For $L=sl_2(\mathbb{F})$ i.e. matrices with trace zero, what is the centre i.e.
$Z(L)$= {$x\in L : [x,y]=[y,x] \ \forall\ y \in L$}.
I will have to find matrices $A \in L$ such that $AB=BA$ for all $B \in L$. But how to appoach this considering trace is zero here.
The centre $Z$ of $L=\mathfrak{sl}_2(\mathbb{F})$ is an abelian ideal in $L$, different from $L$ itself. Hence $\dim(Z)\le 2$. Suppose that $\dim(Z)=1$ or $2$. Then $L/Z$ is $1$ or $2$-dimensional, hence solvable. It follows that $L$ is solvable, a contradiction. So the only possibility is that $Z=0$.