Hyers-Ulam Stability for Linear Differences with Time Dependent and Periodic Coefficients: The Case When the Monodromy Matrix Has Simple Eigenvalues

Abstract

Let $q\ge 2$ be a positive integer and let $\left(a_j\right),\left(b_j\right)$ and $\left(c_j\right)$ (with j nonnegative integer) be three given $\mathbb{C}$-valued and q-periodic sequences. Let $A\left(q\right):=A_{q-1}\cdots A_0,$ where $A_j$ is defined below. Assume that the eigenvalues $x,y,z$ of the “monodromy matrix” $A\left(q\right)$ verify the condition $(x-y)(y-z)(z-x)\ne 0.$ We prove that the linear recurrence in $\mathbb{C}$ $x_{n+3}=a_n{x}_{n+2}+b_n{x}_{n+1}+c_n{x}_n,n\in {\mathbb{Z}}_{+}$ is Hyers–Ulam stable if and only if $\left(\right|x|-1)\left(\right|y|-1)\left(\right|z|-1)\ne 0,$ i.e., the spectrum of $A\left(q\right)$ does not intersect the unit circle $\mathsf{\Gamma }:=\{w\in \mathbb{C}:|w|=1\}.$

1. Introduction

Exponential dichotomy and its links with the unconditional stability of differential dynamics systems were first highlighted by O. Perron in 1930 [1]. The reader can find details on the subsequent evolution of this topic in Coppel’s monograph [2]. The history of the Ulam problem (concerning the stability of a functional equation) and of stability in the sense of Hyers–Ulam is well known. In particular, Hyers–Ulam stability for linear recurrences and for systems of linear recurrences is considered in [3,4,5,6,7,8,9,10,11,12,13,14,15,16], and the references therein.

The relationship between exponential stability and Hyers–Ulam stability has been studied in the articles [3,8,9,17,18], and this article continues these studies.

2. Notations and Definitions

By $\mathbb{C}$, we denote the set complex numbers and ${\mathbb{Z}}_{+}$ is the set of all nonnegative integers. Now, ${\mathbb{C}}^m$ (with m a given positive integer) is the set of all vectors $v={({\xi }_1,\cdots ,{\xi }_m)}^T$ with ${\xi }_j\in \mathbb{C}$ for every integers $1\le j\le m;$ here and in as follows ${}^T$ denotes the transposition. The norm on ${\mathbb{C}}^m$ is the well-known Euclidean norm defined by $\parallel v\parallel :=(|{\xi }_1{|}^2+\cdots +|{\xi }_m{{|}^2)}^{\frac{1}{2}}.$ In addition, ${\mathbb{C}}^{m\times n}$ (with m and n given positive integers) denotes the set of all m by n matrices with complex entries. In particular, ${\mathbb{C}}^{m\times m}$ becomes a Banach algebra when it is endowed with the (Euclidean) matrix norm defined by $\parallel M\parallel :={sup}_{\parallel v\parallel \le 1}\parallel Mv\parallel ,v\in {\mathbb{C}}^m,M\in {\mathbb{C}}^{m\times m}.$ As is usual, the rows and columns of a matrix $M\in {\mathbb{C}}^{m\times n}$ are identified by vectors of the corresponding dimensions and in that case its norm is the vector norm. The entry $m_{ij}$ of a matrix M (i.e., the entry in M located at the intersection between the ith row and the jth column) is denoted by ${\left[M\right]}_{ij}.$ As is usual, the uniform norm of a ${\mathbb{C}}^m$-valued and bounded sequence $g=\left(g_n\right)$ is defined and denoted by ${\parallel g\parallel }_{\infty }:={sup}_{j\in {\mathbb{Z}}_{+}}\parallel g_j\parallel .$

Let $\epsilon >0$ be given. We recall (see also [8] for the two-dimensional case) that a scalar valued sequence $\left(y_j\right)$ is an $\epsilon$-approximative solution of the linear recurrence

$$x_{n+3}=a_n{x}_{n+2}+b_n{x}_{n+1}+c_n{x}_n,n\in {\mathbb{Z}}_{+}$$

(1)

if

$$|y_{n+3}-a_n{y}_{n+2}-b_n{y}_{n+1}-c_n{y}_n|\le \epsilon ,\forall n\in {\mathbb{Z}}_{+}.$$

(2)

The recurrence in Equation (1) is Hyers–Ulam stable if there exists a positive constant L such that for every $\epsilon >0$ and every $\epsilon$-approximative solution $y=\left(y_j\right)$ of Equation (1) there exists an exact solution $\theta =\left({\theta }_j\right)$ of Equation (1) such that ${\parallel y-\theta \parallel }_{\infty }\le L\epsilon .$

Remark 1.

Since any ε-approximative solution of the recurrence in Equation (1) can be seen as a solution of the nonhomogeneous equation

$$x_{n+3}-a_n{x}_{n+2}-b_n{x}_{n+1}-c_n{x}_n=f_{n+1},n\in {\mathbb{Z}}_{+},$$

(3)

for some scalar valued sequence $\left(f_n\right)$ with $f_0=0$ and $\parallel \left(f_k\right){\parallel }_{\infty }\le \epsilon ,$ one has that Equation (1) is Hyers–Ulam stable if and only if there exists a positive constant L such that for every $\epsilon >0,$ every sequence as above, and every initial condition $Y_0={(z_0,v_0,w_0)}^T\in {\mathbb{C}}^3$, there exists an initial condition $X_0={(x_0,x_1,x_2)}^T\in {\mathbb{C}}^3$ such that

$$\left|\varphi (n,Y_0,\left(f_k\right))-\varphi (n,X_0,\left(0\right))\right|\le L\epsilon .$$

(4)

Here, and in what follows, $(\varphi (n,Y_0,\left(f_k\right))$ denotes the solution of the nonhomogeneous linear recurrence in Equation (3) initiated from $Y_0.$

Proof. 

See the proof of Proposition 3.1 in [9]. □

3. Background, Previous Results and the Main Result

Proposition 1.

([19]) Let A be a 3 by 3 matrix whose spectrum (i.e., the set of its eigenvalues $\sigma \left(A\right):=\{x,y,z\})$ satisfies the condition

$$(x-y)(x-z)(y-z)\ne 0.$$

(5)

Then, for every nonnegative integer n, one has

$$A^n=x^{n}B+y^{n}C+z^{n}D$$

(6)

where

$$B={\displaystyle \frac{(A-y{I}_3)(A-z{I}_3)}{(x-y)(x-z)}},C={\displaystyle \frac{(A-x{I}_3)(A-z{I}_3)}{(y-x)(y-z)}}$$

(7)

and

$$D={\displaystyle \frac{(A-x{I}_3)(A-y{I}_3)}{(z-x)(z-y)}}.$$

(8)

Remark 2.

(i) The matrices $B,C$ and D in Equation (6) are orthogonal projections, that is

$$BC=BD=CD=0_3; the null matrix of order three,$$

(9)

and

$$B^2=B,C^2=C, and D^2=D.$$

(10)

(ii) In addition, $B,C,$ and D are nonzero matrices.

Proof. 

Under assumption in Equation (5), the characteristic polynomial $P_A$ and the minimal polynomial $m_A$ of A coincide and $P_A\left(\lambda \right)=(\lambda -x)(\lambda -y)(\lambda -z).$ Thus, from the Hamilton–Cayley Theorem we have $P_A\left(A\right)=(A-x{I}_3)(A-y{I}_3)(A-z{I}_3)=0_3,$ and Equation (9) becomes clear.

To prove Equation (10), it is enough to see that

$$B^2-B=\frac{(A-y{I}_3)(A-z{I}_3)(A-x{I}_3)(A-(y+z-x)I_3))}{{(x-y)}^2{(x-z)}^2};$$

the details are clear thus omitted. Then, we apply the Hamilton–Cayley theorem and obtain Equation (10).

Finally, assuming that $B=0_3,$ the polynomial

$$Q\left(\lambda \right)=\frac{(\lambda -y)(\lambda -z)}{(x-y)(x-z)}$$

is annulated by A and its degree is equal 2 and is a contradiction with the minimality of the degree of $m_A.$ □

Let q, $\left(a_j\right)$, $\left(b_j\right),$ $\left(c_j\right)$ be as above. Recall that

$$A\left(q\right):=A_{q-1}\cdots A_0, \mathrm{where} A_j:=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ c_j& b_j& a_j\end{array}\right),j\in {\mathbb{Z}}_{+}.$$

(11)

Our main result reads as follows.

Theorem 1.

Assume that the eigenvalues $x,y,z$ (of $A\left(q\right)$) satisfy the condition in Equation (5). Then, the following two statements are equivalent:

1.The linear recurrence in $\mathbb{C}$

$$x_{n+3}=a_n{x}_{n+2}+b_n{x}_{n+1}+c_n{x}_n,n\in {\mathbb{Z}}_{+}$$

(12)

is Hyers–Ulam stable.

2.The eigenvalues of $A\left(q\right)$ verify the condition

$$\left(\right|x|-1)\left(\right|y|-1)\left(\right|z|-1)\ne 0.$$

(13)

The proof of the implication $\mathbf{2}\Rightarrow \mathbf{1}$ is covered (for the most part) in the existing literature. We present the ideas and complete the details. For unexplained terminology, we refer the reader to [8,9]. The following result is taken directly from the second section of [9].

Let X be a complex, finite dimensional Banach space and let $\mathcal{B}={\left\{B_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ and $\mathcal{P}={\left\{P_n\right\}}_{n\in {\mathbb{Z}}_{+}}$ be two families of linear operators acting on $X.$ Assume that:

[A1] $B_{n+q}=B_n$ and $P_{n+q}=P_n$, for all $n\in {\mathbb{Z}}_{+}$ and some positive integer $q.$

[A2] $P_n^2=P_n$, for all $n\in {\mathbb{Z}}_{+},$ that is, $\mathcal{P}$ is a family of projections.

[A3] $B_n{P}_n=P_{n+1}{B}_n$, for all $n\in {\mathbb{Z}}_{+}.$ In particular, this yields that $B_{n}x\in ker\left(P_{n+1}\right)$ for each $x\in ker\left(P_n\right).$

[A4] For each $n\in {\mathbb{Z}}_{+},$ the map

$$x\mapsto B_{|n}x:=B_{n}x:ker\left(P_n\right)\to ker\left(P_{n+1}\right)$$

is invertible. Denote by ${\left(B_{|n}\right)}^{-1}$ its inverse.

We say that the family $\mathcal{B}$ is $\mathcal{P}$-dichotomic if there exist four positive constants $N_1,$ $N_2,$ ${\nu }_1$ and ${\nu }_2$ such that

(i)

$\parallel U_{\mathcal{B}}(n,k)P_k\parallel \le N_1{e}^{-{\nu }_1(n-k)}$ for all $n\ge k\ge 0.$

(ii)

$\parallel U_{\mathcal{B}}(n,k)(I-P_k)\parallel \le N_2{e}^{{\nu }_2(n-k)}$ for all $0\le n<k.$

Here, $U_{\mathcal{B}}(n,k)=B_{n-1}\cdots B_k$ when $n>k,$ $U_{\mathcal{B}}(k,k)=I$-the identity operator on $X,$ and $U_{\mathcal{B}}(n,k):={\left(B_{|k}\right)}^{-1}·\cdots ·{\left(B_{|n-1}\right)}^{-1}$ when $n<k.$

Theorem 2.

([9]) Assume that the families $\mathcal{B}$ and $\mathcal{P}$ satisfy [A1]–[A4] above. The following four statements are equivalent:

(1) The monodromy operator $B\left(q\right):=B_{q-1}\cdots B_0$ is hyperbolic (that is, the spectrum of $B\left(q\right)$ does not intersect the unit circle $\mathsf{\Gamma }=\{w\in \mathbb{C}:|w|=1\}$, or equivalently (with the terminology in [9]) it possesses a discrete dichotomy.

(2) The family $\mathcal{B}$ is $\mathcal{P}$-dichotomic.

(3) For each bounded sequence ${\left(G_n\right)}_{n\in {\mathbb{Z}}_{+}},G_0=0$ (of X-valued functions) there exists a unique bounded solution (starting from $ker\left(P_0\right))$ of the difference equation.

$$x_{n+1}=B_n{x}_n+G_{n+1},n\in {\mathbb{Z}}_{+}.$$

(4) The family $\mathcal{B}$ is Hyers–Ulam stable.

We mention that the equivalence between (2) and (3) still works when X is an infinite dimensional Banach space (see [20]). We use Theorem 2 to prove 2⇒1 in Theorem 1.

The main ingredient in the proof of the implication $\mathbf{1}\Rightarrow \mathbf{2}$ in Theorem 1 is the following Lemma.

With $\mathcal{A}$ we denote the set of all matrices $A_j$ (with $j\in {\mathbb{Z}}_{+}),$ where $A_j$ is given in Equation (11)) and the matrix $U_{\mathcal{A}}(n,k)$ is defined above.

Lemma 1.

If the spectrum of $A\left(q\right)$ intersects the unit circle then for each $\epsilon >0$ there exists a $\mathbb{C}$-valued sequence ${\left(f_j\right)}_{j\in {\mathbb{Z}}_{+}}$ with $f_0=0$ and $\parallel \left(f_j\right){\parallel }_{\infty }\le \epsilon$ such that for every initial condition $Z_0={(x_0,y_0,z_0)}^T\in {\mathbb{C}}^3,$ the $\mathbb{C}$-valued sequence

$${\left({\left|\left[U_{\mathcal{A}}(n,0)Z_0+{\sum }_{k=1}^n{U}_{\mathcal{A}}(n,k)F_k\right]\right|}_{11}\right)}_{n\in {\mathbb{Z}}_{+}}$$

(14)

(with $F_k={(0,0,f_k)}^T),$ is unbounded.

4. Proofs

Proof of Lemma 1.

We first use Proposition 1 with $A\left(q\right)$ instead of A. Assume that the eigenvalue x has modulus 1. Let $P_x$ be the Riesz projection associated to $A\left(q\right)$ and x; that is

$$P_x=\frac{1}{2\pi i}\underset{C(x,r)}{\int }{\left(w{I}_3-A\left(q\right)\right)}^{-1}dw,$$

where $C(x,r)$ is the circle centered at x of radius r, and r is small enough that y and z are located outside of the circle. Using Dunford calculus (see [21]), it is easy to see that $P_{x}A{\left(q\right)}^n=x^{n}B$, for each $n\in {\mathbb{Z}}_{+}$. Consider the matrix B from Equation (6), of the form:

$$B=\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right).$$

The solution of the system

$$X_{n+1}=A_n{X}_n+F_{n+1},n\in {\mathbb{Z}}_{+},$$

(15)

initiated from $Z_0$, where $X_n:={\left(\begin{array}{ccc}{z}_n& v_n& w_n\end{array}\right)}^T\in {\mathbb{C}}^3,$ $F_n={\left(\begin{array}{ccc}0& 0& f_n\end{array}\right)}^T$ and

$$A_n:=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ c_n& b_n& a_n\end{array}\right)$$

is given by

$${\mathsf{\Phi }}_n:=\mathsf{\Phi }(n,Z_0,\left(F_k\right))=U_{\mathcal{A}}(n,0)Z_0+{\sum }_{k=1}^n{U}_{\mathcal{A}}(n,k)F_k.$$

(16)

Denote by $\left(\phi (n,Z_0,\left(f_k\right)\right)$ the solution of Equation (1). An obvious calculation yields

$${\phi }_n:=\phi (n,Z_0,\left(f_k\right))={\left[U_{\mathcal{A}}(n,0)Z_0+{\sum }_{k=1}^n{U}_{\mathcal{A}}(n,k)F_k\right]}_{11}.$$

(17)

In fact, one has ${\mathsf{\Phi }}_n={\left(\begin{array}{ccc}{\phi }_n& {\phi }_{n+1}& {\phi }_{n+2}\end{array}\right)}^T$.

Case 1.1. Let $b_{13}\ne 0$. Set

$$F_k=\left\{\begin{array}{c}{x}^{k/q}{u}_0, \mathrm{if} k=nq\hfill \\ 0, \mathrm{if}k \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{multiple} \mathrm{of} q,\hfill \end{array}\right.$$

(18)

where $u_0:={\left(\begin{array}{ccc}0& 0& c_0\end{array}\right)}^T$ and $c_0$ is a randomly chosen nonzero complex scalar with $|c_0|<\epsilon .$ Successively, one has

$${\mathsf{\Phi }}_{nq}=U_{\mathcal{A}}(nq,0)Z_0+{\sum }_{k=1}^{nq}{U}_{\mathcal{A}}(nq,k)F_k$$

$$=U_{\mathcal{A}}(nq,0)Z_0+U_{\mathcal{A}}(nq,0)F_0+U_{\mathcal{A}}(nq,q)F_q+\cdots +U_{\mathcal{A}}(nq,nq)F_{nq}$$

$$=U_{\mathcal{A}}(nq,0)Z_0+{\sum }_{j=1}^n{U}_{\mathcal{A}}(nq,jq)F_{jq}=A{\left(q\right)}^n{Z}_0+{\sum }_{j=1}^n{x}^{j}A{\left(q\right)}^{n-j}{u}_0,$$

that yields

$$P_x\left[U_{\mathcal{A}}(nq,0)Z_0+{\sum }_{k=1}^{nq}{U}_{\mathcal{A}}(nq,k)F_k\right]$$

$$=P_{x}A{\left(q\right)}^n{Z}_0+{\sum }_{j=1}^n{x}^j{P}_{x}A{\left(q\right)}^{n-j}{u}_0$$

$$=x^{n}B{Z}_0+{\sum }_{j=1}^n{x}^j{x}^{n-j}B{u}_0.$$

$$=x^{n}B{Z}_0+n{x}^{n}B{u}_0.$$

Since the sequence ${\left(x^{n}B{Z}_0\right)}_n$ is bounded, it is enough to prove that the sequence ${\left({\left[n{x}^{n}B{u}_0\right]}_{11}\right)}_n$ is unbounded, and note

$$|{\left[n{x}^{n}B{u}_0\right]}_{11}|=n\left|b_{13}{c}_0\right|\to \infty \mathrm{as} n\to \infty .$$

Case 1.2. Let $b_{23}\ne 0$. Arguing as above we can show that $\left({\phi }_{n+1}\right)$ is unbounded, that is that $\left({\phi }_n\right)$ is unbounded as well.

Case 1.3. Analogously, we can treat the case $b_{33}\ne 0$.

Case 2.1. Let $b_{13}=b_{23}=b_{33}=0$ and $b_{12}\ne 0$. Set

$$F_k=\left\{\begin{array}{c}{x}^{k/q}{A}_{q-1}{u}_0, \mathrm{if} k=nq\hfill \\ 0, \mathrm{if} k \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{multiple} \mathrm{of} q,\hfill \end{array}\right.$$

(19)

where $u_0$ and $c_0$ are taken as above. We obtain

$$\begin{array}{ccc}{\mathsf{\Phi }}_{nq}& =& A{\left(q\right)}^n{Z}_0+{\sum }_{j=1}^n{x}^{j}A{\left(q\right)}^{n-j}{A}_{q-1}{u}_0\hfill \end{array}$$

which leads to

$$\begin{array}{ccc}\left|{\phi }_{nq}\right|& =& \left|{\left[{\sum }_{j=1}^n{x}^j{P}_{x}A{\left(q\right)}^{n-j}{A}_{q-1}{u}_0\right]}_{11}\right|\hfill \\ & =& \left|{\left[{\sum }_{j=1}^n{x}^j{x}^{n-j}B{A}_{q-1}{u}_0\right]}_{11}\right|\hfill \\ & =& \left|{\sum }_{j=1}^n{x}^n{b}_{12}{c}_0\right|\hfill \\ & =& \left|n{x}^n{b}_{12}{c}_0\right|=n\left|b_{12}{c}_0\right|\to \infty \mathrm{as} n\to \infty .\hfill \end{array}$$

Case 2.2. Let $b_{22}\ne 0$. Similar to the previous case, we can show that $\left({\phi }_{n+1}\right)$ is unbounded, that is that $\left({\phi }_n\right)$ is unbounded as well.

Case 3. Let $b_{12}=b_{13}=0$ and $b_{11}\ne 0$. Then, set

$$F_k=\left\{\begin{array}{c}{x}^{k/q}{A}_{q-2}{A}_{q-1}{u}_0, \mathrm{if} k=nq\hfill \\ 0, \mathrm{if} k \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{multiple} \mathrm{of} q,\hfill \end{array}\right.$$

(20)

with $u_0$ and $c_0$ as above.

As in the previous cases, we obtain

$$\begin{array}{ccc}{\left[{\sum }_{j=1}^n{x}^j{P}_x{A}^{n-j}{A}_{q-2}{A}_{q-1}{u}_0\right]}_{11}& =& \left|n{x}^n{b}_{11}{c}_0\right|=n\left|b_{11}{c}_0\right|\to \infty \mathrm{as} n\to \infty ,\hfill \end{array}$$

therefore $\left({\phi }_n\right)$ is again unbounded.

Finally, we remark that the matrix B cannot be of the form $\left(\begin{array}{ccc}0& 0& 0\\ *& 0& 0\\ *& *& 0\end{array}\right).$ Indeed, if this is the case, all eigenvalues of B are equal to 0 and the Hamilton–Cayley Theorem yields $B^3=0_3.$ Since $B^2=B$ we obtain $B^2=0_3$, that is, $B=0_3$. This contradicts the statement in Remark 2, (ii). □

Proof of Theorem 1.

1⇒2. We argue by contradiction. Suppose that $\sigma \left(A\right(q\left)\right)$ intersects the unit circle. Without loss of generality, assume that x is an eigenvalue of $A\left(q\right)$ and $\left|x\right|=1.$ Let $Y_0$ and $X_0$ be as in the Remark 1. From Lemma 1, it follows that the sequence in Equation (14) with $\left(Y_0-X_0\right)$ instead of $Z_0$ is unbounded and this contradicts Equation (4).

2⇒1. From the assumption and Theorem 2, it follows that the system $X_{n+1}=A_n{X}_n$ is Hyers–Ulam stable. Thus, for a certain positive constant $L,$ every $\epsilon >0,$ every sequence $\left(f_n\right)$, every $Y_0$ and some $X_0$ one has

$$\left|\varphi (n,Y_0,\left(f_k\right))-\varphi (n,X_0,\left(0\right))\right|$$

$$={\left|\left[U_{\mathcal{A}}(n,0)(Y_0-X_0)+{\sum }_{k=1}^n{U}_{\mathcal{A}}(n,k)F_k\right]\right|}_{11}$$

$$\le \left|\left[U_{\mathcal{A}}(n,0)(Y_0-X_0)+{\sum }_{k=1}^n{U}_{\mathcal{A}}(n,k)F_k\right]\right|\le L\epsilon$$

for all $n\in {\mathbb{Z}}_{+}.$ Now, the assertion follows from Remark 1. □

5. An Example

The following example illustrates our theoretical result.

Example 1.

The linear recurrence of order three

$$x_{n+3}=sin\frac{2n\pi }{3}{x}_{n+2}+cos\frac{2n\pi }{3}{x}_{n+1}+c_n{x}_n,n\in \mathbb{Z}$$

(21)

(with

$$c_n=\left\{\begin{array}{cc}1,& if n is a multiple of 3\\ 0,& elsewhere\end{array}\right.)$$

is Hyers–Ulam stable. Indeed, with the above notation one has

$$A_0=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 1& 0\end{array}\right)A_1=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)$$

and

$$A_2=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right).$$

Now, the monodromy matrix associated to Equation (21) is

$$A\left(3\right)=A_2{A}_1{A}_0=\left(\begin{array}{ccc}1& 1& 0\\ \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}& -\frac{1}{2}\\ -\frac{5}{4}& -\frac{5}{4}& \frac{\sqrt{3}}{4}\end{array}\right)$$

The characteristic equation associated to $A\left(3\right)$ is

$${\lambda }^3-\left(1+\frac{3\sqrt{3}}{4}\right){\lambda }^2+\frac{\sqrt{3}-1}{4}\lambda =0$$

and the absolute value of each of its solutions is different to $1.$

Remark 3.

Reading [22], we note that an interesting question is if the spectral condition $\left(\right|x|-1)\left(\right|y|-1)\left(\right|z|-1)\ne 0$ is equivalent to Hyers–Ulam stability of the recurrence in Equation (12) with ${\mathbb{Z}}_{+}$ replaced by $\mathbb{Z}$. We thank the anonymous reviewer who made us aware of the work in [22].