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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10Hint: $\text{tr}(AB) = \text{tr}(BA)$. – achille hui Jan 13 '16 at 12:51
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As others have hinted, the trace of the commutator $AB-BA$ is necessarily zero (unlike $I$). In fact this is the only restriction on the possibilities for $AB-BA$, as explained in answering this previous Question. – hardmath Jan 13 '16 at 12:55
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1More advanced students can investigate matrices over fields of positive characteristic. – GEdgar Jan 13 '16 at 15:18
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@GEdgar : can you please elaborate? – Qwerty Jan 13 '16 at 15:19
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2@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. – Daniel Fischer Jan 13 '16 at 15:21
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3The 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? – GEdgar Jan 13 '16 at 15:24
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@DanielFischer : Therefore I conclude that the $F$ has to be $\Bbb{R}$ in order to looking at only the trace be enough.. Am I right? – Qwerty Jan 13 '16 at 15:25
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1We guess your question is about $\mathbb R$. That is fine. More advanced students may return to this question later, when they have dealt with fields of positive characteristic. – GEdgar Jan 13 '16 at 15:26
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2For any field of characteristic $0$ it's sufficient to look at the trace, that could also be e.g. $\mathbb{Q}$ or $\mathbb{C}(X)$. – Daniel Fischer Jan 13 '16 at 15:27
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4See also this post and other questions linked there. – Martin Sleziak Jan 13 '16 at 16:05
3 Answers
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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.
BigbearZzz
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you can see for example that
$$\mathrm{Tr}(AB) - \mathrm{Tr}(BA) =0\neq \mathrm{Tr}(\mathrm{I}_n)=n$$
Zanzi
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