Jordan–Chevalley decomposition

In mathematics, the Jordan–Chevalley decomposition, named after Camille Jordan and Claude Chevalley, expresses a linear operator as the sum of its commuting semisimple part and its nilpotent parts. The multiplicative decomposition expresses an invertible operator as the product of its commuting semisimple and unipotent parts. The decomposition is easy to describe when the Jordan normal form of the operator is given, but it exists under weaker hypotheses than the existence of a Jordan normal form. Analogues of the Jordan-Chevalley decomposition exist for elements of linear algebraic groups, Lie algebras, and Lie groups, and the decomposition is an important tool in the study of these objects.

Decomposition of a linear operator

Consider linear operators on a finite-dimensional vector space over a field. An operator T is semisimple if every T-invariant subspace has a complementary T-invariant subspace (if the underlying field is algebraically closed, this is the same as the requirement that the operator be diagonalizable). An operator x is nilpotent if some power x^m of it is the zero operator. An operator x is unipotent if x − 1 is nilpotent.

Now, let x be any operator. A Jordan–Chevalley decomposition of x is an expression of it as a sum

x = x_s + x_n,

where x_s is semisimple, x_n is nilpotent, and x_s and x_n commute. Over a perfect field, such a decomposition exists (cf. #Proof of uniqueness and existence), the decomposition is unique, and the 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.

If x is an invertible operator, then a multiplicative Jordan–Chevalley decomposition expresses x as a product

x = x_s · x_u,

where x_s is semisimple, x_u is unipotent, and x_s and x_u commute. Again, over a perfect field, such a decomposition exists, the decomposition is unique, and x_s and x_u are polynomials in x. The multiplicative version of the decomposition follows from the additive one since, as $x_{s}$ is easily seen to be invertible,

$x=x_{s}+x_{n}=x_{s}\left(1+x_{s}^{-1}x_{n}\right)$

and $1+x_{s}^{-1}x_{n}$ is unipotent. (Conversely, by the same type of argument, one can deduce the additive version from the multiplicative one.)

If x is written in Jordan normal form (with respect to some basis) then x_s is the endomorphism whose matrix contains just the diagonal terms of x, and x_n is the endomorphism whose matrix contains just the off-diagonal terms; x_u is the endomorphism whose matrix is obtained from the Jordan normal form by dividing all entries of each Jordan block by its diagonal element.

Proof of uniqueness and existence

The uniqueness follows from the fact $x_{s},x_{n}$ are polynomial in x: if $x=x_{s}'+x_{n}'$ is another decomposition such that $x'_{s}$ and $x'_{n}$ commute, then $x_{s}-x_{s}'=x_{n}'-x_{n}$, and both $x_{s}',x_{n}'$ commute with x, hence with $x_{s},x_{n}$. Now, the sum of commuting semisimple (resp. nilpotent) endomorphisms is again semisimple (resp. nilpotent). Since the only operator which is both semisimple and nilpotent is the zero operator it follows that $x_{s}=x_{s}'$ and $x_{n}=x_{n}'$.

We show the existence. Let V be a finite-dimensional vector space over a perfect field k and $x:V\to V$ an endomorphism.

First assume the base field k is algebraically closed. Then the vector space V has the direct sum decomposition ${\textstyle V=\bigoplus _{i=1}^{r}V_{i}}$ where each $V_{i}$ is the kernel of $(x-\lambda _{i}I)^{m_{i}}$, the generalized eigenspace and x stabilizes $V_{i}$, meaning $x\cdot V_{i}\subset V_{i}$. Now, define $x_{s}:V\to V$ so that, on each $V_{i}$, it is the scalar multiplication by $\lambda _{i}$. Note that, in terms of a basis respecting the direct sum decomposition, $x_{s}$ is a diagonal matrix; hence, it is a semisimple endomorphism. Since $x-x_{s}:V_{i}\to V_{i}$ is then $x-\lambda _{i}I:V_{i}\to V_{i}$ whose $m_{i}$-th power is zero, we also have that $x_{n}:=x-x_{s}$ is nilpotent, establishing the existence of the decomposition.

(Choosing a basis carefully on each $V_{i}$, one can then put x in the Jordan normal form and $x_{s},x_{n}$ are the diagonal and the off-diagonal parts of the normal form. But this is not needed here.)

The fact that $x_{s},x_{n}$ are polynomials in x follows from the Chinese remainder theorem. Indeed, let $f(t)=\operatorname {det} (tI-x)$ be the characteristic polynomial of x. Then it is the product of the characteristic polynomials of $x:V_{i}\to V_{i}$; i.e., ${\textstyle f(t)=\prod _{i=1}^{r}(t-\lambda _{i})^{d_{i}},d_{i}=\dim V_{i}.}$ Also, $d_{i}\geq m_{i}$ (because, in general, a nilpotent matrix is killed when raised to the size of the matrix). Now, the Chinese remainder theorem applied to the polynomial ring $k[t]$ gives a polynomial $p(t)$ satisfying the conditions

$p(t)\equiv 0{\bmod {t}},\,p(t)\equiv \lambda _{i}{\bmod {(}}t-\lambda _{i})^{d_{i}}$ (for all i).

(There is a redundancy in the conditions if some $\lambda _{i}$ is zero but that is not an issue; just remove it from the conditions.)

The condition $p(t)\equiv \lambda _{i}{\bmod {(}}t-\lambda _{i})^{d_{i}}$, when spelled out, means that $p(t)-\lambda _{i}=g_{i}(t)(t-\lambda _{i})^{d_{i}}$ for some polynomial $g_{i}(t)$. Since $(x-\lambda _{i}I)^{d_{i}}$ is the zero map on $V_{i}$, $p(x)$ and $x_{s}$ agree on each $V_{i}$; i.e., $p(x)=x_{s}$. Also then $q(x)=x_{n}$ with $q(t)=t-p(t)$. The condition $p(t)\equiv 0{\bmod {t}}$ ensures that $p(t)$ and $q(t)$ have no constant terms. This completes the proof of the algebraically closed field case.

If k is an arbitrary perfect field, let $\Gamma =\operatorname {Gal} \left({\overline {k}}/k\right)$ be the absolute Galois group of k. By the first part, we can choose polynomials $p,q$ over ${\overline {k}}$ such that $x=p(x)+q(x)$ is the decomposition into the semisimple and nilpotent part. For each $\sigma$ in $\Gamma$,

$x=\sigma (x)=\sigma (p(x))+\sigma (q(x))=p(x)+q(x).$

Now, $\sigma (p(x))=\sigma (p)(x)$ is a polynomial in $x$; so is $\sigma (q(x))$. Thus, $\sigma (p(x))$ and $\sigma (q(x))$ commute. Also, the application of $\sigma$ evidently preserves semisimplicity and nilpotency. Thus, by the uniqueness of decomposition (over ${\overline {k}}$), $\sigma (p(x))=p(x)$ and $\sigma (q(x))=q(x)$. Hence, $x_{s}=p(x),x_{n}=q(x)$ are $\Gamma$-invariant; i.e., they are endomorphisms (represented by matrices) over k. Finally, since $\left\{1,x,x^{2},\dots \right\}$ contains a ${\overline {k}}$-basis that spans the space containing $x_{s},x_{n}$, by the same argument, we also see that $p,q$ have coefficients in k. This completes the proof. $\square$

Short proof using abstract algebra

(Jacobson 1979) proves the existence of a decomposition as a consequence of the Wedderburn principal theorem. (This approach is not only short but also makes the role of the assumption that the base field be perfect clearer.)

Let V be a finite-dimensional vector space over a perfect field k, $x:V\to V$ an endomorphism and $A=k[x]\subset \operatorname {End} (V)$ the subalgebra generated by x. Note that A is a commutative Artinian ring. The Wedderburn principal theorem states: for a finite-dimensional algebra A with the Jacobson radical J, if $A/J$ is separable, then the natural surjection $p:A\to A/J$ splits; i.e., $A$ contains a semisimple subalgebra $B$ such that $p|_{B}:B{\overset {\sim }{\to }}A/J$ is an isomorphism. In the setup here, $A/J$ is separable since the base field is perfect (so the theorem is applicable) and J is also the nilradical of A. There is then the vector-space decomposition $A=B\oplus J$. In particular, the endomorphism x can be written as $x=x_{s}+x_{n}$ where $x_{s}$ is in $B$ and $x_{n}$ in $J$. Now, the image of x generates $A/J\simeq B$; thus $x_{s}$ is semisimple and is a polynomial of x. Also, $x_{n}$ is nilpotent since $J$ is nilpotent and is a polynomial of x since $x_{s}$ is. $\square$

Nilpotency criterion

The Jordan decomposition can be used to characterize nilpotency of an endomorphism. Let k be an algebraically closed field of characteristic zero, $E=\operatorname {End} _{\mathbb {Q} }(k)$ the endomorphism ring of k over rational numbers and V a finite-dimensional vector space over k. Given an endomorphism $x:V\to V$, let $x=s+n$ be the Jordan decomposition. Then $s$ is diagonalizable; i.e., ${\textstyle V=\bigoplus V_{i}}$ where each $V_{i}$ is the eigenspace for eigenvalue $\lambda _{i}$ with multiplicity $m_{i}$. Then for any $\varphi \in E$ let $\varphi (s):V\to V$ be the endomorphism such that $\varphi (s):V_{i}\to V_{i}$ is the multiplication by $\varphi (\lambda _{i})$. Chevalley calls $\varphi (s)$ the replica of $s$ given by $\varphi$. (For example, if $k=\mathbb {C}$, then the complex conjugate of an endomorphism is an example of a replica.) Now,

Nilpotency criterion — $x$ is nilpotent (i.e., $s=0$) if and only if $\operatorname {tr} (x\varphi (s))=0$ for every $\varphi \in E$. Also, if $k=\mathbb {C}$, then it suffices the condition holds for $\varphi =$ complex conjugation.

Proof: First, since $n\varphi (s)$ is nilpotent,

$0=\operatorname {tr} (x\varphi (s))=\sum _{i}\operatorname {tr} \left(s\varphi (s)|V_{i}\right)=\sum _{i}m_{i}\lambda _{i}\varphi (\lambda _{i})$.

If $\varphi$ is the complex conjugation, this implies $\lambda _{i}=0$ for every i. Otherwise, take $\varphi$ to be a $\mathbb {Q}$-linear functional $\varphi :k\to \mathbb {Q}$ followed by $\mathbb {Q} \hookrightarrow k$. Applying that to the above equation, one gets:

$\sum _{i}m_{i}\varphi (\lambda _{i})^{2}=0$

and, since $\varphi (\lambda _{i})$ are all real numbers, $\varphi (\lambda _{i})=0$ for every i. Varying the linear functionals then implies $\lambda _{i}=0$ for every i. $\square$

A typical application of the above criterion is the proof of Cartan's criterion for solvability of a Lie algebra. It says: if ${\mathfrak {g}}\subset {\mathfrak {gl}}(V)$ is a Lie subalgebra over a field k of characteristic zero such that $\operatorname {tr} (xy)=0$ for each $x\in {\mathfrak {g}},y\in D{\mathfrak {g}}=[{\mathfrak {g}},{\mathfrak {g}}]$, then ${\mathfrak {g}}$ is solvable.

Proof: Without loss of generality, assume k is algebraically closed. By Lie's theorem and Engel's theorem, it suffices to show for each $x\in D{\mathfrak {g}}$, $x$ is a nilpotent endomorphism of V. Write ${\textstyle x=\sum _{i}[x_{i},y_{i}]}$. Then we need to show:

$\operatorname {tr} (x\varphi (s))=\sum _{i}\operatorname {tr} ([x_{i},y_{i}]\varphi (s))=\sum _{i}\operatorname {tr} (x_{i}[y_{i},\varphi (s)])$

is zero. Let ${\mathfrak {g}}'={\mathfrak {gl}}(V)$. Note we have: $\operatorname {ad} _{{\mathfrak {g}}'}(x):{\mathfrak {g}}\to D{\mathfrak {g}}$ and, since $\operatorname {ad} _{{\mathfrak {g}}'}(s)$ is the semisimple part of the Jordan decomposition of $\operatorname {ad} _{{\mathfrak {g}}'}(x)$, it follows that $\operatorname {ad} _{{\mathfrak {g}}'}(s)$ is a polynomial without constant term in $\operatorname {ad} _{{\mathfrak {g}}'}(x)$; hence, $\operatorname {ad} _{{\mathfrak {g}}'}(s):{\mathfrak {g}}\to D{\mathfrak {g}}$ and the same is true with $\varphi (s)$ in place of $s$. That is, $[\varphi (s),{\mathfrak {g}}]\subset D{\mathfrak {g}}$, which implies the claim given the assumption. $\square$

Counterexample to existence over an imperfect field

If the ground field is not perfect, then a Jordan–Chevalley decomposition may not exist. Example: Let p be a prime number, let $k$ be imperfect of characteristic $p$, and choose $a$ in $k$ that is not a $p$th power. Let $V=k[X]/\left(X^{p}-a\right)^{2}$, let $x={\overline {X}}$and let $T$ be the $k$-linear operator given by multiplication by $x$ in $V$. This has as its invariant $k$-linear subspaces precisely the ideals of $V$ viewed as a ring, which correspond to the ideals of $k[X]$ containing $\left(X^{p}-a\right)^{2}$. Since $X^{p}-a$ is irreducible in $k[X]$, ideals of V are $0$, $S$ and $J=\left(x^{p}-a\right)V$. Suppose $T=S+N$ for commuting $k$-linear operators $S$ and $N$ that are respectively semisimple (just over $k$, which is weaker than semisimplicity over an algebraic closure of $k$) and nilpotent. Since $S$ and $N$ commute, they each commute with $T=S+N$ and hence each acts $k[x]$-linearly on $V$. Therefore $S$ and $N$ are each given by multiplication by respective members of $V$ $s=S(1)$ and $n=N(1)$, with $s+n=T(1)=x$. Since $N$ is nilpotent, $n$ is nilpotent in $V$, therefore ${\overline {n}}=0$ in $V/J$, for $V/J$ is a field. Hence, $n\in J$, therefore $n=\left(x^{p}-a\right)h(x)$ for some polynomial $h(X)\in k[X]$. Also, we see that $n^{2}=0$. Since $k$ is of characteristic $p$, we have $x^{p}=s^{p}+n^{p}=s^{p}$. Also, since ${\overline {x}}={\overline {s}}$ in $A/J$, we have $h\left({\overline {s}}\right)=h\left({\overline {x}}\right)$, therefore $h(s)-h(x)\in J$ in $V$. Since $\left(x^{p}-a\right)J=0$, we have $\left(x^{p}-a\right)h(x)=\left(x^{p}-a\right)h(s)$. Combining these results we get $x=s+n=s+\left(s^{p}-a\right)h(s)$. This shows that $s$ generates $V$ as a $k$-algebra and thus the $S$-stable $k$-linear subspaces of $V$ are ideals of $V$, i.e. they are $0$, $J$ and $V$. We see that $J$ is an $S$-invariant subspace of $V$ which has no complement $S$-invariant subspace, contrary to the assumption that $S$ is semisimple. Thus, there is no decomposition of $T$ as a sum of commuting $k$-linear operators that are respectively semisimple and nilpotent. Note that minimal polynomial of $T$ is inseparable over $k$ and is a square in $k[X]$. It can be shown that if minimal polynomial of $k$ linear operator $L$ is separable then $L$ has Jordan-Chevalley decomposition and that if this polynomial is product of distinct irreducible polynomials in $k[X]$, then $L$ is semisimple over $k$.

Analogous decompositions

The multiplicative version of the Jordan-Chevalley decomposition generalizes to a decomposition in a linear algebraic group, and the additive version of the decomposition generalizes to a decomposition in a Lie algebra.

Lie algebras

Let ${\mathfrak {gl}}(V)$ denote the Lie algebra of the endomorphisms of a finite-dimensional vector space V over a perfect field. If $x=x_{s}+x_{n}$ is the Jordan decomposition, then $\operatorname {ad} (x)=\operatorname {ad} (x_{s})+\operatorname {ad} (x_{n})$ is the Jordan decomposition of $\operatorname {ad} (x)$ on the vector space ${\mathfrak {gl}}(V)$. Indeed, first, $\operatorname {ad} (x_{s})$ and $\operatorname {ad} (x_{n})$ commute since $[\operatorname {ad} (x_{s}),\operatorname {ad} (x_{n})]=\operatorname {ad} ([x_{s},x_{n}])=0$. Second, in general, for each endomorphism $y\in {\mathfrak {gl}}(V)$, we have:

1. If $y^{m}=0$, then $\operatorname {ad} (y)^{2m-1}=0$, since $\operatorname {ad} (y)$ is the difference of the left and right multiplications by y.
2. If $y$ is semisimple, then $\operatorname {ad} (y)$ is semisimple.

Hence, by uniqueness, $\operatorname {ad} (x)_{s}=\operatorname {ad} (x_{s})$ and $\operatorname {ad} (x)_{n}=\operatorname {ad} (x_{n})$.

If $\pi :{\mathfrak {g}}\to {\mathfrak {gl}}(V)$ is a finite-dimensional representation of a semisimple finite-dimensional complex Lie algebra, then $\pi$ preserves the Jordan decomposition in the sense: if $x=x_{s}+x_{n}$, then $\pi (x_{s})=\pi (x)_{s}$ and $\pi (x_{n})=\pi (x)_{n}$.

Real semisimple Lie algebras

In the formulation of Chevalley and Mostow, the additive decomposition states that an element X in a real semisimple Lie algebra g with Iwasawa decomposition g = k ⊕ a ⊕ n can be written as the sum of three commuting elements of the Lie algebra X = S + D + N, with S, D and N conjugate to elements in k, a and n respectively. In general the terms in the Iwasawa decomposition do not commute.

Linear algebraic groups

Let $G$ be a linear algebraic group over a perfect field. Then, essentially by definition, there is a closed embedding $G\hookrightarrow \mathbf {GL} _{n}$. Now, to each element $g\in G$, by the multiplicative Jordan decomposition, there are a pair of a semisimple element $g_{s}$ and a unipotent element $g_{u}$ a priori in $\mathbf {GL} _{n}$ such that $g=g_{s}g_{u}=g_{u}g_{s}$. But, as it turns out, the elements $g_{s},g_{u}$ can be shown to be in $G$ (i.e., they satisfy the defining equations of G) and that they are independent of the embedding into $\mathbf {GL} _{n}$; i.e., the decomposition is intrinsic.

When G is abelian, $G$ is then the direct product of the closed subgroup of the semisimple elements in G and that of unipotent elements.

Real semisimple Lie groups

The multiplicative decomposition states that if g is an element of the corresponding connected semisimple Lie group G with corresponding Iwasawa decomposition G = KAN, then g can be written as the product of three commuting elements g = sdu with s, d and u conjugate to elements of K, A and N respectively. In general the terms in the Iwasawa decomposition g = kan do not commute.

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