Combined Matrix of a Tridiagonal Toeplitz Matrix

Abstract

In this work, combined matrices of tridiagonal Toeplitz matrices are studied. The combined matrix is known as the Relative Gain Array in control theory. In particular, given a real tridiagonal Toeplitz matrix of order n, the characterization of its combined matrix as a bisymmetric and doubly quasi-stochastic matrix is studied. Furthermore, this paper addresses the inverse problem, that is, given a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix U of order n, determine under what conditions there exists a real tridiagonal Toeplitz matrix A such that its combined matrix is U.

1. Introduction

Toeplitz matrices are a class of special structured matrices with constant entries along each diagonal. Their structure and properties make these matrices useful in various areas of applied mathematics and engineering, such as algebraic differential equations, queuing theory, signal processing, time series analysis, integral equations, and control theory.

Tridiagonal Toeplitz matrices and their low-rank perturbations play a role in the numerical solution of ordinary and partial differential equations, time series analysis, and as regularization matrices in Tikhonov regularization for discrete ill-posed problems. These matrices are widely studied due to their predictable structure, spectral properties, and the existence of efficient algorithms for their manipulation [1,2,3]. Moreover, the characterization of boundary value problems for second-order linear difference equations has significant implications for computing the inverse of generalized Jacobi matrices [4]. Finally, in [5], the author explores how Toeplitz matrices arise in solving a class of difference equations encountered in enumerative combinatorics, discrete-time dynamic models, and approximate solutions to differential equations.

In control theory, the concept of the combined matrix of a real nonsingular matrix was introduced by Bristol [6], where it is known as the Relative Gain Array (RGA). The RGA is a widely used method for determining the best input–output pairing in multivariable process control systems. It has been considered in many practical open-loop and closed-loop control applications and in the analysis of many fundamental steady-state closed-loop system properties, such as stability and robustness. Specifically, in some chemical engineering applications, a diagonal of RGA with entries close to 1 is used to determine the pairing of inputs and outputs for further design analysis (see, for instance, [7,8]), where many applications in control theory of the RGA matrices are for small sizes of order $n=2,3$ or 4 (see [9,10]).

While the concept of combined matrices has been extensively explored in control theory, their structural properties have also attracted attention in mathematics. In this context, the combined matrix of a nonsingular matrix A is the matrix $\mathcal{C}\left(A\right)=A\circ A^{-t}$ where ∘ represents the Hadamard (entrywise) product. The Hadamard product itself is a fundamental operation in matrix analysis, with applications in various branches of mathematics. For example, it has been studied in the context of multivariate function theory and hypergeometric series [11,12] for foundational results in this direction. Combined matrices possess notable structural features: the sum of the entries of each row and each column is 1. When the combined matrix has nonnegative real numbers, it is a doubly stochastic matrix, and for real numbers, the combined matrices are doubly quasi-stochastic [13,14].

The problem of characterizing when three real numbers are the diagonal entries of a combined matrix can be found in [15,16,17,18]. In [19], doubly stochastic combined matrices are studied, and real matrices of order 3 whose combined matrix is a given doubly stochastic matrix are characterized. In [14], doubly quasi-stochastic matrices are introduced as a natural generalization of doubly stochastic matrices. Moreover, they characterize the conditions that the entries of a specific doubly quasi-stochastic matrix of order 3 must satisfy for there to exist a real matrix whose combined matrix is that one. Two cases are considered: when all entries of the doubly quasi-stochastic matrix are known, or only the main diagonal entries are known.

In this paper, we extend the study of tridiagonal Toeplitz matrices by investigating their combined matrices and characterizing them as bisymmetric, doubly quasi-stochastic matrices. The novelty of the present work lies in extending the inverse problem given in [13,19], for matrices of order 3, to the class of tridiagonal Toeplitz matrices of arbitrary order. Specifically, our contributions are the following: in Section 2, we characterize the structure of the combined matrix $U=\mathcal{C}\left(A\right)$ of a given real tridiagonal Toeplitz matrix A of order n, and show that U is a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix. In Section 3, we analyze the inverse problem: given such a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix U, we establish conditions under which it can be realized as the combined matrix of some real tridiagonal Toeplitz matrix A, and explicitly construct such a matrix when it exists.

2. Combined Matrix of a Tridiagonal Toeplitz Matrix

In this section, we study the combined matrix of a real tridiagonal Toeplitz matrix. From [1], we give a definition of the structure of these matrices.

Definition 1. 

Let ${\mathcal{M}}_n\left(\mathbb{R}\right)$ be the set of real matrices of size $n\times n$ and the sequences ${\left\{a\left(k\right)\right\}}_{k=1}^n\subset \mathbb{R}$, ${\left\{b\left(k\right)\right\}}_{k=1}^{n-1}\subset \mathbb{R}$ and ${\left\{c\left(k\right)\right\}}_{k=1}^{n-1}\subset \mathbb{R}$. The Jacobi matrix associated with these sequences is given by

$$\begin{array}{c}\hfill J=\left(\begin{array}{cccccc}a\left(1\right)& b\left(1\right)& 0& \dots & 0& 0\\ c\left(1\right)& a\left(2\right)& b\left(2\right)& \dots & 0& 0\\ 0& c\left(2\right)& a\left(3\right)& \dots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \dots & a(n-1)& b(n-1)\\ 0& 0& 0& \dots & c(n-1)& a\left(n\right)\end{array}\right)\end{array}$$

A real tridiagonal Toeplitz matrix is a specific case of a tridiagonal Jacobi matrix where $a\left(1\right)=a\left(2\right)=\dots =a\left(n\right)=a$, $b\left(1\right)=b\left(2\right)=\dots =b(n-1)=b$, and $c\left(1\right)=c\left(2\right)=\dots =c(n-1)=c$. It is denoted by $A_n=T(n;c,a,b)$.

In this work, we focus on combined matrices. The concept is defined as follows.

Definition 2. 

The combined matrix of a nonsingular matrix $A=\left(a_{ij}\right)$ is the matrix $\mathcal{C}\left(A\right)=\left(c_{ij}\right)=A\circ A^{-t}$ where ∘ represents the Hadamard (entrywise) product and $A^{-t}$ is the inverse transpose, ${\left(A^t\right)}^{-1}$, of A.

Some fundamental properties of combined matrices are summarized in the following Lemma.

Lemma 1 

([20]). The combined matrix $\mathcal{C}\left(A\right)=\left(c_{ij}\right)$ of a nonsingular matrix $A=\left(a_{ij}\right)$ satisfies the following conditions:

1. 

If $D_1$ and $D_2$ are two nonsingular diagonal matrices, then $\mathcal{C}\left(A\right)=\mathcal{C}\left(D_{1}A{D}_2\right)$.

2. 

If $c_{ij}\ne 0$, then $a_{ij}\ne 0,\forall i,j$.

3. 

The sum of the entries of any row or column of $\mathcal{C}\left(A\right)$ is 1.

4. 

If A is a triangular matrix, then $\mathcal{C}\left(A\right)=I$.

Combined matrices are doubly quasi-stochastic matrices, as defined below.

Definition 3 

([14]). A real matrix $U=\left(u_{ij}\right)$ of order n is called a doubly quasi-stochastic matrix if ${\sum }_{i=1}^n{u}_{ij}=1$, $\forall j=1,2,\dots ,n$ and ${\sum }_{j=1}^n{u}_{ij}=1$, $\forall i=1,2,\dots ,n$. If $u_{ij}\ge 0$, it is called a doubly stochastic matrix.

This section addresses the following question.

Question 1. 

Consider the matrices $A_n=T(n;c,a,b)$ and $U=\mathcal{C}\left(A_n\right)$. What is the structure of U?

If $b=0$ or $c=0$, then $A_n$ is a triangular matrix and by Lemma 1, $\mathcal{C}\left(A_n\right)=I_n$. Otherwise, to answer the question, we distinguish between $a\ne 0$ and $a=0$.

2.1. Case $A_n=T(n;c,a,b)$ with $a\ne 0$

To solve Question 1, we factorize $A_n$ using two nonsingular diagonal matrices to obtain a diagonally equivalent matrix ${\tilde{A}}_n$, which is more useful for our analysis. Although working with $T(n;1,a,1)$ is common, see for example [21], we obtain a different tridiagonal Toeplitz matrix, which we use throughout the paper.

In the following proposition, we establish a relationship between the matrix $A_n=T(n;c,a,b)$ and a simplified form ${\tilde{A}}_n=T(n;1,1,d)$.

Proposition 1. 

Let $A_n=T(n;c,a,b)$.

1. 

$A_n$ is diagonally equivalent to ${\tilde{A}}_n=T(n;1,1,d)$ with $d=bc/a^2$.

2. 

If $A_n$ is a nonsingular matrix, then $\mathcal{C}\left(A_n\right)=\mathcal{C}\left(\widetilde{A_n}\right)$.

Proof. 

• Since $a\ne 0$ and $c\ne 0$, we construct the following nonsingular diagonal matrices,

  $$\begin{array}{cc}\hfill D_1& =\mathrm{diag}\left({\displaystyle \frac{c^{n-1}}{a^n}},{\displaystyle \frac{c^{n-2}}{a^{n-1}}},\dots ,{\displaystyle \frac{c}{a^2}},{\displaystyle \frac{1}{a}}\right)\hfill \\ \hfill D_2& =\mathrm{diag}\left({\displaystyle \frac{a^{n-1}}{c^{n-1}}},{\displaystyle \frac{a^{n-2}}{c^{n-2}}},\dots ,{\displaystyle \frac{a}{c}},1\right).\hfill \end{array}$$

  (1)

  Then,

  $$D_1{A}_n{D}_2=D_{1}T(n;c,a,b)D_2=T(n;1,1,d)={\tilde{A}}_n,$$

  with $d=bc/a^2$.
• If $A_n$ is a nonsingular matrix using $\left(i\right)$, it is diagonally equivalent to $\widetilde{A_n}$, and by Lemma 1, $\mathcal{C}\left(A_n\right)=\mathcal{C}\left(\widetilde{A_n}\right)$.

□

Since $A_n$ and ${\tilde{A}}_n$ are diagonally equivalent matrices, the following results demonstrate some properties of $A_n$ derived from ${\tilde{A}}_n$. In particular, they show the relationships between determinants and eigenvalues of these matrices, some of which are used in the paper.

Proposition 2. 

Let $A_n=T(n;c,a,b)$ and ${\tilde{A}}_n=T(n;1,1,d)$ with $d=bc/a^2$. Then, $det\left(A_n\right)=a^{n}det\left({\tilde{A}}_n\right)$, where

$$\begin{array}{c}\hfill det\left({\tilde{A}}_n\right)=\sum _{j=0}^q{(-1)}^j\left(\begin{array}{c}n-j\\ j\end{array}\right)d^j,\end{array}$$

(2)

with $q=\lfloor n/2\rfloor$, $n=1,2,\dots$, and $det\left({\tilde{A}}_0\right)=1$.

Proof. 

Using (1), we obtain $det\left({\tilde{A}}_n\right)=det\left(D_1\right)det\left(A_n\right)det\left(D_2\right)={\displaystyle \frac{1}{a^n}}det\left(A_n\right).$ Thus, it follows that $det\left(A_n\right)=a^{n}det\left({\tilde{A}}_n\right)$. Now, we prove the Equation (2) by induction. For the base cases,

$$\begin{array}{cc}\hfill det\left({\tilde{A}}_1\right)& =1\hfill \\ \hfill det\left({\tilde{A}}_2\right)& =1-d=\sum _{j=0}^1{(-1)}^j\left(\begin{array}{c}n-j\\ j\end{array}\right)d^j\hfill \\ \hfill det\left({\tilde{A}}_3\right)& =1-2d=\sum _{j=0}^2{(-1)}^j\left(\begin{array}{c}n-j\\ j\end{array}\right)d^j.\hfill \end{array}$$

Now, we assume that (2) holds for $n-1$. Applying Laplace’s method to the matrix ${\tilde{A}}_n$ we obtain the recursive formula,

$$\begin{array}{c}\hfill det\left({\tilde{A}}_n\right)=det\left({\tilde{A}}_{n-1}\right)-ddet\left({\tilde{A}}_{n-2}\right).\end{array}$$

(3)

Substituting the values for $det\left({\tilde{A}}_{n-1}\right)$ and $det\left({\tilde{A}}_{n-2}\right)$ we have

• If $n=2q$

  $$\begin{array}{c}{\displaystyle det\left({\tilde{A}}_n\right)=\sum _{j=0}^{q-1}{(-1)}^j\left(\begin{array}{c}n-1-j\\ j\end{array}\right)d^j-d\left({\displaystyle \sum _{j=0}^{q-1}{(-1)}^j\left(\begin{array}{c}n-2-j\\ j\end{array}\right)d^j}\right)}\hfill \\ \\ =\left(1-\left(\begin{array}{c}n-2\\ 1\end{array}\right)d+\left(\begin{array}{c}n-3\\ 2\end{array}\right)d^2-\left(\begin{array}{c}n-4\\ 3\end{array}\right)d^3+\dots +{(-1)}^{q-1}\left(\begin{array}{c}n-q\\ q-1\end{array}\right)d^{q-1}\right)\hfill \\ \\ -\left(d-\left(\begin{array}{c}n-3\\ 1\end{array}\right)d^2+\left(\begin{array}{c}n-4\\ 2\end{array}\right)d^3-\left(\begin{array}{c}n-5\\ 3\end{array}\right)d^4+\dots +{(-1)}^{q-1}\left(\begin{array}{c}n-1-q\\ q-1\end{array}\right)d^q\right)\hfill \\ \\ =1-\left\{\left(\begin{array}{c}n-2\\ 0\end{array}\right)+\left(\begin{array}{c}n-2\\ 1\end{array}\right)\right\}d+\left\{\left(\begin{array}{c}n-3\\ 1\end{array}\right)+\left(\begin{array}{c}n-3\\ 2\end{array}\right)\right\}{d}^2+\dots +\hfill \\ \\ {(-1)}^{q-1}\left\{\left(\begin{array}{c}n-q\\ q-2\end{array}\right)+\left(\begin{array}{c}n-q\\ q-1\end{array}\right)\right\}{d}^{q-1}+{(-1)}^q\left(\begin{array}{c}n-q-1\\ q-1\end{array}\right)d^q\hfill \\ \\ =1-\left(\begin{array}{c}n-1\\ 1\end{array}\right)d+\left(\begin{array}{c}n-2\\ 2\end{array}\right)d^2+\dots +{(-1)}^{q-1}\left(\begin{array}{c}n-(q-1)\\ q-1\end{array}\right)d^{q-1}+\left(\begin{array}{c}n-q\\ q\end{array}\right)d^q\hfill \\ \\ {\displaystyle =\sum _{j=0}^q{(-1)}^j\left(\begin{array}{c}n-j\\ j\end{array}\right)d^j.}\hfill \end{array}$$
• If $n=2q+1$

  $$\begin{array}{c}\hfill {\displaystyle det\left({\tilde{A}}_n\right)=\sum _{j=0}^q{(-1)}^j\left(\begin{array}{c}n-1-j\\ j\end{array}\right)d^j-d\left({\displaystyle \sum _{j=0}^{q-1}{(-1)}^j\left(\begin{array}{c}n-2-j\\ j\end{array}\right)d^j}\right)}\end{array}$$
  By reasoning analogously to the even case, we obtain

  $$\begin{array}{c}\hfill det\left({\tilde{A}}_n\right)=\sum _{j=0}^q{(-1)}^j\left(\begin{array}{c}n-j\\ j\end{array}\right)d^j.\end{array}$$

Note that $det\left({\tilde{A}}_n\right)$ is a polynomial in d of degree $\lfloor n/2\rfloor$ and the coefficients of this polynomial are combinatorial numbers derived from Tartaglia’s triangle (see Figure 1), by following the marked arrows and considering the starting row. □

The following example illustrates Proposition 2.

Example 1. 

Consider the matrix $A_7=T(7;2,3,9)$. To obtain $det\left(A_7\right)$, we calculate ${\tilde{A}}_7=T(7;1,1,2)$. Using Figure 1, we start from row 7 and follow the marked arrows. Then, we have

$$\begin{array}{c}\hfill det\left({\tilde{A}}_7\right)=\sum _{j=0}^3{(-1)}^j\left(\begin{array}{c}7-j\\ j\end{array}\right)d^j=1-6d+10d^2-4d^3.\end{array}$$

Since $d=2$, we obtain $det\left({\tilde{A}}_7\right)=-3$. Hence, $det\left(A_7\right)=3^{7}det\left({\tilde{A}}_7\right)=-3^8$.

The following proposition establishes relationships between the eigenvalues of ${\tilde{A}}_n$ and $A_n$.

Proposition 3. 

Let $A_n=T(n;c,a,b)$. Then, the following characteristics apply:

1. 

The eigenvalues of $A_n$ are given by

$$\begin{array}{c}\hfill {\mu }_k=a-2\sqrt{bc}cos\left({\displaystyle \frac{k\pi }{n+1}}\right),k=1,2,\dots ,n.\end{array}$$

(4)

2. 

If $bc>0$, we have

(a) 

${\mu }_k\in \mathbb{R}$, $\forall k=1,2,\dots ,n.$

(b) 

If n is odd, $\mu =a$ is an eigenvalue of $A_n$.

(c) 

The eigenvalues are allocated symmetrically with respect to a.

3. 

If $bc<0$ and n is even, then ${\mu }_k\in \mathbb{C}$, $\forall k=1,2,\dots ,n$. If n is odd, $\mu =a$ is an eigenvalue of $A_n$ and the remaining eigenvalues ${\mu }_k$ are complex.

4. 

Since ${\tilde{A}}_n=T(n;1,1,d)$, with $d=bc/a^2$, is the diagonally equivalent matrix to $A_n$, the following is verified:

(a) 

μ is an eigenvalue of $A_n$ if and only if $\lambda =\mu /a$ is an eigenvalue of ${\tilde{A}}_n$. And

$$\begin{array}{c}\hfill {\lambda }_k=1-2\sqrt{d}cos\left({\displaystyle \frac{k\pi }{n+1}}\right),k=1,2,\dots ,n.\end{array}$$

(b) 

If n is odd, $\lambda =1$ is an eigenvalue of ${\tilde{A}}_n$.

(c) 

If $A_n^{\left(1\right)}=T(n;c_1,a_1,b_1)$ and $A_n^{\left(2\right)}=T(n;c_2,a_2,b_2)$ with $b_1{c}_1/a_1^2=b_2{c}_2/a_2^2$, then $A^{\left(1\right)}$ and $A^{\left(2\right)}$ have the same eigenvalues.

Proof. 

• Following the technique given in [21], we obtain the result.

  (a)

  If $bc>0$ the result is straightforward.

  (b)

  If n is odd then, for $k=⌈{\displaystyle \frac{n}{2}}⌉={\displaystyle \frac{n+1}{2}}$ we obtain that

  $$\begin{array}{c}\hfill {\mu }_k=a-2\sqrt{bc}cos\left({\displaystyle \frac{{\displaystyle \frac{n+1}{2}}\pi }{n+1}}\right)=a-2\sqrt{bc}cos\left({\displaystyle \frac{\pi }{2}}\right)=a.\end{array}$$

  (c)

  For $k=1,2,\dots ,⌊{\displaystyle \frac{n}{2}}⌋$, we have

  $$\begin{array}{c}\hfill {\mu }_k=a-\underset{t_k}{\underbrace{2\sqrt{bc}cos\left({\displaystyle \frac{k\pi }{n+1}}\right)}}=a-t_k\end{array}$$

  and

  $$\begin{array}{cc}\hfill {\mu }_{n-(k-1)}& =a-2\sqrt{bc}cos\left({\displaystyle \frac{(n-(k-1\left)\right)\pi }{n+1}}\right)=a-2\sqrt{bc}cos\left(\pi -{\displaystyle \frac{k\pi }{n+1}}\right)\hfill \\ \hfill & =a-2\sqrt{bc}\left(-cos\left({\displaystyle \frac{k\pi }{n+1}}\right)\right)=a+2\sqrt{bc}cos\left({\displaystyle \frac{k\pi }{n+1}}\right)=a+t_k.\hfill \end{array}$$

  Therefore, the eigenvalues of $A_n$ are as follows:

  $$\begin{array}{c}\hfill \begin{array}{c}\mathrm{If} n \mathrm{is} \mathrm{even},\hfill \\ a-t_1,a-t_2,\dots ,a-t_{⌊n/2⌋},a+t_{⌊n/2⌋},\dots ,a+t_2,a+t_1.\hfill \\ \mathrm{If} n \mathrm{is} \mathrm{odd},\hfill \\ a-t_1,a-t_2,\dots ,a-t_{⌊(n-1)/2⌋},a,a+t_{⌊(n-1)/2⌋},\dots ,a+t_2,a+t_1.\hfill \end{array}\end{array}$$

  That is, the eigenvalues are allocated symmetrically with respect to a.

• The result follows directly from the fact that $bc<0$ and n is even.

  (a)

  From (1), $D_1{D}_2={\displaystyle \frac{1}{a}}{I}_n$ and $det\left(D_1{D}_2\right)={\displaystyle \frac{1}{a^n}}$. Thus, the characteristic polynomials of ${\tilde{A}}_n$ and $A_n$ have this relationship,

  $$\begin{array}{cc}\hfill q_{{\tilde{A}}_n}\left(\lambda \right)& =det(\lambda I_n-{\tilde{A}}_n)=det(\lambda I_n-D_1{A}_n{D}_2)=det(\lambda \left(a{D}_1{D}_2\right)-D_1{A}_n{D}_2)\hfill \\ \hfill & =det\left(D_1(a\lambda I_n-A_n)D_2\right)={\displaystyle \frac{1}{a^n}}det(a\lambda I_n-A_n)\hfill \\ \hfill & ={\displaystyle \frac{1}{a^n}}det(\mu I_n-A_n)={\displaystyle \frac{1}{a^n}}{q}_{A_n}\left(\mu \right),\hfill \end{array}$$

  being $\mu =a\lambda$. Therefore,

  $$\begin{array}{c}\hfill q_{{\tilde{A}}_n}\left(\lambda \right)=0\iff q_{A_n}\left(\mu \right)=0.\end{array}$$

  That is, $\mu$ is an eigenvalue of $A_n$ if and only if $\lambda =\mu /a$ is an eigenvalue of ${\tilde{A}}_n$. From (4), we obtain that the eigenvalues of ${\tilde{A}}_n$:

  $$\begin{array}{c}\hfill {\lambda }_k=1-2\sqrt{d}cos\left({\displaystyle \frac{k\pi }{n+1}}\right),k=1,2,\dots ,n.\end{array}$$

  (b)

  If n is odd, we have seen in (2b) that $\mu =a$ is an eigenvalue of $A_n$. By the above result, $\lambda =1$ is an eigenvalue of ${\tilde{A}}_n$.

  (c)

  Since $b_1{c}_1/a_1^2=b_2{c}_2/a_2^2=d$, by Proposition 1, $A^{\left(1\right)}$ and $A^{\left(2\right)}$ are diagonally equivalent to ${\tilde{A}}_n=T(n;1,1,d)$. Applying step 4.(i), we conclude that $A^{\left(1\right)}$ and $A^{\left(2\right)}$ have the same eigenvalues.

□

When ${\tilde{A}}_n$ is nonsingular, $\mathcal{C}\left({\tilde{A}}_n\right)$ is a tridiagonal matrix and it is not necessary to calculate all the entries of ${\tilde{A}}_n^{-t}$ to obtain $\mathcal{C}\left({\tilde{A}}_n\right)$. The following result identifies the specific entries required.

Proposition 4. 

Let ${\tilde{A}}_n=T(n;1,1,d)$. To obtain $U=\left(u_{ij}\right)=\mathcal{C}\left({\tilde{A}}_n\right)$, it is only necessary to calculate the entries $u_{i,i}$ and $u_{i,i+1}$, $i=1,2,\dots ,\lfloor n/2\rfloor$.

Proof. 

Using the structure of ${\tilde{A}}_n$ and, by Lemma 1, $U=\left(u_{ij}\right)=\mathcal{C}\left({\tilde{A}}_n\right)$ is a doubly quasi-stochastic tridiagonal matrix. If $S=\left(s_{ij}\right)={\tilde{A}}_n^{-1}$, we obtain the entries

$$\begin{array}{c}\hfill s_{i,i}={\displaystyle \frac{1}{det\left({\tilde{A}}_n\right)}}{C}_{i,i}\left({\tilde{A}}_n\right),i=1,2,\dots ,n,\end{array}$$

where $C_{i,i}$ is the cofactor of the entry $(i,i)$ of ${\tilde{A}}_n$.

It can be verified that

$$\begin{array}{ccccc}{C}_{1,1}\left({\tilde{A}}_n\right)& =& det\left({\tilde{A}}_{n-1}\right)& =& C_{n,n}\left({\tilde{A}}_n\right)\hfill \\ C_{2,2}\left({\tilde{A}}_n\right)& =& det\left({\tilde{A}}_1\right)det\left({\tilde{A}}_{n-2}\right)& =& C_{n-1,n-1}\left({\tilde{A}}_n\right)\hfill \\ & ⋮& & ⋮& \\ C_{q-1,q-1}\left({\tilde{A}}_n\right)& =& det\left({\tilde{A}}_{q-2}\right)det\left({\tilde{A}}_{n+1-q}\right)& =& C_{n+2-q,n+2-q}\left({\tilde{A}}_n\right)\hfill \\ C_{q,q}\left({\tilde{A}}_n\right)& =& det\left({\tilde{A}}_{q-1}\right)det\left({\tilde{A}}_{n-q}\right)& =& C_{n+1-q,n+1-q}\left({\tilde{A}}_n\right)\hfill \end{array}$$

where $q=\lfloor n/2\rfloor$. From these values, we have, for $i=1,2,\dots ,q$,

$$\begin{array}{c}\hfill u_{i,i}={\displaystyle \frac{C_{i,i}\left({\tilde{A}}_n\right)}{det\left({\tilde{A}}_n\right)}}={\displaystyle \frac{det\left({\tilde{A}}_{i-1}\right)det\left({\tilde{A}}_{n-i}\right)}{det\left({\tilde{A}}_n\right)}}=u_{n+1-i,n+1-i}.\end{array}$$

(5)

Since $U=\mathcal{C}\left({\tilde{A}}_n\right)$ is a doubly quasi-stochastic tridiagonal matrix, we obtain the rest of the entries

$$\begin{array}{c}\hfill u_{2,1}=1-u_{1,1}=u_{1,2}=u_{n-1,n}=u_{n,n-1}\end{array}$$

and, for $i=2,3,\dots ,q$,

$$\begin{array}{c}\hfill u_{i+1,i}=1-u_{i,i}-u_{i,i-1}=u_{i,i+1}=u_{n+1-i,n-i}=u_{n-i,n+1-i}.\end{array}$$

If $n=2q+1$, $q=0,1,2,\dots$, then $u_{q+1,q+1}$ is directly obtained because $U=\mathcal{C}\left({\tilde{A}}_n\right)$ is a doubly quasi-stochastic tridiagonal matrix. □

From Proposition 4, we observe that U is symmetric with respect to both its main and secondary diagonals, that is, U is a bisymmetric matrix. Based on this, we present the following remark.

Remark 1. 

Given the matrix ${\tilde{A}}_n=T(n;1,1,d)$, its combined matrix $U=\mathcal{C}\left({\tilde{A}}_n\right)$ is a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix.

We clarify Proposition 4 with the following example.

Example 2. 

Consider the matrix ${\tilde{A}}_7=T(7;1,1,2)$. To obtain $U=\mathcal{C}\left({\tilde{A}}_7\right)$, we calculate

$$\begin{array}{cc}\hfill C_{1,1}\left({\tilde{A}}_7\right)& =C_{7,7}\left({\tilde{A}}_7\right)=det\left({\tilde{A}}_6\right)=7\hfill \\ \hfill C_{2,2}\left({\tilde{A}}_7\right)& =C_{6,6}\left({\tilde{A}}_7\right)=det\left({\tilde{A}}_1\right)det\left({\tilde{A}}_5\right)=1·5=5\hfill \\ \hfill C_{3,3}\left({\tilde{A}}_7\right)& =C_{5,5}\left({\tilde{A}}_7\right)=det\left({\tilde{A}}_2\right)det\left({\tilde{A}}_4\right)=(-1)·(-1)=1.\hfill \end{array}$$

By Proposition 4, we have

$$\begin{array}{cc}\hfill u_{1,1}& =u_{7,7}={\displaystyle \frac{det\left({\tilde{A}}_6\right)}{det\left({\tilde{A}}_7\right)}}=-{\displaystyle \frac{7}{3}}\hfill \\ \hfill u_{2,2}& =u_{6,6}={\displaystyle \frac{det\left({\tilde{A}}_1\right)det\left({\tilde{A}}_5\right)}{det\left({\tilde{A}}_7\right)}}=-{\displaystyle \frac{5}{3}}\hfill \\ \hfill u_{3,3}& =u_{5,5}={\displaystyle \frac{det\left({\tilde{A}}_2\right)det\left({\tilde{A}}_4\right)}{det\left({\tilde{A}}_7\right)}}=-{\displaystyle \frac{1}{3}}\hfill \end{array}$$

and we obtain the rest of the entries of U,

$$\begin{array}{cc}\hfill u_{1,2}& =u_{2,1}=1-u_{1,1}={\displaystyle \frac{10}{3}},u_{2,3}=u_{3,2}=1-u_{2,1}-u_{2,2}=-{\displaystyle \frac{2}{3}}\hfill \\ \hfill u_{3,4}& =u_{4,3}=1-u_{3,2}-u_{3,3}=2,u_{4,5}=u_{5,4}=u_{3,4}=2\hfill \\ \hfill u_{5,6}& =u_{6,5}=u_{2,3}=-{\displaystyle \frac{2}{3}},u_{6,7}=u_{7,6}=u_{1,2}={\displaystyle \frac{10}{3}}.\hfill \end{array}$$

To obtain $u_{4,4}$, we calculate $u_{4,4}=1-u_{4,3}-u_{4,5}=-3$.

Thus, the matrix U is given by

$$\begin{array}{c}\hfill U={\displaystyle \frac{1}{3}}\left(\begin{array}{ccccccc}-7& 10& 0& 0& 0& 0& 0\\ 10& -5& -2& 0& 0& 0& 0\\ 0& -2& -1& 6& 0& 0& 0\\ 0& 0& 6& -3& 6& 0& 0\\ 0& 0& 0& 6& -1& -2& 0\\ 0& 0& 0& 0& -2& -5& 10\\ 0& 0& 0& 0& 0& 10& -7\end{array}\right).\end{array}$$

As a consequence of the bisymmetry of U, the following result establishes a relationship between the combined matrices of $\widetilde{A_n}=T(n;1,1,d)$ and ${\tilde{A}}_n^t=T(n;d,1,1)$.

Proposition 5. 

Let $\widetilde{A_n}=T(n;1,1,d)$ and ${\tilde{A}}_n^t=T(n;d,1,1)$. Then, $\mathcal{C}\left(\widetilde{A_n}\right)=\mathcal{C}\left({\tilde{A}}_n^t\right)$.

Proof. 

Since $U=\mathcal{C}\left({\tilde{A}}_n\right)$ is a symmetric matrix, we have

$$\begin{array}{cc}\hfill \mathcal{C}\left(T\right(n;1,1,d\left)\right)& =T(n;1,1,d)\circ {\left(T(n;1,1,d)\right)}^{-t}\hfill \\ \hfill {\left(\mathcal{C}\left(T\right(n;1,1,d\left)\right)\right)}^t& ={\left(T(n;1,1,d)\right)}^t\circ {\left({\left(T(n;1,1,d)\right)}^{-t}\right)}^t\hfill \\ \hfill \mathcal{C}\left(T\right(n;1,1,d\left)\right)& =T(n;d,1,1)\circ {\left(T(n;d,1,1)\right)}^{-t}\hfill \\ \hfill \mathcal{C}\left(T\right(n;1,1,d\left)\right)& =\mathcal{C}\left(T\right(n;d,1,1\left)\right).\hfill \end{array}$$

□

Summarizing, we present the following theorem.

Theorem 1. 

Let ${\tilde{A}}_n=T(n;1,1,d)$. Then,

1. 

$U=\mathcal{C}\left({\tilde{A}}_n\right)$ is a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix.

2. 

To construct U it is only necessary to obtain the entries $u_{i,i}$ and $u_{i,i+1}$, $i=1,2,\dots ,\lfloor n/2\rfloor$.

3. 

$\mathcal{C}\left(T\right(n;1,1,d\left)\right)=\mathcal{C}\left(T\right(n;d,1,1\left)\right)$.

Now, we consider the specific case when $bc=a^2$, which corresponds to ${\tilde{A}}_n=T(n;1,1,1)$. In this case, if $n=3q+2$, then $det\left({\tilde{A}}_n\right)=0$ and $\mathcal{C}\left({\tilde{A}}_n\right)$ does not exist. Conversely, if $n\ne 3q+2$, we can directly obtain the combined matrix, without calculations. Therefore, we study the cases when $n=3q$ in Proposition 6 and $n=3q+1$ in Proposition 7.

Proposition 6. 

Let ${\tilde{A}}_{3q}=T(3q;1,1,1)$, $q=1,2,\dots$ Then,

$$\begin{array}{c}\hfill U=\mathcal{C}\left({\tilde{A}}_{3q}\right)=diag(H,H,\dots ,H,H),beingH=\left(\begin{array}{cccc}0& 1& & 0\\ 1& -1& & 1\\ 0& 1& & 0\end{array}\right)\end{array}$$

Proof. 

From Proposition 4, $u_{i,i}={\displaystyle \frac{det\left({\tilde{A}}_{i-1}\right)det\left({\tilde{A}}_{3q-i}\right)}{det\left({\tilde{A}}_{3q}\right)}}$, $i=1,2,\dots ,\lfloor 3q/2\rfloor$. Since $det\left({\tilde{A}}_{3q}\right)=det\left({\tilde{A}}_{3q+1}\right)={(-1)}^q$, for $r=0,1,\dots$

•

if $i=3r+1$, then

$$\begin{array}{cc}\hfill det\left({\tilde{A}}_{i-1}\right)det\left({\tilde{A}}_{3q-i}\right)& =det\left({\tilde{A}}_{3r}\right)det\left({\tilde{A}}_{3(q-r)-1}\right)\hfill \\ \hfill & =det\left({\tilde{A}}_{3r}\right)det\left({\tilde{A}}_{3(q-r-1)+2}\right)={(-1)}^r·0=0\hfill \end{array}$$

•

if $i=3r+2$, then

$$\begin{array}{cc}\hfill det\left({\tilde{A}}_{i-1}\right)det\left({\tilde{A}}_{3q-i}\right)& =det\left({\tilde{A}}_{3r+1}\right)det\left({\tilde{A}}_{3(q-r)-2}\right)\hfill \\ \hfill & =det\left({\tilde{A}}_{3r+1}\right)det\left({\tilde{A}}_{3(q-r-1)+1}\right)\hfill \\ \hfill & ={(-1)}^r·{(-1)}^{q-r-1}={(-1)}^{q-1}\hfill \end{array}$$

•

if $i=3r+3$, then

$$\begin{array}{cc}\hfill det\left({\tilde{A}}_{i-1}\right)det\left({\tilde{A}}_{3q-i}\right)& =det\left({\tilde{A}}_{3r+2}\right)det\left({\tilde{A}}_{3(q-r)-3}\right)\hfill \\ \hfill & =det\left({\tilde{A}}_{3r+2}\right)det\left({\tilde{A}}_{3(q-r-1)}\right)=0·{(-1)}^{q-r-1}=0.\hfill \end{array}$$

Thus, if $i=3r+1$ or $i=3r+3$, then $u_{i,i}=0$ and if $i=3r+2$, then $u_{i,i}={\displaystyle \frac{{(-1)}^{q-1}}{{(-1)}^q}}=-1$.

$U=\left(u_{ij}\right)$ is a doubly quasi-stochastic matrix, hence,

$$\begin{array}{cc}\hfill u_{i,i+1}=u_{i+1,i}=1,& \mathrm{for}i=3r+1,3r+2\hfill \\ \hfill u_{i,i+1}=u_{i+1,i}=0,& fori=3r+3.\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\hfill U=\mathcal{C}\left({\tilde{A}}_{3q}\right)=diag(H,H,\dots ,H,H)beingH=\left(\begin{array}{cccc}0& 1& & 0\\ 1& -1& & 1\\ 0& 1& & 0\end{array}\right).\end{array}$$

□

Reasoning analogously, we arrive at the following result.

Proposition 7. 

Let ${\tilde{A}}_{3q+1}=T(3q+1;1,1,1)$, $q=0,1,2,\dots$. Then,

$$\begin{array}{c}\hfill U=\mathcal{C}\left({\tilde{A}}_{3q+1}\right)=diag(P,P,\dots ,P,P,1)beingP=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right).\end{array}$$

Proof. 

Using the same technique as in Proposition 6, we prove the result. Since $U=\left(u_{ij}\right)$ is a bisymmetric matrix and $u_{1,1}=u_{3q+1,3q+1}$, then $u_{3q+1,3q+1}=1$. □

2.2. Case $A_n=T(n;c,0,b)$

In this subsection, we obtain results analogous to the case when $a\ne 0$, given in Section 2.1, but now considering $a=0$.

Proposition 8. 

Let $A_n=T(n;c,0,b)$. Then, the following conditions are true:

1. 

$A_n$ is diagonally equivalent to ${\tilde{A}}_n=T(n;1,0,g)$ with $g=b/c$.

2. 

$det\left({\tilde{A}}_{2q+1}\right)=0$ and $det\left({\tilde{A}}_{2q}\right)={(-g)}^q$, $q=0,1,2,\dots$

3. 

$U=\mathcal{C}\left({\tilde{A}}_{2q}\right)$ has all entries equal to 0 except $u_{i,i+1}=1$ and $u_{i+1,i}=1$, $i=1,3,\dots ,2q-1$, $q=1,2,\dots$

4. 

If ${\tilde{A}}_{2q}^{\left(1\right)}=T(2q;c_1,0,b_1)$ and ${\tilde{A}}_{2q}^{\left(2\right)}=T(2q;c_2,0,b_2)$ for any $c_1,c_2,b_1$ and $b_2$, then $\mathcal{C}\left({\tilde{A}}_{2q}^{\left(1\right)}\right)=\mathcal{C}\left({\tilde{A}}_{2q}^{\left(2\right)}\right)$.

Proof. 

• Since $c\ne 0$, we construct $D={\displaystyle \frac{1}{c}}{I}_n$ and

  $$\begin{array}{c}\hfill D{A}_n=DT(n;c,0,b)=T(n;1,0,b/c)=T(n;1,0,g)={\tilde{A}}_n.\end{array}$$
  From Lemma 1, if $A_n$ is a nonsingular matrix, then $\mathcal{C}\left(A_n\right)=\mathcal{C}\left({\tilde{A}}_n\right).$
• From (3), since $a=0$, we obtain $det\left({\tilde{A}}_n\right)=-gdet\left({\tilde{A}}_{n-2}\right)$, $n=3,4,\dots$ As $det\left({\tilde{A}}_1\right)=0$, for $n=1,2,\dots$ we have

  $$\begin{array}{ccc}\mathrm{If} n=2q+1\hfill & \to & det\left({\tilde{A}}_{2q+1}\right)={(-g)}^{q}det\left({\tilde{A}}_1\right)=0\hfill \\ \mathrm{If} n=2q\hfill & \to & det\left({\tilde{A}}_{2q}\right)={(-g)}^{q-1}det\left({\tilde{A}}_2\right)={(-g)}^q.\hfill \end{array}$$
  Thus, for $n=2q+1$, $q=0,1,2,\dots$, $A_n$ is singular and $\mathcal{C}\left(A_n\right)$ does not exist.
• From the structure of ${\tilde{A}}_n$, we have $u_{i,i}=0$ for $i=1,2,\dots ,n$. Since U is a doubly quasi-stochastic matrix, then $u_{1,2}=u_{2,1}=1$ and $u_{2,3}=u_{3,2}=0$. Moreover, for $i=3,5,\dots ,n-3$, we have $u_{i,i+1}=u_{i+1,i}=1$ and $u_{i+1,i+2}=u_{i+2,i+1}=0$. Finally, as $u_{n-1,n-1}=u_{n,n}=0$, then $u_{n-1,n}=u_{n,n-1}=1.$ Therefore, the nonzero entries of U are

  $$\begin{array}{c}\hfill u_{i,i+1}=u_{i+1,i}=1,i=1,3,5,\dots ,n-1.\end{array}$$
• Since $U=\mathcal{C}\left({\tilde{A}}_n\right)$ does not depend on g, the result follows immediately.

□

The following result describes the relationships between the eigenvalues of ${\tilde{A}}_n=T(n;1,0,g)$ and $A_n=T(n;c,0,b)$.

Proposition 9. 

Let $A_n=T(n;c,0,b)$ and ${\tilde{A}}_n=T(n;1,0,g)$, with $g=b/c$. Then, the following conditions are true:

1. 

μ is an eigenvalue of $A_n$ if and only if $\lambda =\mu /c$ is an eigenvalue of ${\tilde{A}}_n$. And

$$\begin{array}{c}\hfill {\lambda }_k=-2\sqrt{g}cos\left({\displaystyle \frac{k\pi }{n+1}}\right),k=1,2,\dots ,n\end{array}$$

2. 

If n is odd, $\lambda =0$ is an eigenvalue of ${\tilde{A}}_n.$

3. 

If $g>0$, half of the eigenvalues are positive and the other half are negative, one being the opposite of the other.

4. 

If $A_n^{\left(1\right)}=T(n;c_1,0,b_1)$ and $A_n^{\left(2\right)}=T(n;c_2,0,b_2)$ with $b_1/c_1=b_2/c_2$, then $A^{\left(1\right)}$ and $A^{\left(2\right)}$ have the same eigenvalues.

Proof. 

• Using the characteristic polynomial, we have

  $$\begin{array}{cc}\hfill q_{{\tilde{A}}_n}\left(\lambda \right)& =det(\lambda I_n-{\tilde{A}}_n)=det(\lambda I_n-D{A}_n)\hfill \\ \hfill & =det\left(D(c\lambda I_n-A_n)\right)=det\left(D\right)det(c\lambda I_n-A_n)\hfill \\ \hfill & ={\displaystyle \frac{1}{c^n}}det(\mu I_n-A_n)=q_{A_n}\left(\mu \right).\hfill \end{array}$$
  Thus,

  $$\begin{array}{c}\hfill q_{{\tilde{A}}_n}\left(\lambda \right)=0\iff q_{A_n}\left(\mu \right)=0\end{array}$$

  with $\mu =c\lambda$. Therefore, $\mu$ is an eigenvalue of $A_n$ if and only if $\lambda =\mu /c$ is an eigenvalue of ${\tilde{A}}_n$. From (4), we obtain that the eigenvalues of ${\tilde{A}}_n$ are

  $$\begin{array}{c}\hfill {\lambda }_k=-2\sqrt{g}cos\left({\displaystyle \frac{k\pi }{n+1}}\right),k=1,2,\dots ,n.\end{array}$$
• If n is odd, by Proposition 3, $\mu =0$ is an eigenvalue of $A_n=T(n;c,0,b)$. Then, $\lambda =\mu /c=0$ is an eigenvalue of ${\tilde{A}}_n=T(n;1,0,g).$
• If $g>0$, applying Proposition 3, we obtain that half of the eigenvalues are positive and the other half are negative, one being the opposite of the other.
• Since $b_1/c_1=b_2/c_2=g$, $A^{\left(1\right)}$ and $A^{\left(2\right)}$ are diagonally equivalent to ${\tilde{A}}_n=T(n;1,0,g)$. By applying the previous result, we conclude that $A^{\left(1\right)}$ and $A^{\left(2\right)}$ have the same eigenvalues.

□

We summarize the previous results in the following Theorem.

Theorem 2. 

Let ${\tilde{A}}_n=T(n;1,0,g)$ with $g=b/c$. Then, for $q=0,1,2,\dots$, the following conditions are true:

1. 

If $n=2q+1$, then ${\tilde{A}}_n$ is singular and $\mathcal{C}\left({\tilde{A}}_n\right)$ does not exist.

2. 

If $n=2q$, then $U=\mathcal{C}\left({\tilde{A}}_n\right)$ is a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix, whose nonzero entries are $u_{i,i+1}=1$, $u_{i+1,i}=1$, $i=1,3,5,\dots ,n-1$.

3. 

For any $g_1$ and $g_2$, $\mathcal{C}\left(T(2q;1,0,g_1)\right)=\mathcal{C}\left(T(2q;g_1,0,1)\right)=\mathcal{C}\left(T(2q;1,0,g_2)\right)=\mathcal{C}\left(T(2q;g_2,0,1)\right)$.

3. A Tridiagonal Toeplitz Matrix Whose Combined Matrix Is Given

In this section, we solve the inverse problem summarized by the following question:

Question 2. 

Let U be a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix. Does there exist a matrix $A_n=T(n;c,a,b)$ such that $\mathcal{C}\left(A_n\right)=A_n\circ A_n^{-t}=U$?

Note that if $U=I$, then $A_n=T(n;c,a,b)$ with $b=0$ or $c=0$. Moreover, if the only nonzero entries of U are $u_{i,i+1}=1$ and $u_{i+1,i}=1$, $i=1,3,5,\dots ,n-1$, then $U=\mathcal{C}\left(A_n\right)=\mathcal{C}\left(T(n;c,0,b)\right)$. Therefore, in this section, we consider matrices U corresponding to the other cases. By Proposition 1, $A_n=T(n;c,a,b)$ is diagonally equivalent to ${\tilde{A}}_n=T(n;1,1,d)$ with $d=bc/a^2$. For that reason, we rewrite Question 2.

Question 3. 

Let U be a bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix. Does there exist a matrix ${\tilde{A}}_n=T(n;1,1,d)$, such that $\mathcal{C}\left({\tilde{A}}_n\right)={\tilde{A}}_n\circ {\tilde{A}}_n^{-t}=U$?

To answer this question, the following result provides conditions on the main diagonal entries of U.

Proposition 10. 

Let $U=\left(u_{i,j}\right)$ be a matrix described in Remark 1 of size n. If there exists ${\tilde{A}}_n=T(n;1,1,d)$, such that $\mathcal{C}\left({\tilde{A}}_n\right)=U$, then $u_{i,i}$, $i=1,2,\dots ,q$, $q=1,2,\dots$, is given by

$$\begin{array}{c}\hfill u_{i,i}={\displaystyle \frac{P_k^{\left(i\right)}\left(d\right)}{P_q\left(d\right)}},\end{array}$$

(6)

where $P_q\left(d\right)$ denotes the polynomial in d of degree q and $P_k^{\left(i\right)}\left(d\right)$ denotes the polynomial in d of degree k corresponding to i, with

1. 

$k=q$ if $n=2q+1$ and $i=1$.

2. 

$k=q$ if $n=2q+1$ and $i-1$ is even.

3. 

$k=q-1$ in other cases.

Proof. 

Since $U=\mathcal{C}\left({\tilde{A}}_n\right)$, by (5), we have

$$\begin{array}{c}\hfill u_{i,i}={\displaystyle \frac{det\left({\tilde{A}}_{i-1}\right)det\left({\tilde{A}}_{n-i}\right)}{det\left({\tilde{A}}_n\right)}},i=1,2,3,\dots ,\lfloor n/2\rfloor ,\end{array}$$

with $det\left({\tilde{A}}_0\right)=1$. By Proposition 2, each determinant is a polynomial in d. Thus, $P_q\left(d\right)=det\left({\tilde{A}}_n\right)$ with $q=\lfloor n/2\rfloor$.

If $n=2q$, we have

$$\begin{array}{ccc}i=1\hfill & \to & det\left({\tilde{A}}_{2q-1}\right)=P_{q-1}\left(d\right)\to u_{1,1}={\displaystyle \frac{P_{q-1}^{\left(1\right)}\left(d\right)}{P_q\left(d\right)}}\hfill \\ & & \\ i=2,3,\dots ,q\hfill & \to & \left\{\begin{array}{ccc}i-1\mathrm{even}\hfill & \to & \left\{\begin{array}{c}det\left({\tilde{A}}_{i-1}\right)=P_{{\displaystyle \frac{i-1}{2}}}\left(d\right)\hfill \\ det\left({\tilde{A}}_{2q-i}\right)=P_{{\displaystyle \frac{2q-(i+1)}{2}}}\left(d\right)\hfill \end{array}\right.\to u_{i,i}={\displaystyle \frac{P_{q-1}^{\left(i\right)}\left(d\right)}{P_q\left(d\right)}}\hfill \\ & & \\ i-1\mathrm{odd}\hfill & \to & \left\{\begin{array}{c}det\left({\tilde{A}}_{i-1}\right)=P_{{\displaystyle \frac{i-2}{2}}}\left(d\right)\hfill \\ det\left({\tilde{A}}_{2q-i}\right)=P_{{\displaystyle \frac{2q-i}{2}}}\left(d\right)\hfill \end{array}\right.\to u_{i,i}={\displaystyle \frac{P_{q-1}^{\left(i\right)}\left(d\right)}{P_q\left(d\right)}}\hfill \end{array}\right.\hfill \end{array}$$

If $n=2q+1$, we have

$$\begin{array}{ccc}i=1\hfill & \to & det\left({\tilde{A}}_{2q}\right)=P_q\left(d\right)\to u_{1,1}={\displaystyle \frac{P_q^{\left(1\right)}\left(d\right)}{P_q\left(d\right)}}\hfill \\ & & \\ i=2,3,\dots ,q\hfill & \to & \left\{\begin{array}{ccc}i-1\mathrm{even}\hfill & \to & \left\{\begin{array}{c}det\left({\tilde{A}}_{i-1}\right)=P_{{\displaystyle \frac{i-1}{2}}}\left(d\right)\hfill \\ det\left({\tilde{A}}_{2q+1-i}\right)=P_{{\displaystyle \frac{2q+1-i}{2}}}\left(d\right)\hfill \end{array}\right.\to u_{i,i}={\displaystyle \frac{P_q^{\left(i\right)}\left(d\right)}{P_q\left(d\right)}}\hfill \\ & & \\ i-1\mathrm{odd}\hfill & \to & \left\{\begin{array}{c}det\left({\tilde{A}}_{i-1}\right)=P_{{\displaystyle \frac{i-2}{2}}}\left(d\right)\hfill \\ det\left({\tilde{A}}_{2q+1-i}\right)=P_{{\displaystyle \frac{2q-i}{2}}}\left(d\right)\hfill \end{array}\right.\to u_{i,i}={\displaystyle \frac{P_{q-1}^{\left(i\right)}\left(d\right)}{P_q\left(d\right)}}\hfill \end{array}\right.\hfill \end{array}$$

Therefore, the degree of the polynomial $P_k^{\left(i\right)}\left(d\right)$ is as follows:

• $k=q$ if $n=2q+1$ and $i=1$.
• $k=q$ if $n=2q+1$ and $i-1$ is even.
• $k=q-1$ in other cases.

□

The following result answers Question 3 when n is even, that is, $n=2q$.

Theorem 3. 

Let $U=\left(u_{i,j}\right)$ be a matrix described in Remark 1 of size $n=2q$, $q=1,2,\dots$ If $d\in \mathbb{R}$ satisfies the Equation (6) for $i=1,2,\dots ,q$, then there exists ${\tilde{A}}_n$ such that $\mathcal{C}\left({\tilde{A}}_n\right)=U.$

Proof. 

Using Proposition 10, if $n=2q$, $q=1,2,\dots$, the polynomials $P_q\left(d\right)$ and $P_{q-1}^i\left(d\right)$ are given by

$$\begin{array}{cc}\hfill P_q\left(d\right)& =1-s_{1}d+s_2{d}^2+\dots +{(-1)}^q{d}^q\hfill \\ \hfill P_{q-1}^i\left(d\right)& =1-a_1^{\left(i\right)}d+a_2^{\left(i\right)}{d}^2+\dots +{(-1)}^{q-1}{a}_{q-1}^{\left(i\right)}{d}^{q-1},i=1,2,\dots ,q.\hfill \end{array}$$

Then, using (6) for each i, we obtain a system of q polynomials of degree q in d.

$$\begin{array}{c}{u}_{1,1}-1+\left(a_1^{\left(1\right)}-s_1{u}_{1,1}\right)d+\left(s_2{u}_{1,1}-a_2^{\left(1\right)}\right)d^2+\dots +{(-1)}^{q-1}\left(s_{q-1}{u}_{1,1}-a_{q-1}^{\left(1\right)}\right)d^{q-1}+{(-1)}^q{u}_{1,1}{d}^q=0\hfill \\ u_{2,2}-1+\left(a_1^{\left(2\right)}-s_1{u}_{2,2}\right)d+\left(s_2{u}_{2,2}-a_2^{\left(2\right)}\right)d^2+\dots +{(-1)}^{q-1}\left(s_{q-1}{u}_{2,2}-a_{q-1}^{\left(2\right)}\right)d^{q-1}+{(-1)}^q{u}_{2,2}{d}^q=0\hfill \\ ⋮\hfill \\ u_{q,q}-1+\left(a_1^{\left(q\right)}-s_1{u}_{q,q}\right)d+\left(s_2{u}_{q,q}-a_2^{\left(q\right)}\right)d^2+\dots +{(-1)}^{q-1}\left(s_{q-1}{u}_{q,q}-a_{q-1}^{\left(q\right)}\right)d^{q-1}+{(-1)}^q{u}_{q,q}{d}^q=0\hfill \end{array}$$

(7)

To solve the nonlinear system (7), we apply a change of variables by setting $x_j=d^j$. For $j=1,2,\dots ,q$,

$$\begin{array}{c}\left(a_1^{\left(1\right)}-s_1{u}_{1,1}\right)x_1+\left(s_2{u}_{1,1}-a_2^{\left(1\right)}\right)x_2+\dots +{(-1)}^{q-1}\left(s_{q-1}{u}_{1,1}-a_{q-1}^{\left(1\right)}\right)x_{q-1}+{(-1)}^q{u}_{1,1}{x}_q=1-u_{1,1}\hfill \\ \left(a_1^{\left(2\right)}-s_1{u}_{2,2}\right)x_1+\left(s_2{u}_{2,2}-a_2^{\left(2\right)}\right)x_2+\dots +{(-1)}^{q-1}\left(s_{q-1}{u}_{2,2}-a_{q-1}^{\left(2\right)}\right)x_{q-1}+{(-1)}^q{u}_{2,2}{x}_q=1-u_{2,2}\hfill \\ ⋮\hfill \\ \left(a_1^{\left(q\right)}-s_1{u}_{q,q}\right)x_1+\left(s_2{u}_{q,q}-a_2^{\left(q\right)}\right)x_2+\dots +{(-1)}^{q-1}\left(s_{q-1}{u}_{q,q}-a_{q-1}^{\left(q\right)}\right)x_{q-1}+{(-1)}^q{u}_{q,q}{x}_q=1-u_{q,q}\hfill \end{array}$$

(8)

If the system given in (7) is consistent, then the system given in (8) is also consistent. Thus, if (8) is inconsistent, then (7) is also inconsistent and there does not exist a matrix ${\tilde{A}}_n$ such that $\mathcal{C}\left({\tilde{A}}_n\right)=U$.

If (8) is consistent and at least one solution satisfies that $x_j=x_1^j$, for $j=2,3,\dots ,q$, then $d=x_1$ is a solution of (7) and we obtain ${\tilde{A}}_n$ such that $\mathcal{C}\left({\tilde{A}}_n\right)=U$. If there is no d satisfying this condition, then (7) is not consistent and there does not exist a matrix ${\tilde{A}}_n$ such that $\mathcal{C}\left({\tilde{A}}_n\right)=U$. □

We now rewrite the system in (7) in matrix form, leading to the following remark.

Remark 2. 

If $v={(1,d,\dots ,d^q)}^t$, then the system given in (7) can be expressed as

$$\begin{array}{c}\hfill \left(Z_n^{\left(1\right)}-Z_n^{\left(2\right)}\right)v=0,\end{array}$$

(9)

where $Z_n^{\left(1\right)}v={[u_{11},u_{22},\dots ,u_{qq}]}^t{P}_q\left(d\right)$ of size $q\times (q+1)$ and

$$\begin{array}{c}\hfill Z_n^{\left(2\right)}v=\left(\begin{array}{c}det\left({\tilde{A}}_0\right)det\left({\tilde{A}}_{n-1}\right)\\ det\left({\tilde{A}}_1\right)det\left({\tilde{A}}_{n-2}\right)\\ ⋮\\ det\left({\tilde{A}}_q\right)det\left({\tilde{A}}_{n-q-1}\right)\\ det\left({\tilde{A}}_{q-1}\right)det\left({\tilde{A}}_{n-q}\right)\end{array}\right).\end{array}$$

(10)

The following example illustrates Theorem 3.

Example 3. 

Consider the matrix U of size $n=8$, with the properties described in Remark 1,

$$\begin{array}{c}\hfill U={\displaystyle \frac{1}{17}}\left(\begin{array}{cccccccc}3& 14& 0& 0& 0& 0& 0& 0\\ 14& -7& 10& 0& 0& 0& 0& 0\\ 0& 10& 5& 2& 0& 0& 0& 0\\ 0& 0& 2& -3& 18& 0& 0& 0\\ 0& 0& 0& 18& -3& 2& 0& 0\\ 0& 0& 0& 0& 2& 5& 10& 0\\ 0& 0& 0& 0& 0& 10& -7& 14\\ 0& 0& 0& 0& 0& 0& 14& 3\end{array}\right).\end{array}$$

Using (6), $u_{ii}{P}_4\left(d\right)=P_k^i\left(d\right)$ for $i=1,\dots ,4$. That is,

$$\begin{array}{ccc}\hfill u_{1,1}det\left({\tilde{A}}_8\right)=det\left({\tilde{A}}_7\right)& \to & \phantom{-}{\displaystyle \frac{3}{17}}(1-7d+15d^2-10d^3+d^4)=1-6d+10d^2-4d^3\hfill \\ \hfill u_{2,2}det\left({\tilde{A}}_8\right)=det\left({\tilde{A}}_1\right)det\left({\tilde{A}}_6\right)& \to & -{\displaystyle \frac{7}{17}}(1-7d+15d^2-10d^3+d^4)=1-5d+6d^2-d^3\hfill \\ \hfill u_{3,3}det\left({\tilde{A}}_8\right)=det\left({\tilde{A}}_2\right)det\left({\tilde{A}}_5\right)& \to & \phantom{-}{\displaystyle \frac{5}{17}}(1-7d+15d^2-10d^3+d^4)=1-5d+7d^2-3d^3\hfill \\ \\ \hfill u_{4,4}det\left({\tilde{A}}_8\right)=det\left({\tilde{A}}_3\right)det\left({\tilde{A}}_4\right)& \to & -{\displaystyle \frac{3}{17}}(1-7d+15d^2-10d^3+d^4)=1-5d+7d^2-2d^3.\hfill \end{array}$$

We rewrite the system as $Z_8^{\left(1\right)}v=Z_8^{\left(2\right)}v$, being

$$\begin{array}{cc}\hfill Z_8^{\left(1\right)}& ={\displaystyle \frac{1}{17}}\left(\begin{array}{ccccc}3& -21& 45& -30& 3\\ 7& -49& 105& -70& 7\\ 5& -35& 75& -50& 5\\ 3& -21& 45& -30& 3\end{array}\right),Z_8^{\left(2\right)}=\left(\begin{array}{ccccc}1& -6& 10& -4& 0\\ 1& -5& 6& -1& 0\\ 1& -5& 7& -3& 0\\ 1& -5& 7& -2& 0\end{array}\right)\hfill \\ \hfill v& =\left[\begin{array}{c}1\\ d\\ d^2\\ d^3\\ d^4\end{array}\right].\hfill \end{array}$$

To obtain the solution of the system (9), we apply a change of variables by setting $x_1=d$, $x_2=d^2$, $x_3=d^3$, and $x_4=d^4$, and write the obtained linear system as follows:

$$\left\{\begin{array}{cc}{\displaystyle \frac{81}{17}}{x}_1-{\displaystyle \frac{125}{17}}{x}_2+{\displaystyle \frac{38}{17}}{x}_3+{\displaystyle \frac{3}{17}}{x}_4\hfill & ={\displaystyle \frac{14}{17}}\hfill \\ \hfill & \\ {\displaystyle \frac{134}{17}}{x}_1-{\displaystyle \frac{207}{17}}{x}_2+{\displaystyle \frac{87}{17}}{x}_3-{\displaystyle \frac{7}{17}}{x}_4\hfill & ={\displaystyle \frac{24}{17}}\hfill \\ \hfill & \\ {\displaystyle \frac{50}{17}}{x}_1-{\displaystyle \frac{44}{17}}{x}_2+{\displaystyle \frac{1}{17}}{x}_3+{\displaystyle \frac{5}{17}}{x}_4\hfill & ={\displaystyle \frac{12}{17}}\hfill \\ \hfill & \\ {\displaystyle \frac{106}{17}}{x}_1-{\displaystyle \frac{164}{17}}{x}_2+{\displaystyle \frac{64}{17}}{x}_3-{\displaystyle \frac{3}{17}}{x}_4\hfill & ={\displaystyle \frac{20}{17}}\hfill \end{array}\right.$$

whose solution is

$$\begin{array}{c}\hfill x_1=2,x_2=4,x_3=8,x_4=16.\end{array}$$

Since $x_2=x_1^2=4$, $x_3=x_1^3=8$, and $x_4=x_1^4=16$, we assure that $d=x_1=2$ is a solution to the nonlinear system. Thus, we obtain ${\tilde{A}}_8=T(8;1,1,2)$ such that $\mathcal{C}\left({\tilde{A}}_8\right)=U$. Moreover, any matrix $A_8=T(8;c,a,b)$ satisfying that $bc/a^2=2$ has the same combined matrix.

The following result answers Question 3 when n is odd, that is, $n=2q+1$.

Theorem 4. 

Let $U=\left(u_{i,j}\right)$ be a matrix described in Remark 1 of size $n=2q+1$, $q=0,1,2,\dots$ If $d\in \mathbb{R}$ is a solution of the system given in (6), then there exists ${\tilde{A}}_n$ such that $\mathcal{C}\left({\tilde{A}}_n\right)=U.$

Proof. 

The proof is straightforward using the same technique as in Theorem 3. □

The following example applies Theorem 4 in a concrete case.

Example 4. 

Consider the matrix U described in Remark 1 of size $n=7$,

$$\begin{array}{c}\hfill U={\displaystyle \frac{1}{3}}\left(\begin{array}{ccccccc}2& 1& 0& 0& 0& 0& 0\\ 1& 4& -2& 0& 0& 0& 0\\ 0& -2& 1& 4& 0& 0& 0\\ 0& 0& 4& -5& 4& 0& 0\\ 0& 0& 0& 4& 1& -2& 0\\ 0& 0& 0& 0& -2& 4& 1\\ 0& 0& 0& 0& 0& 1& 2\end{array}\right).\end{array}$$

In this case, the matrices of (9) are

$$\begin{array}{c}\hfill Z_7^{\left(1\right)}={\displaystyle \frac{1}{3}}\left(\begin{array}{cccc}2& -12& 20& -8\\ 4& -24& 40& -16\\ 1& -6& 10& -4\end{array}\right),Z_7^{\left(2\right)}=\left(\begin{array}{cccc}1& -5& 6& -1\\ 1& -4& 3& 0\\ 1& -4& 4& -1\end{array}\right).\end{array}$$

We apply a change of variables by setting $x_1=d$, $x_2=d^2$, and $x_3=d^3$, and the linear system has the solution $x_1=19/45$, $x_2=1/5$, and $x_3=2/15$. Since $x_2\ne x_1^2$, then the system has no solution. Thus, there does not exist a matrix ${\tilde{A}}_7$ such that $\mathcal{C}\left({\tilde{A}}_n\right)=U.$

Note that, as observed from Remark 2 and Examples 3 and 4, $Z_n^{\left(1\right)}v$ depends on the values of $u_{ii}$, for $i=1,2,\dots ,q$, $q=1,2,\dots$ and is relatively easy to obtain. In contrast, $Z_n^{\left(2\right)}v$ depends only on the size of U and becomes difficult to determine for large n. In the following result, we present a simpler technique for constructing $Z_n^{\left(2\right)}v$.

Proposition 11. 

Let $Z_n^{\left(2\right)}v$ be the matrix given in (10). Then, $Z_n^{\left(2\right)}v$ can be expressed as $Z_n^{\left(2\right)}v=\mathrm{tril}\left(\mathrm{ones}(q,q)\right){\tilde{Z}}_n^{\left(2\right)}v$, where $\mathrm{tril}\left(\mathrm{ones}\right(q,q\left)\right)$ denotes a $q\times q$ lower triangular matrix of ones and

$$\begin{array}{c}\hfill {\tilde{Z}}_n^{\left(2\right)}v=\left(\begin{array}{c}det\left({\tilde{A}}_{n-1}\right)\\ det\left({\tilde{A}}_{n-3}\right)d\\ det\left({\tilde{A}}_{n-5}\right)d^2\\ ⋮\\ det\left({\tilde{A}}_{n-2q+3}\right)d^{q-2}\\ det\left({\tilde{A}}_{n-2q+1}\right)d^{q-1}\end{array}\right).\end{array}$$

Proof. 

The matrix $Z_n^{\left(2\right)}$ is a matrix of size $q\times (q+1)$, being $q=\lfloor n/2\rfloor$. Since $Z_n^{\left(2\right)}v$ has the following structure,

$$\begin{array}{c}\hfill Z_n^{\left(2\right)}v=\left(\begin{array}{c}det\left({\tilde{A}}_0\right)det\left({\tilde{A}}_{n-1}\right)\\ det\left({\tilde{A}}_1\right)det\left({\tilde{A}}_{n-2}\right)\\ ⋮\\ det\left({\tilde{A}}_q\right)det\left({\tilde{A}}_{n-q-1}\right)\\ det\left({\tilde{A}}_{q-1}\right)det\left({\tilde{A}}_{n-q}\right)\end{array}\right),\end{array}$$

applying the first iteration of Neville’s elimination, we obtain

$$\begin{array}{c}\hfill {\tilde{Z}}_n^{\left(2\right)}v=\left(\begin{array}{c}det\left({\tilde{A}}_{n-1}\right)\\ det\left({\tilde{A}}_1\right)det\left({\tilde{A}}_{n-2}\right)-det\left({\tilde{A}}_0\right)det\left({\tilde{A}}_{n-1}\right)\\ ⋮\\ det\left({\tilde{A}}_q\right)det\left({\tilde{A}}_{n-q-1}\right)-det\left({\tilde{A}}_{q+1}\right)det\left({\tilde{A}}_{n-q-2}\right)\\ det\left({\tilde{A}}_{q-1}\right)det\left({\tilde{A}}_{n-q}\right)-det\left({\tilde{A}}_q\right)det\left({\tilde{A}}_{n-q-1}\right)\end{array}\right).\end{array}$$

To obtain the rows of ${\tilde{Z}}_n^{\left(2\right)}v$, we use the following decompositions, applying Laplace’s formula given in (3), first on $det\left({\tilde{A}}_{n-i}\right)$ and later on $det\left({\tilde{A}}_{i-1}\right)$ or conversely. In the first case, we have proceeded as follows:

$$\begin{array}{cc}\hfill \left(a\right)det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)& =\left(det\left({\tilde{A}}_{n-i-1}\right)-det\left({\tilde{A}}_{n-i-2}\right)d\right)det\left({\tilde{A}}_{i-1}\right)\hfill \\ \hfill & =det\left({\tilde{A}}_{n-i-1}\right)det\left({\tilde{A}}_{i-1}\right)-det\left({\tilde{A}}_{n-i-2}\right)det\left({\tilde{A}}_{i-1}\right)d\hfill \\ \hfill & =det\left({\tilde{A}}_{n-i-1}\right)det\left({\tilde{A}}_{i-1}\right)-det\left({\tilde{A}}_{n-i-2}\right)\left(det\left({\tilde{A}}_{i-2}\right)-det\left({\tilde{A}}_{i-3}\right)d\right)d\hfill \\ \hfill & =det\left({\tilde{A}}_{n-i-1}\right)det\left({\tilde{A}}_{i-1}\right)-det\left({\tilde{A}}_{n-i-2}\right)det\left({\tilde{A}}_{i-2}\right)d\hfill \\ \hfill & +det\left({\tilde{A}}_{n-i-2}\right)det\left({\tilde{A}}_{i-3}\right)d^2.\hfill \end{array}$$

And, in the second case, we have

$$\begin{array}{cc}\hfill \left(b\right)det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)& =det\left({\tilde{A}}_{n-i}\right)\left(det\left({\tilde{A}}_{i-2}\right)-det\left({\tilde{A}}_{i-3}\right)d\right)\hfill \\ \hfill & =det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-2}\right)-det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-3}\right)d\hfill \\ \hfill & =det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-2}\right)-\left(det\left({\tilde{A}}_{n-i-1}\right)-det\left({\tilde{A}}_{n-i-2}\right)d\right)det\left({\tilde{A}}_{i-3}\right)d\hfill \\ \hfill & =det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-2}\right)-det\left({\tilde{A}}_{n-i-1}\right)det\left({\tilde{A}}_{i-3}\right)d\hfill \\ \hfill & +det\left({\tilde{A}}_{n-i-2}\right)det\left({\tilde{A}}_{i-3}\right)d^2.\hfill \end{array}$$

The process stops when it reaches $det\left({\tilde{A}}_0\right)=1$. Note that, except at the beginning when the determinant is decomposed, the last two terms are repeated. Therefore, depending on whether $i-1$, $i=2,3,\dots ,q$ is odd or even, we obtain the following expressions:

(a)

If, in the first case, we decompose $det\left({\tilde{A}}_{n-i}\right)$, we obtain the following:

(a.1)

if $i-1$ is odd,

$$\begin{array}{c}\hfill det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)=\sum _{s=0}^{i-1}det\left({\tilde{A}}_{n-i-(s+1)}\right)det\left({\tilde{A}}_{i-1-s}\right){(-d)}^s\end{array}$$

(a.2)

if $i-1$ is even,

$$\begin{array}{cc}\hfill det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)& =\sum _{s=0}^{i-1}det\left({\tilde{A}}_{n-i-(s+1)}\right)det\left({\tilde{A}}_{i-1-s}\right){(-d)}^s\hfill \\ \hfill & +det\left({\tilde{A}}_{n-i-(i+1)}\right)det\left({\tilde{A}}_0\right){(-d)}^i.\hfill \end{array}$$

(b)

If, in the first case, we decompose $det\left({\tilde{A}}_{i-1}\right)$, we obtain the following:

(b.1)

if $i-1$ is odd,

$$\begin{array}{c}\hfill det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)=\sum _{s=0}^{i-2}det\left({\tilde{A}}_{n-i-s}\right)det\left({\tilde{A}}_{i-1-(s+1)}\right){(-d)}^s\end{array}$$

(b.2)

if $i-1$ is even,

$$\begin{array}{cc}\hfill det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)& =\sum _{s=0}^{i-2}det\left({\tilde{A}}_{n-i-s}\right)det\left({\tilde{A}}_{i-1-(s+1)}\right){(-d)}^s\hfill \\ \hfill & +det\left({\tilde{A}}_{n-i-(i-1)}\right)det\left({\tilde{A}}_0\right){(-d)}^{i-1}.\hfill \end{array}$$

We apply decomposition (a) or (b) depending on whether the row $R_i$ is the subtrahend (the row being subtracted) or the minuend (the row from which we subtract). For example, when subtracting the rows $i+1$ and i, for $i=2,3,\dots ,q-1$,

if $i-1$ is odd, we obtain

$$\begin{array}{cc}\hfill R_{i+1}-R_i& =det\left({\tilde{A}}_{n-(i+1)}\right)det\left({\tilde{A}}_i\right)-det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)\hfill \\ \hfill & =\sum _{s=0}^{i-1}det\left({\tilde{A}}_{n-(i+1)-s}\right)det\left({\tilde{A}}_{i-(s+1)}\right){(-d)}^s+det\left({\tilde{A}}_{n-(i+1)-i}\right)d^i\hfill \\ \hfill & -\sum _{s=0}^{i-1}det\left({\tilde{A}}_{n-i-(s+1)}\right)det\left({\tilde{A}}_{i-1-s}\right){(-d)}^s=det\left({\tilde{A}}_{n-(2i+1)}\right)d^i,\hfill \end{array}$$

but if $i-1$ is even, we obtain

$$\begin{array}{cc}\hfill R_{i+1}-R_i& =det\left({\tilde{A}}_{n-(i+1)}\right)det\left({\tilde{A}}_i\right)-det\left({\tilde{A}}_{n-i}\right)det\left({\tilde{A}}_{i-1}\right)\hfill \\ \hfill & =\sum _{s=0}^{i-1}det\left({\tilde{A}}_{n-(i+1)-s}\right)det\left({\tilde{A}}_{i-(s+1)}\right){(-d)}^s\hfill \\ \hfill & -\sum _{s=0}^{i-1}det\left({\tilde{A}}_{n-i-(s+1)}\right)det\left({\tilde{A}}_{i-1-s}\right){(-d)}^s+det\left({\tilde{A}}_{n-i-(i+1)}\right)det\left({\tilde{A}}_0\right){(-d)}^i\hfill \\ \hfill & =det\left({\tilde{A}}_{n-(2i+1)}\right)d^i.\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill {\tilde{Z}}_n^{\left(2\right)}v=\left(\begin{array}{c}det\left({\tilde{A}}_{n-1}\right)\\ det\left({\tilde{A}}_{n-3}\right)d\\ det\left({\tilde{A}}_{n-5}\right)d^2\\ ⋮\\ det\left({\tilde{A}}_{n-2q+3}\right)d^{q-2}\\ det\left({\tilde{A}}_{n-2q+1}\right)d^{q-1}\end{array}\right).\end{array}$$

(11)

For $i=1,2,\dots ,q,$ if we denote by $W_i=0$ when n is even and $W_i={(-1)}^{q+1-i}$ when n is odd, and define as $C_{i,j}$ the combinatorial number $\left(\begin{array}{c}i\\ j\end{array}\right)$, the matrix ${\tilde{Z}}_n^{\left(2\right)}$ is given by

$$\begin{array}{c}\hfill {\tilde{Z}}_n^{\left(2\right)}=\left(\begin{array}{cccccccc}1& -C_{n-2,1}& \phantom{-}{C}_{n-3,2}& -C_{n-4,3}& \dots & {(-1)}^{q-2}{C}_{n-(q-1),q-2}& {(-1)}^{q-1}{C}_{n-q,q-1}& W_1\\ 0& 1& -C_{n-4,1}& \phantom{-}{C}_{n-5,2}& \dots & {(-1)}^{q-3}{C}_{n-q,q-3}& {(-1)}^{q-2}{C}_{n-(q+1),q-2}& W_2\\ 0& 0& 1& -C_{n-6,1}& \dots & {(-1)}^{q-4}{C}_{n-(q+1),q-4}& {(-1)}^{q-3}{C}_{n-(q+2),q-3}& W_3\\ ⋮& ⋮& ⋮& ⋮& \dots & ⋮& ⋮& ⋮\\ 0& 0& 0& 0& \dots & \dots & {(-1)}^2{C}_{n-(q+q-3),(q-(q-2\left)\right)}& W_{q-2}\\ 0& 0& 0& 0& \dots & 1& -C_{n-(q+q-2),(q-(q-1\left)\right)}& W_{q-1}\\ 0& 0& 0& 0& \dots & 0& 1& W_q\end{array}\right).\end{array}$$

(12)

Therefore, $Z_n^{\left(2\right)}=\mathrm{tril}\left(\mathrm{ones}(q,q)\right){\tilde{Z}}_n^{\left(2\right)}$. □

To illustrate Proposition 11, we present the following example.

Example 5. 

Consider any bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix U of size $n=14$. The size of $Z_{14}^{\left(2\right)}$ is $7\times 8$ and, since $n=14$ is even, we have $W_i=0$, $i=1,2,\dots ,q$.

Using (11), we obtain

$$\begin{array}{cc}\hfill {\tilde{Z}}_{14}^{\left(2\right)}v& =\left(\begin{array}{c}det\left({\tilde{A}}_{13}\right)\\ det\left({\tilde{A}}_{11}\right)d\\ det\left({\tilde{A}}_9\right)d^2\\ det\left({\tilde{A}}_7\right)d^3\\ det\left({\tilde{A}}_5\right)d^4\\ det\left({\tilde{A}}_3\right)d^5\\ det\left({\tilde{A}}_1\right)d^6\end{array}\right)=\left(\begin{array}{c}1-12d+55d^2-120d^3+126d^4-56d^5+7d^6\\ (1-10d+36d^2-56d^3+35d^4-6d^5)d\\ (1-8d+21d^2-20d^3+5d^4)d^2\\ (1-6d+10d^2-4d^3)d^3\\ (1-4d+3d^2)d^4\\ (1-2d)d^5\\ d^6\end{array}\right)=\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =\underset{{\tilde{Z}}_{14}^{\left(2\right)}}{\underbrace{\left(\begin{array}{cccccccc}1& -12& 55& -120& 126& -56& 7& 0\\ 0& 1& -10& 36& -56& 35& -6& 0\\ 0& 0& 1& -8& 21& -20& 5& 0\\ 0& 0& 0& 1& -6& 10& -4& 0\\ 0& 0& 0& 0& 1& -4& 3& 0\\ 0& 0& 0& 0& 0& 1& -2& 0\\ 0& 0& 0& 0& 0& 0& 1& 0\end{array}\right)}}\underset{v}{\underbrace{\left(\begin{array}{c}1\\ d\\ d^2\\ d^3\\ d^4\\ d^5\\ d^6\\ d^7\end{array}\right)}}\hfill \end{array}$$

From (12) we obtain directly the matrix ${\tilde{Z}}_{14}^{\left(2\right)}$,

$$\begin{array}{cc}\hfill {\tilde{Z}}_{14}^{\left(2\right)}& =\left(\begin{array}{cccccccc}1& -C_{12,1}& C_{11,2}& -C_{10,3}& C_{9,4}& -C_{8,5}& C_{7,6}& 0\\ 0& 1& -C_{10,1}& C_{9,2}& -C_{8,3}& C_{7,4}& -C_{6,5}& 0\\ 0& 0& 1& -C_{8,1}& C_{7,2}& -C_{6,3}& C_{5,4}& 0\\ 0& 0& 0& 1& -C_{6,1}& C_{5,2}& -C_{4,3}& 0\\ 0& 0& 0& 0& 1& -C_{4,1}& C_{3,2}& 0\\ 0& 0& 0& 0& 0& 1& -C_{2,1}& 0\\ 0& 0& 0& 0& 0& 0& 1& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cccccccc}1& -12& 55& -120& 126& -56& 7& 0\\ 0& 1& -10& 36& -56& 35& -6& 0\\ 0& 0& 1& -8& 21& -20& 5& 0\\ 0& 0& 0& 1& -6& 10& -4& 0\\ 0& 0& 0& 0& 1& -4& 3& 0\\ 0& 0& 0& 0& 0& 1& -2& 0\\ 0& 0& 0& 0& 0& 0& 1& 0\end{array}\right).\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill Z_{14}^{\left(2\right)}=\mathrm{tril}\left(\mathrm{ones}(7,7)\right){\tilde{Z}}_{14}^{\left(2\right)}=\left(\begin{array}{cccccccc}1& -12& 55& -120& 126& -56& 7& 0\\ 1& -11& 45& -84& 70& -21& 1& 0\\ 1& -11& 46& -92& 91& -41& 6& 0\\ 1& -11& 46& -91& 85& -31& 2& 0\\ 1& -11& 46& -91& 86& -35& 5& 0\\ 1& -11& 46& -91& 86& -34& 3& 0\\ 1& -11& 46& -91& 86& -34& 4& 0\end{array}\right).\end{array}$$

4. Conclusions

This study provides a comprehensive characterization of the combined matrix $\mathcal{C}\left(A\right)=A\circ A^{-T}$, where A is a real, nonsingular tridiagonal Toeplitz matrix and ∘ denotes the Hadamard product. Taking into account that the matrices $A_n=T(n;c,a,b)$ and ${\tilde{A}}_n=T(n;1,1,d)$, where $d=bc/a^2$, are diagonally equivalent matrices, we established relationships between their determinants and eigenvalues. Furthermore, we demonstrated that $\mathcal{C}\left(A_n\right)=\mathcal{C}\left({\tilde{A}}_n\right)$ is a bisymmetric, doubly quasi-stochastic Jacobi matrix.

We also analyzed the specific case ${\tilde{A}}_n=T(n;1,1,1)$, showing that the combined matrix exists for $n=3q$ and $n=3q+1$, $q=1,2,\dots$, and provided explicit forms for these cases. Moreover, we showed that the combined matrix does not exist when $n=3q+2$.

Additionally, we solved the inverse problem by obtaining necessary and sufficient conditions for the existence of a Toeplitz matrix whose combined matrix corresponds to a prescribed bisymmetric, doubly quasi-stochastic tridiagonal Jacobi matrix. This was achieved by deriving and solving a system of polynomial equations based on the matrix’s diagonal entries and using the results to address both even and odd cases. Furthermore, we provide a result for calculating the matrices involved in the system in a simpler manner.

With this work, we have extended the problem from matrices of order 3 to tridiagonal Toeplitz matrices of any order.