A condition on the powers of exp(A) and exp(B) that implies AB=BA

Details A condition on the powers of exp(A) and exp(B) that implies AB=BA A condition on the powers of exp(A) and exp(B) that implies AB=BA

2010-12-20

arxiv.org/abs/1012.4420v3

Mathematics

Rings and Algebras

17 pages (v3)

Scientific paper

Let A and B be matrices of M_n(C). We show that if exp(A)^k exp(B)^l=exp(kA+lB) for every 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 sub-semigroup 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.

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