When will $(\boldsymbol{A} \boldsymbol{B})^k = \boldsymbol{A}^k \boldsymbol{B} ^k = \boldsymbol{B} ^k \boldsymbol{A}^k$, $\forall k \geq 2$ 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!
 A: 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.
