ1(j) and λj(j), j=2,…,n, satisfy (40). From Theorem 3, there exists Aj, j=2,…,s with λ1(j) and λj(j) as its minimal and maximal eigenvalues, respectively. To show the existence of Aj, j=s+1,…,n with the required spectral properties, is equivalent to showing that the system of (15) has real solutions aj and bj-1, with bj-1>0,j=s+1,s+2,…,n. To do this it is enough to show that the determinant of the coefficients matrix
(41)hj=pj-1(λ1(j))ps-1(λj(j))×∏i=s+1j-1(λj(j)-ai)-pj-1(λj(j))ps-1(λ1(j))×∏i=s+1j-1(λ1(j)-ai)is nonzero.

From Lemmas 6 and 9 it follows that h~j=(-1)j-1hj>0. Hence hj≠0 and the system (15) has real and unique solutions:
(42)aj=(∏i=s+1j-1λ1(j)pj-1(λ1(j))ps-1(λj(j))×∏i=s+1j-1(λj(j)-ai)-λj(j)pj-1(λj(j))ps-1(λ1(j))×∏i=s+1j-1(λ1(j)-ai))(hj)-1,bj-12=(λj(j)-λ1(j))pj-1(λ1(j))pj-1(λj(j))hj,
where
(43)(-1)j-1(λj(j)-λ1(j))pj-1(λ1(j))pj-1(λj(j))>0.
Then it is clear that bj-12>0. Therefore, the bj-1 can be chosen positive and then there exists a unique matrix Aj with the required spectral properties. The necessity of the result comes from Lemma 9.

4. Examples

Now we give an algorithm to construct the solution A of Problem 1.

Algorithm.

Input a positive integer s and real numbers λ1(j) and λj(j), j=1,2,…,n;

let a1=λ1(1). p1(λ)=λ-a1;

for j=2,…,n, calculate pj(λ) according to (5);

compute aj and bj-1 according to (17).

Example 1.

The following numbers [2]
(44)λ1(5)λ1(4)λ1(3)λ1(2)λ1(1)λ2(2)-11.2369-11.1921-10.9106-8.7760-6.0043-2.6295λ3(3)λ4(4)λ5(5)1.85328.426610.4020
satisfy the necessary and sufficient condition (40) of Theorem 10. Then the doubly arrow matrix with bi>0 and s=3 is
(45)A=(-6.00433.05845.24533.0584-5.40115.2453-2.33574.47471.88804.47476.14141.88809.8361).
From the above 5×5 doubly arrow matrix A, we recompute the spectrum σ(Aj) of its submatrix Aj by MATLAB 7.0, j=1,2,…,5, and get(46)σ(A1)λ1(1)¯=-6.0043,σ(A2)λ1(2)¯=-8.7759λ2(2)¯=-2.6295,σ(A3)λ1(3)¯=-10.9106λ2(3)¯=-4.6837λ3(3)¯=1.8532,σ(A4)λ1(4)¯=-11.1921λ2(4)¯=-5.0836λ3(4)¯=0.2494λ4(4)¯=8.4266,σ(A5)λ1(5)¯=-11.2369λ2(5)¯=-5.1390λ3(5)¯=0.1384λ4(5)¯=8.0719λ5(5)¯=10.4020.

Example 2.

We modify the previous example, that some given eigenvalues become equal to [2]
(47)λ1(5)λ1(4)λ1(3)λ1(2)λ1(1)λ2(2)-11.2369-10.9106-10.9106-8.7760-6.0043-6.0043λ3(3)λ4(4)λ5(5)1.85328.426610.4020.
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=(-6.004306.20900-8.77606.2090-3.053104.056208.42664.05628.9205).
Recomputing the spectrum of A, we have(49)σ(A1)λ1(1)¯=-6.0043,σ(A2)λ1(2)¯=-8.7760λ2(2)¯=-6.0043,σ(A3)λ1(3)¯=-10.9106λ2(3)¯=-8.7760λ3(3)¯=1.8532,σ(A4)λ1(4)¯=-10.9106λ2(4)¯=-8.7760λ3(4)¯=1.8532λ4(4)¯=8.4266,σ(A5)λ1(5)¯=-11.2369λ2(5)¯=-8.7760λ3(5)¯=0.6980λ4(5)¯=8.4266λ5(5)¯=10.4020.

Conflict of Interests

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

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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