Research on the Yang-Baxter-Like Matrix Equation for a Diagonalizable Complex Matrix with Three Different Nonzero Eigenvalues

1. Introduction

Matrix theory is of great significance in numerous fields(see [1,2,3,4,5,6,7,8]). When matrix theory is utilized to solve practical problems, a large number of matrix equations have appeared. In this paper, we research the following equation:

$$AXA=XAX,$$

(1)

where A is a given $n\times n$ complex matrix and X is an unknown matrix. Equation (1) is often known as Yang-Baxter-like matrix equation, or YBME for short, as it bears a strong resemblance to the classical Yang-Baxter equation

$$A\left(u\right)B(u+v)A\left(v\right)=B\left(v\right)A(u+v)B\left(u\right),$$

where both A and B are parametric functions of variables u and v. When A and B are independent of u and $v,$ we can get YBME (1). The Yang-Baxter equation originates from the research on the many-body problem studied by Nobel Prize winner Yang Chen-Ning and a lattice model studied by Baxter[9,10]. It has wide applications in many fields such as quantum groups, conformal field theory and non-commutative geometry[11,12].

Equation (1) has drawn a great deal of attention on account of its extensive applications. Research on Equation (1) can be found in references [13,14,15,16,17,18], all of which presuppose that the coefficient matrix A possesses a specific structure. For example, when the coefficient matrix A is a quadratic idempotent matrix[19], a cubic idempotent matrix[20], a quartic idempotent matrix[21], or an involutory matrix[22], explicit expressions for solutions of Equation (1) are given.

Additionally, there has been extensive research on the properties of solutions of the Yang-Baxter-like matrix equation for certain special coefficient matrices. For example, when A is a matrix with nonsingular Jordan blocks, some properties of solutions of the Yang-Baxter-like matrix equation equation are given in the reference[23]

However, a general expression for solutions of Equation (1) has not been found yet. After all, solving Equation (1) is equivalent to solving $n^2$ quadratic equations with $n^2$ unknowns.

When A is a diagonalizable complex matrix which has two different eigenvalues, the explicit solutions of Equation (1) has been derived in the reference[24].

When A is a singular diagonalizable complex matrix which has three different eigenvalues, the explicit solutions of Equation (1) has been derived in the reference[?]. However, when A is a non-singular diagonalizable complex matrix which has three different nonzero eigenvalues, finding the explicit solutions of Equation (1) is very difficult.

On this basis, we research the solutions of Equation (1) when A is an $n\times n$ diagonalizable complex matrix which has three different nonzero eigenvalues and get some distinctive properties of solutions of Equation (1).

We organize the full text in the following manner. In the second section, we present the main results of this paper. In the final section, we provide a summary of the research work conducted in this paper.

2. Results

Theorem 1.

Provided that A is an $n\times n$ complex matrix. Suppose $A=SJ{S}^{-1},$ where S is an $n\times n$ invertible complex matrix and $J=\left(\begin{array}{ccc}\lambda I_k& & \\ & \mu I_l& \\ & & \beta I_m\end{array}\right),\lambda \ne 0,\mu \ne 0,\beta \ne 0,\lambda \ne \mu ,\lambda \ne \beta ,\mu \ne \beta ,k+l+m=n.$ $I_k$ represents the k-order unit matrix,$I_l$ represents the l-order unit matrix,$I_m$ represents the m-order unit matrix. The solutions of Equation (1) are $X=SY{S}^{-1},$ where $Y=\left(\begin{array}{ccc}B& C& D\\ F& G& H\\ P& Q& W\end{array}\right),B\in {\mathbb{C}}^{k\times k},C\in {\mathbb{C}}^{k\times l},D\in {\mathbb{C}}^{k\times m},F\in {\mathbb{C}}^{l\times k},G\in {\mathbb{C}}^{l\times l},H\in {\mathbb{C}}^{l\times m},P\in {\mathbb{C}}^{m\times k},Q\in {\mathbb{C}}^{m\times l},W\in {\mathbb{C}}^{m\times m}.$ And

$$\lambda \mu \beta P+\lambda (\lambda -\mu )PB+\beta (\beta -\mu )WP=0,$$

(2)

$$\lambda \mu \beta D+\lambda (\lambda -\mu )BD+\beta (\beta -\mu )DW=0,$$

(3)

$$\lambda \mu \beta F+\lambda (\lambda -\beta )FB+\mu (\mu -\beta )GF=0,$$

(4)

$$\lambda \mu \beta C+\lambda (\lambda -\beta )BC+\mu (\mu -\beta )CG=0.$$

(5)

$$\lambda \mu \beta Q+\mu (\mu -\lambda )QG+\beta (\beta -\lambda )WQ=0,$$

(6)

$$\lambda \mu \beta H+\mu (\mu -\lambda )GH+\beta (\beta -\lambda )HW=0,$$

(7)

Proof. 

Let $Y=S^{-1}XS$. Then solving Equation (1) is equivalent to solving equation $JYJ=YJY$. We suppose $Y=\left(\begin{array}{ccc}B& C& D\\ F& G& H\\ P& Q& W\end{array}\right),B\in {\mathbb{C}}^{k\times k},C\in {\mathbb{C}}^{k\times l},D\in {\mathbb{C}}^{k\times m},F\in {\mathbb{C}}^{l\times k},G\in {\mathbb{C}}^{l\times l},H\in {\mathbb{C}}^{l\times m},P\in {\mathbb{C}}^{m\times k},Q\in {\mathbb{C}}^{m\times l},W\in {\mathbb{C}}^{m\times m}.$ Then

$$\begin{array}{ccc}& & JYJ\hfill \\ & & =\left(\begin{array}{ccc}\lambda I_k& & \\ & \mu I_l& \\ & & \beta I_m\end{array}\right)\left(\begin{array}{ccc}B& C& D\\ F& G& H\\ P& Q& W\end{array}\right)\left(\begin{array}{ccc}\lambda I_k& & \\ & \mu I_l& \\ & & \beta I_m\end{array}\right)\hfill \\ & & =\left(\begin{array}{ccc}{\lambda }^{2}B& \lambda \mu C& \lambda \beta D\\ \lambda \mu F& {\mu }^{2}G& \mu \beta H\\ \lambda \beta P& \mu \beta Q& {\beta }^{2}W\end{array}\right),\hfill \end{array}$$

$$\begin{array}{ccc}& & YJY\hfill \\ & & =\left(\begin{array}{ccc}B& C& D\\ F& G& H\\ P& Q& W\end{array}\right)\left(\begin{array}{ccc}\lambda I_k& & \\ & \mu I_l& \\ & & \beta I_m\end{array}\right)\left(\begin{array}{ccc}B& C& D\\ F& G& H\\ P& Q& W\end{array}\right)\hfill \\ & & =\left(\begin{array}{ccc}\lambda B^2+\mu CF+\beta DP& \lambda BC+\mu CG+\beta DQ& \lambda BD+\mu CH+\beta DW\\ \lambda FB+\mu GF+\beta HP& \lambda FC+\mu G^2+\beta HQ& \lambda FD+\mu GH+\beta HW\\ \lambda PB+\mu QF+\beta WP& \lambda PC+\mu QG+\beta WQ& \lambda PD+\mu QH+\beta W^2\end{array}\right).\hfill \end{array}$$

As $JYJ=YJY,$ we can get that

$${\lambda }^{2}B=\lambda B^2+\mu CF+\beta DP,$$

(8)

$$\lambda \mu C=\lambda BC+\mu CG+\beta DQ,$$

(9)

$$\lambda \beta D=\lambda BD+\mu CH+\beta DW,$$

(10)

$$\lambda \mu F=\lambda FB+\mu GF+\beta HP,$$

(11)

$${\mu }^{2}G=\lambda FC+\mu G^2+\beta HQ,$$

(12)

$$\mu \beta H=\lambda FD+\mu GH+\beta HW,$$

(13)

$$\lambda \beta P=\lambda PB+\mu QF+\beta WP,$$

(14)

$$\mu \beta Q=\lambda PC+\mu QG+\beta WQ,$$

(15)

$${\beta }^{2}W=\lambda PD+\mu QH+\beta W^2.$$

(16)

(a) According to Equation (16), we know

$$\beta W^2={\beta }^{2}W-\lambda PD-\mu QH.$$

Further

$$\beta W^{2}P={\beta }^{2}WP-\lambda PDP-\mu QHP.$$

According to Equation (14), we know

$$\beta WP=\lambda \beta P-\lambda PB-\mu QF.$$

Further

$$\beta W^{2}P=\lambda \beta WP-\lambda WPB-\mu WQF.$$

So

$${\beta }^{2}WP-\lambda PDP-\mu QHP=\lambda \beta WP-\lambda WPB-\mu WQF.$$

In other words,

$$\beta (\beta -\lambda )WP-\lambda PDP-\mu QHP+\lambda WPB+\mu WQF=0.$$

Further

$${\beta }^2(\beta -\lambda )WP-\lambda \beta PDP-\mu \beta QHP+\lambda \beta WPB+\mu \beta WQF=0.$$

(17)

According to Equation (8), we know

$$\beta DP={\lambda }^{2}B-\lambda B^2-\mu CF.$$

Further

$$-\lambda \beta PDP=-{\lambda }^{3}PB+{\lambda }^{2}P{B}^2+\lambda \mu PCF.$$

(18)

According to Equation (11), we know

$$\beta HP=\lambda \mu F-\lambda FB-\mu GF.$$

Further

$$-\mu \beta QHP=-\lambda {\mu }^{2}QF+\lambda \mu QFB+{\mu }^{2}QGF.$$

(19)

According to equation(14), we know

$$\beta WP=\lambda \beta P-\lambda PB-\mu QF.$$

Further

$$\lambda \beta WPB={\lambda }^2\beta PB-{\lambda }^{2}P{B}^2-\lambda \mu QFB.$$

(20)

According to Equation (15), we know

$$\beta WQ=\mu \beta Q-\lambda PC-\mu QG.$$

Further

$$\mu \beta WQF={\mu }^2\beta QF-\lambda \mu PCF-{\mu }^{2}QGF.$$

(21)

Substituting Equations (18), (19), (20) and (21) into Equation (17), we are able to get

$${\beta }^2(\beta -\lambda )WP+{\mu }^2(\beta -\lambda )QF+{\lambda }^2(\beta -\lambda )PB=0.$$

As $\beta \ne \lambda ,$ we can get that

$${\beta }^{2}WP+{\mu }^{2}QF+{\lambda }^{2}PB=0.$$

According to Equation(14), we know

$$\mu QF=\lambda \beta P-\lambda PB-\beta WP.$$

So

$${\beta }^{2}WP+\mu (\lambda \beta P-\lambda PB-\beta WP)+{\lambda }^{2}PB=0.$$

That is to say,

$$\lambda \mu \beta P+\lambda (\lambda -\mu )PB+\beta (\beta -\mu )WP=0.$$

(b) According to Equation (8), we know

$$\lambda B^2={\lambda }^{2}B-\mu CF-\beta DP.$$

Further

$$\lambda B^{2}D={\lambda }^{2}BD-\mu CFD-\beta DPD.$$

According to Equation (10), we know

$$\lambda BD=\lambda \beta D-\mu CH-\beta DW.$$

Further

$$\lambda B^{2}D=\lambda \beta BD-\mu BCH-\beta BDW.$$

So

$${\lambda }^{2}BD-\mu CFD-\beta DPD=\lambda \beta BD-\mu BCH-\beta BDW.$$

In other words,

$$\lambda (\lambda -\beta )BD-\mu CFD-\beta DPD+\mu BCH+\beta BDW=0.$$

Further

$${\lambda }^2(\lambda -\beta )BD-\lambda \mu CFD-\lambda \beta DPD+\lambda \mu BCH+\lambda \beta BDW=0.$$

(22)

According to Equation (13), we know

$$\lambda FD=\mu \beta H-\mu GH-\beta HW.$$

Further

$$-\lambda \mu CFD=-{\mu }^2\beta CH+{\mu }^{2}CGH+\mu \beta CHW.$$

(23)

According to Equation (16), we know

$$\lambda PD={\beta }^{2}W-\mu QH-\beta W^2.$$

Further

$$-\lambda \beta DPD=-{\beta }^{3}DW+\mu \beta DQH+{\beta }^{2}D{W}^2.$$

(24)

According to Equation (9), we know

$$\lambda BC=\lambda \mu C-\mu CG-\beta DQ.$$

Further

$$\lambda \mu BCH=\lambda {\mu }^{2}CH-{\mu }^{2}CGH-\mu \beta DQH.$$

(25)

According to Equation (10), we know

$$\lambda BD=\lambda \beta D-\mu CH-\beta DW.$$

Further

$$\lambda \beta BDW=\lambda {\beta }^{2}DW-\mu \beta CHW-{\beta }^{2}D{W}^2.$$

(26)

Substituting Equations (23), (24), (25) and (26) into Equation (22), we are able to get

$${\lambda }^2(\lambda -\beta )BD+{\mu }^2(\lambda -\beta )CH+{\beta }^2(\lambda -\beta )DW=0.$$

As $\lambda \ne \beta ,$ we can get that

$${\lambda }^{2}BD+{\mu }^{2}CH+{\beta }^{2}DW=0.$$

According to equation(10), we know

$$\mu CH=\lambda \beta D-\lambda BD-\beta DW.$$

So

$${\lambda }^{2}BD+\mu (\lambda \beta D-\lambda BD-\beta DW)+{\beta }^{2}DW=0.$$

That is to say,

$$\lambda \mu \beta D+\lambda (\lambda -\mu )BD+\beta (\beta -\mu )DW=0.$$

(c) According to Equation (12), we know

$$\mu G^2={\mu }^{2}G-\lambda FC-\beta HQ.$$

Further

$$\mu G^{2}F={\mu }^{2}GF-\lambda FCF-\beta HQF.$$

According to Equation (11), we know

$$\mu GF=\lambda \mu F-\lambda FB-\beta HP.$$

Further

$$\mu G^{2}F=\lambda \mu GF-\lambda GFB-\beta GHP.$$

So

$${\mu }^{2}GF-\lambda FCF-\beta HQF=\lambda \mu GF-\lambda GFB-\beta GHP.$$

In other words,

$$\mu (\mu -\lambda )GF-\lambda FCF-\beta HQF+\lambda GFB+\beta GHP=0.$$

Further

$${\mu }^2(\mu -\lambda )GF-\lambda \mu FCF-\mu \beta HQF+\lambda \mu GFB+\mu \beta GHP=0.$$

(27)

According to Equation (8), we know

$$\mu CF={\lambda }^{2}B-\lambda B^2-\beta DP.$$

Further

$$-\lambda \mu FCF=-{\lambda }^{3}FB+{\lambda }^{2}F{B}^2+\lambda \beta FDP.$$

(28)

According to Equation (14), we know

$$\mu QF=\lambda \beta P-\lambda PB-\beta WP.$$

Further

$$-\mu \beta HQF=-\lambda {\beta }^{2}HP+\lambda \beta HPB+{\beta }^{2}HWP.$$

(29)

According to Equation (13), we know

$$\mu GH=\mu \beta H-\lambda FD-\beta HW.$$

Further

$$\mu \beta GHP=\mu {\beta }^{2}HP-\lambda \beta FDP-{\beta }^{2}HWP.$$

(30)

According to Equation (11), we know

$$\mu GF=\lambda \mu F-\lambda FB-\beta HP.$$

Further

$$\lambda \mu GFB={\lambda }^2\mu FB-{\lambda }^{2}F{B}^2-\lambda \beta HPB.$$

(31)

Substituting Equations (28), (29), (30) and (31) into Equation (27), we are able to get

$${\lambda }^2(\lambda -\mu )FB+{\mu }^2(\lambda -\mu )GF+{\beta }^2(\lambda -\mu )HP=0.$$

As $\lambda \ne \mu ,$ we can get that

$${\lambda }^{2}FB+{\mu }^{2}GF+{\beta }^{2}HP=0.$$

According to equation(11), we know

$$\beta HP=\lambda \mu F-\lambda FB-\mu GF.$$

So

$${\lambda }^{2}FB+{\mu }^{2}GF+\beta (\lambda \mu F-\lambda FB-\mu GF)=0.$$

That is to say,

$$\lambda \mu \beta F+\lambda (\lambda -\beta )FB+\mu (\mu -\beta )GF=0.$$

(d) According to Equation (8), we know

$$\lambda B^2={\lambda }^{2}B-\mu CF-\beta DP.$$

Further

$$\lambda B^{2}C={\lambda }^{2}BC-\mu CFC-\beta DPC.$$

According to Equation (9), we know

$$\lambda BC=\lambda \mu C-\mu CG-\beta DQ.$$

Further

$$\lambda B^{2}C=\lambda \mu BC-\mu BCG-\beta BDQ.$$

So

$${\lambda }^{2}BC-\mu CFC-\beta DPC=\lambda \mu BC-\mu BCG-\beta BDQ.$$

In other words,

$$\lambda (\lambda -\mu )BC-\mu CFC-\beta DPC+\mu BCG+\beta BDQ=0.$$

Further

$${\lambda }^2(\lambda -\mu )BC-\lambda \mu CFC-\lambda \beta DPC+\lambda \mu BCG+\lambda \beta BDQ=0.$$

(32)

According to Equation (12), we know

$$\lambda FC={\mu }^{2}G-\mu G^2-\beta HQ.$$

Further

$$-\lambda \mu CFC=-{\mu }^{3}CG+{\mu }^{2}C{G}^2+\mu \beta CHQ.$$

(33)

According to Equation (15), we know

$$\lambda PC=\mu \beta Q-\mu QG-\beta WQ.$$

Further

$$-\lambda \beta DPC=-\mu {\beta }^{2}DQ+\mu \beta DQG+{\beta }^{2}DWQ.$$

(34)

According to Equation (9), we know

$$\lambda BC=\lambda \mu C-\mu CG-\beta DQ.$$

Further

$$\lambda \mu BCG=\lambda {\mu }^{2}CG-{\mu }^{2}C{G}^2-\mu \beta DQG.$$

(35)

According to Equation (10), we know

$$\lambda BD=\lambda \beta D-\mu CH-\beta DW.$$

Further

$$\lambda \beta BDQ=\lambda {\beta }^{2}DQ-\mu \beta CHQ-{\beta }^{2}DWQ.$$

(36)

Substituting Equations (33), (34), (35) and (36) into Equation (32), we are able to get

$${\lambda }^2(\lambda -\mu )BC+{\mu }^2(\lambda -\mu )CG+{\beta }^2(\lambda -\mu )DQ=0.$$

As $\lambda \ne \mu ,$ we can get that

$${\lambda }^{2}BC+{\mu }^{2}CG+{\beta }^{2}DQ=0.$$

According to equation(9), we know

$$\beta DQ=\lambda \mu C-\lambda BC-\mu CG.$$

So

$${\lambda }^{2}BC+{\mu }^{2}CG+\beta (\lambda \mu C-\lambda BC-\mu CG)=0.$$

That is to say,

$$\lambda \mu \beta C+\lambda (\lambda -\beta )BC+\mu (\mu -\beta )CG=0.$$

(e) According to Equation (16), we know

$$\beta W^2={\beta }^{2}W-\lambda PD-\mu QH.$$

Further

$$\beta W^{2}Q={\beta }^{2}WQ-\lambda PDQ-\mu QHQ.$$

According to Equation (15), we know

$$\beta WQ=\mu \beta Q-\lambda PC-\mu QG.$$

Further

$$\beta W^{2}Q=\mu \beta WQ-\lambda WPC-\mu WQG.$$

So

$${\beta }^{2}WQ-\lambda PDQ-\mu QHQ=\mu \beta WQ-\lambda WPC-\mu WQG.$$

In other words,

$$\beta (\beta -\mu )WQ-\lambda PDQ-\mu QHQ+\lambda WPC+\mu WQG=0.$$

Further

$${\beta }^2(\beta -\mu )WQ-\lambda \beta PDQ-\mu \beta QHQ+\lambda \beta WPC+\mu \beta WQG=0.$$

(37)

According to Equation (9), we know

$$\beta DQ=\lambda \mu C-\lambda BC-\mu CG.$$

Further

$$-\lambda \beta PDQ=-{\lambda }^2\mu PC+{\lambda }^{2}PBC+\lambda \mu PCG.$$

(38)

According to Equation (12), we know

$$\beta HQ={\mu }^{2}G-\lambda FC-\mu G^2.$$

Further

$$-\mu \beta QHQ=-{\mu }^{3}QG+\lambda \mu QFC+{\mu }^{2}Q{G}^2.$$

(39)

According to Equation (14), we know

$$\beta WP=\lambda \beta P-\lambda PB-\mu QF.$$

Further

$$\lambda \beta WPC={\lambda }^2\beta PC-{\lambda }^{2}PBC-\lambda \mu QFC.$$

(40)

According to Equation (15), we know

$$\beta WQ=\mu \beta Q-\lambda PC-\mu QG.$$

Further

$$\mu \beta WQG={\mu }^2\beta QG-\lambda \mu PCG-{\mu }^{2}Q{G}^2.$$

(41)

Substituting Equations (38), (39), (40) and (41) into Equation (37), we are able to get

$${\lambda }^2(\beta -\mu )PC+{\mu }^2(\beta -\mu )QG+{\beta }^2(\beta -\mu )WQ=0.$$

As $\beta \ne \mu ,$ we can get that

$${\lambda }^{2}PC+{\mu }^{2}QG+{\beta }^{2}WQ=0.$$

According to equation(15), we know

$$\lambda PC=\mu \beta Q-\mu QG-\beta WQ.$$

So

$$\lambda (\mu \beta Q-\mu QG-\beta WQ)+{\mu }^{2}QG+{\beta }^{2}WQ=0.$$

That is to say,

$$\lambda \mu \beta Q+\mu (\mu -\lambda )QG+\beta (\beta -\lambda )WQ=0.$$

(f) According to Equation (12), we know

$$\mu G^2={\mu }^{2}G-\lambda FC-\beta HQ.$$

Further

$$\mu G^{2}H={\mu }^{2}GH-\lambda FCH-\beta HQH.$$

According to Equation (13), we know

$$\mu GH=\mu \beta H-\lambda FD-\beta HW.$$

Further

$$\mu G^{2}H=\mu \beta GH-\lambda GFD-\beta GHW.$$

So

$${\mu }^{2}GH-\lambda FCH-\beta HQH=\mu \beta GH-\lambda GFD-\beta GHW.$$

In other words,

$$\mu (\mu -\beta )GH-\lambda FCH-\beta HQH+\lambda GFD+\beta GHW=0.$$

Further

$${\mu }^2(\mu -\beta )GH-\lambda \mu FCH-\mu \beta HQH+\lambda \mu GFD+\mu \beta GHW=0.$$

(42)

According to Equation (10), we know

$$\mu CH=\lambda \beta D-\lambda BD-\beta DW.$$

Further

$$-\lambda \mu FCH=-{\lambda }^2\beta FD+{\lambda }^{2}FBD+\lambda \beta FDW.$$

(43)

According to Equation (16), we know

$$\mu QH={\beta }^{2}W-\lambda PD-\beta W^2.$$

Further

$$-\mu \beta HQH=-{\beta }^{3}HW+\lambda \beta HPD+{\beta }^{2}H{W}^2.$$

(44)

According to Equation (11), we know

$$\mu GF=\lambda \mu F-\lambda FB-\beta HP.$$

Further

$$\lambda \mu GFD={\lambda }^2\mu FD-{\lambda }^{2}FBD-\lambda \beta HPD.$$

(45)

According to Equation (13), we know

$$\mu GH=\mu \beta H-\lambda FD-\beta HW.$$

Further

$$\mu \beta GHW=\mu {\beta }^{2}HW-\lambda \beta FDW-{\beta }^{2}H{W}^2.$$

(46)

Substituting Equations (43), (44), (45) and (46) into Equation (42), we are able to get

$${\lambda }^2(\mu -\beta )FD+{\mu }^2(\mu -\beta )GH+{\beta }^2(\mu -\beta )HW=0.$$

As $\mu \ne \beta ,$ we can get that

$${\lambda }^{2}FD+{\mu }^{2}GH+{\beta }^{2}HW=0.$$

According to equation(13), we know

$$\lambda FD=\mu \beta H-\mu GH-\beta HW.$$

So

$$\lambda (\mu \beta H-\mu GH-\beta HW)+{\mu }^{2}GH+{\beta }^{2}HW=0.$$

That is to say,

$$\lambda \mu \beta H+\mu (\mu -\lambda )GH+\beta (\beta -\lambda )HW=0.$$

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3. Conclusions

When A is an $n\times n$ complex matrix which is diagonalizable and has three different nonzero eigenvalues, we get some distinctive properties of solutions of Equation (1). However, solving Equation (1) remains a significant challenge, necessitating further research.