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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?

Mochammad Zulfikar Aditya
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2 Answers2

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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 $[A,B]=\pmatrix{0&2\\ 2&0}\ne0$.

user1551
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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.

SagarM
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