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I have just started linear functionals when I faced the following problem:

If $A$ and $B$ are $n \times n$ complex matrices, show $AB - BA=\Bbb{I}$ is impossible.

Can someone help me?

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    $\begingroup$ Hint: $\text{tr}(AB) = \text{tr}(BA)$. $\endgroup$ Jan 13, 2016 at 12:51
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    $\begingroup$ @Qwerty If $n$ is a multiple of the characteristic of $F$, then the $n\times n$ identity matrix over $F$ has trace $n\cdot 1_F = 0$. So in that case, looking at the trace doesn't settle the issue. $\endgroup$ Jan 13, 2016 at 15:21
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    $\begingroup$ The accepted answer gets contradiction $n=0$. But what if $n$ is zero in the field? (For example, $2=0$ in the field with two elements.) Then $n=0$ is not a contradiction. Is $AB-BA=I$ still impossible? $\endgroup$
    – GEdgar
    Jan 13, 2016 at 15:24
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    $\begingroup$ For any field of characteristic $0$ it's sufficient to look at the trace, that could also be e.g. $\mathbb{Q}$ or $\mathbb{C}(X)$. $\endgroup$ Jan 13, 2016 at 15:27
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    $\begingroup$ See also this post and other questions linked there. $\endgroup$ Jan 13, 2016 at 16:05

3 Answers 3

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For a matrix $A=[a_{ij}]$ of size $n\times n$, its trace $Tr(A)$ is defined by $$ Tr(A)=\sum_{i=1}^n a_{ii} $$ . You can verify it yourself that $$ Tr(AB)=Tr(BA)$$ and that $$ Tr(A+B)=Tr(A)+Tr(B) $$

Therefore if $AB-BA = \Bbb I$, then we have $$n=Tr(\Bbb I)= Tr(AB-BA)= Tr(AB)-Tr(BA) = 0 $$ which is impossible.

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Look at the trace of $AB$ and the trace of $BA$

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you can see for example that

$$\mathrm{Tr}(AB) - \mathrm{Tr}(BA) =0\neq \mathrm{Tr}(\mathrm{I}_n)=n$$

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