# 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 important in the study of algebraic groups. 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.

## Statement

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 xm 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 = xs + xn

where xs is semisimple, xn is nilpotent, and xs and xn commute. Over a perfect field, such a decomposition exists (cf. #Proof of existence),[1][2] it is unique, and the xs and xn are polynomials in x with no constant terms. The uniqueness is easy to see: if ${\displaystyle x=x_{s}'+x_{n}'}$ is another decomposition, then ${\displaystyle x_{s}-x_{s}'=x_{n}'-x_{n}}$. Now, the sum of commuting semisimple (resp. nilpotent) endomorphisms is again semisimple (resp. nilpotent). Thus, this is possible only if ${\displaystyle x_{s}=x_{s}'}$ and ${\displaystyle x_{n}=x_{n}'}$.

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

x = xs · xu,

where xs is semisimple, xu is unipotent, and xs and xu commute. Again, over a perfect field, such a decomposition exists, it is unique and xs and xu are polynomials in x. Note that multiplicative version follows from the additive one since, as ${\displaystyle x_{s}}$ is invertible (nilpotent operators don't spoil invertibility),

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

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

If x is in the Jordan normal form (with respect to some basis), then xs is the endomorphism whose matrix contains just the diagonal terms of x, and xn is the endomorphism whose matrix contains just the off-diagonal terms; xu 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 existence

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

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

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

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

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

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

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

${\displaystyle x=\sigma (x)=\sigma (p(x))+\sigma (q(x))=p(x)+q(x).}$

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

## Decomposition in a Lie algebra

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

1. If ${\displaystyle y^{m}=0}$, then ${\displaystyle \operatorname {ad} (y)^{2m-1}=0}$, since ${\displaystyle \operatorname {ad} (y)}$ is the difference of the left and right multiplications by y.
2. If ${\displaystyle y}$ is semisimple, then ${\displaystyle \operatorname {ad} (y)}$ is semisimple.[3]

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

If ${\displaystyle \pi :{\mathfrak {g}}\to {\mathfrak {gl}}(V)}$ is a finite-dimensional representation of a semisimple finite-dimensional complex Lie algebra, then ${\displaystyle \pi }$ preserves the Jordan decomposition in the sense: if ${\displaystyle x=x_{s}+x_{n}}$, then ${\displaystyle \pi (x_{s})=\pi (x)_{s}}$ and ${\displaystyle \pi (x_{n})=\pi (x)_{n}}$.[4]

## Decomposition in a linear algebraic group

Let ${\displaystyle G}$ be a linear algebraic group over a perfect field. Then, essentially by definition, there is a closed embedding ${\displaystyle G\hookrightarrow \mathbf {GL} _{n}}$. Now, to each element ${\displaystyle g\in G}$, by the multiplicative Jordan decomposition, there are a pair of a semisimple element ${\displaystyle g_{s}}$ and a unipotent element ${\displaystyle g_{u}}$ a priori in ${\displaystyle \mathbf {GL} _{n}}$ such that ${\displaystyle g=g_{s}g_{u}=g_{u}g_{s}}$. But, as it turns out,[5] the elements ${\displaystyle g_{s},g_{u}}$ can be shown to be in ${\displaystyle G}$ (i.e., they satisfy the defining equations of G) and that they are independent of the embedding into ${\displaystyle \mathbf {GL} _{n}}$; i.e., the decomposition is intrinsic.

When G is abelian, ${\displaystyle G}$ is then the direct product of the closed subgroup of the semisimple elements in G and that of unipotent elements.[6]

## Decomposition in a real semisimple Lie algebra

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 = kan 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.

## Decomposition in a real semisimple Lie group

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.

## Counterexample

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 pth power. Let V = k[x]/(xp-a)2, and let T be the k-linear operator given by multiplication by x on V. This has as its stable k-linear subspaces precisely the ideals of V viewed as a ring. 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. Thus, each preserves the unique nonzero proper k[x]-submodule J=(xp-a)V in V. But by semisimplicity of S, there would have to be an S-stable k-linear complement to J. However, by k[x]-linearity, S and N are each given by multiplication against the respective polynomials s = S(1) and n =N(1) whose induced effects on the quotient V/(xp-a) must be respectively x and 0 since this quotient is a field. Hence, s = x + (xp-a)h(x) for some polynomial h(x) (which only matters modulo (xp-a)), so it is easily seen that s generates V as a k-algebra and thus the S-stable k-linear subspaces of V are precisely the k[x]-submodules. It follows that an S-stable complement to J is also a k[x]-submodule of V, contradicting that J is the only nonzero proper k[x]-submodule of V. Thus, there is no decomposition of T as a sum of commuting k-linear operators that are respectively semisimple and nilpotent.

## References

1. Humphreys 1972, Prop. 4.2, p. 17 for the algebraically closed field case.
2. Waterhouse, Ch. 9, Exercise 1.
3. This is not easy to see but is shown in the proof of (Jacobson, Ch. III, § 7, Theorem 11.). Editorial note: we need to add a discussion of this matter to “semisimple operator”.
4. Fulton, Harris & Theorem 9.20.
5. Waterhouse, Theorem 9.2.
6. Waterhouse, Theorem 9.3.
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