Jordan normal form

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 (on the superdiagonal), 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 representing the operator has the required form exists if and only if all eigenvalues of the operator 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 (for instance, if it is the field of complex numbers). The diagonal entries of the normal form are the eigenvalues (of the operator), 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.

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.

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:

\(A = \left[\begin{array}{*{20}{r}} 5 & 4 & 2 & 1 \\[2pt] 0 & 1 & -1 & -1 \\[2pt] -1 & -1 & 3 & 0 \\[2pt] 1 & 1 & -1 & 2 \end{array}\right].\)

Including multiplicity, the eigenvalues of A are λ = 1, 2, 4, 4. The dimension of the eigenspace corresponding to the eigenvalue 4 is 1 (and not 2), so A is not diagonalizable. However, there is an invertible matrix P such that J = PAP, where

\(J = \begin{bmatrix} 1 & 0 & 0 & 0 \\[2pt] 0 & 2 & 0 & 0 \\[2pt] 0 & 0 & 4 & 1 \\[2pt] 0 & 0 & 0 & 4 \end{bmatrix}.\)

The matrix \(J\) is almost diagonal. This is the Jordan normal form of A. The section Example below fills in the details of the computation.

In general, a square complex matrix A is similar to a block diagonal matrix

\(J = \begin{bmatrix} J_1 & \; & \; \\ \; & \ddots & \; \\ \; & \; & J_p\end{bmatrix}\)

where each block J_i is a square matrix of the form

\(J_i = \begin{bmatrix} \lambda_i & 1 & \; & \; \\ \; & \lambda_i & \ddots & \; \\ \; & \; & \ddots & 1 \\ \; & \; & \; & \lambda_i \end{bmatrix}.\)

So there exists an invertible matrix P such that PAP = 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 ker(A − λ_i I), where I 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 1 × 1 matrices; that is, scalars.
• The Jordan block corresponding to λ is of the form λI + N, where N is a nilpotent matrix defined as N_ij = δ_i,j−1 (where δ is the Kronecker delta). The nilpotency of N can be exploited when calculating f(A) where f is a complex analytic function. For example, in principle the Jordan form could give a closed-form expression for the exponential exp(A).
• The number of Jordan blocks corresponding to λ_i of size at least j is dim ker(A − λ_iI) − dim ker(A − λ_iI). Thus, the number of Jordan blocks of size j is

  \(2 \dim \ker (A - \lambda_i I)^j - \dim \ker (A - \lambda_i I)^{j+1} - \dim \ker (A - \lambda_i I)^{j-1}\)

• Given an eigenvalue λ_i, its multiplicity in the minimal polynomial is the size of its largest Jordan block.

Consider the matrix \(A\) from the example in the previous section. The Jordan normal form is obtained by some similarity transformation:

\(P^{-1}AP = J;\) that is, \(AP = PJ.\)

Let \(P\) have column vectors \(p_i\), \(i = 1, \ldots, 4\), then

\(A \begin{bmatrix} p_1 & p_2 & p_3 & p_4 \end{bmatrix} = \begin{bmatrix} p_1 & p_2 & p_3 & p_4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} p_1 & 2p_2 & 4p_3 & p_3+4p_4 \end{bmatrix}.\)

We see that

\((A - 1 I) p_1 = 0\)

\((A - 2 I) p_2 = 0\)

\((A - 4 I) p_3 = 0\)

\((A - 4 I) p_4 = p_3.\)

For \(i = 1,2,3\) we have \(p_i \in \ker(A-\lambda_{i} I)\), that is, \(p_i\) is an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda_i\). For \(i=4\), multiplying both sides by \((A-4I)\) gives

\((A-4I)^2 p_4 = (A-4I) p_3.\)

But \((A-4I)p_3 = 0\), so

\((A-4I)^2 p_4 = 0.\)

Thus, \(p_4 \in \ker(A-4 I)^2.\)

Vectors such as \(p_4\) are called generalized eigenvectors of A.

This example shows how to calculate the Jordan normal form of a given matrix.

Consider the matrix

\(A = \left[ \begin{array}{rrrr} 5 & 4 & 2 & 1 \\ 0 & 1 & -1 & -1 \\ -1 & -1 & 3 & 0 \\ 1 & 1 & -1 & 2 \end{array} \right]\)

which is mentioned in the beginning of the article.

The characteristic polynomial of A is

\(\begin{aligned} \chi(\lambda) & = \det(\lambda I - A) \\ & = \lambda^4 - 11 \lambda^3 + 42 \lambda^2 - 64 \lambda + 32 \\ & = (\lambda-1)(\lambda-2)(\lambda-4)^2. \, \end{aligned}\)

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 Av = 1v. It is spanned by the column vector v = (−1, 1, 0, 0). Similarly, the eigenspace corresponding to the eigenvalue 2 is spanned by w = (1, −1, 0, 1). Finally, the eigenspace corresponding to the eigenvalue 4 is also one-dimensional (even though this is a double eigenvalue) and is spanned by x = (1, 0, −1, 1). So, the geometric multiplicity (that is, the dimension of the eigenspace of the given eigenvalue) 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

\(J = J_1(1) \oplus J_1(2) \oplus J_2(4) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix}.\)

There are three Jordan chains. Two have length one: {v} and {w}, 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

\(\ker(A-4I)^2 = \operatorname{span} \, \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \left[ \begin{array}{r} 1 \\ 0 \\ -1 \\ 1 \end{array} \right] \right\}\)

where I is the 4 × 4 identity matrix. Pick a vector in the above span that is not in the kernel of A − 4I; for example, y = (1,0,0,0). Now, (A − 4I)y = x and (A − 4I)x = 0, so {y, x} is a chain of length two corresponding to the eigenvalue 4.

\(P = \left[\begin{array}{c|c|c|c} v & w & x & y \end{array}\right] = \left[ \begin{array}{rrrr} -1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array} \right].\)

\(P^{-1}AP=J=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix}.\)

Condensed: the full section is in Wikipedia.

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 ⁠\((A - \lambda I)^b p_b = 0\)⁠. The vector ⁠\(p_1 = (A - \lambda I)^{b-1} p_b\)⁠ is an ordinary eigenvector corresponding to λ. In general, p_i is a preimage of p_i−1 under ⁠\(A - \lambda I\)⁠. So the lead vector generates the chain via multiplication by ⁠\(A - \lambda I\)⁠. 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.

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 ⁠\(A - \lambda I\)⁠, denoted by ⁠\(\operatorname{Ran}(A - \lambda I)\)⁠, is an invariant subspace of A. Also, since λ is an eigenvalue of A, the dimension of ⁠\(\operatorname{Ran}(A - \lambda I)\)⁠, r, is strictly less than n, so, by the inductive hypothesis, ⁠\(\operatorname{Ran}(A - \lambda I)\)⁠ has a basis {p₁, ..., p_r} composed of Jordan chains.

Next consider the kernel, that is, the subspace ⁠\(\ker(A - \lambda I)\)⁠. If

\(\operatorname{Ran}(A - \lambda I) \cap \ker(A - \lambda I) = \{0\},\)

the desired result follows immediately from the rank, nullity theorem. (This would be the case, for example, if A were Hermitian.)

Otherwise, if

\(Q = \operatorname{Ran}(A - \lambda I) \cap \ker(A - \lambda I) \neq \{0\},\)

let the dimension of Q be s ≤ r. Each vector in Q is an eigenvector, so ⁠\(\operatorname{Ran}(A - \lambda I)\)⁠ must contain s Jordan chains corresponding to s linearly independent eigenvectors. Therefore the basis {p₁, ..., p_r} must contain s vectors, say {p₁, ..., p_s}, that are lead vectors of these Jordan chains. We can "extend the chains" by taking the preimages of these lead vectors. (This is the key step.) Let q_i be such that

\(\; (A - \lambda I) q_i = p_i \mbox{ for } i = 1, \ldots,s.\)

Finally, we can pick any basis for

\(\ker(A - \lambda I) / Q\)

and then lift to vectors {z₁, ..., z_t} in ⁠\(\ker(A - \lambda I)\)⁠. Each z_i forms a Jordan chain of length 1. We just need to show that the union of {p₁, ..., p_r}, {z₁, ..., z_t}, and {q₁, ..., q_s} forms a basis for the vector space.

Condensed: the full section is in Wikipedia.

It can be shown that the Jordan normal form of a given matrix A is unique up to the order of the Jordan blocks.

Knowing the algebraic and geometric multiplicities of the eigenvalues is not sufficient to determine the Jordan normal form of A. Assuming the algebraic multiplicity m(λ) of an eigenvalue λ is known, the structure of the Jordan form can be ascertained by analyzing the ranks of the powers (A − λI). To see this, suppose an n × n matrix A has only one eigenvalue λ. So m(λ) = n. The smallest integer k₁ such that

\((A - \lambda I)^{k_1} = 0\)

is the size of the largest Jordan block in the Jordan form of A. (This number k₁ is also called the index of λ. See discussion in a following section.) The rank of

\((A - \lambda I)^{k_1 - 1}\)

is the number of Jordan blocks of size k₁. Similarly, the rank of

\((A - \lambda I)^{k_1 - 2}\)

is twice the number of Jordan blocks of size k₁ plus the number of Jordan blocks of size k₁ − 1. The general case is similar.

This can be used to show the uniqueness of the Jordan form. Let J₁ and J₂ be two Jordan normal forms of A. Then J₁ and J₂ are similar and have the same spectrum, including algebraic multiplicities of the eigenvalues. The procedure outlined in the previous paragraph can be used to determine the structure of these matrices. Since the rank of a matrix is preserved by similarity transformation, there is a bijection between the Jordan blocks of J₁ and J₂. This proves the uniqueness part of the statement.

If A is a real matrix, its Jordan form can still be non-real. Instead of representing it with complex eigenvalues and ones on the superdiagonal, as discussed above, there exists a real invertible matrix P such that PAP = J is a real block diagonal matrix with each block being a real Jordan block. A real Jordan block is either identical to a complex Jordan block (if the corresponding eigenvalue \(\lambda_i\) is real), or is a block matrix itself, consisting of 2×2 blocks (for non-real eigenvalue \(\lambda_i = a_i+ib_i\) with given algebraic multiplicity) of the form

\(C_i = \left[ \begin{array}{rr} a_i & -b_i \\ b_i & a_i \\ \end{array} \right]\)

and describe multiplication by \(\lambda_i\) in the complex plane. The superdiagonal blocks are 2×2 identity matrices and hence in this representation the matrix dimensions are larger than the complex Jordan form. The full real Jordan block is given by

\(J_i = \begin{bmatrix} C_i & I & & \\ & C_i & \ddots & \\ & & \ddots & I \\ & & & C_i \end{bmatrix}.\)

This real Jordan form is a consequence of the complex Jordan form. For a real matrix the nonreal eigenvectors and generalized eigenvectors can always be chosen to form complex conjugate pairs. Taking the real and imaginary part (linear combination of the vector and its conjugate), the matrix has this form with respect to the new basis.

Jordan reduction can be extended to any square matrix M whose entries lie in a field K. The result states that any M can be written as a sum D + N where D is semisimple, N is nilpotent, and DN = ND. This is called the Jordan-Chevalley decomposition. Whenever K contains the eigenvalues of M, in particular when K is algebraically closed, the normal form can be expressed explicitly as the direct sum of Jordan blocks.

Similar to the case when K is the complex numbers, knowing the dimensions of the kernels of (M − λI) for 1 ≤ k ≤ m, where m is the algebraic multiplicity of the eigenvalue λ, allows one to determine the Jordan form of M. We may view the underlying vector space V as a K[x]-module by regarding the action of x on V as application of M and extending by K-linearity. Then the polynomials (x − λ) are the elementary divisors of M, and the Jordan normal form is concerned with representing M in terms of blocks associated to the elementary divisors.

The proof of the Jordan normal form is usually carried out as an application to the ring K[x] of the structure theorem for finitely generated modules over a principal ideal domain, of which it is a corollary.

One can see that the Jordan normal form is essentially a classification result for square matrices, and as such several important results from linear algebra can be viewed as its consequences.

Using the Jordan normal form, direct calculation gives a spectral mapping theorem for the polynomial functional calculus: Let A be an n × n matrix with eigenvalues λ₁, ..., λ_n, then for any polynomial p, p(A) has eigenvalues p(λ₁), ..., p(λ_n).

The characteristic polynomial of A is \(p_A(\lambda)=\det (\lambda I-A)\). Similar matrices have the same characteristic polynomial. Therefore, \(p_A(\lambda)=p_J(\lambda)=\prod_i (\lambda-\lambda_i)^{m_i}\), where \(\lambda_i\) is the ith root of \(p_J\) and \(m_i\) is its multiplicity, because this is clearly the characteristic polynomial of the Jordan form of A.