1. Introduction

The Jordan decomposition theorem for nilpotent matrices is treated in simple way. While the result is known, the interest of our proofs lies in their simplicity. Note that the usual proofs are mostly based on module theory and/or quotient spaces.

Definition 1 A nilpotent Jordan block of size n, denoted, is a square matrix of the form:

Definition 2 A nilpotent Jordan matrix is a block diagonal matrix of the form:

where each is a nilpotent Jordan block.

Theorem 1 Let E be a vector space over a field K, of finite dimension n, and f a linear operator on E, nilpotent of index p. There exists some basis B, in which the matrix representing f in B is a Jordan matrix.

Theorem 1 can be expressed in matrix form as follows:

Theorem 2 Every nilpotent matrix N of index p is similar to an Jordan matrix J in which the size of the largest Jordan block is where k is the rank of N.

A. Galperin and Z. Waksman [1] used elementary concepts to show that is similar to Jordan matrix. Gohberg and Goldberg [2] gave an algorithm that builds Jordan form of an operator A on an n-dimensional space if the Jordan form restricted to an n1 dimensional invariant subspace is known. In what follows, we give two proofs of theorem 2. The first by using elementary operations on matrices, and the second by using a decomposition of E into direct sums of subspaces.

2. Method 1 - Elementary Operations

We shall prove theorem 2 by induction. The following two lemmas are first proved:

Lemma 1 There exists a matrix A representing f having the following form:

Proof 1 As f is nilpotent of index p, there exists such that The family is then linearly independent Suppose the contrary, then there exist constants not all zero such that

Now let

therefore

and

i.e. and, which is a contradiction

By the incomplete basis theorem, there exist vectors such that the family

spans E. The matrix representing f in that basis has the needed property.

The lemma 2 below is the key to prove our theorem. We shall prove (again in two ways!) that the bloc matrix B found in lemma 1 is in fact the zero matrix.

Lemma 2 There exists a matrix representing f having the form:

Proof 2 method 1: The first proof is based on elementary matrix calculations. For this let the triangular matrix defined as follows:

We can easily check that

Define the matrix A' by clearly:

Let and be the i-th rows of X and B respectively, then

Now choose and for , we obtain a matrix A' , similar to A , of the form

A simple calculation yields:

As then

Therefore L=0, and hence

method 2: Let such that the family is linearly independent. Complete the basis of E by vectors

i.e. the family

is basis of E. Let x be in E, then

And define the linear operator:

We now state and prove the following three properties about the linear form l: • property 1: The family is linearly independent in the dual of E

Suppose the contrary, then there exist scalars not all zeros, with

Denote

then

and

implies that which contradicts our hypothesis

Denote now

and let G be the set of all such that

• property 2: The subspace G of E is stable by f

- for all and all , and ; we have then is therefore a subspace of E

- If then and Therefore,

And finally,

• property 3:

- If, then there exist not all zeros, such that by letting

then and which contradicts the hypothesis

- Let such that

is a basis of and let be a basis of E whose dual basis is

, and;

- If

hence

Thus

hence and Consequently

Is it now simple to see that the matrix representing f in the basis

of is of the form:

The first proof of our theorem can now be completed. For and the result is obvious; Assume the result holds up to

By lemma 2, there exists an invertible matrix such that

By the induction hypothesis, the exists an invertible matrix P such that

Let n₁=p, and then

this completes the proof.

3. Method 2 - Decomposition of E into Direct Sums

The second proof was suggested by Rached Mneimné [4] during my visit to the department of Mathematics at Université Diderot in April 2015. The following theorem is on the decomposition of E into direct.

Theorem 3 Let E be a vector space over a field K, of finite dimension n, and f is a linear operator on E, nilpotent of index p. There exists and subspaces such that:

a)

b) is stable by f

c) the operator , restriction of over , is nilpotent of index

Proof 3 The proof is by induction. It is obviously true for and Now suppose that the result holds up to

As f is nilpotent, then and then by the induction hypothesis, there exists and subspaces such that:

a)

b) is stable by f

c) the operator , restriction of , is nilpotent of index

As is nilpotent of order , then by lemma 1, there exists such that

Moreover , then there exists such that Denote

and

Six properties for the subspaces are stated and proved:

1.

In fact, if then and Thus

2. is a basis of

If , then i.e. and as the family is linearly independent, then and, and Therefore

3.

then and

if, then and Therefore

4.

Let with then

As and then By the previous property, there exists such that , and Therefore

5.

If with then by the previous property there exists such that As the subspaces are in direct sum, then and therefore

6.

The proof of theorem 3 can now be completed as follows: Let now H be a complement of in then:

(1)

Thus, are in direct sum, and:

Therefore,

And

with is a basis of H.

The second proof of theorem can be obtained. We can now check that the matrix representing in the basis is:

ACKNOWLEDGEMENTS

The authors are thankful to Salim Kobeissi, Professeur aggregé at Université Pierre Mendès France, Grenoble II, and to Rached Mneimné, Associate Professor at Université Diderot, Paris 7, for their valuable comments.