Jordan–Chevalley decomposition

Statement

Let $F$ be a perfect field, for example a field of characteristic $0$ or an algebraically closed field. Let $V$ be a finite-dimensional vector space over $F$, and let $T \colon V \to V$ be a linear map. Then there are unique linear maps $S, N \colon V \to V$ with these properties:

1. $T = S + N$.
2. $S$ is semisimple and $N$ is nilpotent.
3. $SN = NS$.

Remarks

Moreover, $S$ and $N$ are polynomials in $T$ without constant term. Over $\C$, $S$ is diagonalisable; in a Jordan basis, $S$ is the diagonal part of the Jordan form. For $J_2(\lambda)$, the decomposition is $S = \lambda I_2$ and $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.

Explanations

L1 human

Every linear map over $\C$ splits into a diagonalisable part and a nilpotent part. The two parts commute with each other. This splitting is unique, and it makes functions of matrices easier to compute.