"Prove that for any $n \times n$ matrix $A$ and $B$, $AB-BA \neq I$" So this is a exercise from the course compendium for a matrix course I'm currently taking.
"Prove that for any $n \times n$ matrix $A$ and $B$, $AB-BA \neq I$"
Is the proof that I have constructed a valid one? 
$tr(AB-BA)_{ki}\neq tr(I)\\\leftrightarrow\sum_{i=1}^{n}\sum_{k=1}^{n}(A_{ik}B_{ki}-B_{ki}A_{ik})\neq tr(I) \\\leftrightarrow\sum_{i=1}^{n}\sum_{k=1}^{n}(A_{ik}B_{ki})-\sum_{i=1}^{n}\sum_{k=1}^{n}(B_{ki}A_{ik}) \neq tr(I)\\\leftrightarrow\sum_{i=1}^{n}\sum_{k=1}^{n}(A_{ik}B_{ki})-\sum_{k=1}^{n}\sum_{i=1}^{n}(A_{ik}B_{ki}) \neq tr(I)\\\leftrightarrow\sum_{i=1}^{n}\sum_{k=1}^{n}(A_{ik}B_{ki}-A_{ik}B_{ki})\neq tr(I)\\\leftrightarrow\sum_{i=1}^{n}\sum_{k=1}^{n}(0)\neq tr(I)\\\leftrightarrow 0 \neq tr(I) \\0 \neq n, n > 0 \\\blacksquare$ 
Thanks in advance.
 A: Light answer : I assume that the matrices are with real or complex coefficients. Then you're almost right : $\textrm{Tr}(AB) = \textrm{Tr}(BA)$ so that $\textrm{Tr}(AB-BA) = 0$, and $\textrm{Tr}(I_n) = n \not= 0$ except if $n = 0$. So if you add the hypothesis $n >0$, your argument is correct.
General answer, as you did not specify in which ring the coefficients of $A$ and $B$ belong : $\textrm{Tr}(AB) = \textrm{Tr}(BA)$ so that $\textrm{Tr}(AB-BA) = 0$, whereas $\textrm{Tr}(I_n) = n$. Let $R$ be the (commutative) ring which the coefficients of $A$ and $B$ belong, and $p$ the caracteristic of $A$. Then what you say (concludind thanks to one trace being zero and the other being $\not=0$) is right as soon as $p$ does not divide $n$. For $R = \mathbf{R}$ or $\mathbf{R}$ you argument works if $n>0$, as these field are of caracteristic $0$. But if $R = \mathbf{F}_p = \mathbf{Z}/p \mathbf{Z}$ and if $n = p$ for instance, your argument doesn't work anymore.
Light remark : What matrices of dimension $0$ are, and that they have zero trace, must be clear.
General remark : Finally, what could we say of the equation $AB = BA = I$ for a general $A$ ? If $A = k$ is a field, this is intimately related to Weyl algebras, and I'm really happy to give you this inside reference, which dispenses me to write anything more. ;-) If $A$ is not a field, I don't know.
