Jordan-Chevalley decomposition
Encyclopedia
In mathematics, the Jordan–Chevalley decomposition, named after Camille Jordan
and Claude Chevalley
(also known as Dunford decomposition, named after Nelson Dunford
, as well as SN decomposition), 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 group
s. 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.
over a perfect field
. An operator is semisimple if the roots (in an extension of the ground field) of its minimal polynomial are all distinct (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:
where xss is semisimple, xn is nilpotent, and xss and xn commute. If such a decomposition exists it is unique, and xss and xn are in fact expressible as polynomials in x, .
If x is an invertible operator, then a multiplicative Jordan–Chevalley decomposition expresses x as a product:
where xss is semisimple, xu is unipotent, and xss and xu commute. Again, if such a decomposition exists it is unique, and xss and xu are expressible as polynomials in x.
For endomorphisms of a finite dimensional vector space whose characteristic polynomial splits into linear factors over the ground field
(which always happens if that is an algebraically closed field), the Jordan–Chevalley decomposition exists and has a simple description in terms of the Jordan normal form
. If x is in the Jordan normal form, then xss is the endomorphism whose matrix on the same basis contains just the diagonal terms of x, and xn is the endomorphism whose matrix on that basis 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.
If the ground field is not perfect
, then a Jordan–Chevalley decomposition may not exist. Example: Let p be a prime number, let F = Fp (Xp), let V = Fp (X). This is an extension field of vector space dimension p. Let x be the multiplication by the indeterminate X; it is an endomorphism of V. It is easy to see that the minimal polynomial of x is (tp − Xp), which has X as the only root in Fp (X) and so x is not semisimple. Therefore, if a Jordan–Chevalley decomposition would exist, the nilpotent part would have to be nonzero. But since the nilpotent part is also a polynomial in x, it belongs to the field Fp (X) and therefore must be zero, which gives a contradiction. Hence no Jordan–Chevalley decomposition exists in this example.
Camille Jordan
Marie Ennemond Camille Jordan was a French mathematician, known both for his foundational work in group theory and for his influential Cours d'analyse. He was born in Lyon and educated at the École polytechnique...
and Claude Chevalley
Claude Chevalley
Claude Chevalley was a French mathematician who made important contributions to number theory, algebraic geometry, class field theory, finite group theory, and the theory of algebraic groups...
(also known as Dunford decomposition, named after Nelson Dunford
Nelson Dunford
Nelson Dunford was an American mathematician, known for his work in functional analysis, namely integration of vector valued functions, ergodic theory, and linear operators. The Dunford decomposition, Dunford–Pettis property, and Dunford-Schwartz theorem bear his name.He studied mathematics at the...
, as well as SN decomposition), expresses a linear operator as the sum of its commuting semisimple
Semi-simple operator
In mathematics, a linear operator T on a finite-dimensional vector space is semi-simple if every T-invariant subspace has a complementary T-invariant subspace....
part and its nilpotent
Nilpotent
In mathematics, an element x of a ring R is called nilpotent if there exists some positive integer n such that xn = 0....
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 group
Algebraic group
In algebraic geometry, an algebraic group is a group that is an algebraic variety, such that the multiplication and inverse are given by regular functions on the variety...
s. The decomposition is easy to describe when the Jordan normal form
Jordan normal form
In linear algebra, a Jordan normal form of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called Jordan matrix, representing the operator on some basis...
of the operator is given, but it exists under weaker hypotheses than the existence of a Jordan normal form.
Linear operators
Consider linear operators on a finite-dimensional vector spaceVector space
A vector space is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied by numbers, called scalars in this context. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex...
over a perfect field
Perfect field
In algebra, a field k is said to be perfect if any one of the following equivalent conditions holds:* Every irreducible polynomial over k has distinct roots.* Every polynomial over k is separable.* Every finite extension of k is separable...
. An operator is semisimple if the roots (in an extension of the ground field) of its minimal polynomial are all distinct (if the underlying field is algebraically closed, this is the same as the requirement that the operator be diagonalizable
Diagonalizable matrix
In linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P −1AP is a diagonal matrix...
). 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 = xss + xn,
where xss is semisimple, xn is nilpotent, and xss and xn commute. If such a decomposition exists it is unique, and xss and xn are in fact expressible as polynomials in x, .
If x is an invertible operator, then a multiplicative Jordan–Chevalley decomposition expresses x as a product:
- x = xss · xu,
where xss is semisimple, xu is unipotent, and xss and xu commute. Again, if such a decomposition exists it is unique, and xss and xu are expressible as polynomials in x.
For endomorphisms of a finite dimensional vector space whose characteristic polynomial splits into linear factors over the ground field
Ground field
In mathematics, a ground field is a field K fixed at the beginning of the discussion. It is used in various areas of algebra: for example in linear algebra where the concept of a vector space may be developed over any field; and in algebraic geometry, where in the foundational developments of André...
(which always happens if that is an algebraically closed field), the Jordan–Chevalley decomposition exists and has a simple description in terms of the Jordan normal form
Jordan normal form
In linear algebra, a Jordan normal form of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called Jordan matrix, representing the operator on some basis...
. If x is in the Jordan normal form, then xss is the endomorphism whose matrix on the same basis contains just the diagonal terms of x, and xn is the endomorphism whose matrix on that basis 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.
If the ground field is not perfect
Perfect field
In algebra, a field k is said to be perfect if any one of the following equivalent conditions holds:* Every irreducible polynomial over k has distinct roots.* Every polynomial over k is separable.* Every finite extension of k is separable...
, then a Jordan–Chevalley decomposition may not exist. Example: Let p be a prime number, let F = Fp (Xp), let V = Fp (X). This is an extension field of vector space dimension p. Let x be the multiplication by the indeterminate X; it is an endomorphism of V. It is easy to see that the minimal polynomial of x is (tp − Xp), which has X as the only root in Fp (X) and so x is not semisimple. Therefore, if a Jordan–Chevalley decomposition would exist, the nilpotent part would have to be nonzero. But since the nilpotent part is also a polynomial in x, it belongs to the field Fp (X) and therefore must be zero, which gives a contradiction. Hence no Jordan–Chevalley decomposition exists in this example.