# The Matrix EquationXA-AX=Xαg(X) over Fields or Rings

Authors: Gerald Bourgeois
Journal: Algebra
 (2014)

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## Abstract

Letn,α∈N≥2 and let K be an algebraically closed field with characteristic 0 or greater than n. We show that if f∈K[X] and A,B∈Mn(K) satisfy [A,B]=f(A), then A,B are simultaneously triangularizable. Let R be a reduced ring such that n! is not a zero divisor and let A be a generic matrix over R; we show that X=0 is the sole solution of AX-XA=Xα. Let R be a commutative ring with unity; let A be similar to diag(λ1In1,…,λrInr) such that, for every i≠j,λi-λj is not a zero divisor. If X is a nilpotent solution of XA-AX=Xαg(X) where g∈R[X], then AX=XA.


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## Body

## 1. Introduction

Letn be an integer at least 2. In the first part, we consider n×n matrices with entries in K, a field such that its characteristic char(K) is 0 or greater than n, and K¯ denotes its algebraic closure. In the second part, the entries of the matrices are elements of R, a commutative ring with unity 1≠0; then, Mn(R) is the ring of endomorphisms of the free R-module Rn.Letα∈N≥2 and let g be a polynomial with coefficients in the complex field C such that g(0)≠0. In [1, 2], the matrix A∈Mn(C) is given and the authors study the matrix equations in the unknown X∈Mn(C):
(1)XA-AX=Xα,(2)XA-AX=Xαg(X),whereXisnilpotent.In the present paper, we extend some results obtained in [1, 2] when we replace C with K or R.Two matricesA, B∈Mn(K) are said to be simultaneously triangularizable (abbreviated to ST) over K if there exists P∈GLn(K) such that P-1AP and P-1BP are upper triangular matrices.The following result is a slight improvement of [1, Theorem 1] or of [3, Theorem 11′].Theorem 1.
One assumes thatchar(K)>n or is 0 and that f is a polynomial with coefficients in K:
(3)LetA,B∈Mn(K)besuchthat[A,B]=f(A).
Then,A, B are ST over K¯.Remark 2.
Note that the previous result fails whenchar(K)=n.We now consider (1) and (2).LetR be a commutative ring with unity 1≠0 and let A=[aij] be a n×n matrix where the {aij}1≤i,j≤n are commuting indeterminates. Let R~ be the ring of the polynomials in the indeterminates {aij}1≤i,j≤n and with coefficients in R. Thus, the algebra generated by A is included in Mn(R~). In particular, there are no polynomial relations, with coefficients in R, linking the {aij}1≤i,j≤n. We say that A is ageneric matrix over R. When R is reduced (for every u∈R, u2=0 implies u=0), we obtain a precise result.Proposition 3.
Letn≥2, and let R be a reduced ring such that n! is not a zero divisor. Let A∈Mn(R~) be a generic matrix. Then, X=0 is the sole solution of (1).Otherwise, we only obtain a partial result.Proposition 4.
LetR be a commutative ring with unity, α≥2, and let g be a polynomial with coefficients in R such that g(0)≠0. Let X∈Mn(R) be a nilpotent solution of (2). Then, all elements of the two-sided ideal, in Mn(R), generated by XA-AX are nilpotent.Yet, ifA is diagonalizable and its spectrum is “good,” then we obtain a complete solution.Theorem 5.
LetA∈Mn(R) be similar over R to diag⁡(λ1In1,…,λrInr) with n1+⋯+nr=n such that, for every i≠j, λi-λj is not a zero divisor. Let X be a nilpotent solution of (2). (i)
Then, there isP∈GLn(R) such that
(4)A=Pdiag⁡(λ1In1,…,λrInr)P-1,X=Pdiag⁡(X1,…,Xr)P-1,
where, for every i, Xi∈Mni(R) and Xiαg(Xi)=0.(ii)
If moreoverg(0) is a unit, then Xα=0.


## 2. Equation (1) Over a Field

In this section,K is a field with characteristic not 2. Let Jn denote the nilpotent Jordan-block of dimension n. If A is a square matrix, then χA denotes its characteristic polynomial and C(A) denotes its commutant.We prove Theorem1.Proof.
LetV be the vector space spanned by {B,In,A,…,An-1}. One checks easily by induction:
(5)∀i≥1,AiB-BAi=iAi-1f(A).
By Cayley-Hamilton’s Theorem (that is valid on a commutative ring with unity),AiB-BAi belongs to V, and V is Lie’s algebra. The derived series of V is
(6)V1=[V,V]⊂K[A],V2=[V1,V1]={0}.
Thus,V is solvable. According to Lie’s Theorem (that is valid when char(K)>n or is 0, cf. [4, page 38]), V is triangularizable; that is, A, B are ST.Remark 6.
The hypothesis aboutchar(K) is necessary; indeed, if n=3 and char(K)=3, then
(7)A=(010002-100),B=(00000-1100)
satisfy[A,B]=A2; yet, A3=I3 is not nilpotent.Example 7.
Consider the following equation:(8)A2-2AB+B2=P(A-B),wherePisapolynomial.
It is equivalent to(9)N2-P(N)=[N,B],whereN=A-B.
According to Theorem1, A,B are ST over K¯ and A-B is nilpotent.Proposition 8.
Letα∈⟦2,n-1⟧ and let  A∈Mn(K) be such that its eigenvalues in K¯ have multiplicity at most α. Then, the solutions of (1) commute with A.Proof.
According to [2, Proposition 2.9], if X∈Mn(K¯) is a solution of (1), then each generalized eigenspace of A∈Mn(K¯) is X-invariant. Thus, we may assume that A has dimension at most α. Therefore, Xα=0 and we are done.We consider the relation:(10)AB-BA=Aα,whereα∈N≥2.Proposition 9.
One assumes thatchar(K)>3 or is 0. Let A,B∈M3(K) satisfying (10) such that AB≠BA. Then α=2, A is nilpotent and there is λ∈K and a nilpotent matrix N such that B=λI3+N; moreover, AN-NA=A2. Conversely, there exist such solutions.Proof.
Ifα≥3, then, according to Proposition 8, AB=BA that is contradictory. Thus, α=2 and B has a triple eigenvalue λ=3-1trace(B)∈K. An instance of such a solution is
(11)A=J3,N=(0-2000-1000).Letα∈⟦2,n-1⟧. According to [2, Proposition 2.9], if X∈Mn(K¯) is a solution of (1), then each generalized eigenspace of A∈Mn(K¯) is X-invariant. Thus, we may assume that A is nilpotent.We consider the algebraic variety:(12)Sn,α={(A,B)∈Mn(K¯)2∣Bisnilpotentss(K¯)2and(A,B)satisfies(10)}.Letdim⁡(Sn,α) denote the dimension of the maximal component of Sn,α. Recall that the algebraic variety Nn of nilpotent matrices in Mn(K¯) has dimension n2-n and is irreducible (cf. [5, Section:The nilpotent cone]). Note that the algebraic variety
(13)Wn={(A,B)∈Mn(K¯)2∣A,BarenilpotentandAB=BA(K¯)2}has dimensiondim⁡(Nn)+(n-1)=n2-1 and is irreducible when char(K)>n (cf. [6]).Proposition 10.
Letα∈⟦2,n-1⟧. Then, dim⁡(Sn,α)≥n2-1.Proof.
A generic nilpotent matrixB is similar to Jn. Put B=Jn and consider the equation XJn-JnX=Xα. According to [2, Remark 3.4], X is strictly upper triangular and we can express the entries {xij}1≤i,j≤n of X as functions of the free variables x1,2,…,x1,n (the only condition occurs when n=2 and is x1,2∉{-1/k∣k∈N*}). Then, the algebraic variety Yn,α={X∣XJn-JnX=Xα} has dimension n-1. We deduce that dim⁡(Sn,α)≥dim⁡(Nn)+dim⁡(Yn,α)=n2-1.Remark 11.
We show thatdim⁡(S3,2)=8 (cf. Proposition 9).Case 1.
B
=
0. Then, A2=0; that is, A=0 or A is similar to U=diag⁡(0,J2); we obtain a variety of dimension n2-dim⁡(C(U))=9-5=4.Case 2.
B is similar to U; that is, B goes through a variety of dimension  4. From the Gröbner basis method (implemented in Maple), we deduce that the Hilbert dimension of {X∣XU-UX=X2} is 3. Thus, the dimension of the component of Sn,α associated to this choice of B is 7.Computational evidence leads us to conjecture thatdim⁡(Sn,α)=n2-1, an expression that does not depend on α.We consider the equation:(14)A3-3A2B+3AB2-B3=0n.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 (ker⁡(A-λI)k)k is ultimately stationary. Moreover, the properties about the generalized eigenspaces are similar; A and B have same spectrum (λi)i over K¯; moreover, for every i, the generalized eigenspaces Eλi(A) and Eλi(B) are equal. Thus, to study the solutions (A,B) of (14) can be reduced to study the restrictions of A,B to a generalized eigenspace Eλ(A)=Eλ(B). Moreover, if (A,B) is a solution of (14), then, for every μ∈K, (A-μIn,B-μIn) is also a solution of (14). Finally, it suffices to solve (14) when A, B are assumed to be nilpotent matrices.Let(A,B) be a solution of (14). We may wonder whether A, B are ST. The answer is no, as we can see with the following solution of (14) when n=4 and char(K)≥5 or =0:
(15)((043-13-11034-3410001000),J4).ClearlyA, B are nilpotent and [A,B] is invertible. We say that a pair (U,V) has property L (cf. [8]) if there are orderings of the eigenvalues (λi),(μi) of U, V such that, for every a∈K, the eigenvalues of U+aV are (λi+aμi); if U, V are ST, then they have property L. In our instance, (A,B) has not property L because, if a≠0, then A+aB is not nilpotent.Letα,n be integers ≥2.Proposition 12.
LetK be a field such that char(K)>n or is 0. If A∈Mn(K) has n distinct eigenvalues in K¯, then X=0 is the sole solution of (1).Proof.
Note thatA satisfies the property:
(16)P:foreveryY∈Mn(K),{AY=YA,Yα=0}implyY=0.
Indeed, we may assume thatA is a diagonal matrix over K¯. Since Y commutes with A, Y is also diagonal and clearly Y=0. According to Theorem 1, [A,X] is nilpotent and X too; assume that k, the nilindex of X, is greater than α. According to (5), [A,Xk-α+1]=0; by the property P and X(k-α+1)α=0, we deduce Xk-α+1=0, that is contradictory and therefore k≤α. Thus, Xα=0 and AX=XA; by Property P, we conclude that X=0.Remark 13.
The previous result is shown, whenK is a field of characteristic 0, in [2, Proposition 2.5].


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

(i)
Let(Ri)i∈I be commutative rings with unity. Their ring subdirect product R is defined if there is f:R→Πi∈IRi an injective ring homomorphism such that, for every j∈I, the projection of f on Rj is onto.
(ii)
A commutative ringR with unity is reduced if, for every u∈R, u2=0 implies u=0. That is equivalent to say that R 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, R=Z×Z is a reduced ring with char(R)=0. More generally, char(R) is 0 or a product of distinct primes. Note that R=Z/3Z×Z is reduced with char(R)=0 and yet, 3=(0,3) is a zero divisor.We show Proposition3.Proof.
Note that, ifR is reduced, then R~ also is. Since R~ is a subring of a direct product of algebraic closed fields Πi∈IKi, we may assume R~=Πi∈IKi where, for every i∈I, Ki is a field such that char(Ki)>n or is 0. Let X=(Xi)i and A=(Ai)i be the associated decompositions. Thus, for any i∈I, XiAi-AiXi=Xiα, where the ith component Ai∈Mn(Ki) of A is generic; then, for every i, the discriminant of χAi is not 0 and the matrix Ai has n distinct eigenvalues. According to Proposition 12, for every i∈I, Xi=0 and consequently X=0.Proposition 14.
LetR be a commutative ring with unity such that n! is not a zero divisor and let A∈Mn(R). If X∈Mn(R) is a solution of (1), then X is a nilpotent matrix.Proof.
Note that[A,X] and X commute and that the Cayley-Hamilton theorem is true over R. 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), n![A,X]2n-1=0 that implies [A,X]2n-1=0 and we are done.IfA is generic over R, then we have a more precise result for small n,α.Proposition 15.
Letn=2, 2≤α≤4. Let R be a commutative ring with unity such that, if α=2, 3, or 4, then 2, 3! or 5! is not a zero divisor. Let A∈M2(R~) be a generic matrix. If X=[xij]∈M2(R~) is a solution of (1) then
(17)AX-XA=Xα=0,forevery(i,j),xij2α-1=0.Proof.
The4 parameters are the (aij). We have a system of 4 equations in the 4 unknowns (xij). Using Gröbner basis theory in any specified characteristic, we obtain the required result.Whenn=3,4,5 and 2≤α≤4, the calculations have great complexity; thus, we carry out specializations of the (aij) in the ring R. Then, we randomly choose the matrix A (in order to simulate the generic nature of the matrix) and we formally solve (1) in the n2 unknowns (xij). Numerical experiments, again using Gröbner basis theory in characteristic great enough, lead to the following result: Xα=0n and for every (i,j), xij(α-1)n+1=0; for instance, if n=5, α=4, then the supplementary condition is “13! is not a zero divisor.” Therefore, we conjecture the following.Conjecture. Let n≥2, R be a commutative ring with unity satisfying a condition in the form: “the integer ϕ(n,α)! is not a zero divisor.” Let A∈Mn(R~) be a generic matrix. If X=[xij]∈Mn(R~) is a solution of (1), then AX-XA=Xα=0n and for every (i,j), xij(α-1)n+1=0.Remark 16.
(i) The instanceR=Z/27Z, U=3I2 shows that if U is a nilpotent n×n matrix, then we have not necessarily Un=0.
(ii) In the previous conjecture, note that the exponent(α-1)n+1 is very special; indeed, if Uα=0n, then we have
(18)(trace(U))(α-1)n+1=0,
and we cannot do better (cf. [12]).LetR be a commutative ring with unity and A∈Mn(R). We look for thenilpotent solutions X∈Mn(R) of (2), where α≥2 and g is a polynomial in X, with coefficients in R, such that g(0)≠0. Then, according to (5), for every i,
(19)XiA-AXi=iXα+i-1g(X).Letval(T) denote the valuation of the polynomial T, with the following convention: val(0)=+∞. In the sequel, X is a nilpotent solution of (2).Lemma 17.
Letu be a polynomial in X. Then, for every l, uAl=∑i=0l‍Aivi, where for every i, vi is a polynomial in X such that val(vi)≥val(u).Proof.
For everyi, XiA=AXi+v0(X) with val(v0)≥i. Then uA=Au+v1(X) with val(v1)≥val(u). In the same way, uA2=A2u+2Av1+v2 and uA3=A3u+3A2v1+3Av2+v3 with val(v3)≥val(v2)≥val(v1), and so on.Lemma 18.
Letk≥2 and (vi)i be polynomials in X with, for every i, val(vi)≥α. Then
(20)(∑i‍Aivi)kisintheform∑i‍Aiwi,
where for everyi, wi is a polynomial in X with val(wi)≥kα.Proof.
We use a reasoning by recurrence. Let(21)E=(∑i‍Aivi)k=(∑i‍Aizi)(∑j‍Ajvj),
where for everyi, zi is a polynomial in X such that val(zi)≥(k-1)α. Using Lemma 17, E=∑i,j‍Ai(ziAj)vj=∑i,j‍Ai∑r≤j‍Aryrivj, where val(yri)≥(k-1)α.Lemma 19.
LetP,Q be polynomials in X,A. Then, for every k,
(22)(P(AX-XA)Q)kisintheform∑i‍Aivi,
where for everyi, vi is a polynomial in X such that val(vi)≥kα.Proof.
LetE=(P(AX-XA)Q)k=(PXαgQ)k. Then E=∑‍Πi=1k(PiXαgQi), where for every i, Pi,Qi are monomials in the form Ai1Xj1Ai2Xj2…. Assume, for instance, that Pi=Ai1Xj1Ai2Xj2Ai3Xj3. By Lemma 17,
(23)Pi=Ai1Xj1Ai2(∑r‍Arur)Xj3=∑‍Ai1Xj1Ai2+rurXj3=∑‍∑‍Ai1AswsusXj3
and finallyPi, Qi are in the form ∑j‍Ajpj, ∑j‍Ajqj, where pj, qj are polynomials in X. Now
(24)E=∑‍Πi=1k(∑j‍Ajpj)Xαg(∑j‍Ajqj)=∑‍(Ai1p1XαgAj1q1)⋯(AikpkXαgAjkqk)=∑‍(Ai1v~1)⋯(Aikv~k),
where for everyi, val(v~i)≥α. Using Lemma 18, E=∑i‍Aivi, where for every i, val(vi)≥kα.We deduce Proposition4.Proof.
Use Lemmas18 and 19 and the fact that X is a nilpotent matrix.Remark 20.
(i) WhenR is an arbitrary algebraically closed field, the previous result is equivalent to say that A and X are ST over R (cf. [13] and compare with Theorem 1).
(ii) WhenR is a ring, McCoy, in [14], gave an equivalent condition that, unfortunately, seems almost useless. In fact, if AB=BA and A, B are triangularizable over R, then they have not necessarily a common eigenvector; the following example, for n=2, is due to J. Starr and uses the ring of dual numbers:
(25)R=C[ϵ]〈ϵ2〉,A=(0ϵ00),B=(00ϵ0).
Moreover, letA=(-ϵ2ϵ+105ϵ); then, unlike the case where the entries of the matrices are in a field, the equation XA-AX=X2 admits a solution X=(4ϵ-30ϵ) that does not commute with A; yet X3=0.Note that we can reduce (theoretically) the resolution of (10) to the case α=2. Indeed, let (A,B) be satisfying (10). According to (5), Aα-1B-BAα-1=(α-1)A2α-2. If α-1 is a unit, then put B=(α-1)B1, A1=Aα-1; therefore, A1B1-B1A1=A12.We have a more precise result whenA is diagonalizable and its spectrum is “good.”Lemma 21.
LetA∈Mn(R) be similar to diag⁡(λ1,…,λn) and B∈Mn(R) such that AB=BA. (i)
Assume that, for everyi≠j, λi-λj is not a zero divisor in R. Then A and B are simultaneously diagonalizable.(ii)
Assume that, for everyi≠j, λi-λj is a unit. Then B is a polynomial in A of degree at most n-1 and with coefficients in R.Proof.
We may assume thatA=diag⁡(λ1,…,λn). (i)
IfB=[bij], then AB-BA=[cij] with cij=(λi-λj)bij; if i≠j, then bij=0.(ii)
According to (i), we may assume thatB=diag⁡(μ1,…,μn). We must solve the linear system, in the unknowns (αi)0≤i<n:
(26)foreveryj,μj=∑i‍αiλji.
Since the determinant of the associated Vandermonde matrix is a unit, we are done.Proposition 22.
LetA∈Mn(R) be similar to diag⁡(λ1,…,λn) such that, for every i≠j, λi-λj is not a zero divisor and B∈Mn(R). If A and [A,B] commute, then AB=BA.Proof.
We may assume thatA=diag⁡(λ1,…,λn) and put B=[bij], D=[di,j]=A[A,B]-[A,B]A. We obtain di,j=(λi-λj)2bij=0; therefore, if i≠j, then bij=0.The proof of the following is interesting in itself; yet, Theorem5 is a stronger result.Proposition 23.
LetA∈Mn(R) be similar in R to diag⁡(λ1,…,λn) such that, for every i≠j, λi-λj is not a zero divisor. If n! is not a zero divisor and X∈Mn(R) is a solution of (1), then there are  P∈GLn(R) and {μi}1≤i≤n∈Rn such that
(27)A=Pdiag⁡(λ1,…,λn)P-1;X=Pdiag⁡(μ1,…,μn)P-1;foreveryi,μiα=0.Proof.
According to Proposition14, X is nilpotent. Assume that the nilindex of X is i+α-1. According to (5), Xi and A commute. Since Xi-α+1A-AXi-α+1=(i-α+1)Xi, [Xi-(α-1),A] and A commute. According to Proposition 22, [Xi-(α-1),A]=0. Then we obtain a finite sequence of matrices
(28)(Xi-k(α-1))0≤k<i/(α-1)
that commute withA and where, at each step, the exponent decreases by α-1. Finally, we obtain a matrix Xβ, with β∈⟦1,α-1⟧, that commutes with A. Since Xβ+α-1=0, [Xα,A]=αXβ+α-1Xα-β=0. Finally [X,A] and A commute that implies [A,X]=0.We show Theorem5.Proof.
We may assume thatA=diag⁡(λ1In1,…,λrInr)=diag⁡(μ1,…,μn).
(i)(a)
For everyk≥1, ker⁡((A-λ1I)k)=[e1,…,en1], where (ei)1≤i≤n is the canonical basis of Rn. Indeed, if x=[x1,…,xn]T and (A-λ1I)kx=0, then, for every i, (μi-λ1)kxi=0. Thus, for every i>n1, xi=0.(b)
X(ker⁡(A-λ1I))⊂ker⁡(A-λ1I). Indeed, this comes from the proof of the following.Theorem  24 (cf. [1, Theorem 3]). LetK be an algebraically closed field with characteristic 0 and A∈Mn(K). Assume that the matrix X∈Mn(K) satisfies (2) and that X is a nilpotent matrix. Then, the generalized eigenspaces of A are X-invariant.The formal proof remains valid whenK is replaced with a ring and the generalized eigenspaces simply are eigenspaces. Therefore, AX=XA.(c)
In the same way as in the proof of Lemma21(i), we obtain that X has the required form.(ii) InMn(R), g(0)In is a unit and g(X)-g(0)In is nilpotent; therefore, g(X) is a unit and we are done.Remark 25.
In the previous proposition, consider a matrixXi. According to [15, Theorem 8.54], det⁡(g(Xi)) is the resultant Res(χXi,g). Therefore, if Res(χXi,g) is a unit, then Xiα is zero again; in general, it is not, as we see in the following instance. Let τ∈R such that τ2≠0, τ3=0, Xi=τI2, g(x)=τ, and α=2. Then, Xi2≠0, Xi3=0, and Xi2g(Xi)=0.



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