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Determine all real numbers $x$ for which there exists $3 \times 3$ real matrices $A$ and $B$ such that

$$AB-BA = \begin{pmatrix} 1 & 0 &0 \\ 0 & x& 0\\ 0&0& -1 \end{pmatrix}$$

Mr. Proof
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2 Answers2

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Since $tr(AB) = tr(BA)$, you see that $tr(AB-BA) = 0$. So, the trace of the rhs must also be zero, i.e. $x = 0$.

We have proven so far that, if the equality holds, we must have $x=0$. To complete the proof we must check that when $x=0$, it is possible to find such matrices. We can take for instance $$ A=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right), \quad B = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{array} \right). $$

PierreCarre
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  • It is not if and only if! – Mr. Proof Dec 09 '20 at 04:58
  • @TheMathematician I did not say it was if and only if. I am just guarantying that there exists at least one $x$ is those conditions. The first part of the proof just says that the only $x$ that could fit is $x=0$, but does not establish that it actually does. – PierreCarre Dec 09 '20 at 08:13
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    Oh, I see. Thank you for your comment – Mr. Proof Dec 09 '20 at 08:15
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Using this fact, we have $$\begin{align}\text{tr}(AB - BA) &= \text{tr}(AB) - \text{tr}(BA)\\ &= \text{tr}(AB) - \text{tr}(AB) \\&= 0 \end{align}$$ But $$\text{tr}\begin{pmatrix} 1 & 0 &0 \\ 0 & x& 0\\ 0&0& -1 \end{pmatrix} = 1+x-1 = x $$ Hence, $x = 0$

VIVID
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  • You just show that for such matrices to exist $x$ must be zero. The proof will be complete if you show that for $x=0$ you can in fact find $A,B$ in those conditions. – PierreCarre Dec 05 '20 at 11:56