I am trying to understand why for a semisimple Lie algebra $L$, we have $L = [L,L]$.
here are my thoughts.
Let $I:=[L,L]$ be a proper ideal in $L$, i.e., $I \ne L$, we then consider its orthogonal complement $I^\perp:=\{x \in L: \varkappa(x,y) = 0 \mbox{ for all $y\in I$}\}$ with respect to the Killing form $\varkappa$. Since $\varkappa$ is associative then $I^\perp$ is also an ideal.
On the other hand, since $y \in [L,L]$ then $y = \sum \alpha_{i,j}[y_i,y_j]$, $y_i, y_j \in L$, and $\alpha_{i,j}$ are scalars. We then get $\varkappa(x,y) = \sum\alpha_{i,j}\varkappa(x,[y_i,y_j]) = \sum \alpha_{i,j}\varkappa([x,y_i],y_j)$. And I then want to say that $I^\perp$ can be defiend as a kernal of $\varkappa$, because of $[x,y_i]$ can be also considered as an arbitrary element of $L$, and we thus get a conradiction because $L$ is assumed to be semisimple, but I am not sure that it is ok.