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Endomorphism/Jordan normal form/Introduction/Section

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Let be a field and . A Jordan matrix (to the eigenvalue ) is a square matrix of the form[1]

If we consider such a Jordan matrix as a linear mapping on the standard space , then

In particular, is a eigenvector to the eigenvalue . A simple observation shows that there is no further eigenvector linearly independent to (see exercise). The property on the right is equivalent with the condition[2]

for . The eigenvector is a generating element of the kernel of the mapping , and the other standard vectors arise successively as preimages of under .


A square matrix of the form

where each is a Jordan matrix, is called a matrix in

Jordan normal form.

The Jordan matrices occurring here are called the Jordan blocks of the matrix. Their eigenvalues might be different or equal. In the matrix

there are three Jordan blocks,

with eigenvalues and again .

We state and prove now the theorem about the Jordan normal form for trigonalizable endomorphisms.


For every trigonalizable endomorphism

on a finite-dimensional -vector space , there exists a basis, such that the describing matrix is in

Jordan normal form.

Since is trigonalizable, we can apply fact. Hence, there exists a direct sum decomposition

where the generalized eigenspaces are -invariant. Looking at the situation for each generalized eigenspace, we may assume that has only one eigenvalue , and that

holds. Then,

is nilpotent. Therefore, because of fact, there exists a basis such that is described by a matrix of the form

where the equal or equal . With respect to this basis,

has the form


Hence, every upper triangular matrix is similar to a matrix in Jordan normal form. Over the complex numbers, every matrix can be brought to Jordan normal form. If a matrix has Jordan normal form, then we can read of directly the diagonalizable and the nilpotent part in the sense of fact: The diagonal yields the diagonalizable part, and the entries which are strictly above the diagonal, yield the nilpotent part (in general, this is not true for upper triangular matrices).


We describe how to find to a linear trigonalizable mapping a basis, such that describing matrix with respect to this basis is in Jordan normal form. For this, we determine, for every eigenvalue , the minimal exponent with

This kernel is the generalized eigenspace to . We set

for . This yields the chain

Now, we choose a vector from . The vectors

form a basis for a Jordan block. If this basis generates the generalized eigenspace, then we are done. Otherwise, we look in for another vector which is linearly independent to and to . Again, we add this vector and all its successive images. If is exhausted, then we look whether is already covered, and so on. If the generalized eigenspace to is covered, then we continue with the next eigenvalue.

Under certain circumstances, we can also start with a basis of the eigenspace. If, for example, the eigenspace to is one-dimensional, then we can choose an eigenvector for , and we can find successively preimages under of the vectors, that is, we have to solve the equation

then

etc.

If, for example, the eigenspace is -dimensional and the generalized eigenspace is -dimensional, then we only have to find a preimage for one eigenvector under .


We consider the matrix

and want to bring it to Jordan normal form. bringen. The vector is an eigenvector to the eigenvalue . We have

so that there exists no further linearly independent eigenvector. We look at the linear system . This imples (looking at the second row) and so (we can choose freely to be ). Hence, we set . Finally, we need a solution for . This yields the equation . The matrix acts as

so that the mapping is described with respect to the basis by

This matrix is a Jordan matrix and, in particular, in Jordan normal form.


We consider the matrix

and want to bring it to Jordan normal form. The vectors and are linearly independent eigenvectors to the eigenvalue . We have

so that and span this eigenspace. An eigenvector must be the image of some vector under the matrix . In fact, the linear system

has the solution . Therefore, the matrix acts in the following way

Hence, the mapping is described, with respect to the basis , by the matrix

This matrix is in Jordan normal form with the Jordan blocks and .


We consider the matrix

and want to bring in in Jordan normal for. Here, we have two eigenvalues and, therefore, two two-dimensional generalized eigenspaces, which we treat separately. We have

therefore, belongs to the kernel. The determinant of the upper right submatrix is not , so the rank of the matrix is , and its kernel is one-dimensional. The second power is

a new element of the kernel is . Thus, we have

Because of

we can use the vectors and to establish the first Jordan block.

We have

therefore, belongs to the kernel. The rank of this matrix is again , and the kernel has dimension one. The second power is

a new element in the kernel is . Thus, we have

Because of

we can use the vectors and to establish the second Jordan block. Altogether, the linear mapping defined by has, with respect the basis

the Jordan normal form

  1. Some authors define a Jordan matrix one where the are below the diagonal.
  2. In the context of trigonalizable mappings and in order to find the Jordan normal form, it is useful to work with instead of .