Let $A$ and $B$ be Hermitian matrices of the same size. If $AB − BA$ and $A − B$ commute, show that $A$ and $B$ commute.
I'm not sure where to start with this. Any help is appreciated.
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If $A-B$ commute would't: $A-B=B-A \Rightarrow A-B+A=B \Rightarrow 2A-B=B \Rightarrow 2A=2B \Rightarrow A=B$ ? In witch case $AB = AA = BB $ that obviously commute, is there an error in the question ? – Felipe Dilho Nov 25 '17 at 17:54
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1It seems to me that the question is really trying to say: Let $M_1 = AB-BA$ and let $M_2 = A-B$. Then show $M_1M_2 = M_2M_1$. – Zach Boyd Nov 25 '17 at 18:03
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@ Zach Boyd , Yes, that is correct. – Math-Data Nov 25 '17 at 18:54
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@ZachBoyd. Isn't it rather: show that if $M_1 M_2 = M_2 M_1$ then $AB = BA$? – md2perpe Nov 25 '17 at 20:48
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Yes, that is correct. – Zach Boyd Nov 25 '17 at 21:00
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Let $C=AB-BA.$ Then for $m\geq 1$
$$C^m=C^{m-1}(AB-BA)=C^{m-1}A(B-A)+C^{m-1}(A-B)A.$$ Thus \begin{align*} \text{trace} (C^m)=& \text{trace} (C^{m-1}A(B-A))+ \text{trace} (C^{m-1}(A-B)A) \\=&-\text{trace}(A-B)C^{m-1}A + \text{trace} (C^{m-1}(A-B)A)\\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (\text{as trace}(ST)=\text{trace (TS)})\\ =& - \text{trace} (C^{m-1}(A-B)A)+ \text{trace} (C^{m-1}(A-B)A)\\& ~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~(\text{as}~ C(A-B)=(A-B)C))\\&=0. \end{align*} Since $\text{trace} (C^m)=0$ for all $m,$ $~C$ is nilpotent.(The link of the proof of this is given in the comment)
Also $C^*=(AB-BA)^*=(AB)^*-(BA)^*=B^*A^*-A^*B^*=BA-AB=-C.$
Thus $(iC)^*=\bar i C^*=-i(-C)=iC.$
Thus $iC$ is a Hermitian nilpotent matrix. Hence it must be $0,$ that is $C=0.$
Hence $AB=BA.$
Black-horse
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This is the link https://math.stackexchange.com/questions/159167/traces-of-all-positive-powers-of-a-matrix-are-zero-implies-it-is-nilpotent – Black-horse Nov 25 '17 at 19:18