On commuting matrices and exponentials

Let A and B be matrices of M_n(C). We show that if exp(A)^k exp(B)^l=exp(kA+lB) for all integers k and l, then AB=BA. We also show that if exp(A)^k exp(B)=exp(B)exp(A)^k=exp(kA+B)$ for every positive integer k, then the pair (A,B) has property L of Motzkin and Taussky. As a consequence, if G is a subgroup of (M_n(C),+) and M ->exp(M) is a homomorphism from G to (GL_n(C),x), then G consists of commuting matrices. If S is a subsemigroup of (M_n(C),+) and M ->exp(M) is a homomorphism from S to (GL_n(C),x), then the linear subspace Span(S) of M_n(C) has property L of Motzkin and Taussky.


Notation and definition
iii) The n × n complex matrices A, B are said to be simultaneously triangularizable if there exists an invertible matrix P such that P −1 AP and P −1 BP are upper triangular.
iv) A pair (A, B) of complex n × n matrices is said to have property L if for a special ordering (λ i ) 1≤i≤n , (µ i ) 1≤i≤n of the eigenvalues of A, B, the eigenvalues of xA+yB are (xλ i +yµ i ) 1≤i≤n for all values of the complex numbers x, y.

The problem
It is well known that the exponential is not a group homomorphism from (M n (C), +) to (GL n (C), ×) if n ≥ 2. Nevertheless, when A and B are commuting matrices of M n (C), one has e A+B = e A e B = e B e A .
However (1) is not a sufficient condition for the commutativity of A with B, nor even for A and B to be simultaneously triangularizable. Still, if ∀t ∈ R, e tA e tB = e tB e tA , or ∀t ∈ R, e t(A+B) = e tA e tB , then a power series expansion at t = 0 shows that AB = BA. In the 1950s, pairs of matrices (A, B) of small size such that e A+B = e A e B have been under extensive scrutiny [3,4,6,7,9,10]. More recently, Wermuth [16,17] and Schmoeger [14,15] studied the problem of adding extra conditions on the matrices A and B for the commutativity of e A with e B to imply the commutativity of A with B. A few years ago, Bourgeois (see [1]) investigated, for small n, the pairs (A, B) ∈ M n (C) 2 that satisfy ∀k ∈ N, e kA+B = e kA e B = e B e kA .
The main interest in this condition lies in the fact that, contrary to conditions (2) and (3), it is not possible to use it to obtain information on A and B based only on the local behavior of the exponential around 0. Bourgeois showed that Condition (4) implies that A and B are simultaneously triangularizable if n = 2, and produced a proof that this also holds when n = 3. This last result is however false, as the following counterexample (communicated to us by Jean-Louis Tu) shows: consider the matrices Notice that A 1 and B 1 are not simultaneously triangularizable since they share no eigenvector (indeed, the eigenspaces of A 1 are the lines spanned by the three vectors of the canonical basis, and none of them is stabilized by B 1 ). However, for every t ∈ C, a straightforward computation shows that the characteristic polynomial of tA 1 + B 1 is Then for every t ∈ N, the matrix tA 1 + B 1 has three distinct eigenvalues in 2iπZ, hence is diagonalizable with e tA 1 +B 1 = I 3 . In particular e B 1 = I 3 , and on the other hand e A 1 = I 3 . This shows that Condition (4) holds.
It then appears that one should strengthen Bourgeois' condition as follows in order to obtain at least the simultaneous triangularizability of A and B: Notice immediately that this condition implies that e A and e B commute. Indeed, if Condition (5) holds, then Therefore Condition (5) is equivalent to ∀(k, l) ∈ Z 2 , e kA+lB = e kA e lB = e lB e kA .
Here is our main result.
Theorem 1. Let (A, B) ∈ M n (C) 2 be such that, for all (k, l) ∈ Z 2 , e kA+lB = e kA e lB . Then AB = BA.
The following corollary is straightforward.
Theorem 2. Let G be a subgroup of (M n (C), +) and assume that M → exp(M ) is a homomorphism from (G, +) to (GL n (C), ×). Then, for all (A, B) ∈ G 2 , AB = BA.
The key of the proof of Theorem 1 is Assume that, for every (k, l) ∈ Z 2 , the matrix kA + lB is diagonalizable and Sp(kA + lB) ⊂ Z. Then AB = BA.
One may however wonder whether a subsemigroup S on which the exponential is a homomorphism must be simultaneously triangularizable. Obviously the additive semigroup generated by the matrices A 1 and B 1 above is a counterexample. Nevertheless, we will prove a weaker result, which rectifies and generalizes Bourgeois' results [1]. Note that the converse is obviously false.
The proofs of Theorem 1 and of Proposition 4 have largely similar parts, so they will be tackled simultaneously. There are three main steps.
• We will prove Proposition 4 in the special case where Sp(A) ⊂ 2iπZ and Sp(B) ⊂ 2iπZ. This will involve a study of the matrix pencil z → A + zB. We will then easily derive Proposition 3 using a refinement of the Motzkin-Taussky theorem.
• We will handle the more general case Sp(A) ⊂ 2iπZ and Sp(B) ⊂ 2iπZ in Theorem 1 by using the Jordan-Chevalley decompositions of A and B together with Proposition 3.
• In the general case, we will use an induction to reduce the situation to the previous one, both for Theorem 1 and Proposition 4.
In the last section, we will prove a sort a generalized version of Proposition 4 for additive semigroups of matrices (see Theorem 15).
ii) For M ∈ M n (C), we denote by χ M (X) ∈ C[X] its characteristic polynomial, and we set iii) Given an integer N ≥ 1, we set U N (z) := ζ ∈ C : ζ N = z . . . , f n on C 2 such that

Definition
Using the fact that the eigenvalues are continuous functions of the coefficients, it is obvious that a pair (A, B) ∈ M n (C) 2 has property L if and only if there are affine maps f 1 , . . . , f n from C to C such that

Property L for pairs of matrices with an integral spectrum
We denote by K(C) the quotient field of the integral domain H(C) of entire functions (i.e. analytic functions from C to C). Considering id C as an element of K(C), we may view A + id C B as a matrix of M n K(C) . We define the generic number p of eigenvalues of the pencil z → A + zB as the number of the distinct eigenvalues of A + id C B in an algebraic closure of K(C). A complex number z is called regular when A + zB has exactly p distinct eigenvalues, and exceptional otherwise. In a neighborhood of 0, the spectrum of A + zB may be classically described with Puiseux series as follows (see [2, chapter 7]): there exists a radius r > 0, an integer q ∈ {1, . . . , n}, positive integers d 1 , . . . , d q such that n = d 1 +· · ·+d q , and analytic functions f 1 , . . . , f q defined on a neighborhood of 0 such that We may now prove the following result.
Proof. With the above notation, we prove that f 1 , . . . , f q are polynomial functions. For instance, consider f 1 and its power series expansion is an eigenvalue of kA + B: hence it is an integer. Therefore one has: for every integer k ≥ k 0 , For every integer k ≥ k 0 , the following equality holds: Assume that a j = 0 for some j ≥ 1 with j = N , and define s as the smallest such j. On the one hand, one has for every integer j ∈ N, On the other hand, when k → +∞, one has +∞ j=s+N +1 It follows that to 0 and is not ultimately zero. This is a contradiction. Therefore ∀j ∈ N {0, N }, a j = 0. In the same way, one shows that, for every k ∈ {1, . . . , q}, there Therefore we found affine maps g 1 , . . . , g n from C to C such that, in a neighborhood of 0, The coefficients of these polynomials are polynomial functions of z that coincide on a neighborhood of 0; therefore ∀z ∈ C, χ A+zB (X) = n k=1 X − g k (z) .
The pair (A, B) has property L, and Proposition 6 is proven.

Commutativity for subgroups of diagonalizable matrices with an integral spectrum
Given a matrix M ∈ M n (C) and an eigenvalue λ of it, recall that the eigenprojection of M associated to λ is the projection onto Ker(M − λ I n ) n alongside Here, we derive Proposition 3 from Proposition 6. We start by explaining how Kato's proof [8, p.85 Theorem 2.6] of the Motzkin-Taussky theorem [12] leads to the following refinement.  , we deduce that each eigenprojection of B is sum of some projections chosen among the Π i (0) i≤p . As B is diagonalizable, it is a linear combination of the Π i (0) i≤p , which all commute with A + zB for any regular z. Therefore AB = BA.
We now turn to the proof of Proposition 3.
We now deduce the following special case of Theorem 1.  Note in particular that e N and e N ′ commute. Since N is nilpotent, we have That shows that N is a polynomial in e N . Similarly N ′ is a polynomial in e N ′ . Therefore, The above condition yields ∀(k, l) ∈ Z 2 , e kA+lB = e kN +lN ′ .
For any (k, l) ∈ Z 2 , kN + lN ′ is nilpotent since N and N ′ are commuting nilpotent matrices. Hence kN + lN ′ is a polynomial in e kN +lN ′ . Since kA + lB commutes with e kA+lB , it commutes with kN + lN ′ . Therefore e kD+lD ′ = e kA+lB e −kN −lN ′ = I n .
By Lemma 8, the matrices D and D ′ commute. In particular (D, D ′ ) has property L, which yields affine maps f 1 , . . . , f n from C to C such that The set is clearly finite. We may choose two distinct elements a and b in Z E. The following equivalence holds: Since D and D ′ are simultaneously diagonalizable, it easily follows that D + aD ′ is a polynomial in D + bD ′ and conversely D + bD ′ is a polynomial in D + aD ′ . Hence N + aN ′ and N + bN ′ both commute with D + aD ′ and D + bD ′ . Since N +aN ′ and N +bN ′ both commute with one another, we deduce that A+aB = (D + aD ′ ) + (N + aN ′ ) commutes with A + bB = (D + bD ′ ) + (N + bN ′ ). Since a = b, we conclude that AB = BA. ii) In the sequel, we consider, for k ∈ N {0}, the function iii) For λ ∈ C, we denote by C λ (M ) the characteristic subspace of M with respect to λ, i.e. C λ (M ) = Ker(M − λI n ) n .

Lemma 11. Assume that A satisfies Condition
Then there exists k ∈ N {0} such that γ k is one-to-one.
Lemma 12. Assume that γ 1 is one-to-one and that (A, B) satisfies Equality (5) (resp. Equality (4)). Then the characteristic subspaces of e A and e B are stabilized by A and B.
Proof. Notice that A + B commutes with e A+B , hence commutes with e A e B . It thus stabilizes the characteristic subspaces of e A e B . Let us show that • Since e B and e A commute, e A stabilizes the characteristic subspaces of e B . Considering the characteristic subspaces of the endomorphism of C µ (e B ) induced by e A , we find • Let (λ, µ) ∈ Sp(e A ) × Sp(e B ). Since e A and e B commute, they both stabilize C λ (e A ) ∩ C µ (e B ) and induce simultaneously triangularizable endomorphisms of C λ (e A ) ∩ C µ (e B ) each with a sole eigenvalue, respectively λ and µ: it follows that • Finally, the application (λ, µ) → λµ is one-to-one on Sp(e A ) × Sp(e B ). Therefore for all distinct pairs (λ, µ) and (λ ′ , µ ′ ) in Sp(e A ) × Sp(e B ). One has and C n is the sum of all the characteristic subspaces of e A e B . We deduce that This gives Equality (8).
We deduce that A+B stabilizes every characteristic subspace of e B . However this is also true of B since it commutes with e B . Hence both A and B stabilize the characteristic subspaces of e B . Symmetrically, every characteristic subspace of e A is stabilized by both A and B.
Proof of Theorem 1 and Proposition 4. We use an induction on n. Both Theorem 1 and Proposition 4 obviously hold for n = 1, so we fix n ≥ 2 and assume that they hold for any pair (A, B) ∈ M k (C) 2 with k ∈ {1, . . . , n − 1}. Let (A, B) ∈ M n (C) 2 satisfying Equality (5) (resp. Equality (4)). Assume first that (A, B) is decomposable. Then there exists p ∈ {1, . . . , n − 1}, a non-singular matrix P ∈ GL n (C) and square matrices A 1 , B 1 , A 2 , B 2 respectively in M p (C), in M p (C), in M n−p (C) and in M n−p (C) such that Since the pair (A, B) satisfies Equality (5) (resp. Equality (4)), it easily follows that this is also the case of (A 1 , B 1 ) and (A 2 , B 2 ); hence the induction hypothesis yields that (A 1 , B 1 ) and (A 2 , B 2 ) are commuting pairs (resp. have property L). Therefore (A, B) is also a commuting pair (resp. has property L).
From that point on, we assume that (A, B) is indecomposable. We may also assume that A satisfies Condition (7). Indeed, consider in general the finite set Since its elements are rational numbers, we may find some integer p > 0 such that pE ⊂ Z. Replacing A with pA, we notice that (pA, B) still satisfies Equality (5) (resp. Equality (4)) and that it is a commuting pair (resp. satisfies property L) if and only if (A, B) is a commuting pair (resp. satisfies property L).
Assume now that A satisfies Condition (7) as well as all the previous assumptions, i.e. (A, B) is indecomposable and satisfies Equality (5) (resp. Equality (4)). By Lemma 11, we may choose k ∈ N {0} such that γ k is one-to-one. Replacing A with kA, we lose no generality assuming that γ 1 is one-to-one.
We can conclude: if e B has several eigenvalues, Lemma 12 contradicts the assumption that (A, B) is indecomposable. It follows that e B has a sole eigenvalue, and for the same reason this is also true of e A . Choosing (α, β) ∈ C 2 such that Sp(e A ) = {e α } and Sp(e B ) = {e β }, we find that exp(A − α I n ) and exp(B−β I n ) both have 1 as sole eigenvalue. We deduce that Sp(A−α I n ) ⊂ 2iπZ and Sp(B − β I n ) ⊂ 2iπZ. Set A ′ := A − α I n and B ′ := B − β I n . We now conclude the proofs of Theorem 1 and Proposition 4 by considering the two cases separately.
Thus Theorem 1 and Proposition 4 are proven. 5 Additive semigroups on which the exponential is a homomorphism Notation 13. We denote by Q + the set of non-negative rational numbers. In this short section, we prove the following result.
Theorem 15. Let S be a subsemigroup of (M n (C), +) and assume that M → exp(M ) is a homomorphism from (S, +) to (GL n (C), ×). Then Span(S) has property L.
By Proposition 4, it suffices to establish the following lemma. Since, for every (p 1 , . . . , p r ) ∈ N r , the pair j−1 k=1 p k A k , A j has property L for every j ∈ {2, . . . , r}, by induction we obtain a list (σ 1 , . . . , σ r ) ∈ (Σ n ) r such that OSp r j=1 p j A j = r j=1 p j a (j) σ j (k) 1≤k≤n .