If n=1, then the Jordan block coincides with the given matrix and formula (9) is true. Let us suppose that the Jordan form of the matrix A is found by applying the Jordan block construction formula (9) if the order of the matrix A is smaller than n. we will use mathematical induction.
I step. Assuming that A is singular, 
Considering the corresponding 
matrix we find that in this case the construction based on formulas
(9) is realizable. Namely, in the space 
there are r independent
vectors 
such that the following relations take place
II step. Let us suppose that 
Every vector of the null space
is an eigenvector of the matrix A corresponding to the eigenvalue
of the matrix A 
Therefore, there must be p chains on the I step which begin with
the eigenvectors corresponding
to the eigenvalue 0. We are
interested in the last vector of each such chain. Since the vectors
belonging to the subspace 
must also belong to the space 
then they have to be the linear combinations of the column vectors of the
matrix A
with some 
.
Therefore, the vector
follows the vector
in the chain corresponding to the eigenvalue
.
III step. Since 
there must be n-r-p more linearly
independent vectors 
of the space
in the orthogonal complement of the subspace .
Proposition 5.3.1.
The algorithm of Filipov defines r vectors 
p vectors
and n-r-p vectors 
which determine the Jordan chains. These
vectors are linearly
independent and suit to be the column-vectors of the matrix X,
and J=X-1AX.
Proof. Look Strang
(1988, p. 457). 
Example 5.3.1. Let us find the Jordan
normal form of the matrix
using the algorithm of Filipov.
I step. It comes out from the form of the matrix
that 
and 
Hence r=1 and there is a vector 
from this subspace
satisfying the condition (10).
II step. Let us find the basis
of the null space
of the matrix A:
The vector 
belongs to the subspace
and 
We solve the system
III step. We take for the vector 
the vector 
and form the matrix X:
Now we find the inverse matrix
and the Jordan matrix
The software package ``Maple'' gives for the Jordan
decomposition:
Since the matrix X in the Jordan decomposition of the matrix A is not uniquely defined, then for many problems it is of interest to choose the matrix X so that the conditional number k(X) were the least. Such a problem arose also in example 1.2.9.4.
Problem 5.3.1. Find the
Jordan decomposition of the matrix
Problem 5.3.2.* Find the Jordan
decomposition of the matrix
Problem 5.3.3.* Find the
Jordan decomposition of the matrix
Problem 5.3.4. Let the Jordan
decomposition of the matrix 
be A=MJM-1. Show that 