Existence and uniqueness of Jordan-Chevalley decomposition

DEFINITION 4.1   Let $A$ be a square matrix over a field $k$ . A Jordan-Chevalley decomposition (also called SN-decomposition) of $A$ is a decomposition of $A$

$$A=S+N$$

which satisfies the following conditions.

1. $S$ is semisimple (that means, the minimal polynomial of $S$ has only simple roots.)
2. $N$ is nilpotent.
3. $SN=NS$
4. $S,N\in \overline{k}[A]$

A main objective of this section is to prove the following proposition.

PROPOSITION 4.2   For any square matrix $A$ over a field $k$ , there exists a unique Jordan-Chevalley decomposition.

To prove it, we need some basic facts from linear algebra.

LEMMA 4.3   Let $A$ be a square matrix over a field $k$ . Let $m_A(X)$ be the minimal polynomial of $A$ over $k$ . If $m_A$ is decomposed into two coprime polynomials, that means,

$$m_A=m_1 m_2 \qquad (m_1, m_2) =1,$$

then $A$ is similar to a direct sum

$$A_1\oplus A_2 = \begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}$$

where $m_1(A_1)=0$ , $m_2(A_2)=0$ .

PROOF.. Since $k[X]$ is an Euclidean domain, it is a principal ideal domain. thus we see that there exists a polynomial $l_1(X),l_2(X)\in k[X]$ such that

$$l_1(X)m_1(X)+l_2(X)m_2(X)=1$$

holds. We put

$$E_j=l_j(A)m_j(A) \qquad(j=1,2).$$

Then we see easily that $E_1,E_2$ are mutually orthogonal projection. That means, we have

$$E_1^2=E_1, E_2^2=E_2, E_1+E_2=1, E_1 E_2=0.$$

It is also easy to see that both $E_1$ and $E_2$ commute with $A$ . Now putting $A_1=A\vert _{\operatorname{Range}{E_2}}$ and $A_2=A\vert _{\operatorname{Range}E_1}$ we see that

$$A=A E_2+ A E_1=A_1 \oplus A_2.$$

with $A_1$ and $A_2$ satisfying the required property. $\qedsymbol$

COROLLARY 4.4   Every square matrix $A$ over a field $k$ is similar to a direct sum of square matrices $A_1,A_2,\dots, A_s$ with each minimal polynomial $m_{A_j}$ equals to a power $f_j^{e_j}$ of a irreducible polynomial $f_j$ over $k$ .

$\qedsymbol$

COROLLARY 4.5   When $k$ is algebraically closed, every square matrix $A$ over a field $k$ is similar to a direct sum of square matrices $A_1,A_2,\dots, A_s$ with each minimal polynomial $m_{A_j}$ equals to a power $(X-c_j)^{e_j}$ of a polynomial of degree $1$ over $k$ .

$\qedsymbol$

COROLLARY 4.6   A square matrix over a field $k$ is semisimple if and only if it is diagonalizable (similar to a diagonal matrix) over $\overline{k}$ .

$\qedsymbol$

COROLLARY 4.7   Let $S_1,S_2$ be semisimple square matrices of the same size over $k$ . if $S_1$ and $S_2$ commute, then both $S_1+S_2$ and $S_1 S_2$ are also semisimple.

PROOF.. Using commutativity of $S_1$ and $S_2$ , we may easily see that $S_1$ and $S_2$ are simultaneously diagonalizable over $\overline{k}$ . $\qedsymbol$

COROLLARY 4.8   Let $k$ be a field. Let $C$ be a commutative subalgebra of $M_n(k)$ . If $C$ is generated by semisimple elements, then every element of $C$ is also semisimple.

$\qedsymbol$

On the other hand we have

LEMMA 4.9   Let $k$ be a field. Let $C$ be a commutative subalgebra of $M_n(k)$ . If $C$ is generated by nilpotent elements, then every element of $C$ is also nilpotent.

PROOF.. Easy. $\qedsymbol$

COROLLARY 4.10   A Jordan-Chevalley decomposition (if there exists) of a square matrix $A$ is unique.

PROOF.. Let

$$A=S+N=S'+N'$$

be two Jordan-Chevalley decompositions. Then $S-S'=N'-N$ is a semisimple nilpotent element. Thus $S-S'=N'-N=0$ . $\qedsymbol$

PROOF.. (of Proposition 4.2.) It now remains to prove that Jordan-Chevalley decomposition of a square matrix exists. By definition we may assume that $k$ is algebraically closed. In view of Corollary 4.5, we may then assume that the minimal polynomial $m_A$ of $A$ is of the form $(X-c)^e$ for some $c\in k$ and $e \in \mathbb{Z}_{>0}$ . Then

$$A=c + (A-c)$$

gives the required Jordan-Chevalley decomposition.

$\qedsymbol$

DEFINITION 4.11   Let $k$ be a field. For any square matrix $x \in M_n(k)$ , we denote by $x_s$ (respectively, $x_n$ ) the semisimple (respectively, nilpotent) part of $x$ in the Jordan-Chevalley decomposition of $x$ .

LEMMA 4.12   Let $k$ be a field. Let $x \in M_n(k)$ be a square matrix. then we have

$$(\operatorname{ad}(x))_s=\operatorname{ad}(x_s) ,\quad (\operatorname{ad}(x))_n=\operatorname{ad}(x_n)$$

PROOF.. Follows easily from the uniqueness of the Jordan-Chevalley decomposition. $\qedsymbol$

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