Finding Solutions to the Yang–Baxter-like Matrix Equation for Diagonalizable Coefficient Matrix

Abstract

Let A be a diagonalizable complex matrix. In this paper, we discuss finding solutions to the Yang–Baxter-like matrix equation $AXA=XAX.$ We then present a concrete example to illustrate the validity of the results obtained.

1. Introduction

In past decades, much effort has been put into solving the following Yang–Baxter-like matrix equation

$$AXA=XAX,$$

(1)

where A is a given square matrix, and X is the unknown complex matrix. Recent representative articles relevant to this problem can be referenced from [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22]. Regarding the background, this equation has also been called a star–triangle-like equation in statistical mechanics for many years, and, to the best of our knowledge, it originates from the classical Yang–Baxter equation, which was proposed first by C.N. Yang in December 1967, in his article on simple one-dimensional multi-body problems [21] and, subsequently, in 1972, by R.J. Baxte, who independently discussed this equation when studying some classical two-dimensional statistical mechanics problems [23]. Since then, the Yang–Baxter equation has been transformed into the so-called (simpler) Yang–Baxter-like matrix equation through the appropriate imposition of some restrictions. The Yang–Baxter-like matrix equation is actually the nonparametric form of the classical Yang–Baxter equation in matrix theory. Moreover, in terms of the matrix algebra, finding the relations between A and X turns out to be interesting because it provides information about the commutability and idempotency of matrices [24].

The Yang–Baxter-like matrix equation appears to be simple, but it is generally not easy to find all of its solutions because of the nonlinearity of the equation. Even in the case of a lower-order situation, all of the solutions found thus far have been only for very special cases, and only basic and partial answers have been discovered.

A systematic study of the Yang–Baxter-like matrix equation from the perspective of matrix theory has basically started in the last decade. Ding and Rhee first eastablished a series of important results. For example, when A is nonsingular, and its inverse matrix is a stochastic matrix, the existence of the solution is proven by using Brouwer’s fixed-point theorem. They also obtained some numerical solutions to the equation by making use of the mean ergodic theorem for matrices and the direct iterative method [6]. We then see that, when using the spectral projection theorem in matrix analysis and the generalized eigen-subspace technique, a finite number of spectral solutions to the equation are constructed in [7]. In the case that the eigenvalues of the coefficient matrix are semi-simple, and their multiplicity is at least two, Ding and Zhang claimed that all spectral solutions to the equation can be constructed [9], and, in [8], they discussed finding solutions that satisfy $AX=XA=X^2$. The above solutions are partial, special or commutative, that is, the solutions X satisfy the commutative condition $AX=XA$. Some recent results on the commutative solutions to the equation can be seen in [10,12,13,17,18], and more general discussion in finding the general non-commutative solutions and all solutions of Yang–Baxter like matrix equation are found in [1,2,4,5,16,19,20,22]. When A is an idempotent matrix, all of its solutions were obtained in [4,16], where the idea is based on the property of the diagonalization of A. When A has rank one or rank two, or if $A^{-1}=A$, $A^3=A$, all solutions have been obtained (e.g., see [1,2,20,22]). The construction of all solutions corresponding to diagonalizable matrices with a spectral set $\{1,\alpha ,0\}$ is discussed in [5], and, in [19], the general expressions of X are established for the diagonalizable matrix A with two different eigenvalues.

We notice that all previous discussions are under the assumptions that their coefficient matrices A are either diagonalizable matrices whose number of distinct eigenvalues is no more than three or that the matrices possess special kinds of Jordan blocks. For a more general coefficient matrix, an approach to seek for all of its solutions has not been seen. With this motivation, in this paper, we focus on constructing all of the solutions to the equation when A is a diagonalizable matrix, thereby extending the existing results.

2. All Solutions to the Equation $\mathit{AXA}=\mathit{XAX}$ for Any Diagonalizable Matrices $\mathit{A}$

Let $A,T\in C^{m\times m}$ and $A=TJ{T}^{-1}$, where J is the Jordan canonical form of A, and let $Z=T^{-1}XT.$ Then, the matrix equation $AXA=XAX$ is equivalent to $JZJ=ZJZ$, and, if A is diagonalizable, the orders of all Jordan blocks for A are 1.

Recently, in [19], Equation (1) was discussed for the diagonalizable matrix A with two distinct eigenvalues ${\lambda }_1$ and ${\lambda }_2$, that is, $A=TJ{T}^{-1}$, where $J=diag({\lambda }_1{I}_n,{\lambda }_2{I}_{m-n})$. The main results of [19] are given as follows:

Theorem 1

([19]). Suppose that $A\in {\mathbf{C}}^{m\times m}$ is a diagonalizable matrix with two different eigenvalues ${\lambda }_1$ and ${\lambda }_2$$({\lambda }_1{\lambda }_2\ne 0)$, that is, $A=TJ{T}^{-1}$, in which T is nonsingular, and $J=diag({\lambda }_1{I}_n,{\lambda }_2{I}_{m-n})$. Then, all solutions of the Yang–Baxter-like matrix equation $AXA=XAX$ can be expressed as $X=T\left(\begin{array}{cc}{Y}_1& Y_2\\ Y_3& Y_4\end{array}\right)T^{-1}:$

(I) when ${\lambda }_1^2-{\lambda }_1{\lambda }_2+{\lambda }_2^2=0$, $Y_1$ and $Y_4$ have the forms

$$Y_1=P\left(\begin{array}{cc}{\lambda }_1{I}_{t_1}& 0\\ 0& 0\end{array}\right)P^{-1},Y_4=Q\left(\begin{array}{cc}{\lambda }_2{I}_{t_2}& 0\\ 0& 0\end{array}\right)Q^{-1},$$

P,Q are invertible matrices of appropriate size. $Y_2$ and $Y_3$ have the forms

$$Y_2=P\left(\begin{array}{cc}{Y}_2^{\left(1\right)}& 0\\ 0& 0\end{array}\right)Q^{-1},Y_3=Q\left(\begin{array}{cc}{Y}_3^{\left(1\right)}& 0\\ 0& 0\end{array}\right)P^{-1},$$

in which $0\le t_1\le n,0\le t_2\le m-n$, $Y_2^{\left(1\right)}$ is an arbitrary $t_1\times t_2$ matrix, and $Y_3^{\left(1\right)}=(I-{\left(Y_2^{\left(1\right)}\right)}^{†}{Y}_2^{\left(1\right)})W(I-Y_2^{\left(1\right)}{\left(Y_2^{\left(1\right)}\right)}^{†})$, W is an arbitrary $t_2\times t_1$ matrix.

(II) when ${\lambda }_1^2-{\lambda }_1{\lambda }_2+{\lambda }_2^2\ne 0$, for suitable invertible matrices P,Q with appropriate sizes and the given integer $r,0\le r\le min\{n,m-n\}$, $Y_1$, $Y_2$ and $Y_3$, $Y_4$ have the forms

$$Y_1=P\left(\begin{array}{ccc}\widetilde{{\lambda }_1}{I}_r& 0& 0\\ 0& {\lambda }_1{I}_{\mu }& 0\\ 0& 0& 0\end{array}\right)P^{-1},Y_4=Q\left(\begin{array}{ccc}\widetilde{{\lambda }_2}{I}_r& 0& 0\\ 0& {\lambda }_2{I}_{\nu }& 0\\ 0& 0& 0\end{array}\right)Q^{-1},$$

$$Y_2=P\left(\begin{array}{ccc}{Y}_2^{\left(1\right)}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)Q^{-1},Y_3=Q\left(\begin{array}{ccc}{Y}_4^{\left(1\right)}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)P^{-1},$$

in which $0\le \mu \le n-r,0\le \nu \le m-n-r$, $\widetilde{{\lambda }_1}=\frac{{\lambda }_2^2}{{\lambda }_2-{\lambda }_1},\widetilde{{\lambda }_2}=\frac{{\lambda }_1^2}{{\lambda }_1-{\lambda }_2}$.

Furthermore, $Y_2^{\left(1\right)}$ is an arbitrary $r\times r$ invertible matrix, and

$$Y_3^{\left(1\right)}=\frac{-{\lambda }_1{\lambda }_2({\lambda }_1^2-{\lambda }_1{\lambda }_2+{\lambda }_2^2)}{{({\lambda }_1-{\lambda }_2)}^2}{\left(Y_2^{\left(1\right)}\right)}^{-1}.$$

From the result above, it is readily seen that the structure of the solutions is complicated. When we continue to look for the solutions under the condition that A is diagonalizable and has three different nonzero eigenvalues, it can be seen that the method will appear to be invalid. However, for a general diagonalizable matrix A, by extending the idea and technique introduced in our previous work, we find that setting up the general expression of the solutions turns out to be possible.

Lemma 1.

Suppose that $ZJZ=JZJ$ and $J=diag({\lambda }_1{I}_{n_1},{\lambda }_2{I}_{n_2},\cdots ,{\lambda }_t{I}_{n_t}),$ with ${\lambda }_1{\lambda }_2\cdots {\lambda }_t\ne 0$, $n_1+n_2+\cdots +n_t=m$. Then, $\left(I\right)$ Z is diagonalizable; $\left(II\right)$ any nonzero eigenvalue σ of Z satisfies $\sigma \in \{{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_t\}$, and, if $\sigma ={\lambda }_i,i\in \{1,\cdots ,t\}$, the algebraic multiplicity of σ is no more than $n_i$.

Proof. 

Let $Rank\left(Z\right)=s,0\le s\le m$, and let $Z_{p_i},i=1,\cdots ,s$ be the linearly independent column vectors of Z. From

$$ZJ{Z}_{p_i}=ZJZ{e}_{p_i}=JZJ{e}_{p_i}={\lambda }_{q_i}J{Z}_{p_i},$$

where ${\lambda }_{q_i}$ belongs to the set $\{{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_t\}$. We see that $J{Z}_{p_i},i=1,\cdots ,s$ are the linearly independent eigenvectors of Z corresponding to the eigenvalues ${\lambda }_{q_i}$, $i=1,\cdots ,s.$ If we write the Jordan decomposition of Z as

$$\left(\begin{array}{ccccccc}{J}_1\left({\sigma }_1\right)& & & & & & \\ & J_2\left({\sigma }_2\right)& & & & & \\ & & \ddots & & & & \\ & & & J_j\left({\sigma }_j\right)& & & \\ & & & & J_{j+1}\left(0\right)& & \\ & & & & & \ddots & \\ & & & & & & J_k\left(0\right)\end{array}\right)$$

where ${\sigma }_i\ne 0,i=1,\cdots ,j$, then Z has at most j linearly independent eigenvectors corresponding to the nonzero eigenvalues, i.e., $s\le j$. On the other hand, $s=rank\left(Z\right)\ge j$. Therefore, there must be $s=j$, and every Jordan block of Z has an order of one. Meanwhile, $\{{\sigma }_i\ne 0,i=1,\cdots ,s\}=\{{\lambda }_{q_i},i=1,\cdots ,s\}$, and the algebraic multiplicity of ${\sigma }_i={\lambda }_{q_j}$ is no more than $n_{q_j}$.  □

Theorem 2.

Given $A\in {\mathbf{C}}^{l\times l},$ if $A=Tdiag(J_m,0)T^{-1}$ for some nonsingular matrices, T and $J_m=diag({\lambda }_1{I}_{n_1},{\lambda }_2{I}_{n_2},\cdots ,{\lambda }_t{I}_{n_t})$, then the general solution X to the Yang–Baxter-like matrix equation $XAX=AXA$ is given by

$$X=T\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)T^{-1},Z_4\in {\mathbf{C}}^{(l-m)\times (l-m)}isarbitrary,$$

where $Z_1=P\left(\begin{array}{cc}0& 0\\ 0& \mathrm{\Lambda }\end{array}\right)P^{-1}\in {\mathbf{C}}^{m\times m}$ for a nonsingular matrix $P,\mathrm{\Lambda }$ is a diagonal matrix. $Z_2=0$, and $Z_3=0$ when $Z_1$ is nonsingular. Otherwise, $Z_2=J_m^{-1}P\left(\begin{array}{c}Q\\ 0\end{array}\right)$, $Z_3=(W,0)P^{-1}{J}_m^{-1}$, and $Q\in C^{(m-s)\times (l-m)}$ is arbitrary. $W\in C^{(l-m)\times (m-s)}$ is any matrix satisfying $W\tilde{P}Q=0$ where $\tilde{P}$ is the $m-s$ order leading principle submatrix of $P^{-1}{J}_m^{-1}P$.

Proof. 

Suppose $Z,J\in {\mathbf{C}}^{l\times l}$ satisfies $JZJ=ZJZ$, in which $J=diag(J_m,0)$ ($m\le l$). Let Z be partitioned conformally with J as

$$Z=\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right),Z_1\in {\mathbf{C}}^{m\times m},Z_2\in {\mathbf{C}}^{m\times (l-m)}.$$

(2)

Then, comparing the two sides of the $2\times 2$ block matrix equation

$$\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)\left(\begin{array}{cc}{J}_m& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)=\left(\begin{array}{cc}{J}_m& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)\left(\begin{array}{cc}{J}_m& 0\\ 0& 0\end{array}\right)$$

yields the following system of four equations

$$Z_1{J}_m{Z}_1=J_m{Z}_1{J}_m,Z_1{J}_m{Z}_2=0,Z_3{J}_m{Z}_1=0,Z_3{J}_m{Z}_2=0.$$

(3)

Notice that $Z_4\in {\mathbf{C}}^{(l-m)\times (l-m)}$ can be any matrix.

Therefore, if $Z_1$, $Z_2$, and $Z_3$ are solved, then the solutions X to Equation (1) will be given by the equation $X=T\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)T^{-1},$ where $Z_1\in {\mathbf{C}}^{m\times m},$ and $Z_2\in {\mathbf{C}}^{m\times (l-m)}.$

Since $J_m=diag({\lambda }_1{I}_{n_1},{\lambda }_2{I}_{n_2},\cdots ,{\lambda }_t{I}_{n_t}),$ the matrix $Z_1$ in the equation $Z_1{J}_m{Z}_1=J_m{Z}_1{J}_m$ of (3) can be determined from Lemma 1.

Below, we discuss solving the matrices $Z_2$ and $Z_3$ in (3). Obviously, $Z_2=0$, and $Z_3=0$ when $Z_1$ is nonsingular. Otherwise, if $Z_1$ is singular, there exists a nonsingular matrix P such that $Z_1=P\left(\begin{array}{cc}0& 0\\ 0& \mathrm{\Lambda }\end{array}\right)P^{-1},\mathrm{\Lambda }\in C^{s\times s},0\le s<m,$ where $\mathrm{\Lambda }$ is a nonsingular diagonal matrix. By the second equation of (3), we obtain $Z_2=J_m^{-1}P\left(\begin{array}{c}Q\\ 0\end{array}\right)$, and $Q\in C^{(m-s)\times (l-m)}$ is arbitrary. By the third equation of (3), $Z_3=(W,0)P^{-1}{J}_m^{-1}$ where $W\in C^{(l-m)\times (m-s)}$ is arbitrary. However, because of the last equation of (3), we find $(W,0)P^{-1}{J}_m^{-1}P\left(\begin{array}{c}Q\\ 0\end{array}\right)=0$. If $\tilde{P}$ represents the $m-s$ order leading principle submatrix of $P^{-1}{J}_m^{-1}P$, we may have $W\tilde{P}Q=0.$ This means that for $\forall Q\in C^{(m-s)\times (l-m)}$, we can derive $W\in C^{(l-m)\times (m-s)}$ with it.  □

Next, we present one numerical example to illustrate our results.

Example 1.

Find the general solution X to the equation $AXA=XAX$, where $A=TJ{T}^{-1}$, and $J=diag(3,3,2,5,0,0)$.

By Theorem 2, $l=6,m=4$, $J_4=diag(3,3,2,5)$, $Z_1$ is the solution of $Z_1{J}_4{Z}_1=J_4{Z}_1{J}_4$. From Lemma 1, there exists a nonsingular matrix P such that $P^{-1}{Z}_{1}P=\mathrm{\Lambda }$, in which $\mathrm{\Lambda }$ is one of the following:

$$\begin{array}{}\left(\begin{array}{cccc}3& & & \\ & 3& & \\ & & 2& \\ & & & 5\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 3& & \\ & & 2& \\ & & & 5\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 3& & \\ & & 3& \\ & & & 2\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 3& & \\ & & 3& \\ & & & 5\end{array}\right),\\ \left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 3& \\ & & & 2\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 3& \\ & & & 5\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 2& \\ & & & 5\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 3& \\ & & & 3\end{array}\right),\\ \left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 0& \\ & & & 3\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 0& \\ & & & 2\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 0& \\ & & & 5\end{array}\right),\left(\begin{array}{cccc}0& & & \\ & 0& & \\ & & 0& \\ & & & 0\end{array}\right).\end{array}$$

when $\mathrm{\Lambda }=\left(\begin{array}{cccc}3& & & \\ & 3& & \\ & & 2& \\ & & & 5\end{array}\right)$, $Z_1$ is nonsingular, and hence, $Z_2=0,Z_3=0$. Otherwise, $Z_2=diag(\frac{1}{3},\frac{1}{3},\frac{1}{2},\frac{1}{5})P\left(\begin{array}{c}Q\\ 0\end{array}\right),$ $Z_3=(W,0)P^{-1}diag(\frac{1}{3},\frac{1}{3},\frac{1}{2},\frac{1}{5})$, where $Q\in C^{(4-s)\times 2},W\in C^{2\times (4-s)}$ $(s=Rank(\mathrm{\Lambda }\left)\right)$ are any matrices satisfying $W\tilde{P}Q=0$, and $\tilde{P}$ is the $4-s$ order leading principle submatrix of $P^{-1}diag(\frac{1}{3},\frac{1}{3},\frac{1}{2},\frac{1}{5})P$. Therefore, for $\forall Q\in C^{(4-s)\times 2}$, we can obtain $W\in C^{2\times (4-s)}$ by solving the homogeneous equation $W\tilde{P}Q=0$, and vice versa. We obtain all solutions to (1).

By direct calculation, it can be verified that, when $\mathrm{\Lambda }$ is singular,

$$\begin{array}{cc}\hfill XAX=& T\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)·T^{-1}AT·\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)T^{-1}\hfill \\ \hfill =& T\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)\left(\begin{array}{cc}{J}_4& \\ & 0\end{array}\right)\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)T^{-1}\hfill \\ \hfill =& T\left(\begin{array}{cc}{Z}_1& J_4^{-1}P\left(\begin{array}{c}Q\\ 0\end{array}\right)\\ (W,0)P^{-1}{J}_4^{-1}& Z_4\end{array}\right)\left(\begin{array}{cc}{J}_4& \\ & 0\end{array}\right)\left(\begin{array}{cc}{Z}_1& J_4^{-1}P\left(\begin{array}{c}Q\\ 0\end{array}\right)\\ (W,0)P^{-1}{J}_4^{-1}& Z_4\end{array}\right)T^{-1}\hfill \\ \hfill =& T\left(\begin{array}{cc}{Z}_1{J}_4& 0\\ (W,0)P^{-1}& 0\end{array}\right)\left(\begin{array}{cc}{Z}_1& J_4^{-1}P\left(\begin{array}{c}Q\\ 0\end{array}\right)\\ (W,0)P^{-1}{J}_4^{-1}& Z_4\end{array}\right)T^{-1}\hfill \end{array}$$

$$\begin{array}{cc}\hfill =& T\left(\begin{array}{cc}{Z}_1{J}_4{Z}_1& P\mathrm{\Lambda }\left(\begin{array}{c}Q\\ 0\end{array}\right)\\ (W,0)\mathrm{\Lambda }{P}^{-1}& (W,0)P^{-1}{J}_4^{-1}P\left(\begin{array}{c}Q\\ 0\end{array}\right)\end{array}\right)T^{-1}\hfill \\ \hfill =& T\left(\begin{array}{cc}{Z}_1{J}_4{Z}_1& 0\\ 0& 0\end{array}\right)T^{-1}=T\left(\begin{array}{cc}{J}_4{Z}_1{J}_4& 0\\ 0& 0\end{array}\right)T^{-1}\hfill \\ \hfill =& T\left(\begin{array}{cc}{J}_4& \\ & 0\end{array}\right)T^{-1}·T\left(\begin{array}{cc}{Z}_1& Z_2\\ Z_3& Z_4\end{array}\right)T^{-1}·T\left(\begin{array}{cc}{J}_4& \\ & 0\end{array}\right)T^{-1}=AXA\hfill \end{array}$$

When $\mathrm{\Lambda }$ is nonsingular, we also have $XAX=AXA$.

3. Conclusions

In this paper, we have discussed finding all of the solutions to the Yang–Baxter-like matrix equation $AXA=XAX$ when the orders of all Jordan blocks for A are one. Research on the non-commuting solutions is interesting, but it seems to be hard to find. In the future, we hope to be able to attack the non-commuting solutions to $AXA=XAX$ when the order of Jordan blocks of A is more general.