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Are there matrices $A,B$ (of dimension $n$), that give \begin{equation} AB-BA=I \end{equation}

I have tried getting a result in small scale by using $2\times 2$ matrices and got a false equation $(0=1)$. But I cannot find a way to show that it is true for all $n$. I tried showing it by definition but that didn't get me anywhere. Any ideas?

yoknapatawpha
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yinon eliraz
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1 Answers1

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Hint: Take the trace on both sides.

Harald Hanche-Olsen
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  • I figured out the the trace of AB is the same as BA, but that doesnt mean that each part of A is equal to the same part in B. Am I missing something? – yinon eliraz Jan 13 '15 at 18:14
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    @yinoneliraz If $AB-BA=I$ had a solution, $0=\frac{\text{trace of the LHS}}{n}=\frac{\text{trace of the RHS}}{n}=1$, which would give your "false equation". – Algebraic Pavel Jan 13 '15 at 18:40