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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.

Patric Gustafsson
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    What is $tr(AB-BA)_{ki}$? – Janko Bracic Jan 19 '15 at 14:35
  • This is neither a formal proof (not the right kind of formulas for that) nor an informal proof (which should actually contain words). It is not clear where your reasoning starts or ends. – Marc van Leeuwen Jan 19 '15 at 14:36
  • it may be easier to use properties $tr(AB) = tr(BA), tr(A+\pm B) = tr(A) + \pm tr(B).$ – abel Jan 19 '15 at 14:37
  • @MarcvanLeeuwen It looks like the reasoning starts at the end of the proof (and works its way backward). Since the trace of $I$ is nonzero, that is equivalent to the statement above, which is equivalent to the statement above, etc... – layman Jan 19 '15 at 14:38
  • your proof is essentially correct. you can clean it up a bit now that you know the main idea. – abel Jan 19 '15 at 14:39
  • Also I could not find $n\geq1$ among the hypotheses. – Marc van Leeuwen Jan 19 '15 at 14:40
  • Nor the ring in which the matrix have their coefficients – Olórin Jan 19 '15 at 14:41
  • @MarcvanLeeuwen The OP first states that for each natural number greater than or equal to $1$, it is not equal to $0$. Tehn, since $I$ is an $n \times n$ matrix for some $n \geq 1$, it has trace $n$, which, as was established, must be non-zero. – layman Jan 19 '15 at 14:52
  • @user46944 You're right, even if hypothesis are usual stated upfront ;-) – Olórin Jan 19 '15 at 14:56
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    @RobertGreen Yeah, I agree that the way the proof is presented is very weird. The ideas are all there, but in reverse order. The OP should have written this proof in reverse order, and I hope he makes a note of that. – layman Jan 19 '15 at 14:57
  • @user46944 Definitely. – Olórin Jan 19 '15 at 14:58
  • @user46944 Yes I'm new to writing proofs so any advice is welcome. And I fixed the $n \geq 1$, that was just a weird typo. I guess $n > 0$, should be on the first row instead of the last, or perhaps in the hypotheses itself? – Patric Gustafsson Jan 19 '15 at 15:05

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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.

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