If $(A+B)^n$ is binomial for some $n$, does that imply $AB = BA$?

Question

We say that a matrix power $(A+B)^n$ is binomial iff it satisfies the matrix equality $$(A+B)^n = \sum \limits_{j\,=\,0}^n \binom{n}{j}A^jB^{n-j}.$$ If two matrices have a binomial power for some $n$, does that imply that $AB = BA$?

My approach was the most obvious (I think): I tried to find a non-commutative expression for $(A+B)^n$ to finally get all those ugly permutations with the same degree equal to something that adds up to the term with the respective degree on the other side, but they could "chaotically" add up to that without being commutative, so it's not interesting. I also tried to use a "inductive-like" thinking to get the simplest case of commutativity, i.e., the case where $n=2$, so, by the Euclidean algorithm, $n = 2q + r$. I didn't manage to advance much, though.

Answer 1

BIG EDITION: Added the general case, and expanded the computation of the solution for $n=2$

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The answer is yes for $n=2$ and no in general.

If $n=3$, you can rearrange $$A^3+3A^2B+3AB^2+B^3=(A+B)^3=A^3+A^2B+ABA+BA^2+B^2A+BAB+AB^2$$ to get $$(2A+B)[A,B]+[A,B](A+2B)=0 \;\;\;\; (1),$$ where $[A,B]:=AB-BA$, by taking into account that $[A^2,B]=A[A,B]+[A,B]A$.

Observe that equation (1), via $B':=2A+B$, $A':=A+2B$, since $[A,B']=[A,2A+B]=[A,B]$, is equivalent to $$B'[A,B']+[A,B']A'=0,$$ and since $[A',B']=[A+2B,2A+B]=[A,B]+4[B,A]=[A,B]-4[A,B]=-3[A,B]=-3[A,B']$, by multiplying by $-3$ we get the equivalent equation $$B[A,B]+[A,B]A=0,$$ (I have changed $A',B'$ to $A,B$ by abuse of notation) which, after swap of $A,B$ and a change of sign amounts to $$A[A,B]+[A,B]B=0 \;\;\;\; (2).$$

Equation (2) has solutions with $[A,B]\neq0$ already for $2×2$ matrices. To see this, take into account that $[$M$_2(K),$M$_2(K)]$ is the subspace of trace $0$ matrices. We can find $A,B$ such that $$[A,B]=\begin{pmatrix}0 & c \\ 0 & 0\end{pmatrix}\neq0,$$ and in fact we can do so in such a way that $A[A,B]=0=[A,B]B$; it is enough to pick $$A=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}, B=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix},$$ which give $[A,B]=\begin{pmatrix}0 & -1 \\ 0 & 0\end{pmatrix}.$

Going backwards we can find now an interesting solution for (1): $$A=\begin{pmatrix} 0 & 0 \\ 2 & -1\end{pmatrix}, B=\begin{pmatrix} 0 & 0 \\ -1& 2\end{pmatrix}, [A,B]=\begin{pmatrix} 0 & 0 \\ -3& 0\end{pmatrix}.$$

Observe that if $k>2$ then M$_k(K)$ contains M$_2(K)$ as a subring.

Now, if $n\geq4$:

Consider the free (noncommutative) algebra $K\langle X,Y\rangle$ generated by two elements $X,Y$. Since when $A,B$ commute we have $(A+B)^n=\sum_{i=0}^n\binom{n}i A^iB^{n-i}$ in any $K$-algebra, this means that the element $$p_n(X,Y)=(X+Y)^n-\sum_{i=0}^n\binom{n}i X^iY^{n-i}$$ belongs to the ideal generated by $[X,Y]$ in $K\langle X,Y\rangle$. Therefore for each $n$ we can write $$p_n(X,Y)=a(X,Y)[X,Y]+[X,Y]b(X,Y)+\sum_i c_i(X,Y)[X,Y]d_i(X,Y),$$ where $a,b,c_i,d_i$ are (noncommutative) polynomials in $X,Y$.

E.g., if I did not get it wrong, for $n=4$ we have $$p_4(A,B)=(3A^2+2AB+B^2)[A,B]+[A,B](4B^2+BA+A^2)+A[A,B](A+2B)+(A+2B)[A,B]B.$$ I have found this by pairing monomials of the same degrees in $A$ and $B$. For example: $$A^3B-A^2BA=A^2[A,B], A^3B-ABA^2=A[A^2,B]=A^2[A,B]+A[A,B]A,$$ and the difficult one, $$A^2B^2-BABA=A^2B^2-B[A,B]A-B^2A^2=[A^2,B^2]-B[A,B]A=AB[A,B]+A[A,B]B+[A,B]BA.$$

Now observe that since $p_n$ is homogeneous and $n\geq3$, by a degree argument we must have $a,b,c_i,d_i$ either $0$ or of degree at least $1$, i.e., we cannot have a term of the form $\alpha[X,Y]$ with $0\neq\alpha\in K$. This implies that touching the $[X,Y]$ factor of each term there is either an $X$ or an $Y$ either at its left or at its right.

Therefore, to get $p_n(A,B)=0$ it is enough to have $$A[A,B]=[A,B]A=B[A,B]=[A,B]B=0.$$ This can be done with $[A,B]\neq0$ for matrices of size $3$ or more (just pick sparse matrices with the nonzero elements not in the diagonal, in order to get $[A,B]$ with just one nonzero element and $0$ trace, as in the explanation above for $2\times 2$ matrices).

(Note that the general argument of this section is also valid for $n=3$; the previous section shows that in this particular case we can easily find examples also of size $2$.)

Answer 2

When $n=2$, the answer is obviously yes.

When $n\ge3$, the answer is no in general. Here is a counterexample that works over any field. Let $$ A=\pmatrix{0&1&0\\ 0&0&0\\ 0&0&0}\oplus0 \ \text{ and }\ B=\pmatrix{0&0&0\\ 0&0&1\\ 0&0&0}\oplus0. $$ Then $AB=E_{13}\ne0=BA$, but $A+B$ is nilpotent and all trigrams generated by $A$ and $B$ (i.e. $AAA,AAB,ABA,BAA,ABB,BAB,BBA$ and $BBB$) are zero. Thus $(A+B)^n=\sum_{k=0}^n\binom{n}{k}A^kB^{n-k}=0$ when $n\ge3$.