Research Article | Open Access

# The Matrix Equation over Fields or Rings

**Academic Editor:**Zhongshan Li

#### Abstract

Let and let be an algebraically closed field with characteristic 0 or greater than . We show that if and satisfy , then are simultaneously triangularizable. Let be a reduced ring such that is not a zero divisor and let be a generic matrix over ; we show that is the sole solution of . Let be a commutative ring with unity; let be similar to such that, for every is not a zero divisor. If is a nilpotent solution of where , then .

#### 1. Introduction

Let be an integer at least . In the first part, we consider matrices with entries in , a field such that its characteristic is or greater than , and denotes its algebraic closure. In the second part, the entries of the matrices are elements of , a commutative ring with unity ; then, is the ring of endomorphisms of the free -module .

Let and let be a polynomial with coefficients in the complex field such that . In [1, 2], the matrix is given and the authors study the matrix equations in the unknown :

In the present paper, we extend some results obtained in [1, 2] when we replace with or .

Two matrices , are said to be simultaneously triangularizable (abbreviated to ) over if there exists such that and are upper triangular matrices.

The following result is a slight improvement of [1, Theorem 1] or of [3, Theorem 11′].

Theorem 1. *One assumes that or is and that is a polynomial with coefficients in :
**
Then, , are over .*

*Remark 2. *Note that the previous result fails when .

Let be a commutative ring with unity and let be a matrix where the are commuting indeterminates. Let be the ring of the polynomials in the indeterminates and with coefficients in . Thus, the algebra generated by is included in . In particular, there are no polynomial relations, with coefficients in , linking the . We say that is a* generic* matrix over . When is reduced (for every , implies ), we obtain a precise result.

Proposition 3. *Let , and let be a reduced ring such that is not a zero divisor. Let be a generic matrix. Then, is the sole solution of (1).*

Otherwise, we only obtain a partial result.

Proposition 4. *Let be a commutative ring with unity, , and let be a polynomial with coefficients in such that . Let be a nilpotent solution of (2). Then, all elements of the two-sided ideal, in , generated by are nilpotent.*

Yet, if is diagonalizable and its spectrum is “good,” then we obtain a complete solution.

Theorem 5. *Let be similar over to with such that, for every , is not a zero divisor. Let be a nilpotent solution of (2). *(i)*Then, there is such that
where, for every , and .*(ii)*If moreover is a unit, then .*

#### 2. Equation (1) Over a Field

In this section, is a field with characteristic not . Let denote the nilpotent Jordan-block of dimension . If is a square matrix, then denotes its characteristic polynomial and denotes its commutant.

We prove Theorem 1.

*Proof. *Let be the vector space spanned by . One checks easily by induction:

By Cayley-Hamilton’s Theorem (that is valid on a commutative ring with unity), belongs to , and is Lie’s algebra. The derived series of is

Thus, is solvable. According to Lie’s Theorem (that is valid when or is , cf. [4, page 38]), is triangularizable; that is, , are .

*Remark 6. *The hypothesis about is necessary; indeed, if and , then

satisfy ; yet, is not nilpotent.

*Example 7. *Consider the following equation:

It is equivalent to

According to Theorem 1, are over and is nilpotent.

Proposition 8. *Let and let be such that its eigenvalues in have multiplicity at most . Then, the solutions of (1) commute with .*

*Proof. *According to [2, Proposition 2.9], if is a solution of (1), then each generalized eigenspace of is -invariant. Thus, we may assume that has dimension at most . Therefore, and we are done.

We consider the relation:

Proposition 9. *One assumes that or is . Let satisfying (10) such that . Then , is nilpotent and there is and a nilpotent matrix such that ; moreover, . Conversely, there exist such solutions.*

*Proof. *If , then, according to Proposition 8, that is contradictory. Thus, and has a triple eigenvalue . An instance of such a solution is

Let . According to [2, Proposition 2.9], if is a solution of (1), then each generalized eigenspace of is -invariant. Thus, we may assume that is nilpotent.

We consider the algebraic variety:

Let denote the dimension of the maximal component of . Recall that the algebraic variety of nilpotent matrices in has dimension and is irreducible (cf. [5, Section:* The nilpotent cone*]). Note that the algebraic variety

has dimension and is irreducible when (cf. [6]).

Proposition 10. *Let . Then, .*

*Proof. *A generic nilpotent matrix is similar to . Put and consider the equation . According to [2, Remark 3.4], is strictly upper triangular and we can express the entries of as functions of the free variables (the only condition occurs when and is ). Then, the algebraic variety has dimension . We deduce that .

*Remark 11. *We show that (cf. Proposition 9).

*Case 1. *. Then, ; that is, or is similar to ; we obtain a variety of dimension .

*Case 2. * is similar to ; that is, goes through a variety of dimension . From the Gröbner basis method (implemented in Maple), we deduce that the Hilbert dimension of is . Thus, the dimension of the component of associated to this choice of is .

Computational evidence leads us to conjecture that , an expression that does not depend on .

We consider the equation:

A family of such equations is studied in [7]. As in [1, 2], an essential argument used by the authors is that the sequence of iterated kernels is ultimately stationary. Moreover, the properties about the generalized eigenspaces are similar; and have same spectrum over ; moreover, for every , the generalized eigenspaces and are equal. Thus, to study the solutions of (14) can be reduced to study the restrictions of to a generalized eigenspace . Moreover, if is a solution of (14), then, for every , is also a solution of (14). Finally, it suffices to solve (14) when , are assumed to be nilpotent matrices.

Let be a solution of (14). We may wonder whether , are . The answer is no, as we can see with the following solution of (14) when and or :

Clearly , are nilpotent and is invertible. We say that a pair has property (cf. [8]) if there are orderings of the eigenvalues of , such that, for every , the eigenvalues of are ; if , are , then they have property . In our instance, has not property because, if , then is not nilpotent.

Let be integers .

Proposition 12. *Let be a field such that or is . If has distinct eigenvalues in , then is the sole solution of (1).*

*Proof. *Note that satisfies the property:

Indeed, we may assume that is a diagonal matrix over . Since commutes with , is also diagonal and clearly . According to Theorem 1, is nilpotent and too; assume that , the nilindex of , is greater than . According to (5), ; by the property and , we deduce , that is contradictory and therefore . Thus, and ; by Property , we conclude that .

*Remark 13. *The previous result is shown, when is a field of characteristic , in [2, Proposition 2.5].

#### 3. Equations (1) and (2) Over a Ring

(i)Let be commutative rings with unity. Their ring subdirect product is defined if there is an injective ring homomorphism such that, for every , the projection of on is onto.(ii)A commutative ring with unity is reduced if, for every , implies . That is equivalent to say that is isomorphic to a subring of a direct product of fields or isomorphic to a subdirect product of domains (cf. [9, Theorem 11.6.7]). For instance, is a reduced ring with . More generally, is or a product of distinct primes. Note that is reduced with and yet, is a zero divisor.

We show Proposition 3.

*Proof. *Note that, if is reduced, then also is. Since is a subring of a direct product of algebraic closed fields , we may assume where, for every , is a field such that or is . Let and be the associated decompositions. Thus, for any , , where the th component of is generic; then, for every , the discriminant of is not and the matrix has distinct eigenvalues. According to Proposition 12, for every , and consequently .

Proposition 14. *Let be a commutative ring with unity such that is not a zero divisor and let . If is a solution of (1), then is a nilpotent matrix.*

*Proof. *Note that and commute and that the Cayley-Hamilton theorem is true over . According to the proof of Jacobson lemma (cf. introduction of [10] and also [11] where an improvement of this result is stated within the framework of the algebraic operators on a complex Banach space), that implies and we are done.

If is generic over , then we have a more precise result for small .

Proposition 15. *Let , . Let be a commutative ring with unity such that, if , , or , then , or is not a zero divisor. Let be a generic matrix. If is a solution of (1) then
*

*Proof. *The parameters are the . We have a system of equations in the unknowns . Using Gröbner basis theory in any specified characteristic, we obtain the required result.

When and , the calculations have great complexity; thus, we carry out specializations of the in the ring . Then, we randomly choose the matrix (in order to simulate the generic nature of the matrix) and we formally solve (1) in the unknowns . Numerical experiments, again using Gröbner basis theory in characteristic great enough, lead to the following result: and for every , ; for instance, if , , then the supplementary condition is “ is not a zero divisor.” Therefore, we conjecture the following.

*Conjecture.* Let , be a commutative ring with unity satisfying a condition in the form: “the integer is not a zero divisor.” Let be a generic matrix. If is a solution of (1), then and for every , .

*Remark 16. *(i) The instance , shows that if is a nilpotent matrix, then we have not necessarily .

(ii) In the previous conjecture, note that the exponent is very special; indeed, if , then we have

and we cannot do better (cf. [12]).

Let be a commutative ring with unity and . We look for the* nilpotent* solutions of (2), where and is a polynomial in , with coefficients in , such that . Then, according to (5), for every ,

Let denote the valuation of the polynomial , with the following convention: . In the sequel, is a nilpotent solution of (2).

Lemma 17. *Let be a polynomial in . Then, for every , , where for every , is a polynomial in such that .*

*Proof. *For every , with . Then with . In the same way, and with , and so on.

Lemma 18. *Let and be polynomials in with, for every , . Then
**
where for every , is a polynomial in with .*

*Proof. *We use a reasoning by recurrence. Let

where for every , is a polynomial in such that . Using Lemma 17, , where .

Lemma 19. *Let be polynomials in . Then, for every ,
**
where for every , is a polynomial in such that .*

*Proof. *Let . Then , where for every , are monomials in the form . Assume, for instance, that . By Lemma 17,

and finally , are in the form , , where , are polynomials in . Now

where for every , . Using Lemma 18, , where for every , .

We deduce Proposition 4.

*Proof. *Use Lemmas 18 and 19 and the fact that is a nilpotent matrix.

*Remark 20. *(i) When is an arbitrary algebraically closed field, the previous result is equivalent to say that and are over (cf. [13] and compare with Theorem 1).

(ii) When is a ring, McCoy, in [14], gave an equivalent condition that, unfortunately, seems almost useless. In fact, if and , are triangularizable over , then they have not necessarily a common eigenvector; the following example, for , is due to J. Starr and uses the ring of dual numbers:

Moreover, let ; then, unlike the case where the entries of the matrices are in a field, the equation admits a solution that does not commute with ; yet .

Note that we can reduce (theoretically) the resolution of (10) to the case . Indeed, let be satisfying (10). According to (5), . If is a unit, then put , ; therefore, .

We have a more precise result when is diagonalizable and its spectrum is “good.”

Lemma 21. *Let be similar to and such that . *(i)*Assume that, for every , is not a zero divisor in . Then and are simultaneously diagonalizable.*(ii)*Assume that, for every , is a unit. Then is a polynomial in of degree at most and with coefficients in .*

*Proof. *We may assume that . (i)If , then with ; if , then .(ii)According to (i), we may assume that . We must solve the linear system, in the unknowns :

Since the determinant of the associated Vandermonde matrix is a unit, we are done.

Proposition 22. *Let be similar to such that, for every , is not a zero divisor and . If and commute, then .*

*Proof. *We may assume that and put , . We obtain ; therefore, if , then .

The proof of the following is interesting in itself; yet, Theorem 5 is a stronger result.

Proposition 23. *Let be similar in to such that, for every , is not a zero divisor. If is not a zero divisor and is a solution of (1), then there are and such that
*

*Proof. *According to Proposition 14, is nilpotent. Assume that the nilindex of is . According to (5), and commute. Since , and commute. According to Proposition 22, . Then we obtain a finite sequence of matrices

that commute with and where, at each step, the exponent decreases by . Finally, we obtain a matrix , with , that commutes with . Since , . Finally and commute that implies .

We show Theorem 5.

*Proof. *We may assume that .

(i)(a)For every , , where is the canonical basis of . Indeed, if and , then, for every , . Thus, for every , .(b). Indeed, this comes from the proof of the following.

Theorem 24 (cf. [1, Theorem 3]). *Let be an algebraically closed field with characteristic and . Assume that the matrix satisfies (2) and that is a nilpotent matrix. Then, the generalized eigenspaces of are -invariant.*

The formal proof remains valid when is replaced with a ring and the generalized eigenspaces simply are eigenspaces. Therefore, .(c)In the same way as in the proof of Lemma 21(i), we obtain that has the required form.

(ii) In , is a unit and is nilpotent; therefore, is a unit and we are done.

*Remark 25. *In the previous proposition, consider a matrix . According to [15, Theorem 8.54], is the resultant . Therefore, if is a unit, then is zero again; in general, it is not, as we see in the following instance. Let such that , , , , and . Then, , , and .

#### Disclosure

The 2010 Mathematic Subject Classification: Primary 15A24, 13Axx and Secondary 13P10.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The author thanks J. Bračič for helpful discussions. The author wishes to thank the referees for helpful comments.

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#### Copyright

Copyright © 2014 Gerald Bourgeois. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.