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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
up vote 9 down vote accepted

$2A(B-A)=A+B$ 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 $AB=BA.$

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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
hi sir this question seems is better than my question how did you want prove it? – Maisam Hedyelloo Jan 30 '13 at 19:34

As this question is tagged as "contest-math", I believe that ehsanmo's answer is what the question setter had in mind. Still, it's interesting to try other approaches.

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.

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