A new class of spectrally arbitrary complex sign pattern

Abstract

Assume that S is an nth-order complex sign pattern. If for every nth degree complex coefficient polynomial f(λ) with a leading coefficient of 1, there exists a complex matrix C\in Q\left(S\right) such that the characteristic polynomial of C is f(λ), then S is called a spectrally arbitrary complex sign pattern. That is, if the spectrum of nth-order complex sign pattern S is a set comprised of all spectra of nth-order complex matrices, then S is called a spectrally arbitrary complex sign pattern. This paper presents a class of spectrally arbitrary complex sign pattern with only 3n nonzero elements by adopting the method of Schur complement and row reduction.

Keywords

Complex sign pattern / potentially nilpotent / spectrally arbitrary

1 Introduction

Assume that A = (a_kl) and B = (b_kl) are nth-order sign patterns. Define S=A+\mathrm{i}B as an nth-order complex sign pattern, where i² = −1 and (k, l)-element of S is given by s_kl = a_kl + ib_kl, k, l=1,2,\dots , n. Then sign patterns A and B are referred to as the real part matrix and the imaginary part matrix of the complex pattern S, respectively. Define the total number of nonzero elements in A and B as the number of nonzero elements in complex pattern S.

If there exists a complex matrix C\in Q(S) with a characteristic polynomial f_C\left(\lambda \right)={\lambda }^n, then S is considered potentially nilpotent and C a nilpotent complex matrix. If for every nth-degree complex coefficient polynomial f(x) with a leading coefficient of 1, there exists a complex matrix M\in Q( S), such that f_M\left(\lambda \right)=f\left(x\right), then S is called a spectrally arbitrary complex sign pattern. If S is a spectrally arbitrary complex sign pattern and there is no spectrally arbitrary proper subpattern of S, then S is called a minimally spectrally arbitrary complex sign pattern. Please see references [2-7, 9, 10] for other terms and symbols in this paper.

In reference [4], Gao et al. extended the Nilpotent-Jacobian method [1] that could determine a sign pattern as spectrally arbitrary to a complex sign pattern and used it to demonstrate a class of nth-order irreducible spectrally arbitrary complex sign patterns. They also proposed a conjecture that an nth-order irreducible spectrally arbitrary complex sign pattern contains at least 3n nonzero elements. The Nilpotent-Jacobian method, which obtains the spectral arbitrariness of the studied pattern mainly by judging the non-singularity of corresponding Jacobian matrix, has been widely applied in many mathematical papers, such as [3‒5, 8‒10].

Using the Nilpotent-Jacobian method of complex sign pattern, this paper presents a new class of spectrally arbitrary complex sign patterns containing only 3n nonzero elements.

Assume that

\begin{array}{rl}{S}_n& =A+\mathrm{i}B=\left(\begin{array}{cccccc}-1& 0& -1& & & \\ 0& 1+\mathrm{i}& 1& & & \\ 1& & 0& 1& & \\ ⋮& & & \ddots & \ddots & \\ 1& & & & 0& 1\\ 1-\mathrm{i}& -\mathrm{i}& -\mathrm{i}& \dots & -\mathrm{i}& -\mathrm{i}\end{array}\right)\\ & =\left(\begin{array}{cccccc}-1& 0& -1& & & \\ 0& 1& 1& & & \\ 1& & 0& 1& & \\ ⋮& & & \ddots & \ddots & \\ 1& & & & 0& 1\\ 1& & & & & 0\end{array}\right)+\mathrm{i}\left(\begin{array}{cccccc}0& & & & & \\ & 1& & & & \\ & & 0& & & \\ & & & \ddots & & \\ & & & & 0& \\ -1& -1& -1& \dots & -1& -1\end{array}\right),\end{array}

where n\ge 4.

2 Some Lemmas

For the sake of convenience, and without loss of generality, let

(2.1)C_n= \left(\begin{array}{cccccc}-a_1& 0& -1& & & \\ 0& a_2+\mathrm{i}{b}_0& 1& & & \\ a_3& & 0& 1& & \\ ⋮& & & \ddots & \ddots & \\ a_{n-1}& & & & 0& 1\\ a_n-\mathrm{i}{b}_n& -\mathrm{i}{b}_{n-1}& -\mathrm{i}{b}_{n-2}& \dots & -\mathrm{i}{b}_2& -\mathrm{i}{b}_1\end{array}\right),n\ge 4 ,

where a_k>0, b_l>0, k=1,2,\dots ,n, l=0,1,2,\dots ,n. Then C_n\in Q\left(S_n\right).

Let

f\left(\lambda \right)=|\lambda I -C_n|=\left|\begin{array}{cccccc}\lambda +a_1& 0& 1& & & \\ 0& \lambda -a_2-\mathrm{i}{b}_0& -1& & & \\ -a_3& & \lambda & -1& & \\ ⋮& & & \ddots & \ddots & \\ -a_{n-1}& & & & \lambda & -1\\ -a_n+\mathrm{i}{b}_n& \mathrm{i}{b}_{n-1}& \mathrm{i}{b}_{n-2}& \dots & \mathrm{i}{b}_2& \lambda +\mathrm{i}{b}_1\end{array}\right|.

Expanding the determinant along the first two rows, we obtain

\begin{array}{rl}f\left(\lambda \right)=& \left|\begin{array}{cc}\lambda +a_1& 0\\ 0& \lambda -a_2-\mathrm{i}{b}_0\end{array}\right|\left|\begin{array}{ccccc}\lambda & -1& & & \\ & \lambda & -1& & \\ & & \ddots & \ddots & \\ & & & \lambda & -1\\ \mathrm{i}{b}_{n-2}& \mathrm{i}{b}_{n-3}& \dots & \mathrm{i}{b}_2& \lambda +\mathrm{i}{b}_1\end{array}\right|\\ & -\left|\begin{array}{cc}\lambda +a_1& 1\\ 0& -1\end{array}\right|\left|\begin{array}{ccccc}0& -1& & & \\ & \lambda & -1& & \\ & & \ddots & \ddots & \\ & & & \lambda & -1\\ \mathrm{i}{b}_{n-1}& \mathrm{i}{b}_{n-3}& \dots & \mathrm{i}{b}_2& \lambda +\mathrm{i}{b}_1\end{array}\right|\\ & +\left|\begin{array}{cc}0& 1\\ \lambda -a_2-\mathrm{i}{b}_0& -1\end{array}\right|\left|\begin{array}{ccccc}-a_3& -1& & & \\ -a_4& \lambda & -1& & \\ ⋮& & \ddots & \ddots & \\ -a_{n-1}& & & \lambda & -1\\ -a_n+\mathrm{i}{b}_n& \mathrm{i}{b}_{n-3}& \cdots & \mathrm{i}{b}_2& \lambda +\mathrm{i}{b}_1\end{array}\right|\\ =& (\lambda +a_1)(\lambda -a_2-\mathrm{i}{b}_0){\mathrm{\Delta }}_{n-2}+\mathrm{i}{b}_{n-1}(\lambda +a_1)+(\lambda -a_2-\mathrm{i}{b}_0)[a_3{\mathrm{\Delta }}_{n-3}\\ & +a_4{\mathrm{\Delta }}_{n-4}+\cdots +a_{n-3}{\mathrm{\Delta }}_3+a_{n-2}({\lambda }^2+\mathrm{i}{b}_1\lambda +\mathrm{i}{b}_2)+a_{n-1}(\lambda +\mathrm{i}{b}_1)+a_n-\mathrm{i}{b}_n]\\ =& (\lambda -a_2-\mathrm{i}{b}_0)\left[\right(\lambda +a_1){\mathrm{\Delta }}_{n-2}+a_3{\mathrm{\Delta }}_{n-3}+a_4{\mathrm{\Delta }}_{n-4}+\cdots +a_{n-3}{\mathrm{\Delta }}_3\\ & +a_{n-2}({\lambda }^2+\mathrm{i}{b}_1\lambda +\mathrm{i}{b}_2)+a_{n-1}(\lambda +\mathrm{i}{b}_1)+a_n-\mathrm{i}{b}_n]+\mathrm{i}{b}_{n-1}(\lambda +a_1),\end{array}

where

\begin{array}{rl}{\mathrm{\Delta }}_k& =\left|\begin{array}{ccccc}\lambda & -1& & & \\ & \lambda & -1& & \\ & & \ddots & \ddots & \\ & & & \lambda & -1\\ \mathrm{i}{b}_k& \mathrm{i}{b}_{k-1}& \cdots & \mathrm{i}{b}_2& \lambda +\mathrm{i}{b}_1\end{array}\right|\\ & ={\lambda }^k+\mathrm{i}{b}_1{\lambda }^{k-1}+\mathrm{i}{b}_2{\lambda }^{k-2}+\cdots +\mathrm{i}{b}_{k-1}\lambda +\mathrm{i}{b}_k,k=3,4,\dots ,n-2.\end{array}

Assume that

(2.2)\begin{array}{rl}f\left(\lambda \right)=& {\lambda }^n+(f_1+\mathrm{i}{g}_1){\lambda }^{n-1}+(f_2+\mathrm{i}{g}_2){\lambda }^{n-2}+\cdots +(f_k+\mathrm{i}{g}_k){\lambda }^{n-k}\\ & +\cdots +(f_{n-1}+\mathrm{i}{g}_{n-1})\lambda +(f_n+\mathrm{i}{g}_n),\end{array}

then

(2.3)\{\begin{array}{l}{f}_1=a_1-a_2,\\ g_1=b_1-b_0,\\ f_2=a_3-a_1{a}_2+b_0{b}_1,\\ g_2=b_2+(a_1-a_2)b_1-a_1{b}_0,\\ f_3=a_4-a_2{a}_3+a_1{b}_1{b}_0+b_2{b}_0,\\ g_3=b_3+(a_1-a_2)b_2+(a_3-a_2{a}_1)b_1-a_3{b}_0,\\ \cdots \\ f_k=a_{k+1}-a_2{a}_k+a_{k-1}{b}_1{b}_0+a_{k-2}{b}_2{b}_0+\cdots +a_3{b}_{k-3}{b}_0+a_1{b}_{k-2}{b}_0+b_{k-1}{b}_0,\\ g_k=b_k+(a_1-a_2)b_{k-1}+(a_3-a_2{a}_1)b_{k-2}+(a_4-a_2{a}_3)b_{k-3}+\cdots \\ +(a_{k-1}-a_2{a}_{k-2})b_2+(a_k-a_2{a}_{k-1})b_1-a_k{b}_0,\\ \cdots \\ f_{n-1}=a_n-a_2{a}_{n-1}+a_{n-2}{b}_1{b}_0+a_{n-3}{b}_2{b}_0+\cdots +a_3{b}_{n-4}{b}_0+a_1{b}_{n-3}{b}_0+b_{n-2}{b}_0,\\ g_{n-1}=b_{n-1}+(a_1-a_2)b_{n-2}+(a_3-a_2{a}_1)b_{n-3}+(a_4-a_2{a}_3)b_{n-4}+\cdots \\ +(a_{n-1}-a_2{a}_{n-2})b_1-a_{n-1}{b}_0-b_n,\\ f_n=-a_2{a}_n+a_{n-1}{b}_1{b}_0+a_{n-2}{b}_2{b}_0+\cdots +a_3{b}_{n-3}{b}_0+a_1{b}_{n-2}{b}_0-b_n{b}_0,\\ g_n=a_2{b}_n+a_1{b}_{n-1}-a_2{a}_1{b}_{n-2}-a_2{a}_3{b}_{n-3}-a_2{a}_4{b}_{n-4}-\cdots -a_2{a}_{n-3}{b}_3-a_2{a}_{n-2}{b}_2\\ -a_2{a}_{n-1}{b}_1-a_n{b}_0.\end{array}

Lemma 2.1　 When m\ge 4, complex mode S_nis potentially nilpotent.

Proof　Assume that C_n is matrix (2.1), then (2.2) is the characteristic polymonial of C_n.

Let f_k= 0, g_k = 0, k=1,2,\dots ,n, then it follows from (2.3) that

(2.4)\{\begin{array}{l}{a}_1=a_2,\\ b_1=b_0,\\ a_3=a_1{a}_2-b_0{b}_1=a_2^2-b_1^2,\\ b_2=a_2{b}_1,\\ a_4=a_2{a}_3-a_1{b}_1{b}_0-b_2{b}_0=a_2^3-3a_2{b}_1^2,\\ b_3=(a_2{a}_1-a_3)b_1+a_3{b}_0=a_2{b}_1=a_2{b}_1\\ b_4=(a_2{a}_1-a_3)b_2+(a_2{a}_3-a_4)b_1+a_4{b}_0=a_2{b}_3,\\ \cdots \\ a_{k+1}=(a_2{a}_1-a_2)b_1{b}_0-a_{k-2}{b}_2{b}_0-\cdots +a_3{b}_k-3b_0-a_1{b}_{k-2}{b}_0-b_{k-1}{b}_0,\\ b_k=(a_2{a}_1-a_3)b_{k-2}+(a_2{a}_3-a_4)b_{k-3}+\cdots +(a_2{a}_{k-1}-a_k)b_1+a_k{b}_0,\\ \cdots \\ a_n=a_2{a}_{n-1}-a_{n-2}{b}_0-a_{n-3}{b}_0-\cdots -a_{n-3}{b}_0-a_{n-3}{b}_0-a_{n-3}{b}_0-b_{n-2}{b}_0,\\ b_{n-1}=(a_2{a}_1-a_3)b_{n-3}+(a_2{a}_3-a_4)b_0-\cdots +(a_3{a}_{n-2}-a_{n-1})b_1+a_{n-1}{b}_0+b_n,\\ b_n{b}_0=a_{n-1}{b}_1{b}_0+a_{n-2}{b}_2{b}_0+\cdots +a_3{b}_{n-3}{b}_0+a_1{b}_{n-2}{b}_0-a_2{a}_n,\\ a_1{b}_{n-1}=a_2{a}_1{b}_{n-2}+a_2{a}_3{b}_{n-3}+a_2{a}_4{b}_{n-4}+\cdots +a_2{a}_{n-1}{b}_1+a_n{b}_0-a_2{b}_n.\end{array}

It can be deduced by mathematical induction that when k=2,\dots , n-2, b_k=a_1{b}_{k-1}. When k=4,5,\dots ,n, a_k is the (k-1)th-degree monic polymonial of a_2 and a_k=2a_2{a}_{k-1}-( a_2^2+ b_1^2)a_{k-2}. Therefore, when b_1>0, a_2>0 and a_2 is large enough, a_i>0, b_j>0, a_2^{k-1}>a_k, where i=1,2,\dots ,n, j=1,2,\dots ,n-2, k=3,4,\dots , n.

The remaining three formulas are then discussed as follows:

\begin{array}{}\text{(2.5a)}& b_{n-1}=(a_2{a}_1-a_3)b_{n-3}+(a_2{a}_3-a_4)b_{n-4}+\cdots +(a_2{a}_{n-2}-a_{n-1})b_1+a_{n-1}{b}_0+b_n,\end{array}

\begin{array}{}\text{(2.5b)}& b_n{b}_0=a_{n-1}{b}_1{b}_0+a_{n-2}{b}_2{b}_0+\cdots +a_3{b}_{n-3}{b}_0+a_1{b}_{n-2}{b}_0-a_2{a}_n,\end{array}

\begin{array}{}\text{(2.5c)}& a_1{b}_{n-1}=a_2{a}_1{b}_{n-2}+a_2{a}_3{b}_{n-3}+a_2{a}_4{b}_{n-4}+\cdots +a_2{a}_{n-1}{b}_1+a_n{b}_0-a_2{b}_n.\end{array}

Firstly, it follows from (2.5a) that

\begin{array}{rl}{b}_{n-1}& =(a_2{a}_1-a_3)b_{n-3}+(a_2{a}_3-a_4)b_{n-4}+\cdots +(a_2{a}_{n-2}-a_{n-1})b_1+a_{n-1}{b}_0+b_n\\ & =(a_2{a}_1-a_3)b_{n-3}+a_3{a}_2{b}_{n-4}-a_4(b_{n-4}-a_2{b}_{n-5})-\cdots -a_{n-2}(b_2-a_2{b}_1)\\ & +a_{n-1}(b_0-b_1)+b_n\\ & =(a_2{a}_1-a_3)b_{n-3}+a_3{b}_{n-3}+b_n\\ & =a_2{a}_1{b}_{n-3}+b_n=a_1{b}_{n-2}+b_n.\end{array}

Next, the following equation can be obtained through (2.5b) and (2.5c):

(2.6)(a_2^2+b_0^2)a_n= a_1{b}_0{b}_{n-1}.

Then, it can be obtained through substituting (2.6) into (2.5b) that

\begin{array}{rl}(a_2^2+b_0^2)b_n& =(a_2^2+b_0^2)(a_{n-1}{b}_1+a_{n-2}{b}_2+\cdots +a_3{b}_{n-3}+a_1{b}_{n-2})-a_2^2{b}_{n-1}\\ & =(a_2^2+b_0^2)(a_{n-1}{b}_1+a_{n-2}{b}_2+\cdots +a_3{b}_{n-3}+a_1{b}_{n-2})-a_2^2(a_1{b}_{n-2}+b_n).\end{array}

After rearranging, we obtain

(2a_2^2+ b_0^2)b_n=( a_2^2+ b_0^2)(a_{n-1}{b}_1+a_{n-2}{b}_2+ \cdots +a_3{b}_{n-3})+b_0^2{a}_1{b}_{n-2}>0.

Hence, b_n>0, b_{n-1}>0.

Finally, to prove that equation (2.5c) holds, let’s assume that

\begin{array}{rl}h(a_1,b_1)& =a_2{a}_1{b}_{n-2}+a_2{a}_3{b}_{n-3}+a_2{a}_4{b}_{n-4}+\cdots +a_2{a}_{n-1}{b}_1+a_n{b}_0-a_2{b}_n-a_1{b}_{n-1}\\ & =a_2{a}_1{b}_{n-2}+a_2{a}_3{b}_{n-3}+\cdots +a_2{a}_{n-1}{b}_1+a_n{b}_0+a_1(-b_n-(a_1{b}_{n-2}+b_n\left)\right)\\ & =a_2{a}_3{b}_{n-3}+\cdots +a_2{a}_{n-1}{b}_1+a_n{b}_0-2a_1{b}_n.\end{array}

Then

\begin{array}{rl}(a_2^2& +b_0^2)h(a_1,b_1)\\ & =(a_2^2+b_0^2)(a_2{a}_3{b}_{n-3}+\cdots +a_2{a}_{n-2}{b}_2+a_2{a}_{n-1}{b}_1+a_n{b}_0-2a_1{b}_n)\\ & =(a_2^2+b_0^2)(a_2{a}_3{b}_{n-3}+\cdots +a_2{a}_{n-2}{b}_2+a_2{a}_{n-1}{b}_1)+a_1{b}_0^2{b}_{n-1}-2a_1(a_2^2+b_0^2)b_n\\ & =\left(a_2^2+b_0^2\right)\left(a_2{a}_3{b}_{n-3}+\cdots +a_2{a}_{n-2}{b}_2+a_2{a}_{n-1}{b}_1\right)+a_1^2{b}_0^2{b}_{n-2}-\left(2a_2^2+b_0^2\right)a_1{b}_n\\ & =(a_2^2+b_0^2)(a_2{a}_3{b}_{n-3}+\cdots +a_2{a}_{n-2}{b}_2+a_2{a}_{n-1}{b}_1)+a_1^2{b}_0^2{b}_{n-2}\\ & -a_1(a_2^2+b_0^2)(a_{n-1}{b}_1+a_{n-2}{b}_2+\cdots +a_3{b}_{n-3})-b_0^2{a}_1^2{b}_{n-2}\\ & \equiv 0.\end{array}

Based on the above, there exists a_k>0, b_l>0, k=1,2,\dots ,n, l=0,1,2,\dots ,n, such that complex pattern S_n is potentially nilpotent.

Let

\begin{array}{rl}P& =(r_1,r_2,\dots ,r_{2n})\\ & =\left(a_1,a_3,\dots ,a_{n-3},a_{n-2},a_{n-1},a_n,b_n,b_{n-1},a_2,b_1,b_2,\dots ,b_{n-3},b_{n-2}\right).\end{array}

Then P is the nilpotent point of C_n, which can prove the following conclusion.

Lemma 2.2 \begin{array}{c}|J{|}_P=det\frac{\mathrm{\partial }(f_1,f_2,f_3,\dots ,f_{n-2},f_{n-1},f_n,g_n,g_{n-1},g_{n-2},\dots ,g_3,g_2,g_1)}{\mathrm{\partial }(a_1,a_3,\dots ,a_{n-3},a_{n-2},a_{n-1},a_n,b_n,b_{n-1},a_2,b_1,b_2,\dots ,b_{n-3},b_{n-2})}{|}_P\ne 0.\end{array}

Proof　It can be known from the conditions that the Jacobian matrix of C_n at P is

J{|}_P=\left(\begin{array}{cc}{A}_1^{\mathrm{T}}& A_2^{\mathrm{T}}\\ A_3^{\mathrm{T}}& A_4^{\mathrm{T}}\end{array}\right),

where

A_1^{\mathrm{T}}= \left(\begin{array}{ccccccccc}1& & & & & & & & \\ -a_2& 1& & & & & & & \\ b_0{b}_1& -a_2& 1& & & & & & \\ b_0{b}_2& b_0{b}_1& -a_2& 1& & & & & \\ ⋮& ⋮& \ddots & \ddots & \ddots & & & & \\ b_0{b}_{n-4}& b_0{b}_{n-5}& \cdots & \ddots & -a_2& 1& & & \\ b_0{b}_{n-3}& b_0{b}_{n-4}& \cdots & \cdots & b_0{b}_1& -a_2& 1& & \\ b_0{b}_{n-2}& b_0{b}_{n-3}& \cdots & \cdots & b_0{b}_2& b_0{b}_1& -a_2& -b_0& \\ b_n& -b_{n-2}& \cdots & \cdots & -b_3& -b_2& -b_1& a_2& a_1\end{array}\right),

A_2^{\mathrm{T}}= \left(\begin{array}{cccccccc}-1& 0& & & & & & \\ -a_1& b_0& 0& & & & & \\ -a_3& b_0{a}_1& b_0& 0& & & & \\ -a_4& b_0{a}_3& b_0{a}_1& b_0& 0& & & \\ ⋮& ⋮& \ddots & \ddots & \ddots & \ddots & & \\ ⋮& ⋮& ⋮& \ddots & \ddots & \ddots & \ddots & \\ -a_{n-2}& b_0{a}_{n-3}& b_0{a}_{n-4}& \cdots & \ddots & b_0{a}_1& b_0& 0\\ -a_{n-1}& b_0{a}_{n-2}& b_0{a}_{n-3}& \cdots & \cdots & b_0{a}_3& b_0{a}_1& b_0\\ -a_n& b_0{a}_{n-1}& b_0{a}_{n-2}& \cdots & \cdots & b_0{a}_4& b_0{a}_3& b_0{a}_1\\ b_n-{\displaystyle \sum _{k=2}^{n-1}{a}_k{b}_{n-k}}& -a_2{a}_{n-1}& -a_2{a}_{n-2}& \cdots & \cdots & -a_2{a}_4& -a_2{a}_3& -a_2{a}_1\end{array}\right),

A_3^{\mathrm{T}}= {\left(-11\right)}_{\left(n-1\right)\times \left(n+1\right)},

\begin{array}{rl}& A_4^{\mathrm{T}}=\\ & \left(\begin{array}{cccccccc}-b_{n-2}-{\displaystyle \sum _{k=1}^{n-2}{a}_k{b}_{n-k-1}}& a_{n-1}-a_2{a}_{n-2}& a_{n-2}-a_2{a}_{n-3}& \cdots & \cdots & a_4-a_2{a}_3& a_3-a_2{a}_1& 0\\ -b_{n-3}-{\displaystyle \sum _{k=2}^{n-3}{a}_k{b}_{n-k-2}}& a_{n-2}-a_2{a}_{n-3}& a_{n-3}-a_2{a}_{n-4}& \cdots & \cdots & a_3-a_2{a}_1& 0& 1\\ ⋮& ⋮& ⋮& ⋮& ⋮& \ddots & & \\ ⋮& ⋮& ⋮& \ddots & \ddots & \ddots & & \\ -b_3-a_1{b}_2-a_3{b}_1& a_4-a_2{a}_3& a_3-a_2{a}_1& 0& 1& & & \\ -b_2-a_1{b}_1& a_3-a_2{a}_1& 0& 1& & & & \\ -b_1& 0& 1& & & & & \end{array}\right).\end{array}

Let X=(x_1, x_2, \dots ,x_n,y_n{)}^{\mathrm{T}}, Y=(y_{n-1},\dots ,y_1{)}^{\mathrm{T}}. And examine the matrix equation:

\left(\begin{array}{cc}{A}_1& A_3\\ A_2& A_4\end{array}\right) \left(\begin{array}{c}X\\ Y\end{array}\right)=\left(\begin{array}{c}{O}_{n+1}\\ O_{n-1}\end{array}\right) ,

that is

\begin{array}{}\text{(2.7a)}& A_{1}X+A_{3}Y=O,\end{array}

\begin{array}{}\text{(2.7b)}& A_{2}X+A_{4}Y=O.\end{array}

A_{1}X=- A_{3}Y can be obtained through (2.7a). Obviously, A₁ is invertible, and

\begin{array}{rl}& (A_1^{\mathrm{T}}{)}^{-1}=\\ & \left(\begin{array}{ccccccccc}1& & & & & & & & \\ a_2& 1& & & & & & & \\ a_3& a_2& 1& & & & & & \\ a_4& a_3& a_2& 1& & & & & \\ ⋮& ⋮& ⋮& \ddots & \ddots & & & & \\ a_{n-2}& a_{n-3}& a_{n-4}& \cdots & a_2& 1& & & \\ a_{n-1}& a_{n-2}& a_{n-3}& \cdots & a_3& a_2& 1& & \\ -{\displaystyle \frac{a_n}{b_0}}& -{\displaystyle \frac{a_{n-1}}{b_0}}& -{\displaystyle \frac{a_{n-2}}{b_0}}& \cdots & -{\displaystyle \frac{a_4}{b_0}}& -{\displaystyle \frac{a_2}{b_0}}& -{\displaystyle \frac{a_2}{b_0}}& -{\displaystyle \frac{1}{b_0}}& \\ {\displaystyle \frac{2a_n}{b_0}}& {\displaystyle \frac{2a_1{a}_{n-1}-a_n}{a_1{b}_0}}& {\displaystyle \frac{2a_2{a}_{n-2}-a_{n-1}}{a_1{b}_0}}& \cdots & {\displaystyle \frac{2a_2{a}_1-a_4}{a_1{b}_0}}& {\displaystyle \frac{2a_2{a}_3-a_4}{a_1{b}_0}}& {\displaystyle \frac{a_2^2+b_1^2}{b_1{b}_0}}& {\displaystyle \frac{1}{b_0}}& {\displaystyle \frac{1}{a_1}}\end{array}\right),\end{array}

\begin{array}{rl}& A_3^{\mathrm{T}}(A_1^{\mathrm{T}}{)}^{-1}=\\ & \left(\begin{array}{cccccccc}3a_n& {\displaystyle \frac{3a_2{a}_{n-1}-a_n}{a_1{b}_0}}& {\displaystyle \frac{3a_2{a}_{n-2}-a_{n-1}}{a_1{b}_0}}& \cdots & {\displaystyle \frac{3a_2{a}_3-a_4}{a_1{b}_0}}& {\displaystyle \frac{3a_2{a}_1-a_3}{a_1{b}_0}}& {\displaystyle \frac{2}{b_0}}& {\displaystyle \frac{1}{a_1}}\\ 0& 0& 0& \cdots & 0& 0& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 0& 0& 0& 0\end{array}\right),\end{array}

\begin{array}{rl}& (A_2{A}_1^{-1}{A}_3{)}^{\mathrm{T}}=A_3^{\mathrm{T}}(A_1^{\mathrm{T}}{)}^{-1}{A}_2^{\mathrm{T}}\\ & =\left(\begin{array}{cccccccc}{\displaystyle \frac{3a_n}{b_0}}& {\displaystyle \frac{3a_2{a}_{n-1}-a_n}{a_1{b}_0}}& {\displaystyle \frac{3a_2{a}_{n-2}-a_{n-1}}{a_1{b}_0}}& \cdots & {\displaystyle \frac{3a_2{a}_3-a_4}{a_1{b}_0}}& {\displaystyle \frac{2a_0^2+b_0^2}{a_1{b}_0}}& {\displaystyle \frac{2}{b_0}}& {\displaystyle \frac{1}{a_1}}\\ 0& 0& 0& \cdots & 0& 0& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 0& 0& 0& 0\end{array}\right)\\ & \cdot \left(\begin{array}{cccccc}-1& & & & & \\ -a_1& b_0& & & & \\ -a_3& b_0{a}_1& b_0& & & \\ -a_4& b_0{a}_3& b_0{a}_1& b_0& & \\ ⋮& ⋮& ⋮& ⋮& \ddots & \\ -a_{n-1}& b_0{a}_{n-2}& b_0{a}_{n-3}& \cdots & b_0{a}_1& b_0\\ -a_n& b_0{a}_{n-1}& b_0{a}_{n-2}& \cdots & b_0{a}_3& b_0{a}_1\\ b_n-{\displaystyle \sum _{k=2}^{n-1}{a}_k{b}_{n-k}=-{\displaystyle \frac{a_2{a}_n}{b_0}}}& -a_2{a}_{n-1}& -a_2{a}_{n-2}& \cdots & -a_2{a}_3& -a_2{a}_1\end{array}\right)\\ & =\left(\begin{array}{cccccccc}{s}_1& s_2& s_3& \cdots & s_{n-4}& s_{n-3}& s_{n-2}& s_{n-1}\\ 0& 0& 0& \cdots & 0& 0& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 0& 0& 0& 0\end{array}\right),\end{array}

where

\begin{array}{rl}{s}_1=& -\frac{3a_n}{b_0}+\frac{a_n-3a_2{a}_{n-1}}{b_0}+\frac{(a_{n-1}-3a_2{a}_{n-2})a_3}{a_1{b}_0}+\frac{(a_{n-2}-3a_2{a}_{n-3})a_4}{a_1{b}_0}+\cdots \\ & +\frac{(a_4-3a_2{a}_3)a_{n-2}}{a_1{b}_0}+\frac{(a_3-3a_2{a}_1)a_{n-1}}{a_1{b}_0}-\frac{3a_n}{b_0},\end{array}

\begin{array}{rl}{s}_2=& \frac{(3a_2{a}_{n-1}-a_n)b_0}{a_1{b}_0}+\frac{(3a_2{a}_{n-2}-a_{n-1})a_1{b}_0}{a_1{b}_0}+\frac{(3a_2{a}_{n-3}-a_{n-2})a_3{b}_0}{a_1{b}_0}+\cdots \\ & +\frac{(3a_2{a}_3-a_4)a_{n-3}{b}_0}{a_1{b}_0}+\frac{(3a_2{a}_1-a_3)a_{n-2}{b}_0}{a_1{b}_0}+\frac{a_{n-1}{b}_0}{b_0},\end{array}

\begin{array}{rl}{s}_3=& \frac{(3a_2{a}_{n-2}-a_{n-1})b_0}{a_1{b}_0}+\frac{(3a_2{a}_{n-3}-a_{n-2})a_1{b}_0}{a_1{b}_0}+\frac{(3a_2{a}_{n-4}-a_{n-3})a_3{b}_0}{a_1{b}_0}+\cdots \\ & +\frac{(3a_2{a}_3-a_4)a_{n-4}{b}_0}{a_1{b}_0}+\frac{(3a_2{a}_1-a_3)a_{n-3}{b}_0}{a_1{b}_0}+\frac{a_{n-2}{b}_0}{b_0},\end{array}

…

s_{n-2}=\frac{(3a_2{a}_3-a_4)b_0}{a_1{b}_0}+\frac{(3a_2{a}_1-a_3)a_1{b}_0}{a_1{b}_0}+\frac{a_3{b}_0}{b_0},

s_{n-1}=\frac{(3a_2{a}_1-a_3)b_0}{a_1{b}_0}+\frac{a_1{b}_0}{b_0}.

Based on (2.7a), substituting X=-A_1^{-1}{A}_{3}Y into (2.7b), we have ( A_4- A_2{A}_1^{-1}{A}_3)Y= O, that is

\{\begin{array}{l}{b}_1{y}_2+(b_2+a_1{b}_1)y_3+(b_3+a_1{b}_2+a_3{b}_1)y_4+\cdots +\left(b_{n-3}+{\displaystyle \sum _{k=2}^{n-3}{a}_k{b}_{n-k-2}}\right)y_{n-2}\\ +\left(b_{n-2}+{\displaystyle \sum _{k=2}^{n-2}{a}_k{b}_{n-k-1}+s_1}\right)y_{n-1}=0,\\ y_2+(a_3-a_2{a}_1)y_4+\cdots +(a_{n-4}-a_2{a}_{n-5})y_{n-3}+(a_{n-3}-a_2{a}_{n-4})y_{n-2}\\ +(a_{n-2}-a_2{a}_{n-3}-s_3)y_{n-1}=0,\\ \cdots \\ y_{n-4}+(a_3-a_2{a}_1)y_{n-2}+\left(\right(a_4-a_2{a}_2)-s_{n-3})y_{n-1}=0,\\ y_{n-3}+(a_3-a_2{a}_1-s_2)y_{n-1}=0,\\ y_{n-2}=s_{n-1}{y}_{n-1}=0,\\ y_1+(a_3-a_2{a}_1)y_3+(a_4-a_2{a}_3)y_4+\cdots +(a_{n-3}-a_2{a}_{n-4})y_{n-3}\\ +\left(a_{n-2}-a_2{a}_{n-3}\right)y_{n-2}+\left(a_{n-1}-a_2{a}_{n-3}-s_2\right)y_{n-1}=0.\end{array}

Firstly, by successively adding -( i-1)b_{i-1} times the ith equation to the first equation and then rearranging, we obtain

\{\begin{array}{l}\left[\right(n-2)b_{n-2}+s_1+b_1{s}_3+2b_2{s}_4+\cdots +(n-3\left)b_{n-3}{s}_{n-1}\right]y_{n-1}=0,\\ y_2+(a_3-a_2{a}_1)y_4+(a_4-a_2{a}_3)y_5+\cdots \\ +(a_{n-3}-a_2{a}_{n-4})y_{n-2}+(a_{n-2}-a_2{a}_{n-3}-s_3)y_{n-1}=0,\\ \cdots \\ y_{n-4}+(a_3-a_2{a}_1)y_{n-2}+(a_4-a_2{a}_3-s_{n-3})y_{n-1}=0,\\ y_{n-3}+(a_3-a_2{a}_1-s_{n-2})y_{n-1}=0,\\ y_{n-2}-s_{n-1}{y}_{n-1}=0,\\ y_1+(a_3-a_2{a}_1)y_3+(a_4-a_2{a}_3)y_4+\cdots \\ +(a_{n-2}-a_2{a}_{n-3})y_{n-2}+(a_{n-1}-a_2{a}_{n-2}-s_2)y_{n-1}=0.\end{array}

Next, starting from the (n-2)th equation and proceeding upward, we solve each equation successively, yielding

(2.8)\{\begin{array}{l}{y}_{n-2}=s_{n-1}{y}_{n-1},\\ y_{n-3}=\left(s_{n-2}+a_2{a}_1-a_3\right)y_{n-1},\\ y_{n-4}=\left(a_2{a}_1-a_3\right)y_{n-2}+\left(s_{n-3}+a_2{a}_3-a_4\right)y_{n-1},\\ y_{n-5}=\left(a_2{a}_1-a_3\right)y_{n-3}+\left(a_2{a}_3-a_4\right)y_{n-2}+\left(s_{n-4}+a_2{a}_4-a_5\right)y_{n-1},\\ \cdots \\ y_2=\left(a_2{a}_1-a_3\right)y_4+\left(a_2{a}_3-a_4\right)y_5+\cdots +\left(a_2{a}_{n-4}-a_{n-3}\right)y_{n-2}\\ +\left(s_3+a_2{a}_{n-3}-a_{n-2}\right)y_{n-1},\\ y_1=\left(a_2{a}_1-a_3\right)y_3+\left(a_2{a}_3-a_4\right)y_4+\cdots +\left(a_2{a}_{n-3}-a_{n-2}\right)y_{n-2}\\ +\left(s_2+a_2{a}_{n-2}-a_{n-1}\right)y_{n-1},\\ \left[\right(n-2)b_{n-2}+s_1+b_1{s}_3+2b_2{s}_4+\cdots +(n-3\left)b_{n-3}{s}_{n-1}\right]y_{n-1}=0.\end{array}

Obviously, y_1,y_2, \dots ,y_{n-2} are all monomial functions of y_{n-1}.

Let h=(n- 2)b_{n-2}+s_1+b_1{s}_3+2b_2{s}_4+ \cdots +(n -3) b_{n-3}{s}_{n-1}. It can be known from Lemma 2.1 that when k=2,\dots ,n, ; when k=3,\dots ,n- 1, 3a_2{a}_k-a_{k+1}=a_2{a}_k+ (a_2^2+b_1^2)a_{k-1}>0. Then

Therefore, there is only zero solution y_{n-1}=0 to the last equation (2.8), and y_{n-2}=y_{n-3}=\cdots =y_2=y_1=0, namely Y = O, X = O. So, there is only zero solution to the equation \left(\begin{array}{cc}{A}_1& A_3\\ A_2& A_4\end{array}\right)\left(\begin{array}{c}X\\ Y\end{array}\right)= \left(\begin{array}{c}{O}_{n+1}\\ O_{n-1}\end{array}\right). That is, (J{|}_P{)}^{\mathrm{T}}\left(\begin{array}{c}X\\ Y\end{array}\right) has only a zero solution. Consequently, the rank of the Jacobian matrix J{|}_P is 2n. Hence, J{|}_P\ne 0.

Lemma 2.3 [4]　 Assume that an nth-order complex pattern S=A+\mathrm{i}B, n\ge 2, and there exists a nilpotent complex matrix C=A+\mathrm{i}B\in Q(S), where A\in Q(A), B\in Q(B), and A and B contain at least 2n nonzero elements which, without loss of generality, are marked as a_{i_1{j}_1},\dots ,a_{i_{n_1{j}_{n_1}}}, b_{i_{n_1+1}{j}_{n_1+1}},\dots ,b_{i_{2n}{j}_{2n}}. Assume that X is the matrix after replacing these nonzero elements with variables x_1,\dots ,x_{2n}, and the characteristic polynomial is

\begin{array}{rl}|\lambda I-X|=& {\lambda }^n+\left(f_1\right(x_1,x_2,\dots ,x_{2n})+\mathrm{i}{g}_1(x_1,x_2,\dots ,x_{2n}\left)\right){\lambda }^{n-1}+\cdots \\ & +\left(f_{n-1}\right(x_1,x_2,\dots ,x_{2n})+\mathrm{i}{g}_{n-1}(x_1,x_2,\dots ,x_{2n}))\lambda \\ & +(f_n(x_1,x_2,\dots ,x_{2n})+\mathrm{i}{g}_n(x_1,x_2,\dots ,x_{2n})).\end{array}

If the nilpotent point for Jacobian matrix J=\frac{\mathrm{\partial }(f_1,\dots ,f_n,g_1,\dots ,g_n)}{\mathrm{\partial }(x_1,\dots ,x_{2n})} of X

\left(x_1,\dots ,x_{2n}\right)=\left(a_{i_1{j}_1},\dots ,a_{i_{n_1}{j}_{n_1}},b_{i_{n_1+1}{j}_{n_1+1}},\dots ,b_{i_{2n}{j}_{2n}}\right)

is nonsingular, then the complex pattern S is spectrally arbitrary, and any super pattern of S is a spectrally arbitrary complex sign pattern.

From Lemma 2.1−2.3, the following conclusion is readily established.

Theorem 2.1　 When n\ge 4, the complex pattern S_nand any of its super patterns are spectrally arbitrary.

In fact, when n = 3, Theorem 2.1 is also established.

Example 2.1　Complex pattern

S_3= \left(\begin{array}{ccc}-1& 0& -1\\ 0& 1+\mathrm{i}& 1\\ 1-\mathrm{i}& -\mathrm{i}& -\mathrm{i}\end{array}\right)

and any of its super patterns are spectrally arbitrary.

Proof　Let

C_3= \left(\begin{array}{ccc}-a_1& 0& -1\\ 0& a_2+i{b}_0& 1\\ a_3-i{b}_3& -i{b}_2& -i{b}_1\end{array}\right) \in Q_c\left({\mathcal{S}}_3\right).

Then

f\left(\lambda \right)=|\lambda I -C_3|={\lambda }^3+( f_1+ \mathrm{i}{g}_1){\lambda }^2+(f_2+ \mathrm{i}{g}_2)\lambda +(f_3+\mathrm{i}{g}_3),

where

\begin{array}{rl}{f}_1& =a_1-a_2,g_1=b_1-b_0,\\ f_2& =a_3-a_2{a}_1+b_1{b}_0,g_2=b_2+(a_1-a_2)b_1-a_2{b}_1-b_3,\\ f_3& =-a_2{a}_3+a_1{b}_1{b}_0-b_3{b}_0,g_3=a_2{b}_3+a_1{b}_2-a_2{a}_1{b}_1-a_3{b}_0.\end{array}

Hence,

J=\frac{\mathrm{\partial }(f_1,f_2,f_3,g_3,g_2,g_1)}{\mathrm{\partial }(a_1,a_3,b_3,b_2,a_2,b_1)}=\left(\begin{array}{cccccc}1& 0& 0& 0& -1& 0\\ -a_2& 1& 0& 0& -a_1& b_0\\ b_0{b}_1& -a_2& -b_0& 0& -a_3& a_1{b}_0\\ b_2-a_2{b}_1& -b_0& a_2& a_1& b_3-a_1{b}_1& -a_2{a}_1\\ b_1-b_0& 0& -1& 1& -b_1& a_1-a_2\\ 0& 0& 0& 0& 0& 1\end{array}\right).

When b₁ = b₀ = 1, a_1=a_2=\frac{1}{2}\sqrt{3+\sqrt{17}},a_3=\frac{\sqrt{17}-1}{4},b_2=\frac{9-\sqrt{17}}{8}\sqrt{3+\sqrt{17}}, and b_3= \frac{5-\sqrt{17}}{8}\sqrt{3+\sqrt{17}}, C₃ is nilpotent and det\left(J\right){|}_P=-\sqrt{17(3+\sqrt{17})}\ne 0. From Lemma 2.3, it can be seen that S₃ and any of its super patterns are spectrally arbitrary.

References

[1] Bergsma H, Vander Meulen K N, Van Tuyl A. Potentially nilpotent patterns and the nilpotent-Jacobian method. Linear Algebra Appl 2012; 436(12): 4433–4445

[2] Britz T, McDonald J J, Olesky D D, van den Driessche P. Minimal spectrally arbitrary sign patterns. SIAM J Matrix Anal Appl 2004; 26(1): 257–271

[3] Feng X L, Li Z S. New results on sign patterns that allow diagonalizability. J Math Res Appl 2022; 42(2): 111–120

[4] Gao Y B, Shao Y L, Fan Y Z. Spectrally arbitrary complex sign pattern matrices. Electron J Linear Algebra 2009; 18: 674–692

[5] McDonald J J, Stuart J. Spectrally arbitrary ray patterns. Linear Algebra Appl 2008; 429(4): 727–734

[6] McDonald J J, Yielding A A. Complex spectrally arbitrary zero-nonzero patterns. Linear Multilinear Algebra 2012; 60(1): 11–26

[7] Mei Y Z, Gao Y B, Shao Y L, Wang P. The minimum number of nonzeros in a spectrally arbitrary ray pattern. Linear Algebra Appl 2014; 453: 99–109

[8] Mei Y Z, Gao Y B, Shao Y L, Wang P. A new family of spectrally arbitrary ray patterns. Czechoslovak Math J 2016; 66(4): 1049–1058

[9] Shao J Y, Liu Y, Ren L Z. The inverse problems of the determinantal regions of ray pattern and complex sign pattern matrices. Linear Algebra Appl 2006; 416(2/3): 835–843

[10] Shao J Y, Shan H Y. The determinantal regions of complex sign pattern matrices and ray pattern matrices. Linear Algebra Appl 2005; 395: 211–228

RIGHTS & PERMISSIONS

Higher Education Press 2024