111AMS Classification 2020. Primary: 15A24, 16T25

# Solutions to a system of Yang-Baxter matrix equations

Himadri Mukherjee  Askar Ali M  BITS Pilani K. K. Birla Goa Campus, Goa, India
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this article, we take a system, $XAX=BXB$, $XBX=AXA$, of Yang-Baxter type
matrix equations that “generalizes” the matrix Yang-Baxter equation. We
completely characterize the case when $A,B$ are orthogonal idempotents.

###### Key words and phrases:

Yang-Baxter equation, Matrix equation

##  1\. Introduction

For a given pair of square matrices, $A,B$,

(1) $XAX=BXB\text{ and }XBX=AXA$

will be called as a system of Yang-Baxter matrix equations. Yang-Baxter
equations are extremely useful equations that have found multiple usages in
Physics, computer science, and many areas of pure mathematics. This equation
arises in the context of physics as, in knot theory, if we take the following
braids, which are two among the generators of a braid group, and call the
first one $A$ and the second one $B$ as in Figure 1, we see that $ABA=BAB$ as
braids in Figure 2.

$A=$ $B=$

Figure 1.

$BAB=$ $ABA=$

Figure 2.

This is not a coincidence that the braids satisfy the equation $AXA=XAX$; it
was observed by Artin [3] (especially in [2]) in his development of the famous
and pioneering theory of braid groups. As a result of the above relations
among the generators of a braid group, it makes sense to study the matrix
equation that we call YBE.

In the theory of quantum mechanics, the same equations arise in the guise of
“star-triangle relations” [6]. The mathematical abstraction of physics
emphasizes itself in the theory of quantum groups. Let $V$ be a vector space,
and let $R$ be an operator on $V\otimes V$ depending on parameters $t,q$. Let
us denote $R_{12}$ to be the operator on $V\otimes V\otimes V$ as
$R_{12}=R\otimes I_{v}$, where $R$ acts on the first two coordinates and
$I_{V}$ is the identity operator of $V$. Similarly, we define $R_{13}$ and
$R_{23}$ to be the operators, $R_{ij}$, acting on the $i$-th and the $j$-th
coordinates while acting as identity on the remaining coordinate. One has
[15],

(2) $R_{12}(x)R_{13}(xy)R_{23}(y)=R_{23}(y)R_{13}(xy)R_{12}(x).$

Let us define $P_{ij}$ to be the permutation matrix of appropriate size that
permutes the $i,j$-th coordinates and let us denote
$\tilde{R_{ij}}=P_{ij}R_{ij}$ with the parameters suppressed in the notation.
Once we substitute in the above equation 2, we obtain (with appropriate
parameters),

$\tilde{R_{12}}\tilde{R_{23}}\tilde{R_{12}}=\tilde{R_{23}}\tilde{R_{12}}\tilde{R_{23}}$

The above equation, without the parameters, is called the parameter-free
classical Yang-Baxter equation. This also gives ample reason to study the
matrix YBE.

While working on the paper [13], dealing with the Yang-Baxter equation, namely
$AXA=XAX$, the authors observed the usefulness of the results that were proved
for the coefficient matrix ( ie. $A$) for the variable $X$ owing to the
symmetry of the equation. It was observed that, from the pure matrix equation
perspective, it is much better if we deal with the equation $XYX=YXY$, where
both $X,Y$ are treated as variables. This equation has the advantage of being
homogeneous; thus, many results can be simplified for such an equation, and
when specialized, these in turn give the results for the original YBE. As a
result, we will call the equation

$XYX=YXY$

as the homogeneous matrix Yang-Baxter equation. Let $V$ be the set of
solutions of it over the matrix algebra $M_{n}(\mathbb{K})$ ( for a suitable
field $\mathbb{K}$). We observe the following for the space $V(X,Y)$. Owing to
the symmetry of the equations, this has a $\mathbb{Z}_{2}$ action that flips
the coordinates. As a result, instead of working with the whole $V$, it is
much better to work with the $\mathbb{Z}_{2}$ quotient of it; let us denote
the quotient by $\mathfrak{X}$. It will be very interesting to understand the
geometry of this object using various tools such as commutative algebra and
algebraic geometry. We, taking advantage of the homogeneity of the equation,
can look at the projective zero set associated to it, namely:
$\\{(X,Y)\in\mathbb{P}^{2n^{2}-1}\,\mid\,XYX=YXY\\}$. Let us call this variety
as $\mathfrak{Y}$. It is easily observed that this is not an irreducible
subset of the projective space, owing to the trivial components
$\\{(X,O)|X\neq 0\\}$ and $\\{(0,X)|X\neq 0\\}$, where $0$ represents the zero
matrix of the appropriate dimension. So instead, we can look at the projective
algebraic set
$\\{(X,Y)\in\mathbb{P}^{n^{2}-1}\times\mathbb{P}^{n^{2}-1}\,\mid\,XYX=YXY\\}$.
This gets rid of the problems arising out of the trivial solutions of the YBE,
and further if we take a quotient by the action of $\mathbb{Z}_{2}$, denoted
as $\mathfrak{Z}$, we have a much larger chance of asking the question, when
is this an irreducible projective variety?. This question is very interesting,
and the authors believe it has a good chance of getting interesting answers.
The reason for believing so is that in the paper [12], the authors gave the
connected components of the space of solutions in some special cases, which
turn out to have linear structures.

If the solution of the YBE for the coefficient matrix $A$ is denoted by
$V_{A}$, we can make the following observation.

$V\simeq V_{X}\times V_{Y}$

It was also observed that towards obtaining good results in the YBE main
hindrance was the trivial solution $X=A$; this was also observed in the other
articles as well [12]. This trivial solution gives rise to the lack of
simplicity of the solutions of the equation. While the homogeneous version of
the equation does not solve this problem, if we allow a different matrix $Z$
and twist the equation as $XYX=ZXZ$, we have removed the trivial solution that
gives rise to cluttering the attempts at solving the equation. But now we have
lost the beautiful symmetry of the equation, so we have to look at its twin
pair $XZX=YXY$. This system of Yang-Baxter type equation will be called a
system of YBE, the authors understand that it is not really a system of YBE as
the authors in [13] showed that a true system is just equivalent to a single
Yang-Baxter equation, so we have decided to give the honor of being a system
of YBE to this system of equations.

With similar thoughts of generalizing the equation to obtain more freedom to
exploit, we have decided to study the equation,

$AXA=XBX.$

This equation when specialized for $A=B$ gives back the original YBE and has
the freedom to free it of the difficulties regarding the trivial solutions. In
this paper, we will show that, even though we are considering the system, it
is in reality reduces to just one equation. This, to our thoughts, is a very
interesting result, as the symmetry of the original YBE is broken yet it
contains almost all the information regarding a solution.

One first input that we obtain from freeing the equation of the symmetry is
the possibilities of investigating the equation when the pair $A,B$ are not
only idempotents but also orthogonal, a condition that has no non-trivial
parallel in the single Yang-Baxter equation. Also, in the single YBE
situation, it was observed that the nilpotent coefficient matrix situation is
the most interesting and important. A generalization of this for the system is
the case, $A,B$ are zero-divisors (i.e., $AB=BA=0$). So, it makes ample sense
to investigate the system of YBE when the coefficients satisfy

$AB=BA=0$

.

YBEs are extremely useful fundamental equations in mathematics and physics.
This equation has shown up in unexpected form in many parts of mathematics and
mathematical physics. With the advent of quantum computers and quantum
computing in general, the Yang Baxter equations have come into new focus as
they are used to design the quantum gates for a quantum computer. There are
numerous new articles exploring this area of the Yang-Baxter equations,
namely, [14], [11] and [10] in the areas of neural networks.

Matrix varieties are one of the most important fields in the intersection of
linear algebra and algebraic geometry. The most prominent of these matrix
varieties likely are the determinantal varieties, i.e., the algebraic
varieties of matrices with rank less than a given number. In algebraic
geometry, these varieties turn out to be of particular importance. Other than
the determinantal varieties, the variety of commuting matrices plays an
important role in both linear algebra and algebraic geometry. It has been
proved that the variety of commuting nilpotent matrices are irreducible [5,
4]. A number of geometric papers concerning the YBE has come up recently,
namely the papers [8] and [12]. Also, a recent paper investigated the case for
the coefficient matrix $A$ to be a permutation matrix and foud the solutions
in the convex hull of the permutation matrices (ie the stochastic matrices)
[9]. Such geometric understanding of the YBE warrants use of many geometric
and topological tools, such as the fixed point theory. An algebraic geometric
approach to the same has not been taken so far to the best of the authors’
knowledge.

##  2\. Preliminaries

In this section, we recall some of the established results for the Yang-Baxter
type matrix equation $AXA\ =\ XAX$, which will help the reader to draw a
general picture about the solution set for the matrix equation and to
understand the following sections as well.

######  Lemma 2.1.

[13] Let $Sol_{A}$ denote the set of all solutions $X$ to the YBE $AXA=XAX$,
with the coefficient matrix $A$. Then for any $g\in GL_{n}(K)$,
$gSol_{A}g^{-1}=Sol_{gAg^{-1}}$.

######  Theorem 2.2.

[13] If $A=\lambda I+B$ is a Jordan block, then we have the following

  1. (1)

If $\lambda\neq 0$ and $det(X)\neq 0$ then $X\simeq A$.

  2. (2)

If $\lambda\neq 0$ and $det(X)=0$ then $X=0$.

  3. (3)

If $\lambda=0$, then the YBE $AXA=XAX$ do not have any invertible solution X.

######  Theorem 2.3.

[13] Let $A=\begin{pmatrix}A_{1}&0\\\ 0&A_{2}\end{pmatrix}$, with
$A_{i}=\lambda_{i}I_{n_{i}}+B_{i}$ being the Jordan blocks, further, let us
assume $\lambda_{1},\lambda_{2}\neq 0$ and 0 is an eigenvalue of $X$ and
$A_{1},A_{2}$ have the same dimension, then one of the following is true.

  1. (1)

$X=0$

  2. (2)

$X=\begin{pmatrix}0&ZS^{-1}A\\\ 0&X_{2}\end{pmatrix}$ or
$\begin{pmatrix}0&0\\\ 0&X_{2}\end{pmatrix}$, where $X_{2}\in
Sol_{A_{2}}$,$Z=a_{0}I+a_{1}A+...+a_{k}A^{k}$ for some scalars $a_{i}$s and
integer k ($A=A_{1}=A_{2}\mbox{ in that case}$), and for some $S\in GL_{n}(K)$

  3. (3)

$X=\begin{pmatrix}X_{1}&0\\\ ZS^{-1}A&0\end{pmatrix}$ or
$\begin{pmatrix}X_{1}&0\\\ 0&0\end{pmatrix}$, where $X_{1}\in
Sol_{A_{1}}$,$Z=a_{0}I+a_{1}A+...+a_{k}A^{k}$ for some scalars $a_{i}$s and
integer k ($A=A_{1}=A_{2}\mbox{ in that case }$), and for some $S\in
GL_{n}(K)$

######  Lemma 2.4.

[13] If $A$ is invertible and for each $\lambda\in\sigma(A)$ with the
eigenspace $E_{\lambda}$ one dimensional, and X be a solution to the Yang-
Baxter like matrix equation $AXA\ =\ XAX$, such that $XE_{\lambda}\neq 0$,
then $\sigma(A)\subset\sigma(X)$.

######  Lemma 2.5.

[13] When $A=\begin{pmatrix}A_{1}&0&...&0\\\ 0&A_{2}&...&0\\\ .\\\ .\\\ .\\\
0&0&...&A_{m}\end{pmatrix}$ block diagonal matrix with $A_{i}$ being square
matrix of order $n_{i}$, $X=\begin{pmatrix}X_{1}&0&...&0\\\ 0&X_{2}&...&0\\\
.\\\ .\\\ .\\\ 0&0&...&X_{m}\end{pmatrix}$, is a solution to the YBE with
coefficient matrix $A$, where $X_{i}$ is a solution for the YBE
$A_{i}YA_{i}=YA_{i}Y$.

######  Definition 2.1.

[1] Let $T^{n}=\
\\{x^{a}=x_{1}^{a_{1}}x_{2}^{a_{2}}...x_{n}^{a_{n}}/a_{i}\in\mathbb{N}\\}\
\subset\mathbb{K}[x_{1},...,x_{n}]$. We define lexicographical order on
$T^{n}$ with $x_{1}\ >\ x_{2}\ >...>\ x_{n}$ as follows: For

$a\ =\ (a_{1},...,a_{n}),\ b\ =\ (b_{1},...,b_{n})\in\ \mathbb{N}^{n}$ (3)
$x^{a}<x^{b}\Leftrightarrow\begin{cases}\mbox{the first coordinates from the
left }a_{i}\mbox{ and }b_{i}\mbox{ in }a\mbox{ and }b,\\\ \mbox{ which are
different, satisfy }a_{i}<b_{i}.\end{cases}$

So, in the case of two variables $x_{1}$ and $x_{2}$, we have

$1<x_{2}<x_{2}^{2}<x_{2}^{3}<...<x_{1}<x_{2}x_{1}<x_{2}^{2}x_{1}<...<x_{1}^{2}<...$

.

######  Definition 2.2.

[1] A set of non-zero polynomials $G=\\{g_{1},g_{2},...,g_{t}\\}$ contained in
an ideal $I\subset\mathbb{K}[x_{1},...,x_{n}]$, is called a Gr$\ddot{o}$bner
basis for I if and only if for all $f\in I$ such that $f\neq 0$, there exists
$i\in\\{1,...,t\\}$, such that $lp(g_{i})$ divides $lp(f)$, where $lp(f)$ is
the leading power product of $f$.

Using the term order as defined above one proceeds to perform a “multivariable
division” on the set of polynomial generators of the given ideal $I$. At the
same time we pick up any “remainder” that is obtained through this process. A
result of this process is a collection of generators that divides any element
$f$ of the ideal without leaving any nontrivial “remainder”. This set of
generators is called a Gr$\ddot{o}$bner base of the ideal. This
“multivariable” generalization of division algorithm has many a number of
applications all throughout mathematics. One of the many applications of
Gr$\ddot{o}$bner basis is in the field of elimination theory, in which one
“eliminates” a number of variables from a given set of equations. Once the
variables are eliminated, the resulting equations with fewer variables may be
solved easily in some situations.

######  Theorem 2.6.

[7] Let $\mathbb{k}$ be a field, and let I be an ideal in
$\mathbb{k}[x_{1},\ldots,x_{n}]$. Let G be a Gröbner basis for I with respect
to the lexicographic order with
$x_{1}>x_{2}>\textperiodcentered\textperiodcentered\textperiodcentered>x_{n}$.
Then for any $1\leq t\leq n$, $G\ \cap\ \mathbb{k}[x_{t},\ldots,x_{n}]$ is a
Gr öbner basis for the ideal $I\ \cap\ \mathbb{k}[x_{t},\ldots,x_{n}]$.

The above theorem is a key tool in finding solutions in many cases in this
paper. We have used Sage-Math for computing Gröbner basis. Whenever the
Gr$\ddot{o}$bner basis is found in fine size, authors have included that in
this paper.

##  3\. System of Yang-Baxter Type Matrix Equations

We can rewrite (1) in the block form as follows.

$\begin{pmatrix}B&0\\\ 0&A\end{pmatrix}\begin{pmatrix}X&0\\\
0&X\end{pmatrix}\begin{pmatrix}B&0\\\ 0&A\end{pmatrix}=\begin{pmatrix}X&0\\\
0&X\end{pmatrix}\begin{pmatrix}A&0\\\ 0&B\end{pmatrix}\begin{pmatrix}X&0\\\
0&X\end{pmatrix}.$

This gives, solutions for the initial system of equations (1), are some
special solutions for the equation $A^{\prime}XA^{\prime}=XB^{\prime}X$, where
$A^{\prime}=\begin{pmatrix}B&0\\\ 0&A\end{pmatrix}$, and
$B^{\prime}=\begin{pmatrix}A&0\\\ 0&B\end{pmatrix}$. This leads to the
following result.

######  Lemma 3.1.

If the system of equation $XAX=BXB\text{ and }XBX=AXA$ has a non-trivial
solution, then the equation

(4) $A^{\prime}XA^{\prime}=XB^{\prime}X,$

,where $A^{\prime}=\begin{pmatrix}B&0\\\ 0&A\end{pmatrix}$, and
$B^{\prime}=\begin{pmatrix}A&0\\\ 0&B\end{pmatrix}$ , also has a non-trivial
solution.

######  Corollary 3.1.1.

If $A^{\prime}XA^{\prime}=XB^{\prime}X$ has unique solution, then so is for
(1).

######  Theorem 3.2.

The system of equation $AXA=XBX$ and $BXB=XAX$ have a solution if and only if
the equation
$A^{\prime}I^{\prime}YI^{\prime}A^{\prime}-YI^{\prime}A^{\prime}YI^{\prime}=0$
has a solution, where $I^{\prime}=\begin{pmatrix}0&I\\\ I&0\end{pmatrix}$, and
$I$is identity matrix of order n, $Y\ =\ \begin{pmatrix}X&0\\\
0&X\end{pmatrix}$. Further, any non-trivial solution X of (1), should satisfy
$\sigma(A^{\prime}I^{\prime})\cap\sigma(YI^{\prime})\neq\Phi$.

###### Proof.

Let $I^{\prime}=\begin{pmatrix}0&I\\\ I&0\end{pmatrix}$ be the block matrix of
order $2n\times 2n$. Then,

$I^{\prime}B^{\prime}I^{\prime}=\begin{pmatrix}0&I\\\
I&0\end{pmatrix}\begin{pmatrix}A&0\\\ 0&B\end{pmatrix}\begin{pmatrix}0&I\\\
I&0\end{pmatrix}=\begin{pmatrix}B&0\\\ 0&A\end{pmatrix}=A^{\prime}.$

Also, by letting $Y=\begin{pmatrix}X&0\\\ 0&X\end{pmatrix}$, we have

$I^{\prime}YI^{\prime}=\begin{pmatrix}0&I\\\
I&0\end{pmatrix}\begin{pmatrix}X&0\\\ 0&X\end{pmatrix}\begin{pmatrix}0&I\\\
I&0\end{pmatrix}=\begin{pmatrix}X&0\\\ 0&X\end{pmatrix}=Y.$

It can also be noted that $J$ and $Y$ commute. i.e- $YI^{\prime}=I^{\prime}Y$.
Now let us rewrite our equations with the above notations as follows. We have,

(5) $\displaystyle A^{\prime}YA^{\prime}$ $\displaystyle=YB^{\prime}Y$
$\displaystyle\Leftrightarrow A^{\prime}I^{\prime}YI^{\prime}A^{\prime}$
$\displaystyle-YI^{\prime}A^{\prime}I^{\prime}Y=0$
$\displaystyle\Leftrightarrow A^{\prime}I^{\prime}(YI^{\prime}A^{\prime})$
$\displaystyle-(YI^{\prime}A^{\prime})YI^{\prime}=0.$

Now, by taking $Z\ =\ YI^{\prime}A^{\prime}$, the above equation becomes
$A^{\prime}I^{\prime}Z-ZYI^{\prime}=0$, which is similar to Sylvester’s
equation. This gives, the above equation has a non-trivial solution if the
spectrum of $A^{\prime}I^{\prime}$ and $YI^{\prime}$ have a non-empty
intersection. ∎

######  Lemma 3.3.

For a given $A,B$, $X_{0}$ is a solution for (1), and it is a solution to the
YBE $CXC=XCX$, where $C=A+B$, then $AX_{0}B+BX_{0}A=0$.

###### Proof.

If $X_{0}$ satisfy (1), and $XCX=CXC$, where $C=A+B$, then we have
$X_{0}AX_{0}=BX_{0}B$ and $X_{0}BX_{0}=AX_{0}A$. Adding both these equations
will give,

(6) $\displaystyle X_{0}(A+B)X_{0}=(A+B)X_{0}(A+B)-(AX_{0}B+BX_{0}A).$

Since we have $XCX=CXC$, this implies $AX_{0}B+BX_{0}A=\ 0$. ∎

Note that the equation $AX_{0}B+BX_{0}A=0$, in above lemma is similar to
generalized Sylvester’s equation. Which will lead us to the following
necessary condition to be satisfied for the solution of our system.

######  Lemma 3.4.

The equations in (1) and $CXC=\ XCX$, where $C\ =\ A+B$, have a simultaneous
unique solution if the pencils $A+\lambda B$ and $A-\lambda B$ are regular,
and their spectrum has an empty intersection.

The proof is straightforward from the condition of generalized Sylvester’s
equation’s unique solution case.

######  Lemma 3.5.

Let one of the coefficient matrices be non-singular (without loss of
generality, let us assume A is non-singular), and the pencils $A+\lambda B$
and $A-\lambda B$ are regular. Then, for (1) and $CXC=\ XCX$, where $C\ =\
A+B$, have a simultaneous non-trivial solution, it is necessary that $BA^{-1}$
has a pair of eigenvalues $(\mu_{1},\mu_{2})$, such that their sum is zero.

###### Proof.

From the above lemma 3.4, if (1) has a non-trivial solution, then
$\sigma(A+\lambda B)\cap\sigma(A-\lambda B)$ is non-empty. Let’s say
$\lambda_{0}\in\sigma(A+\lambda B)\cap\sigma(A-\lambda B)$. Then,

$\text{det}(A+\lambda_{0}B)=0\text{ and }\text{det}(A-\lambda_{0}B)=0$
$\Rightarrow\text{det}(I+\lambda_{0}BA^{-1})=0\text{ and
}\text{det}(I-\lambda_{0}BA^{-1})=0$

Which means, there exist a $\mu_{1},\mu_{2}\in\sigma(BA^{-1})$, such that
$1+\lambda_{0}\mu_{1}=0$ and $1-\lambda_{0}\mu_{2}=0$. Now, since $A$ is
invertible, $\lambda=0$ can not be an eigenvalue for the pencils $A+\lambda B$
and $A-\lambda B$. Which, along with the last pair of equation will imply
$\mu_{1}+\mu_{2}=0$. ∎

Similar conditions can be drawn when $B$ is invertible, as the equation
$AX_{0}B+BX_{0}A=0$ has symmetry with respect to $A$ and $B$. In that case,
corresponding pencils will be $B+\lambda A$ and $B-\lambda A$. Also, note that
when both $A,B$ are singular, the necessary condition for the existence of a
non-trivial solution is trivially satisfied as $0\in\sigma(A+\lambda
B)\cap\sigma(A-\lambda B)$.

######  Lemma 3.6.

If X is a commutative solution to (1) and the YBE $CXC\ =\ XCX$ with $C\
=A+B$, and both A and B are invertible, also A and B commute with each other,
then $X=0$.

###### Proof.

If X commute with $A,B$, and $A$ and $B$ are invertible and commute with each
other, then by the above lemma, we have

(7) $\displaystyle AXB+BXA$ $\displaystyle=0$ (8) $\displaystyle 2ABX$
$\displaystyle=0$

Which gives, $X=0$. ∎

######  Lemma 3.7.

If both $A,B$ are non-singular, then (1) has a non-singular solution, only if
$det(A^{3})=det(B^{3})$.

###### Proof.

We have,

$det(AXA)=det(XBX)\text{ and }det(BXB)=det(XAX)$ $\Rightarrow
det(A^{2})det(X)=det(X^{2})det(B)$ $\text{ and
}det(B^{2})det(X)=det(X^{2})det(A).$

Now, if $X$ is a non-singular solution, then the above equations give,

$det(X)=\frac{det(A^{2})}{det(B)}=\frac{det(B^{2})}{det(A)}.$

Which gives, $det(A^{3})=det(B^{3})$. ∎

######  Corollary 3.7.1.

With the conditions in the above lemma, if A and B have only real eigenvalues,
then det(A) = det(B).

######  Lemma 3.8.

If one of the coefficient matrices is singular, and the other is non-singular,
then (1) can not have a non-singular solution.

###### Proof.

Without loss of generality, assume that $A$ is singular and $B$ is non-
singular. Let $(0\neq)v\in Ker(A)$. Then, $AXAv=0$, gives $XBXv=0$. Now, if
$Xv=0$, then $v\in Ker(X)$ and X is singular. If $Xv\neq 0$, then as B is non-
singular $BXv\neq 0$. That means, $BXv\in Ker(X)$, and X is singular. ∎

######  Lemma 3.9.

Let A be singular, and B be a non-singular matrix. Then, if $u\in Ker(A)$,
then either $u\in Ker(X)$ or $BXu\in Ker(X)$.

###### Proof.

If $u\in Ker(A)$, then we have $AXAu=0$, which gives $BXBu=0$. Then if
$u\notin Ker(X)$, then $BXu(\neq 0)\in Ker(X)$, as B is invertible. ∎

######  Lemma 3.10.

If A and B are invertible, and $v\in Ker(X)$, then X kills $Bv\text{ and }Av$.

###### Proof.

We have $XAXv=0$, gives $BXBv=0$. Since B is invertible, this gives $XBv=0$.
Similarly, we can show that $XAv=0$. ∎

######  Lemma 3.11.

For matrices C, D and Y, $YC=DY$ implies $YC^{n}=D^{n}Y$.

###### Proof.

Let $YC=DY$, then

$DYC=D^{2}Y$ $\Rightarrow YC^{2}=D^{2}Y.$

Then by induction, we have, $YC^{n}=D^{n}Y$. ∎

######  Theorem 3.12.

Equations in (1) implies, $X^{2n}BX=BX(BA)^{n}$. Further,
$\Phi_{BA}(X^{2})BX=0$, where $\Phi_{BA}$ is characteristic polynomial of
$BA$.

###### Proof.

We have $XAXA=BXBA$, which implies $X^{2}BX=BX(BA)$. Then by previous lemma,
by letting $Y=BX$, we have $X^{2n}BX=BX(BA)^{n}$.
This implies that, for any polynomial $f$, $f(X^{2})BX=BXf(BA)$. Then, for
characteristic polynomial $\Phi_{BA}$, of $BA$, the later equation can be
written as $\Phi_{BA}(X^{2})BX=0$. ∎

######  Corollary 3.12.1.

$AXA\ =\ XBX$ and $BXB\ =\ XAX$, implies $e^{X^{2}}BX\ =\ BXe^{BA}$,
(equivalently, $e^{X^{2}}AX\ =\ AXe^{AB}$).

######  Corollary 3.12.2.

Let A or B be invertible. Then for any solution X of (1), $e^{AB}$ or $e^{BA}$
preserve Ker(X).

###### Proof.

Let B be an invertible matrix, and X be a solution to (1). Now, if $v\in
Ker(X)$, then we have,

$Xv\ =\ 0$ $\Rightarrow e^{X^{2}}BXv\ =\ 0$ $\Rightarrow BXe^{BA}v\ =\ 0$

As B is invertible, the above equation implies, $Xe^{BA}v\ =\ 0$. But
irrespective of A or B being invertible, we have $e^{BA}$ is an invertible
matrix, hence $e^{BA}v\ \neq 0$. That is, $e^{BA}v\in Ker(X)$. ∎

######  Lemma 3.13.

If one of A,B is invertible and let $\lambda\in\sigma(BA)$ and
$\lambda\notin\sigma(X^{2})$. Then X annihilate $E_{\lambda}(BA)$, the eigen
space of BA corresponding to eigenvalue $\lambda$.

###### Proof.

With out loss of generality, let us assume B is invertible. Also, let $v(\neq
0)\ \in\ E_{\lambda}(BA)$. Then we have,

$\displaystyle BAv\ $ $\displaystyle=\ \lambda v$ $\displaystyle BXBAv\ $
$\displaystyle=\ \lambda BXv$ $\displaystyle XAXAv\ $ $\displaystyle=\ \lambda
BXv$ $\displaystyle X^{2}BXv\ $ $\displaystyle=\ \lambda BXv$

Now, since $\lambda\notin\sigma(X^{2})$, $BXv$ must be zero. But, as B is
invertible, this gives $Xv\ =\ 0$. i.e. $XE_{\lambda}(BA)\ =\ 0$. ∎

######  Lemma 3.14.

If one of A,B is invertible and let $\lambda\in\sigma(BA)$ and
$\lambda\notin\sigma(X^{2})$. Then X annihilates the generalized eigenspace of
BA corresponding to the eigenvalue $\lambda$, $P_{\lambda}(BA).$

###### Proof.

Let B be invertible, and $\\{v_{1},v_{2},...,v_{r}\\}$ be the canonical basis
for $P_{\lambda}$. Then we have,

$BAv_{1}\ =\ \lambda v_{1}$ $BAv_{i}\ =\ v_{i-1}+\lambda v_{i},\text{ for
}i=2,3,..,r.$

But, from above Lemma, we have $BAv_{1}\ =\ 0$. Now,

$XBAv_{2}\ =\ Xv_{1}+\lambda Xv_{2}$ $\Rightarrow XBAv_{2}\ =\ \lambda Xv_{2}$
$\Rightarrow BXBAv_{2}\ =\ \lambda BXv_{2}$ $\Rightarrow X^{2}BXv_{2}\ =\
\lambda BXv_{2}$

We have $\lambda\notin\sigma(X^{2})$, gives $Xv_{2}\ =\ 0$. On continuing this
process through induction, we have $XP_{\lambda}(BA)\ =\ 0$. ∎

######  Lemma 3.15.

If both A and B are invertible, then for any non-singular solution X of (1),
$\sigma(X^{2})=\sigma(BA)$, (equivalently, $\sigma(X^{2})=\sigma(AB)$).

###### Proof.

Let A and B be invertible and X be an invertible solution for (1), then for
any $\lambda\in\sigma(X^{2})$, we have a non-zero vector $v\in\mathbb{K}^{n}$,
such that,

$X^{2}v\ =\ \lambda v$ $\Rightarrow XAX^{2}v\ =\ \lambda XAv$ $\Rightarrow
BXBXv\ =\ \lambda XAv$ $\Rightarrow BAXAv\ =\ \lambda XAv$

Since A and X are invertible, we have $XAv\neq 0$, gives
$\lambda\in\sigma(BA)$. Conversely, if $\lambda\in\sigma(BA)$, then for some
non-zero vector $u\in\mathbb{K}^{n}$, we have,

$BAu\ =\ \lambda u$ $\Rightarrow BXBAu\ =\ \lambda BXu$ $\Rightarrow XAXAu\ =\
\lambda BXu$ $\Rightarrow X^{2}BXu\ =\ \lambda BXu$

Now, as $X$ and $B$ are non-singular, we get $BXu\neq 0$. This gives
$\lambda\in\sigma(X^{2})$. ∎

######  Theorem 3.16.

Let A, and B be invertible matrices, and X be an invertible solution to (1),
then $X^{2}$ is similar to $BA$,(equivalently to $AB$ also).

###### Proof.

We have,

$BXB\ =\ XAX$ $\Rightarrow B\ =\ X^{-1}B^{-1}XAX$ $\Rightarrow BA\ =\
X^{-1}B^{-1}XAXA$ $\Rightarrow BA\ =\ X^{-1}B^{-1}X^{2}BX$

∎

######  Lemma 3.17.

If A and B are simultaneously diagonalizable, then (1) can be rewritten as
follows:

$YJ_{A}Y=J_{B}YJ_{B}\text{ and }YJ_{B}Y=J_{A}YJ_{A},$

where $J_{A},J_{B}$ denote Jordan canonical form of A and B respectively.

###### Proof.

Let $S\in Gl_{n}(K)$, such that $SAS^{-1}=J_{A}$ and $SBS^{-1}=J_{B}$. Then,
(1) implies

(9) $\displaystyle XS^{-1}SAS^{-1}SX$ $\displaystyle=BS^{-1}SXS^{-1}SB\text{
and }$ (10) $\displaystyle XS^{-1}SBS^{-1}SX$
$\displaystyle=AS^{-1}SXS^{-1}SA.$

Then by taking $Y=SXS^{-1}$, above equation will become,

$YJ_{A}Y=J_{B}YJ_{B}\text{ and }YJ_{B}Y=J_{A}YJ_{A},$

∎

######  Corollary 3.17.1.

Let A and B are simultaneously diagonalizable and are similar, then (1) will
become a Yang-Baxter equation: $YJ_{A}Y=J_{A}YJ_{A}$.

###### Proof.

When A and B are similar and are simultaneously diagonalizable, we have
$J_{A}=J_{B}$. Then rest is immediately followed by the above lemma. ∎

######  Lemma 3.18.

Let X be a solution to (1), then for any $\alpha\in\mathbb{K}$, $X+\alpha C$
is a solution to (1) for some $C\in M_{n}(\mathbb{K})$, if $AC=CA=0$ and
$BC=CB=0$.

###### Proof.

We have,

$A(X+\alpha C)A=(X+\alpha C)B(X+\alpha C)$ and  $B(X+\alpha C)B=(X+\alpha
C)A(X+\alpha C),$

is true, if $AC=CA=0$ and $BC=CB=0$. ∎

######  Theorem 3.19.

Let $X_{1},X_{2}$ be solution to (1), then $X_{1}+X_{2}$ is a solution to (1),
if and only if $X_{1}AX_{2}+X_{2}AX_{1}\ =\ 0$ and $X_{1}BX_{2}+X_{2}BX_{1}\
=\ 0$.

###### Proof.

$A(X_{1}+X_{2})A=(X_{1}+X_{2})B(X_{1}+X_{2})$ and
$B(X_{1}+X_{2})B=(X_{1}+X_{2})A(X_{1}+X_{2}),$ $\Leftrightarrow
AX_{1}A+AX_{2}A=X_{1}BX_{1}+X_{1}BX_{2}+X_{2}BX_{1}+X_{2}BX_{2}$ and
$BX_{1}B+BX_{2}B=X_{1}AX_{1}+X_{1}AX_{2}+X_{2}AX_{1}+X_{2}AX_{2},$
$\Leftrightarrow X_{1}AX_{2}+X_{2}AX_{1}=0\text{ and }X_{1}BX_{1}+X_{1}BX_{2}\
=0$

∎

The following result is an extension from the theorem given in [8], which
states, for an arbitrary square matrix A, there are no isolated nontrivial
solutions to the YBE $AXA\ =\ XAX$, which do not commute with A.

######  Theorem 3.20.

Let A and B commute each other, then any non-trivial non-commutative solution
to (1) is not isolated.

###### Proof.

Let $X_{0}$ be a non-trivial solution to (1), which does not commute with both
A and B. Then we have, for any $t\in\mathbb{K}$, $e^{-BAt}X_{0}e^{BAt}$ is
also a solution to (1). We have, as A and B commute, A commute with $e^{BAt}$.
This implies,

$\displaystyle A(e^{-BAt}X_{0}e^{BAt})A\ $ $\displaystyle=\
e^{-BAt}AX_{0}Ae^{BAt}$ $\displaystyle=e^{-BAt}X_{0}BX_{0}e^{BAt}$
$\displaystyle=(e^{-BAt}X_{0}e^{BAt})B(e^{-BAt}X_{0}e^{BAt}).$

Similarly, we can see that
$B(e^{-BAt}X_{0}e^{BAt})B=(e^{-BAt}X_{0}e^{BAt})A(e^{-BAt}X_{0}e^{BAt})$. This
gives, $e^{-BAt}X_{0}e^{BAt}$ is a solution to (1). As a consequence, we can
see $X_{0}$ lies in a path of solutions, inside solution space. ∎

##  4\. Idempotent Orthogonal Complements

Two square matrices A and B are called idempotent orthogonal complements, when
$A^{2}=A,B^{2}=B$, and $AB=BA=0$. In this section, we are looking for
solutions to (1) when A and B are idempotent orthogonal complements. As A and
B are idempotent, their spectrum will be a subset of $\\{0,1\\}$, also both
are diagonalizable.
Let $E_{A}(\lambda_{i})$ denote the eigenspace of A with respect to the
eigenvalue $\lambda_{i}$. Similarly, for B, $E_{B}(\lambda_{j})$ denotes the
eigenspace of B with respect to the eigenvalue $\lambda_{j}$. Now, for a non-
zero vector $v\in E_{B}(1)$, we have $Bv=v$. This gives, $ABv=0$, implies
$Av=0$. Similarly, we have any eigenvector for A corresponding to eigenvalue
1, belongs to $Ker(B)$. Hence we have the following inclusions.

$E_{B}(1)\subset E_{A}(0)$ $E_{A}(1)\subset E_{B}(0)$

Analogously, we can see that any column vector of B belongs to $Ker(A)$, and
vice versa. As we discussed above, both A and B are diagonalizable and
commute, we get they are simultaneously diagonalizable. i.e- There exist an
invertible matrix $S\in GL_{n}(\mathbb{K})$, such that $J_{A}=SAS^{-1}$ and
$J_{B}=SBS^{-1}$. Then, as a consequence of Lemma 3.17, without loss of
generality, we can assume both A and B are in their Jordan-canonical form.
Also, since their spectrum is subset of $\\{0,1\\}$, we have, for appropriate
$r$, $A\simeq\begin{pmatrix}I_{r}&0\\\ 0&0\end{pmatrix}$, where $I_{r}$ is
identity matrix of order r, and $0^{\prime}s$ are zero blocks with appropriate
dimensions. To avoid the triviality of equations in (1), let’s take $0<r<n$.
Now, by the above inclusions for eigenspaces of A and B, we have
$nullity(B)\geq rank(A)$. This leads, $B\simeq\begin{pmatrix}0&0\\\
0&I_{m}\end{pmatrix}$ , such that $n-m\geq r$.
Throughout the discussion, $J$ represents the following matrix, with
appropriate dimensions.

$J=\begin{pmatrix}0&1&0&...&0\\\ 0&0&0&...&0\\\ .\\\ .\\\ .\\\
0&0&0&...&0\end{pmatrix}.$

######  Theorem 4.1.

Let A and B be two idempotent matrices, which are orthogonal complements and
are in their Jordan-canonical forms as considered above, such that rank(A) =
nullity(B) = r. Then any solution to (1) has one of the following forms.

  * i-

$X=\begin{pmatrix}0&C\\\ D&0\end{pmatrix}$, such that $CD=0$ and $DC=0$.

  * ii-

$X=\begin{pmatrix}0&C\\\ D&Y_{2}\end{pmatrix}$, where $Y_{2}=UJU^{-1}$, for
any $U\in GL_{n-r}(\mathbb{K})$, and for $Z=U^{-1}C,\ W\ =\ DU$, $ZJ=0,\
JW=0,\ ZW=0,\ WZ=J$.

  * iii-

$X=\begin{pmatrix}Y_{1}&C\\\ D&0\end{pmatrix}$, where $Y_{1}=VJV^{-1}$, for
any $V\in GL_{r}(\mathbb{K})$, and for $Z\ =\ V^{-1}C,\ W\ =\ DV$, $JZ\ =\ 0,\
WJ\ =\ 0,\ ZW\ =\ J,\ WZ\ =\ 0$.

  * iv-

$X=\begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}$, where
$Y_{1}=U_{1}JU_{1}^{-1},Y_{2}=U_{2}JU_{2}^{-1}$ for any $(U_{1},U_{2})\in
GL_{r}(\mathbb{K})\times GL_{n-r}(\mathbb{K})$, and for $Z_{1}\ =\
U_{1}^{-1}C,\ Z_{2}\ =\ CU_{2},\ W_{1}\ =\ DU_{1},\ W_{2}\ =\ U_{2}^{-1}D$,
$JZ_{1}\ =\ 0,\ Z_{2}J\ =\ 0,\ W_{1}J\ =\ 0,\ JW_{2}\ =\ 0,\ Z_{1}W_{1}\ =\
J,\ Z_{2}W_{2}\ =\ J$.

###### Proof.

When $rank(A)=nullity(B)$; i.e $r=n-m$, we have,

$A=\begin{pmatrix}I_{r}&0\\\ 0&0\end{pmatrix}\text{ and
}B=\begin{pmatrix}0&0\\\ 0&I_{n-r}\end{pmatrix}.$

Also, let us partition $X$ corresponding to the block forms of A and B. Take,
$X=\begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}$ , where $Y_{1}$ is $r\times
r$, $Y_{2}$ is $(n-r)\times(n-r)$, C is $r\times n-r$, and D is $r\times n-r$
matrices. Then, (1) implies,

$\begin{pmatrix}I_{r}&0\\\ 0&0\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}I_{r}&0\\\
0&0\end{pmatrix}=\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}0&0\\\
0&I_{n-r}\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}$

and

$\begin{pmatrix}0&0\\\ 0&I_{n-r}\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}0&0\\\
0&I_{n-r}\end{pmatrix}=\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}I_{r}&0\\\
0&0\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}.$

This gives the following system of equations,

(11) $\displaystyle
Y_{1}^{2}=0,Y_{2}^{2}=0,Y_{1}C=0,CY_{2}=0,DY_{1}=0,Y_{2}D=0,CD=Y_{1},DC=Y_{2}.$

This gives us, $Y_{1}$ and $Y_{2}$ are nilpotent matrices with nipotency $\leq
2$. With the choice of $Y_{1}$ and $Y_{2}$, we can determine the remaining
blocks $C,D$.
Let both $Y_{1}$ and $Y_{2}$ be zero matrices, then equations in (11) will
reduce to finding C and D such that $CD=0,DC=0$. Now, if $Y_{1}=0$ and $Y_{2}$
is non-zero, then $Y_{2}$ must be similar to J. Say, $Y_{2}=UJU^{-1}$, for
some $U\in GL_{n-r}(\mathbb{K})$. Then by taking $Z=CU,W=U^{-1}D$, we can
rewrite equations in (11) as the following.

(12) $\displaystyle ZJ\ =\ 0,\ JW\ =\ 0,\ ZW\ =\ 0,\ WZ\ =\ J.$

Note that, here, the first column of Z and second row of W are zero vectors,
and rows of Z are orthogonal to column vectors of W.
Similarly, if $Y_{2}\ =\ 0$ and $Y_{1}$ is non-zero, then $Y_{1}$ must be
similar to $J$. Then for some $V\in GL_{r}(\mathbb{K})$, $Y_{1}\ =\ VJV^{-1}$.
Then by taking $Z\ =\ V^{-1}C,\ W\ =\ DV$, we can rewrite (11) as follows.

$JZ\ =\ 0,\ WJ\ =\ 0,\ ZW\ =\ J,\ WZ\ =\ 0.$

Now, let both $Y_{1}$ and $Y_{2}$ be non-zero. Then, $Y_{1}=U_{1}JU_{1}^{-1}$
and $Y_{2}=U_{2}JU_{2}^{-1}$, for some $(U_{1},U_{2})\in
GL_{r}(\mathbb{K})\times GL_{n-r}(\mathbb{K})$. Then, by taking
$Z_{1}=U_{1}^{-1}C,Z_{2}=CU_{2},W_{1}=DU_{1},W_{2}=U_{2}^{-1}D$, we can
rewrite (11) as the following.

(13) $\displaystyle JZ_{1}=0,\ Z_{2}J=0,\ W_{1}J=0,\ JW_{2}=0,\ Z_{1}W_{1}=J,\
W_{2}Z_{2}=J.$

Here, the first columns of $Z_{2}$ and $W_{1}$ are zero, also the second rows
of $Z_{1}$ and $W_{2}$ are zero.
∎

######  Example 4.1.1.

Let $A\ =\frac{1}{31}\begin{pmatrix}-39&-20&-50\\\ 14&35&10\\\
49&14&66\end{pmatrix}$, and
$B\ =\frac{1}{31}\begin{pmatrix}70&20&50\\\ -14&-4&-10\\\
-49&-14&-35\end{pmatrix}.$ Then clearly, both A and B are idempotent and
orthogonal complements and are simultaneously diagonalizable as follows.

$J_{A}\ =\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&0\end{pmatrix}\ =UAU^{-1}$
$=\begin{pmatrix}3&1&4\\\ 2&3&2\\\
7&2&5\end{pmatrix}\frac{1}{31}\begin{pmatrix}-39&-20&-50\\\ 14&35&10\\\
49&14&66\end{pmatrix}\frac{1}{31}\begin{pmatrix}-11&-3&10\\\ -4&13&-2\\\
17&-1&-7\end{pmatrix}.$

Similarly, for same U, we can see, $J_{B}\ =UBU^{-1}=\begin{pmatrix}0&0&0\\\
0&0&0\\\ 0&0&1\end{pmatrix}$.
Then, a solution X for (1), can be found using the equations,
$YJ_{A}Y=J_{B}YJ_{B}$ and $YJ_{B}Y=J_{A}YJ_{A}$, where $Y\ =\ UXU^{-1}$. For
the convenience, let us rename $J_{A}$ as A, $J_{B}$ as B, and $Y$ as X.
Now, take $X\ =\ \begin{pmatrix}a&b&c\\\ d&e&f\\\ g&h&i\end{pmatrix}$. Then,
$AXA\ =\ XBX$, and $BXB\ =\ XAX$ implies the following equations. The solution
X will be constituted by the zeros of the following system.

(14) $\displaystyle\begin{matrix}f_{1}=-cg+a&f_{2}=-ch+b&f_{3}=-ci\\\
f_{4}=-fg+d&f_{5}=-fh+e&f_{6}=-fi\\\ f_{7}=-gi&f_{8}=-hi&f_{9}=-i^{2}\\\
f_{10}=-a2-bd&f_{11}=-ab-be&f_{12}=-ac-bf\\\ f_{13}=-ad-
de&f_{14}=-bd-e2&f_{15}=-cd-ef\\\ f_{16}=-ag-dh&f_{17}=-bg-eh&f_{18}=-cg-
fh+i\end{matrix}$

Now, if we calculate the Gr$\ddot{o}$bner basis for ideal $\langle
f_{1},f_{2},...,f_{18}\rangle$ in $\mathbb{C}[a,b,c,...,i]$ with respect to
the lexicographic order for $a>b>c>d>...>i$, we get, $\\{a+e-i,\ i^{2},\ hi,\
gi,\ fi,\ ei,\ di,\ ci,\ bi,\ fh-e,\ ch-b,\ fg-d,\ eg-dh,\ cg+e-i,\ bg+eh,\
ce-bf,\ cd+ef,\ bd+e^{2}\\}.$
From the Gr$\ddot{o}$bner basis, we have $i=\ 0$, gives the block $Y_{2}=0$.
From remaining equations, we can see that $Y_{1}^{2}\ =\ \begin{pmatrix}a&b\\\
d&e\end{pmatrix}^{2}=0$. Also,

$CD\ =\ \begin{pmatrix}c\\\ f\end{pmatrix}\begin{pmatrix}g&h\end{pmatrix}\
=\begin{pmatrix}cg&ch\\\ fg&fh\end{pmatrix}\ =\ \begin{pmatrix}a&b\\\
d&e\end{pmatrix}\ =\ Y_{1}.$

Similarly, we can see that $DC=0$, $Y_{1}C\ =0$, $DY_{1}\ =0$. So, the
solution set is precisely, $X\ =\begin{pmatrix}Y_{1}&C\\\ D&0\end{pmatrix}$,
with $Y_{1}^{2}\ =\ 0$, $CD\ =\ Y_{1}$, $DC\ =\ 0$, $Y_{1}C\ =0$, $DY_{1}\
=0$.

######  Theorem 4.2.

Let A and B be two idempotent matrices, which are orthogonal complements and
are in their Jordan-canonical form as considered above, such that
$rank(A)<nullity(B)$. Then any solution to (1) has one of the following forms.

  * i-

$X=\begin{pmatrix}0&C\\\ D&Y_{2}\end{pmatrix}$, such that
$DC=B^{\prime}Y_{2}B^{\prime},~{}CB^{\prime}D=Y_{1},~{}CB^{\prime}Y_{2}=0,~{}Y_{2}B^{\prime}D=0,~{}Y_{2}B^{\prime}Y_{2}=0.$

  * ii-

$X=\begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}$, where $Y_{1}=UJU^{-1}$,
for any $U\in GL_{r}(\mathbb{K})$, and for $Z\ =\ U^{-1}C,\ W\ =\ DU$,
$JZ=0,~{}WJ=0,~{}DC=B^{\prime}Y_{2}B^{\prime},~{}ZB^{\prime}W=J,~{}CB^{\prime}Y_{2}=0,~{}Y_{2}B^{\prime}D=0,~{}Y_{2}B^{\prime}Y_{2}=0.$

###### Proof.

If $nullity(B)>rank(A)$, then we have,

$A=\begin{pmatrix}I_{r}&0\\\ 0&0\end{pmatrix}\text{ and
}B=\begin{pmatrix}0&0\\\ 0&B^{\prime}\end{pmatrix},$

where $B^{\prime}=\begin{pmatrix}I_{s}&0\\\ 0&0\end{pmatrix}$,
$(n-r)\times(n-r)$ matrix, and $s=rank(B)$, such that $1\leq s<n-r$. Also, let
us take X as previously, i.e., $X=\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}$. Then (1) implies,

$\begin{pmatrix}I_{r}&0\\\ 0&0\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}I_{r}&0\\\
0&0\end{pmatrix}=\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}0&0\\\
0&B^{\prime}\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}$

and

$\begin{pmatrix}0&0\\\ 0&B^{\prime}\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}0&0\\\
0&B^{\prime}\end{pmatrix}=\begin{pmatrix}Y_{1}&C\\\
D&Y_{2}\end{pmatrix}\begin{pmatrix}I_{r}&0\\\
0&0\end{pmatrix}\begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}.$

This gives rise to the following equations.

(15) $\displaystyle
Y_{1}^{2}=0,~{}Y_{1}C=0,~{}DY_{1}=0,~{}DC=B^{\prime}Y_{2}B^{\prime},$
$\displaystyle
CB^{\prime}D=Y_{1},~{}CB^{\prime}Y_{2}=0,~{}Y_{2}B^{\prime}D=0,~{}Y_{2}B^{\prime}Y_{2}=0.$

Further, by partitioning C,D, and $Y_{2}$ into blocks with dimensions
corresponding to $B^{\prime}$, we have

$C=\begin{pmatrix}C_{1}&C_{2}\\\
C_{3}&C_{4}\end{pmatrix},D=\begin{pmatrix}D_{1}&D_{2}\\\
D_{3}&D_{4}\end{pmatrix},Y_{2}=\begin{pmatrix}X_{1}&X_{2}\\\
X_{3}&X_{4}\end{pmatrix}.$

Then we have the following implications.

(16) $\displaystyle
Y_{2}B^{\prime}Y_{2}=0\Rightarrow~{}~{}X_{1}^{2}=0,~{}X_{1}X_{2}=0,~{}X_{3}X_{1}=0,~{}X_{3}X_{2}=0$
$\displaystyle
CB^{\prime}Y_{2}=0\Rightarrow~{}~{}C_{1}X_{1}=0,~{}C_{1}X_{2}=0,~{}C_{3}X_{1}=0,~{}C_{3}X_{2}=0$
$\displaystyle
Y_{2}B^{\prime}D=0\Rightarrow~{}~{}X_{1}D_{1}=0,~{}X_{1}D_{2}=0,~{}X_{3}D_{1}=0,~{}X_{3}D_{2}=0$
$\displaystyle
DC=B^{\prime}Y_{2}B^{\prime}\Rightarrow~{}~{}D_{1}C_{1}+D_{2}C_{3}=X_{1},~{}D_{1}C_{2}+D_{2}C_{4}=0,$
$\displaystyle D_{3}C_{1}+D_{4}C_{3}=0,~{}D_{3}C_{2}+D_{4}C_{4}=0$

Now, from (15), it is clear that $Y_{1}$ is a nilpotent matrix with
$nilpotency\leq 2$. Then, $Y_{1}$ is either 0 or similar to J. When $Y_{1}=0$,

$CB^{\prime}D=Y_{1}=0\Rightarrow
C_{1}D_{1}=0,C_{1}D_{2}=0,C_{3}D_{1}=0,C_{3}D_{2}=0.$

If $Y_{1}\neq 0$, then for some $U\in GL_{r}(\mathbb{K})$, $Y_{1}=UJU^{-1}$.
Then by taking $U^{-1}C=Z$ and $DU=W$, we can rewrite (15) as following,

(17)
$JZ=0,WJ=0,DC=B^{\prime}Y_{2}B^{\prime},ZB^{\prime}W=J,CB^{\prime}Y_{2}=0,Y_{2}B^{\prime}D=0,Y_{2}B^{\prime}Y_{2}=0.$

Further, from the first equation of (16), it is clear that $X_{1}$ is
nilpotent with $nilpotency\leq 2$. Then $X_{1}$ is 0 or similar to J. Now, if
$X_{1}=0$, then we can rewrite $CB^{\prime}Y_{2}=0$ as,
$\begin{pmatrix}0&C_{1}X_{2}\\\ 0&C_{3}X_{3}\end{pmatrix}\ =\ Y_{1}$. Also,
$Y_{2}B^{\prime}D=0$ can be rewritten, $\begin{pmatrix}0&0\\\
X_{3}D_{1}&X_{3}D_{2}\end{pmatrix}\ =\ \begin{pmatrix}0&0\\\
0&0\end{pmatrix}$.
Also, when $s\geq 2$, and if $Y_{1}\neq 0$, then by taking
$Z=\begin{pmatrix}Z_{1}&Z_{2}\\\ Z_{3}&Z_{4}\end{pmatrix}$ and
$W=\begin{pmatrix}W_{1}&W_{2}\\\ W_{3}&W_{4}\end{pmatrix}$, $ZB^{\prime}W=J$
implies, $Z_{1}W_{2}=0,Z_{3}W_{1}=0,Z_{3}W_{2}=0$, and $Z_{1}W_{1}=J_{s}$,
where $J_{s}$ is the left-top corner $s\times s$ block of J. ∎

######  Example 4.2.1.

Let $A\ =\ \begin{pmatrix}29/2&24&12&33/2\\\ -1/2&-2&-1&-7/2\\\
-43/2&-34&-17&-41/2\\\ 9/2&8&4&13/2\end{pmatrix}$, and
$B\ =\ \begin{pmatrix}-12&-15&-9&-6\\\ 0&0&0&0\\\ 20&25&15&10\\\
-4&-5&-3&-2\end{pmatrix}$. Then, both A and B are idempotent and orthogonal
complements. Then we can simultaneously diagonalize them using some $U\in
GL_{n}(\mathbb{K})$, as follows.

$J_{A}\ =\ \begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{pmatrix}\ =\ UAU^{-1}$ $=\begin{pmatrix}1&2&1&2\\\ 3&4&2&1\\\
4&5&3&2\\\ 1&6&2&7\end{pmatrix}\begin{pmatrix}29/2&24&12&33/2\\\
-1/2&-2&-1&-7/2\\\ -43/2&-34&-17&-41/2\\\
9/2&8&4&13/2\end{pmatrix}\begin{pmatrix}7&5/2&-3&-3/2&\\\ -2&1/2&0&1/2\\\
-8&-9/2&5&3/2\\\ 3&1/2&-1&-1/2\end{pmatrix}.$

Similarly, for same U, we have $J_{B}\ =\ UBU^{-1}\ =\begin{pmatrix}0&0&0&0\\\
0&0&0&0\\\ 0&0&1&0\\\ 0&0&0&0\end{pmatrix}.$
Then, a solution X for (1), can be found using the equations,
$YJ_{A}Y=J_{B}YJ_{B}$ and $YJ_{B}Y=J_{A}YJ_{A}$, where $Y\ =\ UXU^{-1}$. For
the convenience, let us rename $J_{A}$ as A, $J_{B}$ as B, and $Y$ as X.
Take $X\ =\ \begin{pmatrix}Y_{1}&C\\\ D&Y_{2}\end{pmatrix}\ =\
\begin{pmatrix}a&b&c&d\\\ e&f&g&h\\\ i&j&k&l\\\ m&n&p&q\end{pmatrix}$.
Then the Gr$\ddot{o}$bner basis generated corresponding to the equations from
the system $AXA\ =\ XBX$ and $BXB\ =\ XAX$ is:
$\\{a+f-k,\ lp,\ kp,\ jp,\ ip,\ fp,\ ep,\ bp,\ fm-km-en,\ dm+hn,\ cm+gn,\
bm+fn,\ kl,\ gl,\ fl,\ el,\ cl,\ bl,\ k^{2},\ jk,\ ik,\ gk,\ fk,\ ek,\ ck,\
bk,\ gj-f,\ cj-b,\ gi-e,\ fi-ej,\ di+hj,\ ci+f-k,\ bi+fj,\ df-bh-dk,\ cf-bg,\
de+fh,\ ce+fg,\ be+f^{2},\ hkn,\ dkn,\ dgn-chn,\ hkm,\ hjm-hin,\ ejm-ein,\
bdg-bch\\}$

From above Gr$\ddot{o}$bner basis, we get $k=0$, which gives $a\ =-f$. Then,
we have,

$Y_{2}\ =\ \begin{pmatrix}X_{1}&X_{2}\\\
X_{3}&X_{4}\end{pmatrix}=\begin{pmatrix}k&l\\\
p&q\end{pmatrix}=\begin{pmatrix}0&l\\\ p&q\end{pmatrix},\text{ with $lp=0$}.$

Also, $Y_{1}^{2}\ =\begin{pmatrix}a&b\\\ e&f\end{pmatrix}^{2}\ =0.$ Similarly,
by considering $C\ =\begin{pmatrix}C_{1}&C_{2}\\\ C_{3}&C_{4}\end{pmatrix}\
=\begin{pmatrix}c&d\\\ g&h\end{pmatrix}$ and $D\
=\begin{pmatrix}D_{1}&D_{2}\\\ D_{3}&D_{4}\end{pmatrix}\
=\begin{pmatrix}i&j\\\ m&n\end{pmatrix}$, we can see,

$CB^{\prime}Y_{2}\ =\begin{pmatrix}0&C_{1}X_{2}\\\ 0&C_{3}X_{2}\end{pmatrix}\
=\begin{pmatrix}0&cl\\\ 0&gl\end{pmatrix}.$

In same passion we have, $Y_{2}B^{\prime}D=0,\ Y_{2}B^{\prime}Y_{2}=0,\
CB^{\prime}D=Y_{1},\ DC=B^{\prime}Y_{2}B^{\prime},\ DY_{1}=0,\ Y_{1}C=0$. Now,
if C is chosen in a way it’s non-singular, then $b=n=0$. Also, if D is non-
singular, then $e=h=0$. These equations together give the complete solutions
for the system of Yang-Baxter like matrix equations, for the given A and B.

##  5\. Solutions when coefficient matrices are of size 2

In this section, we discuss explicit solutions of the system of equations in
(1) when A and B are $2\times 2$ matrices, and are in their Jordan-canonical
form. For finding these solutions, we have equipped with the Gr$\ddot{o}$bner
basis techniques.

###  5.1. When A and B are diagonalizable

Let both A and B be diagonalizable and are in Jordan form. Say,

$A=\begin{pmatrix}a&0\\\ 0&b\end{pmatrix},\text{ }B=\begin{pmatrix}c&0\\\
0&d\end{pmatrix},\text{ and }X=\begin{pmatrix}x_{1}&x_{2}\\\
x_{3}&x_{4}\end{pmatrix}.$

Then, solutions to the equations $AXA=XBX$ and $BXB=XAX$, equivalently can be
read as zeros of a system of 8 polynomials, $f_{1},f_{2},...,f_{8}$.

######  Proposition 5.1.

When A,B assume above mentioned form, and $a,b,c,d\neq 0$, then any non-
trivial solution of (1) have the following form:

  * i)

If we assume $x_{1}$ to be 0, then $X=\begin{pmatrix}0&\alpha\\\
0&\frac{b^{2}}{d}\end{pmatrix}$ or $\begin{pmatrix}0&0\\\
\alpha&\frac{b^{2}}{d}\end{pmatrix}$, where $\alpha\in\mathbb{K},a=b\text{ and
}b^{3}=d^{3}$.

  * ii)

If we assume $x_{2}$ to be 0, then $X=\begin{pmatrix}\frac{a^{2}}{c}&0\\\
\alpha&\frac{b^{2}}{d}\end{pmatrix}$ or $\begin{pmatrix}\frac{a^{2}}{c}&0\\\
0&\frac{b^{2}}{d}\end{pmatrix}$, where
$a^{3}=c^{3},b^{3}=d^{3},ab=a^{2}+b^{2},\text{ and }cd=c^{2}+d^{2}$.

  * iii)

If we assume $x_{3}$ to be 0, then $X=\begin{pmatrix}\frac{a^{2}}{c}&\alpha\\\
0&\frac{b^{2}}{d}\end{pmatrix}$ or $\begin{pmatrix}\frac{a^{2}}{c}&0\\\
0&\frac{b^{2}}{d}\end{pmatrix}$, where
$a^{3}=c^{3},b^{3}=d^{3},ab=a^{2}+b^{2},\text{ and }cd=c^{2}+d^{2}$.

  * iv)

If we assume $x_{4}$ to be 0, then $X=\begin{pmatrix}\frac{a^{2}}{c}&\alpha\\\
0&0\end{pmatrix}$ or $\begin{pmatrix}\frac{a^{2}}{c}&0\\\
\alpha&0\end{pmatrix}$, where $\alpha\in\mathbb{K},a^{2}b=c^{2}d$.

  * v)

Any other non-singular solution $X$ of (1), will have the form $X^{2}\ =\
U\begin{pmatrix}ac&0\\\ 0&bd\end{pmatrix}U^{-1}$, for some $U\in
GL_{2}(\mathbb{K})$.

###### Proof.

Equations in (1) give the following polynomials, and a solution will be a
solution for this system of polynomials.

(18) $\displaystyle f_{1}$ $\displaystyle=a^{2}x_{1}-cx_{1}^{2}-dx_{2}x_{3}$
$\displaystyle f_{2}$ $\displaystyle=$
$\displaystyle~{}abx_{2}-cx_{1}x_{2}-dx_{2}x_{4}$ $\displaystyle f_{3}$
$\displaystyle=abx_{3}-cx_{1}x_{3}-dx_{3}x_{4}$ $\displaystyle f_{4}$
$\displaystyle=$ $\displaystyle~{}b^{2}x_{4}-cx_{2}x_{3}-dx_{4}^{2}$
$\displaystyle f_{5}$ $\displaystyle=c^{2}x_{1}-ax_{1}^{2}-bx_{2}x_{3}$
$\displaystyle f_{6}$ $\displaystyle=$
$\displaystyle~{}cdx_{2}-ax_{1}x_{2}-bx_{2}x_{4}$ $\displaystyle f_{7}$
$\displaystyle=cdx_{3}-ax_{1}x_{3}-bx_{3}x_{4}$ $\displaystyle f_{8}$
$\displaystyle=$ $\displaystyle~{}d^{2}x_{4}-ax_{2}x_{3}-bx_{4}^{2}$

Now we can find a Gr$\ddot{o}$bner basis for the ideal $\langle
f_{1},f_{2},...,f_{8}\rangle$ in $\mathbb{C}[a,b,c,d,x_{1},...,x_{4}]$ with
respect to the lexicographic order for $a>b>c>d>x_{1}>...>x_{4}$. But
Gr$\ddot{o}$bner basis for this in general is more complicated and not helping
to solve the equations. So, we have taken some additional assumptions such as
$x_{1}=0$, or $x_{2}=0$, or $x_{3}=0$, or $x_{4}=0$, and solved the system in
each of the cases using Gr$\ddot{o}$bner basis. This gives solutions given in
$(i)$ to $(iv)$. As the Gr$\ddot{o}$bner basis is very large, we have not
included that here. Now, any other non-singular solution will have the form
$X^{2}\ =\ U\begin{pmatrix}ac&0\\\ 0&bd\end{pmatrix}U^{-1}$, for some $U\in
GL_{2}(\mathbb{K})$, which is a consequence of Theorem 3.16. ∎

######  Proposition 5.2.

  * i)

When $a=0,b,c,d\neq 0$, then the only non-trivial solution to (1) is
$X=\begin{pmatrix}0&0\\\ 0&\frac{d^{2}}{b}\end{pmatrix}$, where $b^{3}=d^{3}$.

  * ii)

Similarly, when $b=0,a,c,d\neq 0$, then the only non-trivial solution is
$X=\begin{pmatrix}\frac{c^{2}}{a}&0\\\ 0&0\end{pmatrix}.$

###### Proof.

when $a=0$, we can rewrite the equations in (18), and calculate the
Gr$\ddot{o}$bner basis, which is simpler than the previous case. The
corresponding generators of the ideals are the following.

(19) $\displaystyle f_{1}$ $\displaystyle=cx_{1}^{2}+dx_{2}x_{3},$
$\displaystyle f_{2}$ $\displaystyle=$
$\displaystyle~{}cx_{1}x_{2}+dx_{2}x_{4}$ $\displaystyle f_{3}$
$\displaystyle=cx_{1}x_{3}+dx_{3}x_{4},$ $\displaystyle f_{4}$
$\displaystyle=$ $\displaystyle~{}b^{2}x_{4}-cx_{2}x_{3}-dx_{4}^{2}$
$\displaystyle f_{5}$ $\displaystyle=c^{2}x_{1}-bx_{2}x_{3},$ $\displaystyle
f_{6}$ $\displaystyle=$ $\displaystyle~{}cdx_{2}-bx_{2}x_{4}$ $\displaystyle
f_{7}$ $\displaystyle=cdx_{3}-bx_{3}x_{4},$ $\displaystyle f_{8}$
$\displaystyle=$ $\displaystyle~{}d^{2}x_{4}-bx_{4}^{2}.$

Similar way, we can rewrite the equations in (18) when $b=0$, and $a,c,d\ \neq
0$, and calculate the Gr$\ddot{o}$bner basis, which will lead to our solution.
∎

######  Proposition 5.3.

When $a,c=0$ and $b,d\neq 0$, the solution has the following form.
$X=\begin{pmatrix}\alpha&0\\\ \beta&0\end{pmatrix}$ or
$\begin{pmatrix}\alpha&\beta\\\ 0&0\end{pmatrix}$, or
$\begin{pmatrix}\alpha&0\\\ 0&\frac{b^{2}}{d}\end{pmatrix}$ where
$\alpha,\beta\in\mathbb{K}$ and $b^{3}=d^{3}$.

###### Proof.

When $a,c=0$ and $b,d\neq 0$, the system of equation has a much simpler form,
and easy to find solutions. The corresponding equations are the following.

(20) $\displaystyle f_{1}$ $\displaystyle=dx_{2}x_{3},$ $\displaystyle f_{2}$
$\displaystyle=$ $\displaystyle~{}dx_{2}x_{4}$ $\displaystyle f_{3}$
$\displaystyle=dx_{3}x_{4},$ $\displaystyle f_{4}$ $\displaystyle=$
$\displaystyle~{}b^{2}x_{4}-dx_{4}^{2}$ $\displaystyle f_{5}$
$\displaystyle=bx_{2}x_{3},$ $\displaystyle f_{6}$ $\displaystyle=$
$\displaystyle~{}bx_{2}x_{4}$ $\displaystyle f_{7}$
$\displaystyle=bx_{3}x_{4},$ $\displaystyle f_{8}$ $\displaystyle=$
$\displaystyle~{}d^{2}x_{4}-bx_{4}^{2}.$

Then the corresponding Gr$\ddot{o}$bner basis for the ideal generated by the
above polynomials is,

$\\{x_{3}x_{4}d,\ x_{2}x_{4}d,\ x_{2}x_{3}d,\ x_{4}b^{2}-x_{4}^{2}d,\
x_{4}^{2}b-x_{4}d^{2},\ x_{3}x_{4}b,\ x_{2}x_{4}b,\ x_{2}x_{3}b,\
x_{4}^{3}d-x_{4}bd^{2}\\}$

On finding zeros of these, we get the solution forms as above. ∎

###  5.2. When A is non-diagonalizable and B is diagonalizable

Let $A\ =\ \begin{pmatrix}a&1\\\ 0&a\end{pmatrix}$ and $B\ =\
\begin{pmatrix}b&0\\\ 0&c\end{pmatrix}$, and $X$ as previously
$\begin{pmatrix}x_{1}&x_{2}\\\ x_{3}&x_{4}\end{pmatrix}$.

######  Proposition 5.4.

  * a)

When $a=0$, for a solution to (1), we have the following forms.

    * i

When $b,c\neq 0$, (1) has only trivial solution.

    * ii

If $b=0,c\neq 0$, then the solutions to (1) are $X\ =\
\begin{pmatrix}\alpha&\beta\\\ 0&0\end{pmatrix}$ , where
$\alpha,\beta,\in\mathbb{K}$.

    * iii

If $b\neq 0,c=0$, then the solutions to (1) are $X\ =\
\begin{pmatrix}0&\alpha\\\ 0&\beta\end{pmatrix}$ , where
$\alpha,\beta,\in\mathbb{K}$.

  * b)

When $a\neq 0$, and $b,c\ \in\ \mathbb{K}$, (1) has only trivial solution.

###### Proof.

When $A\ =\ \begin{pmatrix}0&1\\\ 0&0\end{pmatrix}$, $B\ =\
\begin{pmatrix}b&0\\\ 0&c\end{pmatrix}$, and $X\ =\
\begin{pmatrix}x_{1}&x_{1}\\\ x_{3}&x_{4}\end{pmatrix}$, the initial
computation of Gr$\ddot{o}$bner basis correspond to the ideal generated by
component polynomial equations from $AXA\ =\ XBX$ and $BXB\ =\ XAX$ shows that
$x_{3}=0$. After modifying $X$ as $\begin{pmatrix}x_{1}&x_{2}\\\
0&x_{4}\end{pmatrix}$, we have the Gr$\ddot{o}$bner basis as,

$\\{x_{4}c^{2},\ x_{2}bc-x_{1}x_{4},\ x_{4}^{2}c,\ x_{1}b^{2},\
x_{1}x_{2}b+x_{2}x_{4}c,\ x_{1}^{2}b,\ x_{1}x_{4}^{2},\
x_{1}^{2}x_{4},x_{1}x_{2}x_{4}c\\}.$

Then, solving from Gr$\ddot{o}$bner basis gives us the solutions in $(a)$.
Now, when $A\ =\ \begin{pmatrix}a&1\\\ 0&a\end{pmatrix}$, and $B\ =\
\begin{pmatrix}b&0\\\ 0&c\end{pmatrix}$, such that $a\neq 0$, the initial
Gr$\ddot{o}$bner basis to the corresponding ideal shows that $x_{3}=0$,
irrespective the value of $b,\ c$. Then on next iteration, after giving
$x_{3}\ =0$, we will get $x_{1},\ x_{2},\ x_{4}\ =\ 0$. ∎

###  5.3. When A and B are non-diagonalizable

Let $A\ =\ \begin{pmatrix}a&1\\\ 0&a\end{pmatrix}$ and $B\ =\
\begin{pmatrix}b&1\\\ 0&b\end{pmatrix}$, and $X$ as previously
$\begin{pmatrix}x_{1}&x_{2}\\\ x_{3}&x_{4}\end{pmatrix}$.

######  Proposition 5.5.

Let A and B be as above. Then for any non-trivial solution X of (1), we have,

  * i-

When $a,b\neq 0$, X will have one of the following form.
$\begin{pmatrix}a&1\\\
0&a\end{pmatrix},\begin{pmatrix}a+a\sqrt{\alpha}&\alpha\\\
-a^{2}&a-a\sqrt{\alpha}\end{pmatrix}or\begin{pmatrix}a-a\sqrt{\alpha}&\alpha\\\
-a^{2}&a+a\sqrt{\alpha}\end{pmatrix}$ where $\alpha\in K$, and $a=\ b$, in
that case.

  * ii-

When $a,b\neq 0$, and $a\neq b$, then any non-trivial solution X will have the
form $\begin{pmatrix}\alpha&\frac{b^{2}+\alpha^{2}-\alpha(a+b)}{ab}\\\
-ab&(a+b)-\alpha\end{pmatrix}$.

  * iii-

When one of $a,b$ is zero, then (1) has only trivial solution.

###### Proof.

For the given $A,B$, and $X$ as $\begin{pmatrix}x_{1}&x_{2}\\\
x_{3}&x_{4}\end{pmatrix}$, the initial computation of Gr$\ddot{o}$bner basis
for the ideal generated by polynomials appeared in (1) shows that
$x_{3}(a^{2}-b^{2})=0$. Then either $x_{3}=0$ or $a^{2}=b^{2}$. If
$a^{2}=b^{2}$, and $a=b$, then the system of equations becomes single YBE, and
authors have found its solution in [13], which are given in $(i)$ above. Now,
if $a\neq b$, but $a^{2}=b^{2}$, we have $b^{3}+x_{3}a=0$ in Gr$\ddot{o}$bner
basis. This gives $x_{3}=-ab$. After giving $x_{3}$ this value, on the next
iteration, we have $x_{1}ab^{2}+x_{4}ab^{2}-ab^{3}-b^{4}=0$ in the
Gr$\ddot{o}$bner basis. This gives $x_{1}+x_{4}=a+b$. Now, adding
$x_{1}+x_{4}-(a+b)$ to the basis, the Gr$\ddot{o}$bner basis will become:

$\\{x_{1}+x_{4}-a-b,\ a^{2}-b^{2},\ x_{2}ab-x_{4}^{2}+x_{4}a+x_{4}b-b^{2},\
x_{2}b^{3}-x_{4}^{2}a+x_{4}ab+x_{4}b^{2}-ab^{2}\\}.$

Solving this will give the solution in $(ii)$.
When $a\neq 0$ and $b=0$, the Gr$\ddot{o}$bner basis correspond to the system
of equations in (1) will give, $x_{3}^{2}=0,x_{4}^{3}a=0$. Then we have
$x_{3},x_{4}=0$. Adding this to the basis, Gr$\ddot{o}$bner basis will reduce
to $\\{x_{2}a^{2}+x_{1}a,\ x_{1}a^{2},\ x_{1}x_{2}a,\ x_{1}^{2}a\\}$. This
gives us (1) has only trivial solution in this case. ∎

## References

  * [1] W. W. Adams and P. Loustaunau.  An introduction to Gröbner bases, volume 3 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 1994.
  * [2] E. Artin.  Braids and permutations.  Ann. of Math. (2), 48:643–649, 1947.
  * [3] E. Artin.  Theory of braids.  Ann. of Math. (2), 48:101–126, 1947.
  * [4] V. Baranovsky.  The variety of pairs of commuting nilpotent matrices is irreducible.  Transform. Groups, 6(1):3–8, 2001.
  * [5] R. Basili.  On the irreducibility of commuting varieties of nilpotent matrices.  J. Algebra, 268(1):58–80, 2003.
  * [6] R. J. Baxter.  On looking for solutions of the star-triangle relations.  In Quantum field theory, statistical mechanics, quantum groups and topology (Coral Gables, FL, 1991), pages 11–16. World Sci. Publ., River Edge, NJ, 1992.
  * [7] D. A. Cox, J. Little, and D. O’Shea.  Ideals, varieties, and algorithms.  Undergraduate Texts in Mathematics. Springer, Cham, fourth edition, 2015\.  An introduction to computational algebraic geometry and commutative algebra.
  * [8] N. C. Dincic and D. Djordjevic.  On the intrinsic structure of the solution set to the yang–baxter-like matrix equation.  Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A: Matemáticas (RACSAM), 116(2):20, 2022.
  * [9] B. D. Djordjević.  Doubly stochastic and permutation solutions to axa= xax when a is a permutation matrix.  Linear Algebra and its Applications, 2022.
  * [10] H. Huang, Z. Huang, C. Wu, C. Jiang, D. Fu, and C. Lin.  Modified newton integration neural algorithm for solving time-varying yang-baxter-like matrix equation.  Neural Processing Letters, pages 1–15, 2022.
  * [11] R. Iordanescu, F. F. Nichita, and I. M. Nichita.  The yang-baxter equation,(quantum) computers and unifying theories.  Axioms, 3(4):360–368, 2014.
  * [12] L. Lu.  Manifold expressions of all solutions of the yang–baxter-like matrix equation for rank-one matrices.  Applied Mathematics Letters, 132:108175, 2022.
  * [13] H. Mukherjee and A. A. M.  Solutions to the matrix yang-baxter equation, 2022.
  * [14] F. A. Vind, A. Foerster, I. S. Oliveira, R. S. Sarthour, D. d. O. Soares-Pinto, A. M. d. Souza, and I. Roditi.  Experimental realization of the yang-baxter equation via nmr interferometry.  Scientific reports, 6(1):1–8, 2016.
  * [15] H. Wenzl.  Representations of braid groups and the quantum Yang-Baxter equation.  Pacific J. Math., 145(1):153–180, 1990.