Jordan normal form

Example of a matrix in Jordan normal form.
All matrix entries not shown are zero. The
outlined squares are known as "Jordan blocks".
Each Jordan block contains one number λ_i
on its main diagonal, and 1s directly above
the main diagonal. The λ_is are the eigenvalues
of the matrix; they need not be distinct.

In linear algebra, a Jordan normal form, also known as a Jordan canonical form,
is an upper triangular matrix of a particular form called a Jordan matrix representing a linear operator on a finite-dimensional vector space with respect to some basis. Such a matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal, and with identical diagonal entries to the left and below them.
Let V be a vector space over a field K. Then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is algebraically closed. The diagonal entries of the normal form are the eigenvalues, and the number of times each eigenvalue occurs is called the algebraic multiplicity of the eigenvalue.
If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.
The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.
The Jordan normal form is named after Camille Jordan, who first stated the Jordan decomposition theorem in 1870.

Overview

Notation

Some textbooks have the ones on the subdiagonal; that is, immediately below the main diagonal instead of on the superdiagonal. The eigenvalues are still on the main diagonal.

Motivation

An n × n matrix A is diagonalizable if and only if the sum of the dimensions of the eigenspaces is n. Or, equivalently, if and only if A has n linearly independent eigenvectors. Not all matrices are diagonalizable; matrices that are not diagonalizable are called defective matrices. Consider the following matrix:
Including multiplicity, the eigenvalues of A are λ = 1, 2, 4, 4. The dimension of the eigenspace corresponding to the eigenvalue 4 is 1, so A is not diagonalizable. However, there is an invertible matrix P such that J = P⁻¹AP, where
The matrix is almost diagonal. This is the Jordan normal form of A. The section Example below fills in the details of the computation.

Complex matrices

In general, a square complex matrix A is similar to a block diagonal matrix
where each block J_i is a square matrix of the form
So there exists an invertible matrix P such that P⁻¹AP = J is such that the only non-zero entries of J are on the diagonal and the superdiagonal. J is called the Jordan normal form of A. Each J_i is called a Jordan block of A. In a given Jordan block, every entry on the superdiagonal is 1.
Assuming this result, we can deduce the following properties:

Counting multiplicities, the eigenvalues of J, and therefore of A, are the diagonal entries.
Given an eigenvalue λ_i, its geometric multiplicity is the dimension of where is the identity matrix, and it is the number of Jordan blocks corresponding to λ_i.
The sum of the sizes of all Jordan blocks corresponding to an eigenvalue λ_i is its algebraic multiplicity.
A is diagonalizable if and only if, for every eigenvalue λ of A, its geometric and algebraic multiplicities coincide. In particular, the Jordan blocks in this case are matrices; that is, scalars.
The Jordan block corresponding to λ is of the form, where N is a nilpotent matrix defined as N_ij = δ_i_,j−1. The nilpotency of N can be exploited when calculating f where f is a complex analytic function. For example, in principle the Jordan form could give a closed-form expression for the exponential exp.
The number of Jordan blocks corresponding to λ_i of size at least j is Thus, the number of Jordan blocks of size j is
:
Given an eigenvalue λ_i, its multiplicity in the minimal polynomial is the size of its largest Jordan block.
Example

Consider the matrix from the example in the previous section. The Jordan normal form is obtained by some similarity transformation:
Let have column vectors,, then
We see that
For we have, that is, is an eigenvector of corresponding to the eigenvalue. For, multiplying both sides by gives
But, so
Thus,
Vectors such as are called generalized eigenvectors of A.

Example: Obtaining the normal form

This example shows how to calculate the Jordan normal form of a given matrix.
Consider the matrix
which is mentioned in the beginning of the article.
The characteristic polynomial of A is
This shows that the eigenvalues are 1, 2, 4 and 4, according to algebraic multiplicity. The eigenspace corresponding to the eigenvalue 1 can be found by solving the equation It is spanned by the column vector v = ^T. Similarly, the eigenspace corresponding to the eigenvalue 2 is spanned by w = ^T. Finally, the eigenspace corresponding to the eigenvalue 4 is also one-dimensional and is spanned by x = ^T. So, the geometric multiplicity of each of the three eigenvalues is one. Therefore, the two eigenvalues equal to 4 correspond to a single Jordan block, and the Jordan normal form of the matrix A is the direct sum
There are three Jordan chains. Two have length one: and, corresponding to the eigenvalues 1 and 2, respectively. There is one chain of length two corresponding to the eigenvalue 4. To find this chain, calculate
where is the identity matrix. Pick a vector in the above span that is not in the kernel of for example, y = ^T. Now, and, so is a chain of length two corresponding to the eigenvalue 4.
The transition matrix P such that P⁻¹AP = J is formed by putting these vectors next to each other as follows
A computation shows that the equation P⁻¹AP = J indeed holds.
If we had interchanged the order in which the chain vectors appeared, that is, changing the order of v, w and together, the Jordan blocks would be interchanged. However, the Jordan forms are equivalent Jordan forms.

Generalized eigenvectors

Given an eigenvalue λ, every corresponding Jordan block gives rise to a Jordan chain of linearly independent vectors p_i, i = 1,..., b, where b is the size of the Jordan block. The generator, or lead vector, p_b of the chain is a generalized eigenvector such that. The vector is an ordinary eigenvector corresponding to λ. In general, p_i is a preimage of p_i−1 under. So the lead vector generates the chain via multiplication by. Therefore, the statement that every square matrix A can be put in Jordan normal form is equivalent to the claim that the underlying vector space has a basis composed of Jordan chains.

A proof

We give a proof by induction that any complex-valued square matrix A may be put in Jordan normal form. Since the underlying vector space can be shown to be the direct sum of invariant subspaces associated with the eigenvalues, A can be assumed to have just one eigenvalue λ. The 1 × 1 case is trivial. Let A be an n × n matrix. The range of, denoted by, is an invariant subspace of A. Also, since λ is an eigenvalue of A, the dimension of, r, is strictly less than n, so, by the inductive hypothesis, has a basis composed of Jordan chains.
Next consider the kernel, that is, the subspace. If
the desired result follows immediately from the rank–nullity theorem.
Otherwise, if
let the dimension of Q be. Each vector in Q is an eigenvector, so must contain s Jordan chains corresponding to s linearly independent eigenvectors. Therefore the basis must contain s vectors, say, that are lead vectors of these Jordan chains. We can "extend the chains" by taking the preimages of these lead vectors. Let q_i be such that
Finally, we can pick any basis for
and then lift to vectors in. Each z_i forms a Jordan chain of length 1. We just need to show that the union of,, and forms a basis for the vector space.
By the rank-nullity theorem,, so, and so the number of vectors in the potential basis is equal to n. To show linear independence, suppose some linear combination of the vectors is 0. Applying we get some linear combination of p_i, with the q_i becoming lead vectors among the p_i. From linear independence of p_i, it follows that the coefficients of the vectors q_i must be zero. Furthermore, no non-trivial linear combination of the z_i can equal a linear combination of p_i, because then it would belong to and thus, which is impossible by the construction of z_i. Therefore the coefficients of the z_i will also be 0. This leaves in the original linear combination just the p_i terms, which are assumed to be linearly independent, and so their coefficients must be zero too. We have found a basis composed of Jordan chains, and this shows A can be put in Jordan normal form.