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I have known that when $\boldsymbol{A}\boldsymbol{B}= \boldsymbol{B}\boldsymbol{A}$, the equality $(\boldsymbol{A} \boldsymbol{B})^k = \boldsymbol{A}^k \boldsymbol{B} ^k = \boldsymbol{B} ^k \boldsymbol{A}^k$, $\forall k \geq 2$ holds. But the converse is not true, so I'm wondering what condition is necessary and sufficient to this equality? Any comments would be appreciated!

Kris Prokins
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  • It is easy to show the converse holds under the assumption that $A$ and $B$ are both invertible. Idk if there is a weaker condition where the converse holds. – JimmyK4542 Sep 20 '19 at 03:13
  • Your question makes sense in any semigroup, but the tag linear algebra seems to indicate that you are only interested in some particular case (matrices?). Anyway, what is the precise context of your question? – J.-E. Pin Sep 20 '19 at 05:06
  • Sorry, I’ve not learned any theories about groups. So this question would be about matrices. But I am interested in if there’s any general case about it. – Kris Prokins Sep 20 '19 at 05:29

1 Answers1

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I suppose you are talking about conditions that appear in the form of equalities between permuted words comprising of products of $A$ and $B$.

Note that if both $A$ and $B$ are nonsingular, the condition $AABB=ABAB$ already implies that $AB=BA$.

If both $A$ and $B$ are allowed to be singular, the situation is hopeless beyond rescue. Consider $$ A=\pmatrix{0&1&0\\ 0&0&0\\ 0&0&0},\ N=\pmatrix{0&0&0\\ 0&0&1\\ 0&0&0} $$ and $B=N$, so that $BA=0\ne AB$. In this case, conditions like "$(AB)^k=A^kB^k=B^kA^k=0$ for all $k\ge2$" concerning monomials of degree $\ge4$ are useless. In fact, even if you lower the degrees of the monomials involved from $4$ to $3$ or $2$, all such monomials except $AB$ will be equal to zero, because $A^2=B^2=BA=0$,

The remaining possibility is that one of $A,B$ is singular and the other is nonsingular. In the above example, if we put $B=I+N$ (so that $A$ is singular and $B$ is nonsingular), we still have $AB\ne0=BA$ but $(AB)^k=(BA)^k=A^kB^k=B^kA^k=0$ for every $k\ge2$. So, the condition you proposed still doesn't work, but that doesn't exclude the possibility that equalities of other permuted words may be effective.

user1551
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