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Jordan normal form

In linear algebra, the Jordan normal form, also called the Jordan canonical form, named in honor of the 19th and early 20th-century French mathematician Camille Jordan, answers the question, for a given square matrix M over a field K, to what extent M can be simplified into a standard shape by changing basis.

 Contents

Motivation

Consider the situation of matrix diagonalization. A square matrix is diagonalizable if the sum of the dimensions of the eigenspaces is the number of rows or columns of the matrix. Let us examine the following matrix

$A=\begin{pmatrix} 322 & -323 & -323 & 322 \\ 325 & -326 & -325 & 326 \\ -259 & 261 & 261 & -260 \\ -237 & 237 & 238 & -237 \end{pmatrix}$

We have eigenvalues of A being only λ = 5, 5, 5, 5. Now, the dimension of the kernel of A-5I is 1, so A is not diagonalizable. However, we can construct the Jordan form of this matrix. Since the kernel of A-5I is 1, we know that the Jordan form is comprised of only one Jordan block, that is, the Jordan form of A is

$J=J_4(5)=\begin{pmatrix} 5 & 1 & 0 & 0 \\ 0 & 5 & 1 & 0 \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 5 \end{pmatrix}$

Observe that J can be written as 5I+N, where N is a nilpotent matrix. Since we have now A similar to such a simple matrix, we can perform calculations involving A by using the Jordan form, which can ease the calculations in many cases. For example calculating powers of matrices is significantly easier by using the Jordan form.

General case

It is not possible to make all such matrices M diagonal, even when K is algebraically closed: what the Jordan normal form does is to quantify the failure. In abstract terms, any M is written as a sum D + N where D is diagonalizable, N is nilpotent, and D commutes with N.

The way the normal form is usually written is explicitly as the direct sum of block square matrices, known as Jordan blocks. Jordan blocks are of the form λI + N=Jn(λ), where λ is one of the eigenvalues of M, n the number of rows or columns of the Jordan block, and N is a special nilpotent matrix defined as Niji,j+1 (where δ is the Kronecker delta). This form is valid over an algebraically closed field. That is, one typical Jordan block looks like

$\begin{pmatrix} \lambda & 1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & 0 & \cdots & 0 \\ 0 & 0 & \lambda & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & 0 & 0 & \lambda \\\end{pmatrix}$

If one knows the dimensions of the kernels (MI)k for 1 ≤ km, where m is the algebraic multiplicity of the eigenvalue λ, one can determine the Jordan form that exists for M. Calculating the invertible transition matrix P such that P-1MP=J can be done by considering eigenvectors.

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 principal ideal domains, of which it is a corollary.

Algorithms and methods

Let us examine the methods of determining the transition matrix by example.

Example 1

Consider the calculation of the transition matrix for the matrix above. Recall

$A=\begin{pmatrix} 322 & -323 & -323 & 322 \\ 325 & -326 & -325 & 326 \\ -259 & 261 & 261 & -260 \\ -237 & 237 & 238 & -237 \end{pmatrix}$

We concern ourselves with obtaining generalized eigenvectors, that is, solutions to

$(A-\lambda I)^k\mathbf{v} = \mathbf{0}$

which will allow us to calculate "chains" of vectors, whose elements form the columns of the transition matrix.

For A above, we know there is only one Jordan block (see above), so we firstly obtain one generalized eigenvector - since (A-5I)^4 is the zero matrix, ker (A-5I)^4 is the entire space, so we can pick one of the standard basis vectors for the space, v=(1,0,0,0)T, since none of the standard basis vectors are an eigenvector of (A-5I)^3, (A-5I)^2, or A-5I. Then, forming the chain

$\left\{(A-5I)^3\mathbf{v}, (A-5I)^2\mathbf{v}, (A-5I)\mathbf{v}, \mathbf{v}\right\}$
$=\left\{ \begin{pmatrix} 5922 \\ 4230 \\ -3572 \\ -5170 \end{pmatrix}, \begin{pmatrix} 2857 \\ 2363 \\ -1962 \\ -2392 \end{pmatrix}, \begin{pmatrix} 317 \\ 325 \\ -259 \\ -237 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix}\right\}$

so, we can form the transition matrix as

$P=\begin{pmatrix} 5922 & 2857 & 317 & 1 \\ 4230 & 2363 & 325 & 0 \\ -3572 & -1962 & -259 & 0 \\ -5170 & -2392 & -237 & 0 \\ \end{pmatrix}$

Example 2

Say we have

$B = \begin{pmatrix} 5 & 4 & 2 & 1 \\ 0 & 1 & -1 & -1 \\ -1 & -1 & 3 & 0 \\ 1 & 1 & -1 & 2 \\ \end{pmatrix}$

The eigenvalues of B are 4, 4, 2 and 1. Now, we have

$\mathrm{dim}\ \ker{B-I} = 1, \mathrm{dim}\ \ker{B-2I} = 1, \mathrm{dim}\ \ker{B-4I} = 1, \mathrm{dim}\ \ker{B-4I}^2 = 2$

so we can say that the Jordan form of the matrix is

$J=J_1(1)\oplus J_1(2)\oplus J_2(4)$

since vectors in ker B-4I are also in ker (B-4I)2.

We have that

$\ker{(B-4I)}^4 = \mathrm{sp}\ \left\{\begin{pmatrix} 0 \\ 0 \\ -1 \\ 1\end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -1 \\ 1\end{pmatrix}\right\}$

but we pick a vector in the span that is not in any of the kernels of (B-4I)3, (B-4I)2, or B-4I, so choose v=(0,0,-1,1)T since (1,0,-1,1) is in the kernel of B-4I.

Now, there are three chains, {(B-4I)v, v}, {w}, and {x}, where w=(1,-1,0,1)T is the basis vector of the 1-dimensional kernel of B-2I and likewise x=(-1,1,0,0) is the basis vector of the 1-dimensional kernel of B-I. Form the transition matrix from these chain vectors as follows:

$P=\begin{pmatrix} -1 & 0 & 1 & -1\\ 0 & 0 & -1 & 1\\ 1 & -1 & 0 & 0\\ -1 & 1 & 1 & 0\end{pmatrix}=\left((B-4I)\mathbf{v}\left|\mathbf{v}\left|\mathbf{w}\left|\mathbf{x}\right)\right.\right.\right.$

and

$P^-1BP=J=\begin{pmatrix} 4 & 1 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$

If we had interchanged the order of which the chain vectors appeared, that is, changing the order of w, 'x, and {(B-4I)v, v} together, the Jordan blocks would be interchanged, giving equivalent Jordan forms, however.

03-10-2013 05:06:04