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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$

$\displaystyle 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,

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

then $ A$ is similar to a direct sum

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

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

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

holds. We put

$\displaystyle 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

$\displaystyle 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

$\displaystyle 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

$\displaystyle 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

$\displaystyle 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

$\displaystyle (\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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next up previous
Next: -rationality Up: Jordan-Chevalley decomposition of a Previous: Jordan-Chevalley decomposition of a
2007-12-19