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ON THE MATRIX EQUATION XA−AX=X
DIETRICH BURDE
Abstract. We study the matrix equation XA − AX = Xp in M (K) for 1 < p < n. It is
n
shown that every matrix solution X is nilpotent and that the generalized eigenspaces of A are
X-invariant. For A being a full Jordan block we describe how to compute all matrix solutions.
m ℓ ℓ m ℓ
Combinatorial formulas for A X ,X A and (AX) are given. The case p = 2 is a special
case of the algebraic Riccati equation.
1. Introduction
Let p be a positive integer. The matrix equation
XA−AX=Xp
arises from questions in Lie theory. In particular, the quadratic matrix equation XA−AX = X2
plays a role in the study of affine structures on solvable Lie algebras.
An affine structure on a Lie algebra g over a field K is a K-bilinear product g × g → g,
(x,y) 7→ x · y such that
x·(y·z)−(x·y)·z =y·(x·z)−(y·x)·z
[x,y] = x · y − y · x
for all x,y,z ∈ g where [x,y] denotes the Lie bracket of g. Affine structures on Lie algebras
correspond to left-invariant affine structures on Lie groups. They are important for affine
manifolds and for affine crystallographic groups, see [1], [3], [5].
We want to explain how the quadratic matrix equations XA − AX = X2 arise from affine
structures. Let g be a two-step solvable Lie algebra. This means we have an exact sequence of
Lie algebras
ι π
0 →a−→g−→b→0
with the following data: a and b are abelian Lie algebras, ϕ : b 7→ End(a) is a Lie algebra
representation, Ω ∈ Z2(b,a) is a 2-cocycle, and the Lie bracket of g = a × b is given by
[(a,x),(b,y)] := (ϕ(x)b − ϕ(y)a + Ω(x,y),0).
Let ω : b × b → a be a bilinear map and ϕ1, ϕ2 : b 7→ End(a) Lie algebra representations. A
natural choice for a left-symmetric product on g is the bilinear product given by
(a,x)◦(b,y) := (ϕ1(y)a+ϕ2(x)b+ω(x,y),0).
One of the necessary conditions for the product to be left-symmetric is the following:
ϕ1(x)ϕ(y)−ϕ(y)ϕ1(x) = ϕ1(y)ϕ1(x).
Date: January 27, 2006.
1991 Mathematics Subject Classification. Primary 15A24.
Key words and phrases. Algebraic Riccati equation, weighted Stirling numbers.
I thank Joachim Mahnkopf for helpful remarks.
1
2 D. BURDE
Let (e ,...e ) be a basis of b and write X := ϕ (e ) and A := ϕ(e ) for the linear operators.
1 m i 1 i j j
Weobtain the matrix equations
XA −AX =XX
i j j i j i
for all 1 ≤ i,j ≤ m. In particular we have matrix equations of the type XA − AX = X2.
2. General results
Let K be an algebraically closed field of characteristic zero. In general it is quite difficult to
determine the matrix solutions of a nonlinear matrix equation. Even the existence of solutions
is a serious issue as illustrated by the quadratic matrix equation
X2 = 0 1
0 0
which has no solution. On the other hand our equation XA−AX = Xp always has a solution,
for any given A, namely X = 0. However, if A has a multiple eigenvalue, then we have a lot
of nontrivial solutions and there is no easy way to describe the solution set algebraically. A
special set of solutions is obtained by the matrices X satisfying XA−AX = 0 = Xp. First one
can determine the matrices X commuting with A and then pick out those satisfying Xp = 0.
Let E denote the n × n identity. We will assume most of time that p ≥ 2 since for p = 1 we
obtain the linear matrix equation AX +X(E −A) = 0 which is a special case of the Sylvester
matrix equation AX + XB = C. Let S : Mn(K) → Mn(K) with S(X) = AX +XB be the
Sylvester operator. It is well known that the linear operator S is singular if and only if A and
−Bhave a common eigenvalue, see [4]. For B = E −A we obtain the following result.
Proposition 2.1. The matrix equation XA−AX = X has a nonzero solution if and only if
Aand A−E have a common eigenvalue.
The general solution of the matrix equation AX = XB is given in [2]. We have the following
results on the solutions of our general equation.
Proposition 2.2. Let A ∈ M (K). Then every matrix solution X ∈ M (K) of XA−AX = Xp
n n
is nilpotent and hence satisfies Xn = 0.
Proof. We have Xk(XA−AX) = Xk+p for all k ≥ 0. Taking the trace on both sides we obtain
tr(Xk+p) = 0 for all k ≥ 0. Let λ ,...,λ be the pairwise distinct eigenvalues of X. For s ≥ 1
1 r
we have
r
tr(Xs) = Xmiλs.
i
i=1
For s ≥ p we have tr(Xs) = 0 and hence
r
X(miλp)λk =0
i i
i=1
for all k ≥ 0. This is a system of linear equations in the unknowns x = m λp for i = 1,2,...,r.
i i i
The determinant of its coefficients is a Vandermonde determinant. It is nonzero since the λi
are pairwise distinct. Hence it follows m λp = 0 for all i = 1,2,...,r. This means λ = λ =
i i 1 2
· · · = λ = 0 so that X is nilpotent with Xn = 0.
r
XA−AX=Xp 3
Since for p = n our equation reduces to Xn = 0 and the linear matrix equation XA = AX,
we may assume that p < n.
Proposition 2.3. Let K be an algebraically closed field and p be a positive integer. If X,A ∈
M(K)satisfy XA−AX =Xp then X and A can be simultaneously triangularized.
n
Proof. Let V be the vector space generated by A and all Xi. Since X is nilpotent we can choose
a minimal m ∈ N such that Xm = 0. Then V = span{A,X,X2,...,Xm−1}. We define a Lie
bracket on V by taking commutators. Using induction on ℓ we see that for all ℓ ≥ 1
(1) XℓA−AXℓ=ℓXp+ℓ−1
Hence the Lie brackets are defined by
[A,A] = 0
[A,Xi] = AXi −XiA = −iXp+i−1
[Xi,Xj] = 0.
It follows that V is a finite-dimensional Lie algebra. The commutator Lie algebra [V,V] is
abelian and V/[V,V] is 1-dimensional. Hence V is solvable. By Lie’s theorem V is triangular-
izable. Hence there is a basis such that X and A are simultaneously upper triangular.
p i k j
Corollary 2.4. Let X,A ∈ Mn(K) satisfy the matrix equation XA−AX = X . Then A X A
is nilpotent for all k ≥ 1 and i,j ≥ 0. So are linear combinations of such matrices.
Proof. We mayassumethatX andAaresimultaneouslyuppertriangular. SinceX isnilpotent,
Xk is strictly upper triangular. The product of such a matrix with an upper triangular matrix
i j
A or A is again strictly upper triangular. Moreover a linear combination of strictly upper
triangular matrices is again strictly upper triangular.
Proposition 2.5. Let p ≥ 2 and A ∈ Mn(K). If A has no multiple eigenvalue then X = 0 is
the only matrix solution of XA − AX = Xp. Conversely if A has a multiple eigenvalue then
there exists a nontrivial solution X 6= 0.
Proof. Assume first that A has no multiple eigenvalue. Let B = (e ,...,e ) be a basis of Kn
1 n
such that A = (a ) and X = (x ) are upper triangular relative to B. In particular a =0for
ij ij ij
i > j and xij = 0 for i ≥ j. Since all eigenvalues of A are distinct, A is diagonalizable. We
can diagonalize A by a base change of the form e 7→ µ e +µ e +···+µ e which also keeps
i 1 1 2 2 i i
X strictly upper triangular. Hence we may assume that A is diagonal and X is strictly upper
triangular. Then the coefficients of the matrix XA−AX = (cij) satisfy
c =x (a −a ), x =0fori≥j.
ij ij jj ii ij
Consider the lowest nonzero line parallel to the main diagonal in X. Since α −α 6=0for all
jj ii
i 6= j this line stays also nonzero in XA−AX, but not in Xp because of p ≥ 2. It follows that
X=0.
Now assume that A has a multiple eigenvalue. There exists a basis of Kn such that A has
canonical Jordan block form. Each Jordan block is an matrix of the form
λ 1 ... 0 0
0 λ ... 0 0
. . . . .
J(r,λ) = . . .. . . ∈ Mr(K).
. . . .
0 0 . . . λ 1
0 0 . . . 0 λ
4 D. BURDE
For λ = 0 we put J(r) = J(r,0). It is J(r,λ) = J(r) + λE. Consider the matrix equation
XJ(r,λ)−J(r,λ)X =Xp in M (K). It is equivalent to the equation XJ(r)−J(r)X = Xp. If
r
r ≥ 2 it has a nonzero solution, namely the r × r matrix
0 . . . 1
. . .
X= . .. . .
. .
0 . . . 0
Indeed, XJ(r) − J(r)X = 0 = Xp in that case. Since A has a multiple eigenvalue, it has a
Jordan block of size r ≥ 2. After permutation we may assume that this is the first Jordan
block of A. Let X ∈ Mr(K) be the above matrix and extend it to an n×n-matrix by forming
a block matrix with X and the zero matrix in Mn−r(K). This will be a nontrivial solution of
XA−AX=XpinM (K).
n
Lemma 2.6. Let A,X ∈ M (K) and A = SAS−1, X = SXS−1 for some S ∈ GL (K).
n 1 1 n
Then XA−AX =Xp if and only if X A −A X =Xp.
1 1 1 1 1
Proof. The equation Xp = XA−AX is equivalent to
p −1 p p −1
X =(SXS ) =SX S
1
=(SXS−1)(SAS−1)−(SAS−1)(SXS−1)
=XA −AX.
1 1 1 1
The lemma says that we may choose a basis of Kn such that A has canonical Jordan form.
Denote by C(A) = {S ∈ Mn(K) | SA = AS} the centralizer of A ∈ Mn(K). Applying the
lemma with A = SAS−1 = A, where S ∈ C(A)∩GL (K), we obtain the following corollary.
1 n
Corollary 2.7. If X is a matrix solution of XA−AX = Xp then so is X = SX S−1 for any
0 0
S ∈ C(A)∩GLn(K).
Let B = (e ,...,e ) be a basis of Kn such that A has canonical Jordan form. Then A is a
1 n
block matrix
A=diag(A ,A ,...,A )
1 2 k
with A ∈ M (K) and A leaves invariant the corresponding subspaces of Kn. Let X satisfy
i ri
XA−AX = Xp. Does it follow that X is also a block matrix X = diag(X ,...,X ) with
1 r
X ∈M (K)relative to the basis B ? In general this is not the case.
i ri
Example 2.8. The matrices
0 0 0 −1 0 1
A= 0 0 1 , X= −1 0 0
0 0 0 −1 0 1
satisfy XA−AX =X2.
Here A = diag(J(1),J(2)) leaves invariant the subspaces span{e } and span{e ,e } corre-
1 2 3
sponding to the Jordan blocks J(1) and J(2), but X does not. Also the subspace kerA is
not X-invariant. This shows that the eigenspaces E = {x ∈ Kn | Ax = λx} of A need not
λ
be X-invariant. However, we have the following result concerning the generalized eigenspaces
n k
H ={x∈K |(A−λE) x=0forsomek≥0}ofA.
λ
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