Computation of the zeros of a quaternionic polynomial using matrix methods

Abstract

In a recent paper, Ishfaq Dar (2024), worked on the problem of locating the zeros of quaternion polynomials by introducing various matrix techniques. In this paper, we use those newly developed matrix methods to locate the left eigenvalues of the quaternion companion matrix of the polynomial, which in turn yield various results concerning the location of the zeros of quaternionic polynomials including extensions of the results of Rubinstein and A. Aziz to the quaternionic setting as well.

Keywords:

• Companion matrix
• quaternionic polynomials
• right (and left) quaternionic eigenvalues

Mathematics Subject Classification (2020):

• 30A10
• 30C10
• 30C15

1. Introduction

In an effort to expand complex numbers to greater spatial dimensions, the Irish mathematician Sir William Rowan Hamilton (1805-1865) invented quaternions in 1843. Hamilton became obsessed with quaternions and their uses (Hankins, 1980) after inventing them, and he did so for the remainder of his life. However, he probably never imagined that his invention, quaternions, would one day be used to programme video games and steer spacecraft (Turner, 2006). Quaternions are used in many areas of science and engineering, including computer graphics, robotics, and aerospace. In computer graphics they are particularly used in computer vision and machine learning, where they are used to represent and manipulate 3D data. In engineering they are used in the field of augmented reality, where they are used to track the position and orientation of objects in 3D space. Similarly they are used to solve problems in robotics such as inverse kinematics and path planning. Quaternions are denoted by $\mathbb{H}=\{a=a_0+a_{1}i+a_{2}j+a_{3}k:a_0,a_1,a_2,a_3\in \mathbb{R}$ and i, j, k satisfy $i^2=j^2=k^2=\mathit{\text{ijk}}=-1,\mathit{\text{ij}}=-\mathit{\text{ji}}=k,\mathit{\text{jk}}=-\mathit{\text{kj}}=i,\mathit{\text{ki}}=-\mathit{\text{ik}}=j..$ The set of quaternions is a skew-field and because of non-commutative nature, they differ from complex numbers $\mathbb{C}$ and real numbers $\mathbb{R}.$ A number in the quaternions is denoted by q where $q=\alpha +\beta i+\gamma j+\delta k\in \mathbb{H},$ contains one real part $\alpha$ and three imaginary parts $\beta ,$ $\gamma$ and $\delta .$ The conjugate of q, denoted by $\overline{q}$ is a quaternion $q=\alpha -\beta i-\gamma j-\delta k$ and the norm of q is $\left|q\right|=\sqrt{q\overline{q}}=\sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2}.$ The inverse of each non zero element q of $\mathbb{H}$ is given by $q^{-1}=|q{|}^{-2}\overline{q}.$

Depending upon the position of the coefficients, the quaternion polynomial of degree n in indeterminate q is defined as $f(q)=q^n+q^{n-1}{a}_1+\cdots +q{a}_{n-1}+a_n$ or $g(q)=q^n+a_1{q}^{n-1}+\cdots +a_{n-1}q+a_n.$

The quaternion Companion Matrix: The $n\times n$ companion matrix of a monic quaternion polynomial of the form $f(q)=q^n+q^{n-1}{a}_1+\dots +q{a}_{n-1}+a_n,$ is given by $$C_f=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& -a_n\\ 1& 0& 0& \cdots & 0& -a_{n-1}\\ 0& 1& 0& \cdots & 0& -a_{n-2}\\ 0& 0& 1& \cdots & 0& -a_{n-3}\\ ⋮& & & & & \\ 0& 0& 0& \cdots & 1& -a_1\end{array}\right],$$ whereas, the $n\times n$ companion matrix for a monic quaternion polynomial of the form $g(q)=q^n+a_1{q}^{n-1}+\cdots +a_{n-1}q+a_n,$ is given by $$C_g=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& & & & \\ 0& 0& 0& \cdots & 1\\ -a_n& -a_{n-1}& -a_{n-2}& \cdots & -a_1\end{array}\right]$$

Right Eigenvalue: Given an $n\times n$ matrix $A=[a_{\mu \nu }]$ of quaternions, $\lambda \in \mathbb{H}$ is called the right eigenvalue of A, if $\mathit{\text{Ax}}=x\lambda$ for some non zero eigenvector $x={[x_1,x_2,\dots x_n]}^T$ of quaternions.

Left Eigenvalue: Given an $n\times n$ matrix $A=[a_{\mu \nu }]$ of quaternions, $\lambda \in \mathbb{H}$ is called the left eigenvalue of A, if $\mathit{\text{Ax}}=\lambda x$ for some non zero eigenvector $x={[x_1,x_2,\dots ,x_n]}^T$ of quaternions.

For complex case, concerning the location of the eigenvalues, the famous Geršgorin theorem (Marden, 1949) can be stated as;

Theorem A.

All the eigenvalues of a $n\times n$ complex matrix $A=(a_{\mu \nu })$ are contained in the union of n Geršgorin discs defined by $D_{\mu }=\{z\in \mathbb{C}:|z-a_{\mu \mu }|\le {\sum }_{\begin{array}{c}\nu =1\\ \nu \ne \mu \end{array}}^n|a_{\mu \nu }\left|\right\}.$

Recently, Ishfaq Dar (2024) proved the following quaternion version of Geršgorin theorem.

Theorem B.

All the left eigenvalues of a $n\times n$ matrix $A=(a_{\mu \nu })$ of quaternions lie in the union of the n Geršgorin balls defined by $B_{\mu }=\{q\in \mathbb{H}:|q-a_{\mu \mu }|\le {\rho }_{\mu }(A)\},$ where ${\rho }_{\mu }(A)={\sum }_{\begin{array}{c}\nu =1\\ \nu \ne \mu \end{array}}^n\left|a_{\mu \nu }\right|.$

In the same paper, they considered the quaternion polynomial with coefficients on left side and gave connection between its zeros and the left eigen values of its corresponding companion matrix by proving the following result

Theorem C.

Let $P(q)=q^n+a_{n-1}{q}^{n-1}+\dots +a_{1}q+a_0$ be a quaternion polynomial with quaternionic coefficients and q be quaternionic variable, then for any diagonal matrix $D=\mathit{\text{diag}}(d_1,d_2,\dots ,d_{n-1},d_n),$ where $d_1,d_2,\dots ,d_n$ are positive real numbers, the left eigenvalues of $D^{-1}{C}_{P}D$ and the zeros of $P(q)$ are same.

Since variable and the coefficients are quaternions, the question arises, does above theorem holds for polynomials with coefficients on right side, the answer is provided by the following theorem.

Theorem 1.

Let $f(q)=q^n+q^{n-1}{a}_1+\cdots +q{a}_{n-1}+a_n$ be a quaternion polynomial with quaternionic coefficients and q be quaternionic variable, then for any diagonal matrix $T=\mathit{\text{diag}}(d_1,d_2,\dots ,d_{n-1},d_n),$ where $d_1,d_2,\dots ,d_n$ are positive real numbers, the left eigenvalues of $T^{-1}{C}_{f}T$ and the zeros of $f(q)$ are same.

From this theorem, we conclude that the zeros of quaternion polynomial and left eigen values of its corresponding companion matrix are same irrespective of the position of its coefficients.

Remark 1.

It is easy to see that, if we take $d_1=d_2=d_3=\dots =d_n=1,$ then both Theorem C and Theorem 1 reduce to the following result.

Corollary 1.

If $\lambda$ is the left eigenvalue of the companion matrix $C_P$ associated with the quaternionic polynomial $P(q),$ then $\lambda$ is a zero of $P(q).$

In last two years various results were proved by different authors regarding the location of zeros of quaternion polynomials. Recently, Carney, Gardner, Keaton, and Powers (2020) extended Eneström-Kakeya theorem to quaternion settings by proving following result.

Theorem D.

If $p(q)=q^n{a}_n+q^{n-1}{a}_{n-1}+q^{n-2}{a}_{n-2}+\dots +q{a}_1+a_0$ is a polynomial of degree n (where q is a quaternionic variable) with real coefficients satisfying $0\le a_0\le a_1\le \dots \le a_n,$ then all the zeros of p lie in $\left|q\right|\le 1.$

In the same paper, they generalized of Theorem D to the polynomials whose coefficients are monotonic but not necessarily non-negative by establishing the following result.

Theorem E.

If $p(q)=q^n{a}_n+q^{n-1}{a}_{n-1}+q^{n-2}{a}_{n-2}+\dots +q{a}_1+a_0$ is a polynomial of degree n (where q is a quaternionic variable) with real coefficients satisfying $a_0\le a_1\le \dots \le a_n,$ then all the zeros of p lie in $\left|q\right|\le \frac{a_n+\left|a_0\right|-a_0}{\left|a_n\right|}.$

Milovanović, Mir, and Ahmad (2022) generalized Theorem D and Theorem E by proving the following result.

Theorem F.

If $p(q)=q^n{a}_n+q^{n-1}{a}_{n-1}+q^{n-2}{a}_{n-2}+\dots +q{a}_1+a_0$ is a polynomial of degree n (where q is a quaternionic variable) with real coefficients satisfying $a_0\le a_1\le \dots \le a_{\lambda -1}\le a_{\lambda }\ge \dots \ge a_{n-1}\ge a_n,$ where $0\le \lambda \le n,$ then all the zeros of p lie in $\left|q\right|\le \frac{2a_{\lambda }-a_n+\left|a_0\right|-a_0}{\left|a_n\right|}.$

Because of the restriction on the coefficients that they should be real and monotonic, the results discussed above are applicable to a small class of polynomials, so its interesting to look for the results without any restriction on the coefficients and applicable to every quaternionic polynomial with quaternion/complex or real coefficients. In this direction, Ishfaq Dar (2024) proved various results concerning the location of the zeros of quaterionic polynomials with quaternionic coefficients without any restriction on the coefficients and besides this extended Cauchy’s theorem to quaternion settings by proving the following result.

Theorem G.

If $f(q)=q^n+q^{n-1}{a}_1+\cdots +q{a}_{n-1}+a_n$ is a quaternion polynomial with quaternion coefficients and q is quaternionic variable, then all the zeros of $f(q)$ lie inside the ball $\left|q\right|\le 1+{\mathrm{\text{max}}}_{1\le \nu \le n}\left|a_{\nu }\right|.$

Now in view of Theorem B and the fact that the zeros of a quaternion polynomial and the left eigenvalues of corresponding companion matrix are same, here we prove some results concerning the location of the zeros of quaternion polynomials, which extend various well known results to quaternion settings and show by examples that obtained results give better estimation of the zeros as compared to the results already available in the literature. We begin by proving the following refinement of Theorem G.

Theorem 2.

Let $f(q)=q^n+q^{n-1}{a}_1+q^{n-2}{a}_2+\cdots +q{a}_{n-1}+a_n$ be a monic quaternion polynomial of degree n with quaternion coefficients and q be a quaternion variable. If ${\alpha }_2$ $\ge$ ${\alpha }_3$ $\ge$ $\cdots$ $\ge$ ${\alpha }_n$ are ordered positive numbers, (1) $${\alpha }_{\nu }=\frac{\left|a_{\nu }\right|}{r^{\nu }},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\nu =2,3,\dots ,n$$(1) where r is a positive real number. Then all the zeros of f(q) lie in the union of balls $$\left\{q\in H:\left|q\right|\le r(1+{\alpha }_2)\right\}\mathit{\text{and}}\left\{q\in H:|q+a_1|\le r\right\}.$$

Next as an application of Theorem 2, we prove the following result concerning the location of the zeros of the quaternion polynomials with quaternion coefficients.

Theorem 3.

Let $f(q)=q^n+q^{n-1}{a}_1+q^{n-2}{a}_2+\cdots +q{a}_{n-1}+a_n$ be a monic quaternion polynomial with quaternion coefficients and q be a quaternion variables. If ${\alpha }_2$ $\ge$ ${\alpha }_3$ $\ge$ $\cdots$ $\ge$ ${\alpha }_n$ are ordered positive numbers with ${\alpha }_{\nu }$ = $\frac{\left|a_{\nu }\right|}{r^{\nu }},\nu =2,3,\dots ,n,$ where r is a positive real number, then all the zeros of $f(q)$ lie in the union of balls (2) $$\left\{q\in \mathbb{H}:\left|q\right|\le r(1+\beta )\right\}\hspace{0.5em}\mathit{\text{and}}\hspace{0.5em}\left\{q\in \mathbb{H}:|q+a_1|\le r\right\},$$(2) where $\beta ={\alpha }_2-\frac{{\delta }_2}{1+{\alpha }_2}-\frac{{\delta }_3}{{(1+{\alpha }_2)}^2}-\cdots -\frac{{\delta }_n}{{(1+{\alpha }_2)}^{n-1}},$ ${\delta }_{\nu }={\alpha }_{\nu }-{\alpha }_{\nu +1},\hspace{1em}\nu =2,3,\dots ,n$ and ${\alpha }_{n+1}=0.$

Remark 2.

If we assume that the variable q and coefficients $a_1,a_2,\dots ,a_n$ in Theorem 3 are complex, then Theorem 3 reduces to a result of Zalman Rubinstein [Theorem 1, 6].

Lastly, we prove the following result.

Theorem 4.

Let $f(q)=q^n+q^{n-1}{a}_{n-1}+q^{n-2}{a}_{n-2}+\cdots +q{a}_1+a_0$ be a monic quaternion polynomial with quaternion coefficients and q be a quaternion variable. If $\lambda ={\{\underset{0\le j\le n-1}{\mathrm{\text{max}}}|a_j|\}}^{\frac{1}{n}},$ then all the zeros of $f(q)$ lie in the ball (3) $$\{q\in \mathbb{H}:|q|\le \lambda +\mathrm{\text{max}}({\lambda }^2+{\lambda }^3+\cdots +{\lambda }^n)\}.$$(3)

2. Computations and analysis

In this section, we present some examples for which existing Eneström-Kakeya type results are not applicable but the obtained results are applicable and give better information about the location of the zeros than existing results present in the literature. It is worth mentioning that all existing Eneström-Kakeya type results are applicable to a small class of polynomials with real coefficients satisfying monotonicity condition, whereas the results proved in this paper are applicable to a larger class of polynomial with quaternion/complex or real coefficients.

Example 1:

Let $p(q)=q^3+3q^2+q-1.$ It is easy to see that Theorem D and Theorem E are not applicable and on using Theorem F (with $\lambda =2,n=3$), it follows that all the zeros of $p(q)$ lie in the ball $|q+2|\le 7.$ On using Theorem G, it follows that all the zeros of $p(q)$ lie in the ball $\left|q\right|\le 4.$ Whereas, if we use Theorem 2 (with $r=1$), it follows that all the zeros of $p(q)$ lie in the union of the balls $\left\{\right|q|\le 2\}$ and $\left\{\right|q+3|\le 1\}.$ Thus, Theorem 2 gives better bound with a significant improvement.

Example 2:

Let $p(q)=q^3+3q^2+2q-1.$ It is easy to see that Theorem D and Theorem E are not applicable and on using Theorem F (with $\lambda =2,n=3$), it follows that all the zeros of $p(q)$ lie in the ball $|q+2|\le 7.$ On using Theorem G, it follows that all the zeros of $p(q)$ lie in the ball $\left|q\right|\le 4.$ Whereas, if we use Theorem 3 (with $r=1$), it follows that all the zeros of $p(q)$ lie in the union of the balls $\left\{\right|q|\le 2.66\}$ and $\left\{\right|q+3|\le 1\}.$ Thus, Theorem 3 gives better bound with a significant improvement.

3. Proof of the main theorems

Proof of Theorem 1.

Let $C_f$ be the companion matrix of the polynomial $f(q)=q^n+q^{n-1}{a}_1+\cdots +q{a}_{n-1}+a_n$ and $T=\mathit{\text{diag}}(d_1,d_2,\dots ,d_{n-1},d_n)$ be diagonal matrix, where $d_1,d_2,\dots ,d_n$ are positive real numbers, then one can easily see that $$T^{-1}{C}_{f}T=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& \frac{-a_n{d}_n}{d_1}\\ \frac{d_1}{d_2}& 0& 0& \cdots & 0& \frac{-a_{n-1}{d}_n}{d_2}\\ 0& \frac{d_2}{d_3}& 0& \cdots & 0& \frac{-a_{n-2}{d}_n}{d_3}\\ ⋮& & & & & \\ 0& 0& 0& \cdots & \frac{d_{n-1}}{d_n}& -a_1\end{array}\right].$$

Now if $\lambda$ is the left eigenvalue of $T^{-1}{C}_{f}T$ corresponding to the non-zero eigenvector $v={[v_1,v_2,\dots ,v_n]}^T,$ then by definition of left eigenvalues $(T^{-1}{C}_{f}T)v=\lambda v,$ implies $$\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& \frac{-a_n{d}_n}{d_1}\\ \frac{d_1}{d_2}& 0& 0& \cdots & 0& \frac{-a_{n-1}{d}_n}{d_2}\\ 0& \frac{d_2}{d_3}& 0& \cdots & 0& \frac{-a_{n-2}{d}_n}{d_3}\\ ⋮& & & & & \\ 0& 0& 0& \cdots & \frac{d_{n-1}}{d_n}& -a_1\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\\ \cdots \\ \cdots \\ v_n\end{array}\right]=\lambda \left[\begin{array}{c}{v}_1\\ v_2\\ v_3\\ \cdots \\ \cdots \\ v_n\end{array}\right]$$

Which on simplification yields the following system of linear equations $$\begin{array}{l}-a_n\frac{d_n}{d_1}{v}_n=\lambda v_1\\ \frac{d_1}{d_2}{v}_1-a_{n-1}\frac{d_n}{d_2}{v}_n=\lambda v_2\\ ⋮\\ \frac{d_{n-2}}{d_{n-1}}{v}_{n-2}-a_2\frac{d_n}{d_{n-1}}{v}_n=\lambda v_{n-1}\\ \frac{d_{n-1}}{d_n}{v}_{n-1}-a_1{v}_n=\lambda v_n\end{array}$$

Since $v={[v_1,v_2,\dots ,v_n]}^T$ being a non-zero vector, implies that $v_n\ne 0,$ hence from above equations we get $$\begin{array}{l}{a}_n=-\lambda \frac{d_1}{d_n}{v}_1{v}_n^{-1}\\ a_{n-1}=-\lambda \frac{d_2}{d_n}{v}_2{v}_n^{-1}+\frac{d_1}{d_n}{v}_1{v}_n^{-1}\\ ⋮\\ a_2=-\lambda \frac{d_{n-1}}{d_n}{v}_{n-1}{v}_n^{-1}+\frac{d_{n-2}}{d_n}{v}_{n-2}{v}_n^{-1}\\ a_1=\frac{d_{n-1}}{d_n}{v}_{n-1}{v}_n^{-1}-\lambda \end{array}$$

Multiplying above equations respectively by $1,\lambda ,{\lambda }^2,\dots ,{\lambda }^{n-1}$ and then on adding we get $${\lambda }^{n-1}{a}_1+{\lambda }^{n-2}{a}_2+\cdots +\lambda a_{n-1}+a_n=-{\lambda }^n$$ that is, $${\lambda }^n+{\lambda }^{n-1}{a}_1+{\lambda }^{n-2}{a}_2+\cdots +\lambda a_{n-1}+a_n=0$$

This shows that $\lambda$ is a zero of $f(q).$ Since $\lambda$ was chosen arbitrarily, it follows that the left eigenvalues of $T^{-1}{C}_{f}T$ are the zeros of $f(q).$

Proof of Theorem 2.

By definition the companion matrix of the quaternion polynomial $f(q)=q^n+q^{n-1}{a}_1+q^{n-2}{a}_2+\cdots +q{a}_{n-1}+a_n$ is given by $$C_f=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& -a_n\\ 1& 0& 0& \cdots & 0& -a_{n-1}\\ 0& 1& 0& \cdots & 0& -a_{n-2}\\ ⋮& & & & & \\ 0& 0& 0& \cdots & 1& -a_1\end{array}\right].$$

We take a diagonal matrix T = $\mathit{\text{diag}}(r^{n-1},r^{n-2},\dots ,r,1),$ where r is a positive real number, then $$T^{-1}{C}_{f}T=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& \frac{-a_n}{r^{n-1}}\\ r& 0& 0& \cdots & 0& \frac{-a_{n-1}}{r^{n-2}}\\ 0& r& 0& \cdots & 0& \frac{-a_{n-2}}{r^{n-3}}\\ ⋮& & & & & \\ 0& 0& 0& \cdots & r& -a_1\end{array}\right].$$

Applying Theorem B to the matrix $T^{-1}{C}_{f}T,$ it follows that all the left eigenvalues of $T^{-1}{C}_{f}T$ lie in the union of balls $$\begin{array}{l}\left|q\right|\le \left|\frac{a_n}{r^{n-1}}\right|=r\frac{\left|a_n\right|}{r^n}<r+r\frac{\left|a_n\right|}{r^n},\\ \left|q\right|\le r+\left|\frac{a_{n-1}}{r^{n-2}}\right|<r+r\frac{|a_{n-1|}}{r^{n-1}},\\ ⋮\\ \left|q\right|\le r+\left|\frac{a_2}{r}\right|<r+\frac{r\left|a_2\right|}{r^2}\end{array}$$ and $$|q+a_1|\le r.$$

Since by $(1),$ ${\alpha }_{\nu }=\left|\frac{a_{\nu }}{r^{\nu }}\right|,\nu =2,3,4,\dots ..,n.$ Therefore all the left eigenvalues of $T^{-1}{C}_{f}T$ lie in the union of balls $$\begin{array}{l}\left|q\right|<r(1+{\alpha }_n),\\ \left|q\right|\le r(1+{\alpha }_{n-1}),\\ ⋮\\ \left|q\right|\le r(1+{\alpha }_2),\end{array}$$ and $$|q+a_1|\le r.$$

Using the fact that ${\alpha }_2\ge {\alpha }_3\ge {\alpha }_4\ge \cdots \ge {\alpha }_n,$ it follows that all left eigenvalues of $T^{-1}{C}_{f}T$ lie in the union of balls (4) $$\left\{q\in \mathbb{H}:\left|q\right|\le r(1+{\alpha }_2)\right\}\mathrm{\hspace{0.5em}}\text{and}\mathrm{\hspace{0.5em}}\left\{q\in \mathbb{H}:|q+a_1|\le r\right\}.$$(4)

Since T is a diagonal matrix with real positive entries, by Theorem 1, it follows that all the zeros of $f(q)$ lie in the union of the balls given by (Ishfaq Dar, 2024).

This completes the proof of Theorem 2.

Proof of Theorem 3.

Let w be a zero of quaternion polynomial $f(q)=q^n+q^{n-1}{a}_1+q^{n-2}{a}_2+\cdots +q{a}_{n-1}+a_n.$ Since by Theorem 2, all the zeros of quaternion polynomial f(q) lie in the union of balls (5) $$\left\{q\in H:\left|q\right|\le r(1+{\alpha }_2)\right\}\text{\hspace{0.5em}and\hspace{0.5em}}\left\{q\in H:|q+a_1|\le r\right\},$$(5) it follows that $\left|w\right|\le r(1+{\alpha }_2)$ and $|w+a_1|\le r.$ If w lies in the ball $\left\{q\in H:|q+a_1|\le r\right\},$ then clearly w lies in the union of balls defined by (Milovanović et al., 2022). Hence we assume that the zero w does not lie in the ball $\left\{|q+a_1|\le r\right\},$ then clearly from (Marden, 1949), it follows that $\left|w\right|\le r(1+{\alpha }_2).$ Since $(1+{\alpha }_2)\le {(1+{\alpha }_2)}^{\nu },\hspace{1em}\nu =1,2,3,\dots ,n-1,$therefore $$\begin{array}{cc}\frac{{\delta }_2}{(1+{\alpha }_2)}+\frac{{\delta }_3}{{(1+{\alpha }_2)}^2}+\cdots +\frac{{\delta }_n}{{(1+{\alpha }_2)}^{n-1}}\hfill & \le \frac{{\delta }_2+{\delta }_3+\cdots +{\delta }_n}{(1+{\alpha }_2)}\hfill \\ =\frac{({\alpha }_2-{\alpha }_3)+({\alpha }_3-{\alpha }_4)+\cdots +({\alpha }_n-{\alpha }_{n+1)}}{(1+{\alpha }_2)}\hfill & \hfill \\ =\frac{{\alpha }_2}{(1+{\alpha }_2)} \text{where} {\alpha }_{n+1}=0.\hfill & \hfill \end{array}$$

Thus $$\begin{array}{cc}\beta ={\alpha }_2-\left\{\frac{{\delta }_2}{(1+{\alpha }_2)}+\frac{{\delta }_3}{{(1+{\alpha }_2)}^2}+\cdots +\frac{{\delta }_n}{{(1+{\alpha }_2)}^{n-1}}\right\}\hfill & \hfill \\ \ge {\alpha }_2-\frac{{\alpha }_2}{(1+{\alpha }_2)}\hfill & \hfill \\ =\frac{{\alpha }_2^2}{(1+{\alpha }_2)}>0.\hfill & \hfill \end{array}$$

In view of this, if $\left|w\right|\le r,$ then $\left|w\right|\le r(1+\beta ),$ therefore we assume that $\left|w\right|>r,$ that is, $r<\left|w\right|\le r(1+{\alpha }_2).$ So that (6) $$\left|w\right|=\mathit{\text{rt}}\hspace{1em}\text{where}\hspace{1em}1<t\le (1+{\alpha }_2).$$(6)

Now $q(w)=0$ implies that $$\begin{array}{cc}|w^n+w^{n-1}{a}_1|=|-(w^{n-2}{a}_2+\cdots +w{a}_{n-1}+a_n)|\hfill & \hfill \\ \le \left|w{|}^{n-2}\right|a_2|+\cdots +|w\left|\right|a_{n-1}|+|a_n|,\hfill & \hfill \end{array}$$ that is, $$|w{|}^{n-1}\left|w+a_1\right|\le |w{|}^{n-2}\left|a_2\right|+\cdots +\left|w\right|\left|a_{n-1}\right|+\left|a_n\right|.$$

Hence by using (Rubinstein, 1963) and the hypothesis, we get $$\begin{array}{cc}{(\mathit{\text{rt}})}^{n-1}|w+a_1|\le {(\mathit{\text{rt}})}^{n-2}\left|a_2\right|+\cdots +\left|\mathit{\text{rt}}\right|\left|a_{n-1}\right|+\left|a_n\right|\hfill & \hfill \\ =r^n\left\{t^{n-2}{\alpha }_2+\cdots +t{\alpha }_{n-1}+{\alpha }_n\right\}.\hfill & \hfill \end{array}$$

This leads to $$t^{n-1}|w+a_1|\le r\left\{t^{n-2}{\alpha }_2+\cdots +t{\alpha }_{n-1}+{\alpha }_n\right\}.$$

Multiplying both sides by $\frac{(t-1)}{t^{n-1}}$ we get $$\begin{array}{cc}(t-1)|w+a_1|\le \frac{r(t-1){\alpha }_2}{t}+\frac{r(t-1){\alpha }_3}{t^2}+\cdots +\frac{r(t-1){\alpha }_{n-1}}{t^{n-2}}+\frac{r(t-1){\alpha }_{n-1}}{t^{n-1}}\hfill & \hfill \\ =r{\alpha }_2-\frac{r}{t}({\alpha }_2-{\alpha }_3)-\frac{r}{t^2}({\alpha }_3-{\alpha }_4)-\cdots -\frac{r}{t^{n-2}}({\alpha }_{n-1}-{\alpha }_n)-\frac{r}{t^{n-1}}{\alpha }_n\hfill & \hfill \\ =r{\alpha }_2-\frac{r}{t}{\delta }_2-\frac{r}{t^2}{\delta }_3-\cdots -\frac{r}{t^{n-2}}{\delta }_{n-1}-\frac{r}{t^{n-1}}{\delta }_n.\hfill & \hfill \end{array}$$

Since $t\le (1+{\alpha }_2),$ $$(t-1)|w+a_1|<r{\alpha }_2-\frac{r{\delta }_2}{(1+{\alpha }_2)}-\frac{r{\delta }_3}{{(1+{\alpha }_2)}^2}-\cdots -\frac{r{\delta }_{n-1}}{{(1+{\alpha }_2)}^{n-2}}-\frac{r{\delta }_n}{{(1+{\alpha }_2)}^{n-1}}$$ which implies (7) $$(t-1)\left|w+a_1\right|\le r\left\{{\alpha }_2-\frac{{\delta }_2}{(1+{\alpha }_2)}-\frac{{\delta }_3}{{(1+{\alpha }_2)}^2}-\cdots -\frac{{\delta }_n}{{(1+{\alpha }_2)}^{n-1}}\right\}.$$(7)

Using the fact that $|w+a_1|>r,$ and $t>1,$ implies from inequality (Ell, Bihan, & Sangwine, 2014), we obtain $$(t-1)<{\alpha }_2-\frac{{\delta }_2}{(1+{\alpha }_2)}-\frac{{\delta }_3}{{(1+\alpha )}^2}-\cdots -\frac{{\delta }_n}{{(1+{\alpha }_2)}^{n-1}}$$ equivalently $$\begin{array}{c}t<1+{\alpha }_2-\frac{{\delta }_2}{(1+{\alpha }_2)}-\frac{{\delta }_3}{{(1+\alpha )}^2}-\cdots -\frac{{\delta }_n}{{(1+{\alpha }_2)}^{n-1}}\\ =1+\beta .\end{array}$$

Using this in (Rubinstein, 1963), we get $$\left|w\right|=\mathit{\text{rt}}<r(1+\beta )\hspace{1em}\text{where}\hspace{1em}\beta ={\alpha }_2-\frac{{\delta }_2}{(1+{\alpha }_2)}-\frac{{\delta }_3}{{(1+{\alpha }_2)}^2}-\cdots -\frac{{\delta }_n}{{(1+{\alpha }_2)}^{n-1}}$$

Hence, it follows that all the zeros of $f(q)$ which are outside the ball $\{q\in \mathbb{H}:|q+a_1|\le r\}$ lie in the ball $\{q\in \mathbb{H}:|q|<r(1+\beta )\}.$ Therefore, it follows that all the zeros of $f(q)$ lie in the union of balls $\left\{q\in \mathbb{H}:\left|q\right|\le r(1+\beta )\right\}$ and $\left\{q\in \mathbb{H}:|q+a_1|\le r\right\}.$

This completes the proof of Theorem 3.

Proof of Theorem 4.

By definition the companion matrix of the polynomial $f(q)=q^n+q^{n-1}{a}_{n-1}+\cdots +q{a}_1+a_0$ is given by $$C_f=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& -a_0\\ 1& 0& 0& \cdots & 0& -a_1\\ 0& 1& 0& \cdots & 0& -a_2\\ ⋮& & & & & \\ 0& 0& 0& \cdots & 1& -a_{n-1}\end{array}\right].$$

Now for any positive real number $\lambda$ define a diagonal matrix $T=\mathit{\text{diag}}({\lambda }^{n-1},{\lambda }^{n-2},\dots ,\lambda ,1),$ then $$T^{-1}{C}_{f}T=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& \frac{-a_0}{{\lambda }^{n-1}}\\ \lambda & 0& 0& \cdots & 0& \frac{-a_1}{{\lambda }^{n-2}}\\ 0& \lambda & 0& \cdots & 0& \frac{-a_2}{{\lambda }^{n-3}}\\ ⋮& & & & & \\ 0& 0& 0& \cdots & \lambda & -a_{n-1}\end{array}\right].$$

Applying Theorem B to matrix $T^{-1}{C}_{f}T,$ it follows that all the left eigenvalues of $T^{-1}{C}_{f}T$ lie in the union of the balls, $$\begin{array}{l}\left|q\right|\le \frac{\left|a_0\right|}{{\lambda }^{n-1}},\\ \left|q\right|\le \lambda +\frac{\left|a_1\right|}{{\lambda }^{n-2}},\\ ⋮\\ \left|q\right|\le \lambda +\left|a_{n-1}\right|.\end{array}$$ that is, all the left eigenvalues of $T^{-1}{C}_{f}T$ lie in the ball $$\left|q\right|\le \underset{1\le \nu \le n-1}{\mathrm{\text{max}}}\left\{\frac{\left|a_0\right|}{{\lambda }^{n-1}},\lambda +\frac{\left|a_{\nu }\right|}{{\lambda }^{n-\nu -1}}\right\}.$$

Since by hypothesis $${\lambda }^n=\underset{0\le j\le n-1}{\mathrm{\text{max}}}\left|a_j\right|,\lambda \ne 0$$ we have $$\left|a_j\right|\le {\lambda }^n,j=0,1,2,\dots ,n-1,$$

This implies $$\begin{array}{cc}\left|q\right|\le \underset{1\le \nu \le n-1}{\mathrm{\text{max}}}\left\{\lambda ,\lambda +{\lambda }^{\nu +1}\right\}\hfill & \hfill \\ =\mathrm{\text{max}}\left\{\lambda ,\lambda +{\lambda }^2,\lambda +{\lambda }^3,\dots ,\lambda +{\lambda }^n\right\}.\hfill & \hfill \end{array}$$ that is, all the left eigenvalues of the matrix $T^{-1}{C}_{f}T$ lie in the ball (8) $$\begin{array}{c}\left|q\right|\le \lambda +\mathrm{\text{max}}\{{\lambda }^2,{\lambda }^3,\dots ,{\lambda }^n\}.\hfill \end{array}$$(8)

Since T is a diagonal matrix with real positive entries, by Theorem 1, it follows that all the zeros of $f(q)$ lie in the ball given by (Turner, 2006).

This completes the proof of Theorem 4.▪

Author’s contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Acknowledgement

The author’s are highly grateful to the referees for their valuable suggestions and comments which has improved the quality of this work.

Disclosure statement

The authors declare that they have no competing interests.