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This question already has an answer here:

The task is:

Show that the equation $[A,B] :=AB-BA= iqI$, where $q\neq 0$ is a non-zero real constant and $I$ is the identity matrix cannot be satisfied by any finite dimensional Hermitian matrices $A,B$.

What I got so far:

If I set $A=B$ and $q=0$ than the right hand side is $[A,B]=[A,A]=0$ and the left hand side is $iqI=0$. This means the equation is satisfied in this case (counter-example).

Furhter I calculated $[A,B]= AB-BA = AB-(AB)^\dagger$ which I can use for the diagonal elements to show that $([A,B])_{ii} = \sum_k A_{ik}B_{ki}-A^\ast_{ik}B^\ast_{ki} = i \sum_k 2\, \Im(A_{ik}B_{ki})$ where I am stuck.

Do you have any idea how this statement can be proven even though I found a counter expample?

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marked as duplicate by Qmechanic quantum-mechanics Sep 23 '16 at 22:58

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Take the trace of both sides, then use the property of traces.

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