Jordan Decomposition

Question

Suppose that $x$ is a linear map from a finite dimensional vector space to itself, then $x$ has a unique Jordan Decomposition ($x=d+n$, $d$ is diagonalisable, $n$ is nilpotent). It's easy to show that $\mathrm{ad}(x)=\mathrm{ad}(d)+\mathrm{ad}(n)$, $\mathrm{ad}(d)$ is diagonalisable and $\mathrm{ad}(n)$ is nilpotent. But how can I know that this has already been a Jordan decomposition for $\mathrm{ad}(x)$?

Answer 1

Here is the reason:

$ad(.)$ preserve semi-simple and nil-potent properties, also, since $[ad(d),ad(n)]=ad[d,n]=0$, then they commute, hence it is the Jordan decomposition of $ad(x).$

Note that Jordan decomposition is unique.