Jordan–Chevalley decomposition

The uniqueness follows from the fact ${\displaystyle x_{s},x_{n}}$ are polynomial in x: if ${\displaystyle x=x_{s}'+x_{n}'}$ is another decomposition such that ${\displaystyle x'_{s}}$ and ${\displaystyle x'_{n}}$ commute, then ${\displaystyle x_{s}-x_{s}'=x_{n}'-x_{n}}$, and both ${\displaystyle x_{s}',x_{n}'}$ commute with x, hence with ${\displaystyle x_{s},x_{n}}$ since they are polynomials in ${\displaystyle x}$. The sum of commuting nilpotent endomorphisms is nilpotent, and over a perfect field the sum of commuting semisimple endomorphisms is again semisimple. Since the only operator which is both semisimple and nilpotent is the zero operator it follows that ${\displaystyle x_{s}=x_{s}'}$ and ${\displaystyle x_{n}=x_{n}'}$.

We show the 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 ${\textstyle V=\bigoplus _{i=1}^{r}V_{i}}$ 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., ${\textstyle f(t)=\prod _{i=1}^{r}(t-\lambda _{i})^{d_{i}}}$, where ${\displaystyle d_{i}=\dim V_{i}.}$ 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

${\displaystyle p(t)\equiv 0{\bmod {t}},\,p(t)\equiv \lambda _{i}{\bmod {(}}t-\lambda _{i})^{d_{i}}}$ (for all 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 ${\displaystyle x_{s}}$ 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 ${\displaystyle \Gamma =\operatorname {Gal} \left({\overline {k}}/k\right)}$ 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 ${\displaystyle \left\{1,x,x^{2},\dots \right\}}$ 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. Q.E.D.

(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, ${\displaystyle x:V\to V}$ an endomorphism and ${\displaystyle A=k[x]\subset \operatorname {End} (V)}$ 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 ${\displaystyle A/J}$ is separable, then the natural surjection ${\displaystyle p:A\to A/J}$ splits; i.e., ${\displaystyle A}$ contains a semisimple subalgebra ${\displaystyle B}$ such that ${\displaystyle p|_{B}:B{\overset {\sim }{\to }}A/J}$ is an isomorphism.[4] In the setup here, ${\displaystyle A/J}$ 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 ${\displaystyle A=B\oplus J}$. In particular, the endomorphism x can be written as ${\displaystyle x=x_{s}+x_{n}}$ where ${\displaystyle x_{s}}$ is in ${\displaystyle B}$ and ${\displaystyle x_{n}}$ in ${\displaystyle J}$. Now, the image of x generates ${\displaystyle A/J\simeq B}$; thus ${\displaystyle x_{s}}$ is semisimple and is a polynomial of x. Also, ${\displaystyle x_{n}}$ is nilpotent since ${\displaystyle J}$ is nilpotent and is a polynomial of x since ${\displaystyle x_{s}}$ is. ${\displaystyle \square }$

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

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

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

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

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

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

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

Proof:[6] Without loss of generality, assume k is algebraically closed. By Lie's theorem and Engel's theorem, it suffices to show for each ${\displaystyle x\in D{\mathfrak {g}}}$, ${\displaystyle x}$ is a nilpotent endomorphism of V. Write ${\textstyle x=\sum _{i}[x_{i},y_{i}]}$. Then we need to show:

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

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

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

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.[7]

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 ${\displaystyle \pi (x_{n})=\pi (x)_{n}}$.[8]

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.

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,[9] 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.[10]

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.