I'm quite new to Lie Algebras, and so there's a lot of easy stuff that I'm probably missing. Anyway following Kac notes I'm asked to compute the center of $\mathfrak{gl}(n,\Bbb K)$, and I've done it in the usual way (defining a bases for two matrices and working it out with brute force).
Anyway I was wondering: since the kernel of the adjoint representation is the center of the group, i.e. $ker(ad)=Z(\mathfrak{gl}(n,\Bbb K))$, is there any more elegant way to arrive to the result, i.e. $Z(\mathfrak{gl}(n,\Bbb K))= \lambda I$, starting from some proprieties of the adojoint representation?
I also think that the adjoint representation will not give a "more elegant proof". The computation of the kernel amounts again to the problem to show that $[A,B]=AB-BA=0$ for all $B$ implies that $A=\lambda I$. I do not agree that the proof here necessarily needs to be "brute force". There are several proofs given, among them such which try to avoid matrix computations, see these answers. Also, the usual proof is not so bad.