Prove or disprove A,B,C commute Let $A,B,C$ be the real square matrices that satisfy
$$\begin{align}
A(B+C) &= (B+C)A \\
B(C+A) &= (C+A)B \\
C(A+B) &= (A+B)C
\end{align}$$
If $A$ is symmetry and $B^t=C$, how do we prove or disprove that $A$, $B$, and $C$ commute?
 A: Let $H$ and $K$ be respectively the symmetric and skew-symmetric parts of $B$. One may verify that the three given conditions reduce to
$$
[A,H]=0=[A+2H,K],\tag{1}
$$
where $[X,Y]$ denotes the commutator $XY-YX$. Now, if $A$ commutes with $B$, we must have $[A,H]=[A,K]=0$ and hence by $(1)$, we obtain
$$
[H,K]=0.\tag{2}
$$
However, one can pick $A,H$ and $K$ such that $(1)$ is satisfied but $(2)$ isn't, such as by setting $A=-2H$ and by picking $H$ and $K$ such that $[H,K]\ne0$. For instance, suppose
\begin{aligned}
&H=\pmatrix{1&0\\ 0&0},
\ K=\pmatrix{0&-1\\ 1&0},\\
&A=-2H=\pmatrix{-2&0\\ 0&0},
\ B=H+K=\pmatrix{1&-1\\ 1&0},
\ C=B^T=\pmatrix{1&1\\ -1&0}.
\end{aligned}
Then the three given conditions in question are satisfied but $[A,B]=\pmatrix{0&2\\ 2&0}\ne0$.
A: A commutator is defined as 
$$[A,B] = AB-BA$$
Hence, the conditions given in the question are 
$$[A,B]=[C,A]$$
$$[B,C]=[A,B]$$
$$[C,A]=[B,C]$$
Hence,
$$[A,B]=[B,C] = [C,A]$$
In order for them to commute we need to check iff 
$$[A,B]=[B,C] = [C,A] = 0$$
Now, let $$[B,C] = X$$
taking transpose we get, 
$$[C,B] = X^{T}  = X$$
You can take transpose on all commutators and verify that it again gives only symmetry of $X$, hence no condition is forcing $X=0$, so I guess there may exist symmetric $X$ satisfying the above conditions and hence disproving the argument in the question.

EDIT: Correction made in the answer according to the comment. Thanks a lot. The example in @user1551's answer agrees with my guess. 
