Endomorphisms with trace zero

Question

If $L=\mathfrak{gl}(V)$ or $L=\mathfrak{sl}(V)$, and if $g\in GL(V)$ is any invertible endomorphism of $V$, show that $gLg^{-1}=L$.

Suppose $L=\mathfrak{gl}(V)$, and let $l\in L$. Clearly, $glg^{-1}$ is an endomorphism of $V$, so $gLg^{-1}\subseteq L$. Similarly, $g^{-1}Lg\subseteq L$, so $L\subseteq gLg^{-1}$. Combining the two, we have $gLg^{-1}=L$.

Suppose $L=\mathfrak{sl}(V)$, and let $l\in L$. So $l$ is an endomorphism with trace zero. We need to show that $glg^{-1}$ is also an endomorphism with trace zero. Why is that?

Answer 1

Use the fact that, for any square matrices $A$ and $B$, the products $AB$ and $BA$ have the same trace.