Construction of H-Symmetric pentadiagonal matrices by three spectra

ABSTRACT

The set of eigenvalues of a square matrix P is denoted by $\sigma (P)$, and the set of eigenvalues of the submatrix obtained from P by deleting its first i rows and columns of P is denoted by ${\sigma }_i(P)$. It is known that a symmetric pentadiagonal matrix may be constructed from $\sigma ,{\sigma }_1$ and ${\sigma }_2$. The pairs $\sigma ,{\sigma }_1$ and ${\sigma }_1,{\sigma }_2$ must interlace; the construction is not unique. In this paper we solve the inverse eigenvalue problem for H-Symmetric pentadiagonal matrices, not necessarily symmetric. Using $\sigma ,{\sigma }_1,{\sigma }_2$, not necessarily interlacing, and $\sigma (P)$ may contain complex eigenvalues, we construct the solution by modified Lanczos algorithm.

KEYWORDS:

• Inverse eigenvalue problem
• modified Lanczos algorithm
• pentadiagonal matrix
• pseudo-symmetric
• H-Symmetric matrix

MSC:

• 15A18
• 65F18

1. Introduction

Let $H=\mathrm{diag}({\delta }_1,{\delta }_2,\dots ,{\delta }_n)$ be a diagonal matrix such that $H^2=I$, where I is identity matrix. In ${\mathbb{C}}^n$ we define the inner product $[x,y{]}_H=y^{\ast }Hx.$ A square real matrix P is said to be H-Symmetric if $H{P}^{T}H=P.$ The matrix $H{P}^{T}H$ is called $H-$adjoint of P and it is denoted by $P^{\mathrm{♯}}$. Let ${ϵ}_1,{ϵ}_2,\dots ,{ϵ}_{n-1}$ be real numbers such that ${ϵ}_i^2=1$, for $i=1,\dots ,n-1$. It can be easily proved that any pentadiagonal real matrix P of the form (1) $P=\left(\begin{array}{cccccc}{a}_1& {ϵ}_1{b}_1& {ϵ}_1{ϵ}_2{c}_1& & & \\ b_1& a_2& {ϵ}_2{b}_2& {ϵ}_2{ϵ}_3{c}_2& & \\ c_1& b_2& a_3& \cdots & & \\ & \ddots & \ddots & \ddots & \ddots & {ϵ}_{n-2}{ϵ}_{n-1}{c}_{n-2}\\ & & & & & {ϵ}_{n-1}{b}_{n-1}\\ & & & c_{n-2}& b_{n-1}& a_n\end{array}\right),$(1) is a H-Symmetric matrix, where $H=\mathrm{diag}(1,{ϵ}_1,{ϵ}_1{ϵ}_2,\dots ,\prod _{i=1}^{n-1}{ϵ}_i)$. It is proved that [1–3], if P is H-Symmetric and $\sigma (P)$ is real and distinct, then the eigenvectors are H-Orthonormal, i.e. there are eigenvectors $u_1,u_2,\dots ,u_n$ of P such that $H{U}^{T}HU=I,\mathrm{where}U=[u_1,u_2,\dots ,u_n].$ If H = I then P is pentadiagonal symmetric matrix. Pentadiagonal symmetric matrices arise in discrete vibrating beams, see [4,5,8,9] for more details. Inverse eigenvalue problems for symmetric pentadiagonal matrices are studied by many authors, for example see [4–7,9,14]. In most cases the reconstruction procedure requires three interlacing real spectra. Using a finite difference method for discretizing non-smooth beams may lead to a nonsymmetric stiffness matrix, see [12]. H-Symmetric pentadiagonal matrices of the form (1(1) $P=\left(\begin{array}{cccccc}{a}_1& {ϵ}_1{b}_1& {ϵ}_1{ϵ}_2{c}_1& & & \\ b_1& a_2& {ϵ}_2{b}_2& {ϵ}_2{ϵ}_3{c}_2& & \\ c_1& b_2& a_3& \cdots & & \\ & \ddots & \ddots & \ddots & \ddots & {ϵ}_{n-2}{ϵ}_{n-1}{c}_{n-2}\\ & & & & & {ϵ}_{n-1}{b}_{n-1}\\ & & & c_{n-2}& b_{n-1}& a_n\end{array}\right),$(1) ) appear in non-hermitian quantum mechanics [11,15]. Other inverse eigenvalue problems for H-symmetric matrices are studied in [13,16–18]. The objective of this paper is to study the inverse eigenvalue problem for H-Symmetric pentadiagonal matrices. Indeed using three given spectra that may or may not have interlacing property, we construct H-Symmetric pentadiagonal matrices of the form (1(1) $P=\left(\begin{array}{cccccc}{a}_1& {ϵ}_1{b}_1& {ϵ}_1{ϵ}_2{c}_1& & & \\ b_1& a_2& {ϵ}_2{b}_2& {ϵ}_2{ϵ}_3{c}_2& & \\ c_1& b_2& a_3& \cdots & & \\ & \ddots & \ddots & \ddots & \ddots & {ϵ}_{n-2}{ϵ}_{n-1}{c}_{n-2}\\ & & & & & {ϵ}_{n-1}{b}_{n-1}\\ & & & c_{n-2}& b_{n-1}& a_n\end{array}\right),$(1) ) such that $\sigma (P),{\sigma }_1(P)$ and ${\sigma }_2(P)$ are the prescribed spectra. We use the modified form of Lanczos algorithm to construct the solution and we show that the solution is not unique. The solution obtained by this algorithm produces eigenvectors that for large size matrices may not be H-Orthonormal. To resolve this case we use a modified Gram–Schmidt orthogonalization procedure to make eigenvectors to be H-Orthonormal.

2. Construction of the solution

In this section, we state the main inverse eigenvalue problem and construct the solution. We consider conditions on the given data such that this problem has solution.

Inverse Problem.

Given three spectra $\{{\lambda }_i{\}}_{i=1}^n,\{{\mu }_i{\}}_{i=1}^{n-1},\{{\nu }_i{\}}_{i=1}^{n-2},$ construct a H-Symmetric pentadiagonal matrix P of the form (1(1) $P=\left(\begin{array}{cccccc}{a}_1& {ϵ}_1{b}_1& {ϵ}_1{ϵ}_2{c}_1& & & \\ b_1& a_2& {ϵ}_2{b}_2& {ϵ}_2{ϵ}_3{c}_2& & \\ c_1& b_2& a_3& \cdots & & \\ & \ddots & \ddots & \ddots & \ddots & {ϵ}_{n-2}{ϵ}_{n-1}{c}_{n-2}\\ & & & & & {ϵ}_{n-1}{b}_{n-1}\\ & & & c_{n-2}& b_{n-1}& a_n\end{array}\right),$(1) ) such that $\sigma (P)=\{{\lambda }_i{\}}_{i=1}^n,{\sigma }_1(P)=\{{\mu }_i{\}}_{i=1}^{n-1},{\sigma }_2(P)=\{{\nu }_i{\}}_{i=1}^{n-2}.$ We give three theorems that lead to the solution of this inverse problem. During the proof we give the conditions on the data for solvability of the inverse problem.

2.1. Real eigendata

In this subsection, we consider the case that the three given spectra are real. We denote the column vector consisting of ith components of the eigenvectors with $q_i=[q_{1i},q_{2i},\dots ,q_{ni}{]}^T$. Indeed $q_{ki}=u_{ik}$ where $\{u_i{\}}_1^n$ are the eigenvectors. In the following lemma, we prove that the vectors $\{q_i{\}}_1^n$ are also H-Orthonormal.

Lemma 2.1.

Let $\{u_i{\}}_1^n$ be H-Orthonormal eigenvectors of the matrix P then, the vectors $\{q_i{\}}_1^n$ are also H-Orthonormal.

Proof.

Suppose $U=[u_1,u_2,\dots ,u_n]$ and $Q=[q_1,q_2,\dots ,q_n]$, then we have $U^T=Q$. The H-Orthonormality of the eigenvectors $u_i$ implies that $H{U}^{T}HU=I$. This implies that HU is invertible (HU is a square matrix), therefore $HUH{U}^T=I$. Thus $H{Q}^{T}HQ=I$. This means that the vectors $q_i$ are H-Orthonormal which completes the proof.

In the following theorem, we construct H-Symmetric pentadiagonal matrix and H-Orthonormal eigenvectors by using eigenvalues and the first and second components of the eigenvectors. The proposed algorithm is called the modified Lanczos algorithm.

Theorem 2.2.

Let P be a H-Symmetric matrix with $H=\mathrm{diag}({\delta }_1,{\delta }_2,\dots ,{\delta }_n).$ Let $\sigma (P)=\{{\lambda }_i{\}}_{i=1}^n$ and $q_1=[q_{11},q_{21},\dots ,q_{n1}{]}^T,q_2=[q_{12},q_{22},\dots ,q_{n2}{]}^T,$ are the vectors of first and second components of the eigenvectors of P, respectively, such that $[q_1,q_2{]}_H=0,[q_1,q_1{]}_H={\delta }_1,[q_2,q_2{]}_H={\delta }_2.$ Then the entries of P and other components of the eigenvectors can be constructed as follows: $\begin{array}{rl}{a}_i& ={\delta }_i\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{ji}^2,b_i={\epsilon }_i{\delta }_{i+1}\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{ji}{q}_{j,i+1},\\ c_i& =\sqrt{D_i},D_i={\delta }_{i+2}\sum _{j=1}^n{\delta }_j[({\lambda }_j-a_i)q_{ji}-c_{i-2}{q}_{j,i-2}-b_{i-1}{q}_{j,i-1}-{ϵ}_i{b}_i{q}_{j,i+1}{]}^2,\\ q_{j,i+2}& =\frac{1}{{ϵ}_i{ϵ}_{i+1}{c}_i}[({\lambda }_j-a_i)q_{ji}-c_{i-2}{q}_{j,i-2}-b_{i-1}{q}_{j,i-1}-{ϵ}_i{b}_i{q}_{j,i+1}],\end{array}$ where $c_0=c_{-1}=b_0=0$.

Proof.

Suppose that $U=[u_1,u_2,\dots ,u_n]$ is a matrix such that its columns are eigenvectors of P and $\mathrm{\Lambda }=\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$ then, $PU=U\mathrm{\Lambda }$. On transposing we find $U^T{P}^T=\mathrm{\Lambda }{U}^T$, where $U^T=Q=\left(\begin{array}{cccc}{q}_{11}& q_{12}& \cdots & q_{1n}\\ q_{21}& q_{22}& \cdots & q_{2n}\\ ⋮& ⋮& \cdots & ⋮\\ q_{n1}& q_{n2}& \cdots & q_{nn}\end{array}\right).$ Thus the eigenvalue equation is as follows (2) $Q{P}^T=\mathrm{\Lambda }Q.$(2) Equating the first column on both sides of Equation (2(2) $Q{P}^T=\mathrm{\Lambda }Q.$(2) ) implies that (3) $a_1{q}_{j1}+{ϵ}_1{b}_1{q}_{j2}+{ϵ}_1{ϵ}_2{c}_1{q}_{j3}={\lambda }_j{q}_{j1}.$(3) Multiplying both sides of (3(3) $a_1{q}_{j1}+{ϵ}_1{b}_1{q}_{j2}+{ϵ}_1{ϵ}_2{c}_1{q}_{j3}={\lambda }_j{q}_{j1}.$(3) ) by ${\delta }_j{q}_{j1}$ and summing up from j = 1 to j = n and using H-Orthonormal property of the vectors $q_1$ and $q_2$ we find $a_1={\delta }_1\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{j1}^2.$ Similarly, multiplying both sides of (3(3) $a_1{q}_{j1}+{ϵ}_1{b}_1{q}_{j2}+{ϵ}_1{ϵ}_2{c}_1{q}_{j3}={\lambda }_j{q}_{j1}.$(3) ) by ${\delta }_j{q}_{j2}$ and summing up from j = 1 to j = n we find ${ϵ}_1{b}_1{\delta }_2=\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{j1}{q}_{j2},$ which implies that $b_1={ϵ}_1{\delta }_2\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{j1}{q}_{j2}.$ Using Equation (3(3) $a_1{q}_{j1}+{ϵ}_1{b}_1{q}_{j2}+{ϵ}_1{ϵ}_2{c}_1{q}_{j3}={\lambda }_j{q}_{j1}.$(3) ) implies that ${ϵ}_1{ϵ}_2{c}_1{q}_{j3}=({\lambda }_j-a_1)q_{j1}-{ϵ}_1{b}_1{q}_{j2}.$ Taking to the power 2 and multiplying the last equation by ${\delta }_j$ and summing up implies that $c_1^2{\delta }_3=\sum _{j=1}^n{\delta }_j[({\lambda }_j-a_1)q_{j1}-{ϵ}_1{b}_1{q}_{j2}{]}^2,$ which concludes $c_1=\sqrt{D_1}$ with $D_1={\delta }_3\sum _{j=1}^n{\delta }_j[({\lambda }_j-a_1)q_{j1}-{ϵ}_1{b}_1{q}_{j2}{]}^2$.Again using Equation (3(3) $a_1{q}_{j1}+{ϵ}_1{b}_1{q}_{j2}+{ϵ}_1{ϵ}_2{c}_1{q}_{j3}={\lambda }_j{q}_{j1}.$(3) ) we find $q_{j3}=\frac{1}{{ϵ}_1{ϵ}_2{c}_1}[({\lambda }_j-a_1)q_{j1}-{ϵ}_1{b}_1{q}_{j2}].$ Now consider the second column of Equation (2(2) $Q{P}^T=\mathrm{\Lambda }Q.$(2) ) which is (4) $b_1{q}_{j1}+a_2{q}_{j2}+{ϵ}_2{b}_2{q}_{j3}+{ϵ}_2{ϵ}_3{c}_2{q}_{j4}={\lambda }_j{q}_{j2}.$(4) Multiplying both sides by ${\delta }_j{q}_{j2}$ and taking sigma, we find $a_2={\delta }_2\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{j2}^2.$ Multiplying (4(4) $b_1{q}_{j1}+a_2{q}_{j2}+{ϵ}_2{b}_2{q}_{j3}+{ϵ}_2{ϵ}_3{c}_2{q}_{j4}={\lambda }_j{q}_{j2}.$(4) ) by ${\delta }_j{q}_{j3}$ and taking sigma we find $b_2={ϵ}_2{\delta }_3\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{j2}{q}_{j3}.$ We can rewrite (4(4) $b_1{q}_{j1}+a_2{q}_{j2}+{ϵ}_2{b}_2{q}_{j3}+{ϵ}_2{ϵ}_3{c}_2{q}_{j4}={\lambda }_j{q}_{j2}.$(4) ) as follows ${ϵ}_2{ϵ}_3{c}_2{q}_{j4}=({\lambda }_j-a_2)q_{j2}-b_1{q}_{j1}-{ϵ}_2{b}_2{q}_{j3}.$ Squaring both sides of the last equation and multiplying by ${\delta }_j$ and summing up implies that $c_2=\sqrt{D_2}$ with $D_2={\delta }_4\sum _{j=1}^n{\delta }_j[({\lambda }_j-a_2)q_{j2}-b_1{q}_{j1}-{ϵ}_2{b}_2{q}_{j3}{]}^2$, and $q_{j4}=\frac{1}{{ϵ}_2{ϵ}_3{c}_2}[({\lambda }_j-a_2)q_{j2}-b_1{q}_{j1}-{ϵ}_2{b}_2{q}_{j3}].$ In general for $i=3,\dots ,n$ equating ith column in Equation (2(2) $Q{P}^T=\mathrm{\Lambda }Q.$(2) ) we obtain (5) $c_{i-2}{q}_{j,i-2}+b_{i-1}{q}_{j,i-1}+a_i{q}_{ji}+{ϵ}_i{b}_i{q}_{j,i+1}+{ϵ}_i{ϵ}_{i+1}{c}_i{q}_{j,i+2}={\lambda }_j{q}_{ji},$(5) where $q_{i-2}$, $q_{i-1}$, $q_i$, $q_{i+1}$, $b_{i-1}$ and $c_{i-2}$ are computed previously and we suppose that $b_n=c_{n-1}=c_n=0$. Multiplying both sides of (5(5) $c_{i-2}{q}_{j,i-2}+b_{i-1}{q}_{j,i-1}+a_i{q}_{ji}+{ϵ}_i{b}_i{q}_{j,i+1}+{ϵ}_i{ϵ}_{i+1}{c}_i{q}_{j,i+2}={\lambda }_j{q}_{ji},$(5) ) by ${\delta }_j{q}_{ji}$ and taking sum over j we find $a_i={\delta }_i\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{ji}^2.$ Again multiplying both sides of (5(5) $c_{i-2}{q}_{j,i-2}+b_{i-1}{q}_{j,i-1}+a_i{q}_{ji}+{ϵ}_i{b}_i{q}_{j,i+1}+{ϵ}_i{ϵ}_{i+1}{c}_i{q}_{j,i+2}={\lambda }_j{q}_{ji},$(5) ) by ${\delta }_j{q}_{j,i+1}$ and taking sum over j we find $b_i={ϵ}_i{\delta }_{i+1}\sum _{j=1}^n{\lambda }_j{\delta }_j{q}_{ji}{q}_{j,i+1}.$ Equation (5(5) $c_{i-2}{q}_{j,i-2}+b_{i-1}{q}_{j,i-1}+a_i{q}_{ji}+{ϵ}_i{b}_i{q}_{j,i+1}+{ϵ}_i{ϵ}_{i+1}{c}_i{q}_{j,i+2}={\lambda }_j{q}_{ji},$(5) ) can be written as follows: ${ϵ}_i{ϵ}_{i+1}{c}_i{q}_{j,i+2}=({\lambda }_j-a_i)q_{ji}-c_{i-2}{q}_{j,i-2}-b_{i-1}{q}_{j,i-1}-{ϵ}_i{b}_i{q}_{j,i+1}.$ Squaring both sides of the last equation, multiplying by ${\delta }_j$, summing up over j and using H-Orthonormality we can find $c_i$ and $q_{j,i+2}$. Thus all entries of the matrix P and the corresponding H-Orthonormal eigenvectors are constructed. Note that for solvability of the inverse problem by this Theorem the data ${\lambda }_i$, ${\delta }_i$ and $q_1,q_2$ must be chosen such that $D_i$ given by theorem to be nonnegative, otherwise the problem has no solution.

Now we are ready to construct the solution of the inverse problems by three given spectra. First, we compute the first entries of the eigenvectors of P by two given spectra.

Theorem 2.3.

Let P be H-Symmetric and $\sigma (P)=\{{\lambda }_i{\}}_{i=1}^n,{\sigma }_1(P)=\{{\mu }_i{\}}_{i=1}^{n-1},$ are real and distinct such that there exists a permutation of $\sigma (P)$ like $\{{\lambda }_{k_i}{\}}_{i=1}^n$ such that ${\delta }_i\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-1}({\mu }_j-{\lambda }_{k_i})}}{{\displaystyle {\mathrm{\Pi }}_{j=1,j\ne i}^n({\lambda }_{k_j}-{\lambda }_{k_i})}}>0,i=1,2,\dots ,n.$ Then, the first entries of the eigenvectors of P i.e. $u_{1i}$ is computed as follows: (6) $u_{1i}=\sqrt{{\delta }_i\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-1}({\mu }_j-{\lambda }_{k_i})}}{{\displaystyle {\mathrm{\Pi }}_{j=1,j\ne i}^n({\lambda }_{k_j}-{\lambda }_{k_i})}}}.$(6)

Proof.

We modify the proof in [10] for H-symmetric case. Let P be H-Symmetric. By definition of the eigenvalues, we have $u^{T}HPu=\lambda u^{T}Hu.$ That is to say that ${\sigma }_1(P)$ is the set of stationary points of the function $u^{T}HPu$ with condition $u^{T}Hu=\delta$ for $\delta =\pm 1$ and $u_{11}=0.$ Thus $\{{\mu }_i{\}}_{i=1}^{n-1}$ are stationary points of the function $f(u)=u^{T}HPu-\lambda u^{T}Hu-2\nu u^T{e}_1,$ where ν is a lagrange multiplier. Thus the equation $\frac{\mathrm{\partial }f}{\mathrm{\partial }u}=0$, implies $HPu+P^{T}Hu-2\lambda Hu-2\nu e_1=0,$ that is $Pu-\lambda u-\nu e_1=0.$ Setting $u=\sum _{i=1}^n{\alpha }_i{u}_{k_i}$, we find $\sum _{i=1}^n({\lambda }_{k_i}-\lambda ){\alpha }_i{u}_{k_i}=\nu e_1.$ Multiplying both sides by $u_{k_j}^{T}H$ implies that $\sum _{i=1}^n({\lambda }_{k_i}-\lambda ){\alpha }_i{u}_{k_j}^{T}H{u}_{k_i}=\nu u_{k_j}^{T}H{e}_1.$ Therefore ${\alpha }_j=\nu \frac{{\delta }_j{u}_{1k_j}}{({\lambda }_{k_j}-\lambda )}.$ Thus $u=\nu \sum _{j=1}^n\frac{{\delta }_j{u}_{1k_j}}{({\lambda }_{k_j}-\lambda )}{u}_{k_j}.$ Equalizing the first components implies that $u_{11}=\nu \sum _{j=1}^n\frac{{\delta }_j{u}_{1k_j}^2}{({\lambda }_{k_j}-\lambda )}.$ Imposing the condition $u_{11}=0$, implies that $\sum _{j=1}^n\frac{{\delta }_j{u}_{1k_j}^2}{(\lambda -{\lambda }_{k_j})}=0.$ Since ${\sigma }_1(P)$ are the roots of the last equation, thus $\sum _{j=1}^n\frac{{\delta }_j{u}_{1k_j}^2}{(\lambda -{\lambda }_{k_j})}=\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-1}(\lambda -{\mu }_j)}}{{\displaystyle {\mathrm{\Pi }}_{i=1}^n(\lambda -{\lambda }_{k_i})}}.$ Multiplying both sides of the last equation by $(\lambda -{\lambda }_{k_i})$ then taking limit $\lambda \to {\lambda }_{k_i}$ we find the required result.

Example 2.4.

Consider the matrix P as follows $P=\left(\begin{array}{cccccc}-3& -22& 4& 0& 0& 0\\ 22& 92& -22& -4& 0& 0\\ 4& 22& 1& 4& 4& 0\\ 0& 4& 4& 9& 4& 4\\ 0& 0& 4& 4& 9& 4\\ 0& 0& 0& 4& 4& 5\end{array}\right).$ The matrix P is H-Symmetric with $H=diag(1,-1,1,1,1,1)$ and we have $\begin{array}{rl}\sigma (P)& =\{0.0179,80.2501,2.0163,4.3221,6.5181,19.3656\},\\ {\sigma }_1(P)& =\{0.0630,4.8283,6.0585,19.1020,85.9482\}.\end{array}$ Conditions of Theorem 2.2 hold and using (6(6) $u_{1i}=\sqrt{{\delta }_i\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-1}({\mu }_j-{\lambda }_{k_i})}}{{\displaystyle {\mathrm{\Pi }}_{j=1,j\ne i}^n({\lambda }_{k_j}-{\lambda }_{k_i})}}}.$(6) ) we obtain, the first components of the eigenvectors of P as follows $q_1=[0.1488,0.2712,0.9623,0.0319,0.3275,0.1310{]}^T.$

Let $P_1$ be a matrix obtained from P by deleting the first row and column. The following theorem finds the relation between the eigenvalues of matrices P and $P_1$.

Theorem 2.5.

Let $H_1=\mathrm{diag}({\delta }_1^1,{\delta }_2^1,\dots ,{\delta }_{n-1}^1)=\mathrm{diag}(1,{ϵ}_2,{ϵ}_2{ϵ}_3,\dots ,{\mathrm{\Pi }}_{i=2}^{n-1}{ϵ}_i)$ and $P_1$ be $H_1-$Symmetric matrix. Suppose $V=[v_1,v_2,\dots ,v_{n-1}]$ is the matrix with columns consisting the eigenvectors of $P_1$. Then the eigenvalues of P are the roots of the following equation: (7) $(a_1-\lambda )-\sum _{i=1}^{n-1}\frac{{\delta }_i^1{ϵ}_1(b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i}{)}^2}{{\mu }_i-\lambda }=0.$(7)

Proof.

Let $X=[x_1,x_2,\dots ,x_n]$ be an arbitrary eigenvector of P, then this vector can be written as follows: (8) $X=x_1{e}_1+p_1\left[\begin{array}{c}0\\ v_{11}\\ v_{21}\\ ⋮\\ v_{n-1,1}\end{array}\right]+\cdots +p_{n-1}\left[\begin{array}{c}0\\ v_{1,n-1}\\ v_{2,n-1}\\ ⋮\\ v_{n-1,n-1}\end{array}\right]=\left[\begin{array}{cccc}1& & & \\ & & & \\ & & & \\ & & & V\end{array}\right]\left[\begin{array}{c}{x}_1\\ p_1\\ ⋮\\ p_{n-1}\end{array}\right],$(8) where $V^{\mathrm{♯}}V=I$ and $V^{\mathrm{♯}}{P}_{1}V=\mathrm{diag}({\mu }_1,{\mu }_2,\dots ,{\mu }_{n-1})$. The equation $PX=\lambda X$ can be written as follows: (9) $\left[\begin{array}{cccc}{a}_1& & & b^{\ast }\\ & & & \\ & & & \\ b& & & P_1\end{array}\right]\left[\begin{array}{cccc}1& & & \\ & & & \\ & & & \\ & & & V\end{array}\right]\left[\begin{array}{c}{x}_1\\ p_1\\ ⋮\\ p_{n-1}\end{array}\right]=\lambda \left[\begin{array}{cccc}1& & & \\ & & & \\ & & & \\ & & & V\end{array}\right]\left[\begin{array}{c}{x}_1\\ p_1\\ ⋮\\ p_{n-1}\end{array}\right],$(9) where $b=[b_1,c_1,0,\dots ,0{]}^T$ and $b^{\ast }=[{ϵ}_1{b}_1,{ϵ}_1{ϵ}_2{c}_1,0,\dots ,0].$ Multiplying both sides by $\left[\begin{array}{cccc}1& & & \\ & & & \\ & & & \\ & & & V^{\mathrm{♯}}\end{array}\right]$ and simple calculation shows that (10) $\{\begin{array}{l}(a_1-\lambda )x_1+{\displaystyle \sum _{i=1}^{n-1}({ϵ}_1{b}_1{v}_{1i}+{ϵ}_1{ϵ}_2{c}_1{v}_{2i})p_i=0,}\\ ({\delta }_i^1{\delta }_1^1{b}_1{v}_{1i}+{\delta }_2^1{\delta }_i^1{c}_1{v}_{2i})x_1+({\mu }_i-\lambda )p_i=0,i=1,2,\dots ,n-1.\end{array}$(10) Combining the last equations together we find $[(a_1-\lambda )-{\displaystyle \sum _{i=1}^{n-1}\frac{{\delta }_i^1{ϵ}_1(b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i}{)}^2}{{\mu }_i-\lambda }}]x_1=0.$ According to Theorem 2.3 and Equation (6(6) $u_{1i}=\sqrt{{\delta }_i\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-1}({\mu }_j-{\lambda }_{k_i})}}{{\displaystyle {\mathrm{\Pi }}_{j=1,j\ne i}^n({\lambda }_{k_j}-{\lambda }_{k_i})}}}.$(6) ), if the matrices P and $P_1$ have distinct eigenvalues then the first entry of the eigenvectors is not zero, i.e. $x_1\ne 0$, thus $a_1-\lambda -{\displaystyle \sum _{i=1}^{n-1}\frac{{\delta }_i^1{ϵ}_1(b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i}{)}^2}{{\mu }_i-\lambda }=0.}$

Remark 2.6.

Suppose that P and $P_1$ have a common eigenvalue ${\lambda }_j={\mu }_k$, then Equation (6(6) $u_{1i}=\sqrt{{\delta }_i\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-1}({\mu }_j-{\lambda }_{k_i})}}{{\displaystyle {\mathrm{\Pi }}_{j=1,j\ne i}^n({\lambda }_{k_j}-{\lambda }_{k_i})}}}.$(6) ) implies that the first entry of the eigenvector corresponding to $\lambda ={\lambda }_j$ is zero. Substituting $\lambda ={\lambda }_j$ and $x_1=0$ in the system (10(10) $\{\begin{array}{l}(a_1-\lambda )x_1+{\displaystyle \sum _{i=1}^{n-1}({ϵ}_1{b}_1{v}_{1i}+{ϵ}_1{ϵ}_2{c}_1{v}_{2i})p_i=0,}\\ ({\delta }_i^1{\delta }_1^1{b}_1{v}_{1i}+{\delta }_2^1{\delta }_i^1{c}_1{v}_{2i})x_1+({\mu }_i-\lambda )p_i=0,i=1,2,\dots ,n-1.\end{array}$(10) ), we obtain $\begin{array}{r}\{\begin{array}{l}{\displaystyle \sum _{i=1}^{n-1}(b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i})p_i=0,}\\ p_i=0,i=1,2,\dots ,n-1,i\ne k,\end{array}⟹(b_1{v}_{1k}+{ϵ}_2{c}_1{v}_{2k})p_k=0.\end{array}$ Note that $p_k$ is different from zero otherwise the eigenvector X in (8(8) $X=x_1{e}_1+p_1\left[\begin{array}{c}0\\ v_{11}\\ v_{21}\\ ⋮\\ v_{n-1,1}\end{array}\right]+\cdots +p_{n-1}\left[\begin{array}{c}0\\ v_{1,n-1}\\ v_{2,n-1}\\ ⋮\\ v_{n-1,n-1}\end{array}\right]=\left[\begin{array}{cccc}1& & & \\ & & & \\ & & & \\ & & & V\end{array}\right]\left[\begin{array}{c}{x}_1\\ p_1\\ ⋮\\ p_{n-1}\end{array}\right],$(8) ) corresponding to the eigenvalue $\lambda ={\lambda }_j$ will be zero which is not true, thus $b_1{v}_{1k}+{ϵ}_2{c}_1{v}_{2k}=0$. Therefore, if P and $P_1$ have the set of common eigenvalues $S=\{{\mu }_{i_1},{\mu }_{i_2},\dots ,{\mu }_{i_k}\}$ then other eigenvalues of P are the roots of the following equation: $a_1-\lambda -{\displaystyle \sum _{i=1,{\mu }_i\notin S}^{n-1}\frac{{\delta }_i^1{ϵ}_1(b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i}{)}^2}{{\mu }_i-\lambda }=0.}$

Note that since $\{{\lambda }_i\}$ are the eigenvalues of Equation (7(7) $(a_1-\lambda )-\sum _{i=1}^{n-1}\frac{{\delta }_i^1{ϵ}_1(b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i}{)}^2}{{\mu }_i-\lambda }=0.$(7) ) thus $a_1-\lambda -{\displaystyle \sum _{i=1}^{n-1}\frac{{\delta }_i^1{ϵ}_1(b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i}{)}^2}{{\mu }_i-\lambda }=\frac{{\displaystyle {\displaystyle {\mathrm{\Pi }}_{i=1}^n({\lambda }_i-\lambda )}}}{{\displaystyle {\mathrm{\Pi }}_{i=1}^{n-1}({\mu }_i-\lambda )}}.}$ Multiplying both sides by $({\mu }_j-\lambda )$ then replacing $\lambda ={\mu }_j$, we find (11) $(b_1{v}_{1j}+{ϵ}_2{c}_1{v}_{2j}{)}^2=-{\delta }_j^1{ϵ}_1\frac{{\displaystyle {\mathrm{\Pi }}_{i=1}^n({\lambda }_i-{\mu }_j)}}{{\displaystyle {\mathrm{\Pi }}_{i=1,i\ne j}^{n-1}({\mu }_i-{\mu }_j)}},j=1,2,\dots ,n-1.$(11) For existence of the solution the right-hand side of the last equations must be nonnegative. Thus, there must be a permutation $\{{\mu }_{k_1},{\mu }_{k_2},\dots ,{\mu }_{k_{n-1}}\}$ of ${\sigma }_1(P)$ such that the following fractions to be negative: ${\delta }_j^1{ϵ}_1\frac{{\mathrm{\Pi }}_{i=1}^n({\lambda }_i-{\mu }_{k_j})}{{\mathrm{\Pi }}_{i=1,i\ne j}^{n-1}({\mu }_{k_i}-{\mu }_{k_j})}<0,j=1,2,\dots ,n-1.$ Now given three spectra, we find the second entries of the eigenvectors in the following theorem.

Theorem 2.7.

Let P be H-Symmetric and $P_1$ be $H_1$-Symmetric such that ${\sigma }_2(P)=\{{\nu }_i{\}}_{i=1}^{n-2}$. Then $u_{2i}=u_{1i}\sum _{j=1}^{n-1}\frac{{\delta }_j^1(b_1{v}_{1j}+{ϵ}_2{c}_1{v}_{2j})}{({\lambda }_i-{\mu }_j)}{v}_{1j},i=1,2,\dots ,n,$ where $b_1{v}_{1j}+{ϵ}_2{c}_1{v}_{2j}$ is computed using (11(11) $(b_1{v}_{1j}+{ϵ}_2{c}_1{v}_{2j}{)}^2=-{\delta }_j^1{ϵ}_1\frac{{\displaystyle {\mathrm{\Pi }}_{i=1}^n({\lambda }_i-{\mu }_j)}}{{\displaystyle {\mathrm{\Pi }}_{i=1,i\ne j}^{n-1}({\mu }_i-{\mu }_j)}},j=1,2,\dots ,n-1.$(11) ).

Proof.

According to Theorem 2.3, we have $\begin{array}{rl}{u}_{1i}& =\sqrt{{\delta }_i\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-1}({\mu }_j-{\lambda }_i)}}{{\displaystyle {\mathrm{\Pi }}_{j=1,j\ne i}^n({\lambda }_j-{\lambda }_i)}}},i=1,2,\dots ,n,\\ v_{1j}& =\sqrt{{\delta }_i^1\frac{{\displaystyle {\mathrm{\Pi }}_{j=1}^{n-2}({\nu }_j-{\mu }_i)}}{{\displaystyle {\mathrm{\Pi }}_{j=1,j\ne i}^{n-1}({\mu }_j-{\mu }_i)}}},j=1,2,\dots ,n-1.\end{array}$ Now let $u_j=\left[\begin{array}{c}{u}_{1j}\\ y_j\end{array}\right],$ where $y_j\in {\mathbb{R}}^{n-1}$and $u_j$ is eigenvector of P corresponding to the ${\lambda }_j$. Thus we have $\left[\begin{array}{cc}{a}_1& b^{\ast }\\ b& P_1\end{array}\right]\left[\begin{array}{c}{u}_{1j}\\ y_j\end{array}\right]={\lambda }_j\left[\begin{array}{c}{u}_{1j}\\ y_j\end{array}\right],$ therefore $a_1{u}_{1j}+b^{\ast }{y}_j={\lambda }_j{u}_{1j},$ and (12) $b{u}_{1j}+P_1{y}_j={\lambda }_j{y}_j.$(12) Since $P_1$ is $H_1$-Symmetric and its eigenvalues are distinct, thus (13) $V^{\mathrm{♯}}{P}_{1}V=M=\mathrm{diag}({\mu }_1,{\mu }_2,\dots ,{\mu }_{n-1}).$(13) Multiplying both sides of Equation (12(12) $b{u}_{1j}+P_1{y}_j={\lambda }_j{y}_j.$(12) ) by $V^{\mathrm{♯}}=H_1{V}^T{H}_1$ implies that $V^{\mathrm{♯}}b{u}_{1j}+V^{\mathrm{♯}}{P}_1{y}_j={\lambda }_j{V}^{\mathrm{♯}}{y}_j.$ Thus $({\lambda }_{j}I-M)V^{\mathrm{♯}}{y}_j=V^{\mathrm{♯}}b{u}_{1j}.$ Therefore $y_j$ can be computed as follows: $y_j=V({\lambda }_{j}I-M{)}^{-1}\left[\begin{array}{c}{\delta }_1^1{\delta }_1^1{b}_1{v}_{11}+{\delta }_1^1{\delta }_2^1{c}_1{v}_{21}\\ {\delta }_1^1{\delta }_2^1{b}_1{v}_{12}+{\delta }_2^1{\delta }_2^1{c}_1{v}_{22}\\ ⋮\\ {\delta }_1^1{\delta }_{n-1}^1{b}_1{v}_{1,n-1}+{\delta }_2^1{\delta }_{n-1}^1{c}_1{v}_{2,n-1}\end{array}\right]u_{1j}.$ Equalizing the first components of both sides in the last equation and simple calculation completes the proof.

Now given the spectrum $\{{\lambda }_i{\}}_{i=1}^n$ and having the first and second components of eigenvectors, i.e. $u_{1i}$ and $u_{2i}$ we can construct the matrix P using modified Lanczos algorithm (Theorem 2.2).

Remark 2.8.

Lanczos algorithm constructs pentadiagonal matrix P and H-Orthonormal eigenvectors. For large scale matrices the constructed eigenvectors might not be H-Orthonormal. To overcome this difficulty we use the modified Gram–Schmidt orthogonalization. This algorithm transforms the vectors $u_1,u_2,\dots ,u_n$ into H-Orthonormal vectors as follows:

1. Put $u_1=\frac{u_1}{\sqrt{u_1^{T}H{u}_1}}$, for $i=2,3,\dots ,n$ do the steps $(2)$ and $(3)$,

2. Define $S_i=u_i-\sum _{j=1}^{i-1}{\delta }_j[u_i,u_j]u_j,$

3. Set $u_i=\frac{S_i}{\sqrt{\mid [S_i,S_i{]}_H\mid }}.$

Remark 2.9.

Due to the fact that the components of $u_{1i}$ and $v_{1i}$ have signs +, −, moreover $b_1{v}_{1j}+c_1{v}_{2j}$ have signs +, −, therefore the solution matrix is not unique. Therefore, in order to show the efficiency of the modified algorithm in the numerical examples, we compare the prescribed eigenvalues with the eigenvalues of the constructed matrix.

2.2. Complex eigenvalues in spectrum

In this section, we consider an inverse eigenvalues problem for pentadiagonal H-Symmetric matrix P using three spectra consisting of complex eigenvalues. That is we want to construct pentadiagonal matrix P such that $\sigma (P)$ may have some complex numbers. Indeed, we construct a pentadiagonal matrix P by prescribed eigendata $\sigma (P),{\sigma }_1(P),{\sigma }_2(P)$ such that ${\sigma }_1(P),{\sigma }_2(P)$ are real and distinct, but $\sigma (P)$ has some complex eigenvalues. Since P has complex eigenvalues, thus the corresponding eigenvectors do not make a H-Orthonormal matrix. Therefore, we can not use the Lanczos algorithm to construct matrix P. Using (11(11) $(b_1{v}_{1j}+{ϵ}_2{c}_1{v}_{2j}{)}^2=-{\delta }_j^1{ϵ}_1\frac{{\displaystyle {\mathrm{\Pi }}_{i=1}^n({\lambda }_i-{\mu }_j)}}{{\displaystyle {\mathrm{\Pi }}_{i=1,i\ne j}^{n-1}({\mu }_i-{\mu }_j)}},j=1,2,\dots ,n-1.$(11) ) we compute $z_i$, where (14) $z_i=b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i},i=1,2,\dots ,n-1,$(14) On the other hand, the eigenvectors of $P_1$ are $H_1$-Orthonormal, thus simple calculation shows that (15) $b_1={\delta }_1^1{\mathbf{v}}_1^T{H}_1\mathbf{z},{\mathbf{v}}_1=[v_{11},v_{12},\dots ,v_{1n-1}],\mathbf{z}=[z_1,z_2,\dots ,z_{n-1}]$(15) From (14(14) $z_i=b_1{v}_{1i}+{ϵ}_2{c}_1{v}_{2i},i=1,2,\dots ,n-1,$(14) ), we find (16) $c_1^2{v}_{2i}^2=(z_i-b_1{v}_{1i}{)}^2,i=1,2,\dots ,n-1.$(16) Multiplying both sides of (16(16) $c_1^2{v}_{2i}^2=(z_i-b_1{v}_{1i}{)}^2,i=1,2,\dots ,n-1.$(16) ) by ${\delta }_i^1$ and taking sum in i and considering the condition $\sum _{i=1}^{n-1}{\delta }_i^1{v}_{2i}^2={\delta }_2^1$, we obtain (17) $c_1=\sqrt{{\delta }_2^1\sum _{i=1}^{n-1}{\delta }_i^1(z_i-b_1{v}_{1i}{)}^2},$(17) therefore, we find $v_{2i}=\frac{z_i-b_1{v}_{1i}}{{ϵ}_2{c}_1},i=1,2,\dots ,n-1.$ Now using the first and the second components of $P_1$, i.e. $v_{1i},v_{2i},$ and the spectrum ${\sigma }_1(P)$ we may use the Lanczos algorithm to construct $P_1$. The entries $b_1$ and $c_1$ are found by (15(15) $b_1={\delta }_1^1{\mathbf{v}}_1^T{H}_1\mathbf{z},{\mathbf{v}}_1=[v_{11},v_{12},\dots ,v_{1n-1}],\mathbf{z}=[z_1,z_2,\dots ,z_{n-1}]$(15) ) and (17(17) $c_1=\sqrt{{\delta }_2^1\sum _{i=1}^{n-1}{\delta }_i^1(z_i-b_1{v}_{1i}{)}^2},$(17) ). Using trace formula we find $a_1$ as follows: (18) $a_1=\sum _{i=1}^n{\lambda }_i-\sum _{i=1}^{n-1}{\mu }_i.$(18) Therefore, P is constructed completely.

3. Numerical examples

In this section, to display the efficiency of the proposed method, some numerical experiments are considered. The numerical computations are performed using MATLAB R2015a on an Intel(R) Core(TM) i5 system. In order to test and evaluate the obtained numerical results, first we consider P as a known matrix and compute the eigendata $\sigma (P)$, ${\sigma }_1(P)$, ${\sigma }_2(P)$ using eig function in Matlab. Then by using these eigendata and proposed method we reconstruct the matrix and call it $P^{(c)}$. Finally, we compare the spectral data of P and $P^{(c)}$.

Example 3.1.

Consider the H-symmetric pentadiagonal matrix of the form (1(1) $P=\left(\begin{array}{cccccc}{a}_1& {ϵ}_1{b}_1& {ϵ}_1{ϵ}_2{c}_1& & & \\ b_1& a_2& {ϵ}_2{b}_2& {ϵ}_2{ϵ}_3{c}_2& & \\ c_1& b_2& a_3& \cdots & & \\ & \ddots & \ddots & \ddots & \ddots & {ϵ}_{n-2}{ϵ}_{n-1}{c}_{n-2}\\ & & & & & {ϵ}_{n-1}{b}_{n-1}\\ & & & c_{n-2}& b_{n-1}& a_n\end{array}\right),$(1) ) with entries $a_i=\{\begin{array}{l}20,i=1,\\ 6,i=2,\dots ,n-1,\\ 5,i=n,\end{array}{b}_i=-4,c_i=1,{ϵ}_i=\{\begin{array}{l}-1,i=1,\\ 1,i=2,\dots ,n-1.\end{array}$ In Tables 1 and 2, for different values of n, we compared the spectra of computed matrix $P^{(c)}$ with the initial given matrix P.

Table 1. Numerical results for Example 3.1 without modified Gram–Schmidt method.

n	$\Vert \sigma (P^{(c)})-\sigma (P)\Vert$	$\Vert {\sigma }_1(P^{(c)})-{\sigma }_1(P)\Vert$	$\Vert {\sigma }_2(P^{(c)})-{\sigma }_2(P)\Vert$
10	1.41e$-07$	1.37e$-07$	1.45e$-07$
50	165.59	166.77	166.88
100	337.07	338.06	338.20
200	374.86	375.59	375.72

Table 2. Numerical results for Example 3.1 with applying modified Gram–Schmidt method.

n	$\Vert \sigma (P^{(c)})-\sigma (P)\Vert$	$\Vert {\sigma }_1(P^{(c)})-{\sigma }_1(P)\Vert$	$\Vert {\sigma }_2(P^{(c)})-{\sigma }_2(P)\Vert$
10	4.06e−14	1.19e−14	9.74e−15
50	1.35e−13	3.12e−14	3.50e−14
100	1.55e−13	4.12e−14	4.35e−14
200	4.30e−13	5.51e−14	5.71e−14

We consider the following numerical examples where the given matrix P has some complex eigenvalues.

Example 3.2.

Consider the H-symmetric pentadiagonal matrix of the form (1(1) $P=\left(\begin{array}{cccccc}{a}_1& {ϵ}_1{b}_1& {ϵ}_1{ϵ}_2{c}_1& & & \\ b_1& a_2& {ϵ}_2{b}_2& {ϵ}_2{ϵ}_3{c}_2& & \\ c_1& b_2& a_3& \cdots & & \\ & \ddots & \ddots & \ddots & \ddots & {ϵ}_{n-2}{ϵ}_{n-1}{c}_{n-2}\\ & & & & & {ϵ}_{n-1}{b}_{n-1}\\ & & & c_{n-2}& b_{n-1}& a_n\end{array}\right),$(1) ) with entries $a_i=\{\begin{array}{l}-1,i=1,\\ 20,i=2,\\ 9,i=3,\dots ,n-1,\\ 5,i=n,\end{array}{b}_i=c_i=4,{ϵ}_i=\{\begin{array}{l}-1,i=1,\\ 1,i=2,\dots ,n-1.\end{array}$ In Tables 3 and 4, for different values of n, we compared the spectra of computed matrix $P^{(c)}$ with the initial given matrix P.

Table 3. Numerical results for Example 3.2 without Modified Gram–Schmidt method.

n	$\Vert \sigma (P^{(c)})-\sigma (P)\Vert$	$\Vert {\sigma }_1(P^{(c)})-{\sigma }_1(P)\Vert$	$\Vert {\sigma }_2(P^{(c)})-{\sigma }_2(P)\Vert$
10	2.16e−12	2.22e−12	2.33e−12
50	7.0467	5.5257	5.5861
100	52.7912	52.6453	52.7120
200	77.5283	77.4396	77.4848

Table 4. Numerical results for Example 3.2 with applying Modified Gram–Schmidt method.

n	$\Vert \sigma (P^{(c)})-\sigma (P)\Vert$	$\Vert {\sigma }_1(P^{(c)})-{\sigma }_1(P)\Vert$	$\Vert {\sigma }_2(P^{(c)})-{\sigma }_2(P)\Vert$
10	5.48e−14	4.48e−14	3.80e−14
50	2.38e−13	2.00e−13	1.82e−13
100	2.75e−13	3.62e−13	2.68e−13
200	1.05e−12	5.86e−13	5.19e−13

Example 3.3.

Consider the H-symmetric pentadiagonal matrix of the form (1(1) $P=\left(\begin{array}{cccccc}{a}_1& {ϵ}_1{b}_1& {ϵ}_1{ϵ}_2{c}_1& & & \\ b_1& a_2& {ϵ}_2{b}_2& {ϵ}_2{ϵ}_3{c}_2& & \\ c_1& b_2& a_3& \cdots & & \\ & \ddots & \ddots & \ddots & \ddots & {ϵ}_{n-2}{ϵ}_{n-1}{c}_{n-2}\\ & & & & & {ϵ}_{n-1}{b}_{n-1}\\ & & & c_{n-2}& b_{n-1}& a_n\end{array}\right),$(1) ) with entries $\begin{array}{rl}& a_i=\{\begin{array}{l}-1,i=1,\\ 20,i=2,\\ 19,i=3,\dots ,n-1,\\ 18,i=n,\end{array}{b}_i=\{\begin{array}{l}12,i=1,n-1\\ 6,i=2,\dots ,n-2,\end{array}{c}_i=9,\\ & {ϵ}_i=\{\begin{array}{l}-1,i=1,\\ 1,i=2,\dots ,n-1.\end{array}\end{array}$ In Tables 5 and 6, for different values of n, we compared the spectra of computed matrix $P^{(c)}$ with the initial given matrix P.

Table 5. Numerical results for Example 3.3 without Modified Gram–Schmidt method.

n	$\Vert \sigma (P^{(c)})-\sigma (P)\Vert$	$\Vert {\sigma }_1(P^{(c)})-{\sigma }_1(P)\Vert$	$\Vert {\sigma }_2(P^{(c)})-{\sigma }_2(P)\Vert$
10	2.86e−12	2.68e−12	2.17e−12
50	2.57e−04	2.96e−04	2.71e−04
100	37.84	32.82	32.99
200	106.71	104.99	105.15

Table 6. Numerical results for Example 3.3 with applying Modified Gram–Schmidt method.

n	$\Vert \sigma (P^{(c)})-\sigma (P)\Vert$	$\Vert {\sigma }_1(P^{(c)})-{\sigma }_1(P)\Vert$	$\Vert {\sigma }_2(P^{(c)})-{\sigma }_2(P)\Vert$
10	7.87e−14	9.90e−14	1.29e−14
50	3.86e−13	4.99e−13	3.84e−13
100	9.60e−13	6.07e−13	5.95e−13
200	2.05e−12	1.07e−12	1.02e−12

4. Conclusions

In this paper, given three spectra $\{{\lambda }_i{\}}_{i=1}^n,\{{\mu }_i{\}}_{i=1}^{n-1},\{{\nu }_i{\}}_{i=1}^{n-2}$, not necessarily interlacing, we construct H-Symmetric pentadiagonal matrix P that admits the spectrum, i.e. $\sigma (P)=\{{\lambda }_i{\}}_{i=1}^n,{\sigma }_1(P)=\{{\mu }_i{\}}_{i=1}^{n-1},{\sigma }_2(P)=\{{\nu }_i{\}}_{i=1}^{n-2}.$ The construction procedure shows that the solution is not unique. The significance of this paper is to consider the cases where $\sigma (P)$ may contain complex eigenvalues. Moreover, in general the eigenvalues do not need to satisfy interlacing properties.

Acknowledgments

The authors would like to thank three anonymous reviewers for their grateful comments on earlier version of the paper.

Disclosure statement

No potential conflict of interest was reported by the author(s).