Suppose $\forall x \in L$ ad$x$ is diagonalisable. Then $L$ is abelian

Question

Let $L$ be a Lie algebra over an algebraically closed field $k$.

Suppose that $\forall x \in L$ the adjoint map ad$x$ is diagonalisable. How can we show $L$ is abelian?

The first thing I tried was to consider the Cartan decomposition of $L$. Since each ad$x$ is diagonalisable then if $($ad$y - \alpha(y))^n(x) = 0$ then $x$ is an eigenvector of ad$y$ for all $y$ in a Cartan subalgebra.

Hint as to how to continue?

Answer 1

Let $L=\mathfrak{so}_3(\Bbb C)$ with basis $(x,y,z)$ and Lie brackets $$ [x,y]=z,\; [z,x]=y,\; [y,z]=x. $$ Then the operators $ad(x)$ are real skew-symmetric matrices in $M_3(\Bbb C)$, which are diagonalizable since they are normal. However, the Lie algebra $L$ is non-abelian.

References:

is a real skew-symmetric matrix diagonalizable?

Skew-symmetric non-diagonalizable matrix?