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Can someone help me with this problem ?

Show that for every $A$,$B$ $\in {U}$ : $AB-BA \in {U}$

$$U:=\bigl\{A \in \mathbb{R}^{3\times3} \mid \operatorname{Tr}(A)=0 \bigr\} $$

J. W. Tanner
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Kai
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    So you need to show that the trace of $AB-BA$ is $0$. Equivalently (by linearity) $tr(AB)=tr(BA)$. This is a property of the trace function. – Just dropped in Mar 03 '19 at 14:24

1 Answers1

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To write the comment of @JustDroppedIn as an answer,

let $U:=\bigl\{M \in \mathbb{R}^{3\times3} \mid \operatorname{Tr}(M)=0 \bigr\},$ and take any $A,\;B \in U.$

Let $C=AB-BA.$ The trace is a linear map, so Tr$(C)=$Tr$(AB)-$T$r(BA)$.

Furthermore, Tr$(AB)=$Tr$(BA).$

(That can be shown from the component-wise definitions of matrix multiplication and trace.)

Thus, Tr$(C)=0$ and $C \in U$.

Note that we did not need to use the fact that Tr$(A)=$Tr$(B)=0$ to prove that Tr$(AB-BA)=0.$

For any two $n\times n$ matrices, the trace of their commutator vanishes.

J. W. Tanner
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