# let A,B be complex matrics and $2A(B-A)=A+B$ how prove $AB=BA$

let $A,B\in M_n(\mathbb C)$ $\mathbb C$ is complex field such that $$2A(B-A)=A+B$$ how prove $AB=BA$ thanks in advance

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My answer does not lead to a solution. You should un-accept it. –  Joe Johnson 126 Jan 30 '13 at 18:53

$2A(B-A)=A+B$ is in fact $(2A-I)(B-A-I)=I$ so one is the inverse of the other, thus $(B-A-I)(2A-I)=I.$ From these two we get $AB=BA.$
I was thinking of the fact that for $A,B,C\in M(\mathbb C)$, $(AB-BA)C=C(AB-BA)$. And was trying to make a solution for this question, but I couldn't. –  Babak S. Jan 30 '13 at 18:59
One common trick for handling equalities involving commutators or the like is to left-multiply it by a matrix and also right-multiply it by the same matrix, and see what happens. Here we have \begin{align*} 2A(B - A) &= A + B,\\ 2AAB - 2AAA &= AA+AB,\\ 2ABA - 2AAA &= AA+BA. \end{align*} Subtract the third equation from the second one, we get \begin{align*} &2A(AB - BA) = AB-BA,\\ &(2A-I)(AB-BA)=0. \end{align*} So, if we can show that $2A-I$ is nonsingular, we are done. Suppose the contrary. Then $x^T(2A-I)=0$ or $x^T(2A)=x^T$ for some nonzero vector $x$. Therefore $2A(B-A)=A+B$ implies that $x^T(B-A)=x^T(A+B)$, i.e. $x^T(2A)=x^T=0$, which is a contradiction.