Some non-commuting solutions of the Yang-Baxter-like matrix equation

Abstract Let A be a square matrix satisfying A 4 = A {A}^{4}=A . We solve the Yang-Baxter-like matrix equation A X A = X A X AXA=XAX to find some solutions, based on analysis of the characteristic polynomial of A and its eigenvalues. We divide the problem into small cases so that we can find the solution easily. Finally, in order to illustrate the results, two numerical examples are presented.


Introduction
Let A be an × n n complex matrix. The quadratic matrix equation: is often called the Yang-Baxter-like matrix equation since it is similar to the classical parameter-free Yang-Baxter equation in format [1][2][3]. The original Yang-Baxter equation has many applications in statistical mechanics, integrable systems, quantum theory, knot theory, braid group theory, and so on [3][4][5][6].
(1) is difficult, almost all the works so far have been toward constructing commuting solutions of the equation; see, e.g., [7][8][9][10][11][12][13][14][15][16][17] and references therein. Commuting solutions mean that the unknown matrix X satisfies the commutability condition = AX XA. All commuting solutions of (1) have been obtained when A is a matrix with general Jordan structure forms in [7], but finding all non-commuting solutions of (1) is still a challenging task when A is arbitrary.
Up to now, there are only isolated results toward this goal for special classes of the given matrix A, e.g., [18][19][20][21][22][23][24][25][26]. All solutions have been constructed for rank-1 matrices A in [23], rank-2 matrices A in [24,25], non-diagonalizable elementary matrices A in [26] in [21], and diagonalizable matrices A with two different eigenvalues in [22]. In this paper, we try to solve the matrix Eq. (1) to obtain some non-commuting solutions when the given matrix A satisfies = A A 4 . This is an important step to solve more general matrices.
We first provide some preliminary results. Then we study some solutions for the Yang-Baxter-like matrix equation of (1) when the given matrix A satisfies = A A 4 through analysis of the characteristic polynomial of A and its eigenvalues. Finally, we give some numerical experiments to illustrate our results.

Preliminary results
In this section, we give some results for our further discussion. First, we need the following lemmas. is true for any × n k matrix V.
We assume that A is an × n n complex matrix with ≥ n 4. Since and the minimal polynomial ( ) g λ of A, which is the unique annihilator of A with minimal degree and leading coefficient 1, is a factor of ( ) p λ . Therefore, each eigenvalue of A is semi-simple, and then A is the diagonalizable matrix. So there exists a nonsingular matrix S such that , clearly that X is the solution of (1) if and only if Y is the solution of the equation: Moreover, X is a commuting solution if and only if Y is a commuting solution. Thus, in the following we just solve (3) to get solutions of (1). ( ) In this paper, an analysis of all cases except Case 10 and Case 11 is given in the next section. It is too difficult to solve all the solutions of the Yang-Baxter-like matrix Eq. (1) when A is a diagonalizable complex matrix with three distinct nonzero eigenvalues. So Case 10 and Case 11 are still challenging tasks that are not easy to solve in short term and will be further studied in the future.
3 Some solutions of the matrix equation 3 n m n m . By applying Lemma 2.1, we have the following conclusion. with rank m. Then all solutions of (1) are n m n m arbitrary matrix. According to the first condition of (7), we have = − s m the rank of K. Applying this result to the last three conditions of (7), we get According to the block structure of Σ, partition W and H as respectively. Then  In this case, A is idempotent. So A has two eigenvalues 0 and 1. Let (1) in this case has been studied previously by [21]. But for the completeness of the presentation, we summarize the main results of [21] in the following theorem. and its inverse partitioned as Applying Lemma 2.1, we obtain all commuting solutions and non-commuting solutions of (1) in the following theorem.
(2) All the non-commuting solutions of Eq. (1) are where K is any × m m diagonalizable matrix and Z is any ( − ) × ( − ) n m n m diagonalizable matrix such that (i) the nonzero matrices C and D have the same rank r such that (ii) K and Z have eigenvalues − (iv) the other eigenvalues of K and Z belong to { } 0, 1 and − + 0, We get = C 0 and = D 0. If = C 0 and = D 0, it is easy to prove that = JY YJ. Thus, (8) implies that all commuting respectively.
We show that there are no solutions of (8) such that = are not eigenvalues of Z. This is a contradiction. The same results can also be obtained with assumption ≠ C 0 and = D 0. Thus, any solutions of (8) with = C 0 or = D 0 is a commuting one. So all non-commuting solutions of (8) must have ≠ C 0 and ≠ D 0. From the first two equations of (8), we have From the first equation and the third equation of (8), we obtain Combining these results we have Similarly, Substituting (9) into the right of the third equation of (8), we get Substituting (10) into the right of the fourth equation of (8), we have Multiplying C to the first equation of (8) from the right and using (11), we obtain .
From which, we have Hence, ( ) = ( ) r C r CD . Multiplying D to the second equation of (8) from the right and using (10), we get Hence, ( ) = ( ) r D r CD . Therefore, This means that (i) is true. From (10) and (11), we know that all nonzero columns of D and C are eigenvectors of Z and K associated with eigenvalues must be an eigenvalue of K and Z, respectively. Furthermore, they are semi-simple eigenvalues of K and Z, respectively. If and ( ) ( ) The eigenvector u and the generalized eigenvector v must be linearly independent. In fact, if , then via multiplying this equality by ( ) from the left we get ( ) , from which = b 0 and so = a 0. Since 3 3 , we get is a semi-simple eigenvalue of Z. So (ii) and (iii) are true.
, we obtain Applying Lemma 2.2 to the first equation of (8), it follows that According to the second equation of (8) and the fact that the eigenvalues of the square of a matrix are the squares of the eigenvalues of the matrix, the aforementioned identity can be written as are the eigenvalues of K and , for 1, , . , we obtain , the aforementioned identity can be further simplified to Multiplying v to the first equation of (8) from the right, we get Combining this with (11), we obtain This is a contradiction. If 1 is an eigenvalue of K that is not semi-simple, then there exists a vector Multiplying v to the first equation of (8) from the right, we have Combining this with (11) , , , satisfies (i)-(iv). We show that it is a solution of (8). According to (iii), we have (9), (10), (11), and (12). Combining (9) and (11), we have the third equality of (8). Combining (10) and (12), we have the fourth equality of (8). Then from (11) and (ii), we obtain Thus, Since the columns of C  and C  form a basic of m , then , we can obtain , , , is a solution of (8).  (1), we partition Y as  (2) All the non-commuting solutions of (1) are where K is any × m m diagonalizable matrix and Z is any ( − ) × ( − ) n m n m diagonalizable matrix such that (i) the nonzero matrices C and D have the same rank r and satisfy ; (ii) K and Z have eigenvalues (iv) the other eigenvalues of K and Z belong to { } 0, 1 and − − 0, Proof. The proof is similar to the proof of Theorem 3.4 and is omitted here.  I  i  I  diag  1  3  2  ,  1  3  2 .  (2) All the non-commuting solutions of (1) are where K is any × m m diagonalizable matrix and Z is any ( − ) × ( − ) n m n m diagonalizable matrix such that (i) the nonzero matrices C and D have the same rank r and satisfy (ii) K and Z have eigenvalues     Next, we will find all the non-commuting solutions of (1). So Z is an arbitrary ( − ) × ( − ) n m n m matrix for all of its solutions. Next, we can get some results for several special cases as follows. . Suppose that the rank of matrix A is m and the multiplicity of eigenvalue 1 is k. where K is any × k k diagonalizable matrix and T is any ( − ) × ( − ) m k m k diagonalizable matrix such that (i) the nonzero matrices F and E have the same rank r such that ; (ii) K and T have eigenvalues − . Its general solution has been constructed in Theorem 3.4. Thus, K is any × k k diagonalizable matrix and T is any ( − ) × ( − ) m k m k diagonalizable matrix such that (i) the nonzero matrices F and E have the same rank r such that ; (ii) K and T have eigenvalues − , respectively; We solve the remaining three equations of (13) to get C and D for each such obtained solution (K, F, E, T). Then the last three equations of (13) are as follows: The first equation of (14) implies accordingly. We have the following main results in this case. . Suppose that the rank of matrix A is m and the multiplicity of eigenvalue 1 is k.
(1) All commuting solutions of (1) are given by (2) All non-commuting solutions of (1) are given by where K is any × k k diagonalizable matrix and T is any ( − ) × ( − ) m k m k diagonalizable matrix such that (i) the nonzero matrices F and E have the same rank r and satisfy ; (ii) K and T have eigenvalues (iv) the other eigenvalues of K and T belong to { } 0, 1 and − − 0, . Z is an arbitrary ( − ) × ( − ) n m n m matrix.
Proof. The proof is similar to the proof of Theorems 3.7 and 3.9 and is omitted here. □ 3.9 Case 9: , then the minimal polynomial of A is accordingly. We have the following main results in this case. (2) All non-commuting solutions of (1) are given by where K is any × k k diagonalizable matrix and T is any ( − ) × ( − ) m k m k diagonalizable matrix such that (i) the nonzero matrices F and E have the same rank r and satisfy (ii) K and T have eigenvalues , respectively; Proof. The proof is similar to the proof of Theorems 3.7 and 3.9 and is omitted here. □

Numerical examples
We present two numerical examples to illustrate our results.    In this paper, we have found some solutions of the Yang-Baxter-like matrix Eq. (1) when the given matrix A satisfies = A A 4 , which has extended the previous results of [18][19][20][21]. Our approach here is to use the Jordan decomposition of A to obtain a simplified Yang-Baxter-like matrix equation with A replaced by a simple block diagonal matrix, and then we solve a system of several matrix equations for the smaller sized solution blocks. The same idea and technique in this paper can be applied to find all solutions of (1) when A satisfies the condition = − A A 4 or when = A A k for some ∈ k N. Once we obtain all the solutions of (1) when A is a diagonalizable complex matrix with three distinct nonzero eigenvalues, we can solve Cases 10 and 11. However, their commuting solution can be obtained by the same way used for these cases before. Finding all the non-commuting solutions of the Yang-Baxter-like matrix Eq. (1) for a general matrix A is a hard task, which will be further studied in the future.