Centre of $\mathfrak {sl}_{3}\mathbb{C} $ Lie algebra $\mathfrak {sl}_{3}\mathbb{C}$ is the Lie algebra of $3\times3$ matrices with complex entries and trace $0$ and Lie bracket $[X,Y] = XY-YX \hspace{3mm}  \forall \hspace{3mm} X,Y\in \mathfrak  {sl}_{3}\mathbb{C}$. 
The centre of $\mathfrak  {sl}_{3}\mathbb{C}$ is the set of all $Z\in \mathfrak  {sl}_{3}\mathbb{C}$ s.t. $[Z,X]=0 \hspace{2mm} \forall \hspace{2mm} X \in \mathfrak  {sl}_{3}\mathbb{C}$.
I want to show that this centre is $\left\{\mathbf{0}\right\}$. Is there a way to show this without ending up with many equations to solve? I was thinking of representing $X$ in terms of the basis for $ \mathfrak{sl}_{3}\mathbb{C}$ and then it would be enough to show that $[Z,K]=0$ implies $Z=\mathbf{0}$ for every basis element $K$ of $  \mathfrak{sl}_{3}\mathbb{C}.$ But I'm not sure if this is true and it also means solving many equations. 
Thanks
 A: As some people already said in the comments, it is necessary for intuition that you work on the basis $(E_{ij})$ (the matrices with $1$ in one entry and $0$ in all the others). But there are other means of proving that the center $\mathfrak z(\mathfrak{sl}_n\mathbb C)$ of $\mathfrak{sl}_n\mathbb C$ is null as, for example, using the Schur's Lemma.
For a better understanding of which hypothesis are needed, let us generalize the discussion to a finite dimensional vector space $V$ under a algebraically closed field $\mathbb K$ of characteristic $0$ instead of $\mathbb C^n$. 
We say that a subset $\Gamma\subset\mathfrak{gl}(V)$ is irreducible if for all proper subspace W (i.e. $\neq 0$ or $V$) of $V$ there are $v\in W$ and $X\in\Gamma$ such that $Xv\notin W$.
One can state a version of the Schur's Lemma as:

Let $V$ (under the hypothesis above) and $\Gamma$ a irreducible subset of $\mathfrak{gl}(V)$. If $Z$ commutes with every element of $\Gamma$ then $Z=\lambda I$ (where $I$ is the identity function) for some $\lambda\in\mathbb K$.

So, one can check that $\mathfrak{sl}(V)$ is a irreducible set of $\mathfrak{gl}(V)$ (you can do it chosing a basis for $V$). Hence, if $Z\in\mathfrak z(\mathfrak{sl}(V))$ then
$$ZX-XZ=[Z,X]=0,$$
for all $X\in\mathfrak{sl}(V)$, and, by the Schur's Lemma, $Z=\lambda I$, for some $\lambda\in\mathbb K$. Thus, since $\operatorname{tr}Z=0$ for all $Z\in\mathfrak z(\mathfrak{sl}(V))$, we must have that $\mathfrak z(\mathfrak{sl}(V))=0$.
Let us prove the above version of Schur Lemma.
Suppose $\lambda\in\mathbb K$ is an eigenvalue (possibly $0$) of $Z$. Define
$$T:=Z-\lambda I.$$
Since $Z$ and $I$ commute with every element of $\Gamma$ then so does $T$.
The image $\operatorname{Im}(T)$ of $T$ is invariant by $\Gamma$ (i.e. there are no $v\in\operatorname{Im}(T)$ and $X\in\Gamma$ such that $Xv\notin\operatorname{Im}(T)$). In fact, for every $v\in V$,
$$X(Tv)=T(Xv)\in\operatorname{Im}(T).$$
So, since $\Gamma\subset\mathfrak{gl}(V)$ is irreducible, we must have that $\operatorname{Im}(T)=V$ or $0$. Also, $\operatorname{Im}(T)$ can not be $V$ because $T=Z-\lambda I$ has non-null kernel and $\dim V<\infty$. Therefore,
$$Z-\lambda I = T = 0.$$
