Jordan-Chevalley decomposition

Definition

For a linear operator $x$, A Jordan–Chevalley decomposition of $x$ is an expression of it as a sum

$$x=x_s+x_n,$$

where $x_s$ is semi-simple, $x_n$ is nilpotent, and $x_s$ and $x_n$ commute.

Theorem

Over a perfect field, a Jordan-Chevalley decomposition exists and is unique, where $x_s$ and $x_n$ are polynomials in $x$ with no constant terms. In particular, for any such decomposition over a perfect field, an operator that commutes with $x$ also commutes with $x_s$ and $x_n$.

\begin{proof} Here we only prove the case over finite dimensional vector space over $\mathbb{C}$. For other cases, see here.

i) Uniqueness: suppose $x=x_n+x_s$ and $x=x_n^{\prime }+x_s^{\prime }$. Since $x_n,x_s$ and $x_n^{\prime },x_s^{\prime }$ are polynomials in $x$ with no constant terms, they commute. Note that $x_n-x_n^{\prime }=x_s^{\prime }-x_s$. Since the sum of commuting nilpotent endomorphisms is nilpotent, and over a perfect field the sum of commuting semi-simple endomorphisms is again semi-simple, $x_n-x_n^{\prime }$ is both semi-simple and nilpotent. And so $x_n-x_n^{\prime }=0$.

ii) Existence: By the existence of Jordan canonical form.

iii) Polynomials: Let $f(t)=\det (tI-x)={\prod }_{i=1}^r(t-{\lambda }_i{)}^{d_i}$ be the characteristic polynomial of $x$, where $d_i=\dim V_i$. The Chinese remainder theorem gives a polynomial $p(t)$ satisfying the conditions$$ \begin{aligned} p(t)&\equiv 0\mod t\ p(t)&\equiv \lambda_i\mod (t-\lambda_i)^{d_i} \end{aligned}

Then for each $i\in\{1,\cdots,r\}$, $p(x)=\lambda_iI+(x-\lambda_iI)^{d_i}g_i(x)$ for some polynomial $g_i(t)$. So $x_s$ and $p(x)$ agree on each $V_i$, i.e. $p(x)=x_s$. Also then $q(t)=t-p(t)$ satisfies $q(x)=x_n$. The condition $p(t)\equiv 0\pmod t$ ensures that $p(t)$ and $q(t)$ have no constant term. `\end{proof}`